Circuits nilsson 7th solution manual

1Circuit Variables

Assessment Problems

AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year todollars/millisecond. We begin by expressing $10 billion in scientific notation:

$100 billion = $100 × 109

Now we determine the number of milliseconds in one year, again using a product ofratios:

1 year365.25 days

· 1 day24 hours

· 1 hour60 mins

· 1 min60 secs

· 1 sec1000 ms

=1 year

31.5576 × 109 ms

Now we can convert from dollars/year to dollars/millisecond, again with a productof ratios:

$100 × 109

1 year· 1 year31.5576 × 109 ms

=100

31.5576= $3.17/ms

AP 1.2 First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal willtravel in 10−9 s if it is traveling at 80% of the speed of light. Remember that thespeed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108) = 2.4 × 108

m/s. Now, we use a product of ratios to convert from meters/second toinches/nanosecond:

2.4 × 108 m1s

· 1 s109 ns

· 100 cm1 m

· 1 in2.54 cm

=(2.4 × 108)(100)

(109)(2.54)=

9.45 in1 ns

Thus, a signal traveling at 80% of the speed of light will travel 9.45′′ in ananosecond.

1–1

1–2 CHAPTER 1. Circuit Variables

AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or i = dqdt

Inthis problem, we are given the current and asked to find the total charge. To do this,we must integrate Eq. (1.2) to find an expression for charge in terms of current:

q(t) =∫ t

0i(x) dx

We are given the expression for current, i, which can be substituted into the aboveexpression. To find the total charge, we let t → ∞ in the integral. Thus we have

qtotal =∫ ∞

020e−5000x dx =

20−5000

e−5000x

∣∣∣∣∞

0=

20−5000

(e∞ − e0)

=20

−5000(0 − 1) =

205000

= 0.004 C = 4000µC

AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dqdt

. Inthis problem we are given an expression for the charge, and asked to find themaximum current. First we will find an expression for the current using Eq. (1.2):

i =dq

dt=

d

dt

[ 1α2 −

(t

α+

1α2

)e−αt

]

=d

dt

( 1α2

)− d

dt

(t

αe−αt

)− d

dt

( 1α2 e−αt

)

= 0 −( 1

αe−αt − α

t

αe−αt

)−(−α

1α2 e−αt

)

=(− 1

α+ t +

1α

)e−αt

= te−αt

Now that we have an expression for the current, we can find the maximum value ofthe current by setting the first derivative of the current to zero and solving for t:

di

dt=

d

dt(te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0

Since e−αt never equals 0 for a finite value of t, the expression equals 0 only when(1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this valueof t, the current is

i =1α

e−α/α =1α

e−1

Remember in the problem statement, α = 0.03679. Using this value for α,

i =1

0.03679e−1 ∼= 10 A

Problems 1–3

AP 1.5 Start by drawing a picture of the circuit described in the problem statement:

Also sketch the four figures from Fig. 1.6:

[a] Now we have to match the voltage and current shown in the first figure with thepolarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2is the same as 4A of current leaving Terminal 1. We get

(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4 A

(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4 A

[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,p = vi = (−20)(−4) = 80 W. Since the power is greater than 0, the box isabsorbing power.

[c] From the calculation in part (b), the box is absorbing 80 W.

AP 1.6 Applying the passive sign convention to the power equation using the voltage andcurrent polarities shown in Fig. 1.5, p = vi. From Eq. (1.3), we know that power isthe time rate of change of energy, or p = dw

dt. If we know the power, we can find the

energy by integrating Eq. (1.3). To begin, find the expression for power:

p = vi = (10,000e−5000t)(20e−5000t) = 200,000e−10,000t = 2 × 105e−10,000t W

Now find the expression for energy by integrating Eq. (1.3):

w(t) =∫ t

0p(x) dx


Substitute the expression for power, p, above. Note that to find the total energy, welet t → ∞ in the integral. Thus we have

w =∫ ∞

02 × 105e−10,000x dx =

2 × 105

−10,000e−10,000x

∣∣∣∣∣∞

0

=2 × 105

−10,000(e−∞ − e0) =

2 × 105

−10,000(0 − 1) =

2 × 105

10,000= 20 J

AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thusentering the lower terminal where the polarity marking of the voltage is negative.Thus, using the passive sign convention, p = −vi. Substituting the values of voltageand current given in the figure,

p = −(800 × 103)(1.8 × 103) = −1440 × 106 = −1440 MW

Thus, because the power associated with the Oregon end of the line is negative,power is being generated at the Oregon end of the line and transmitted by the line tobe delivered to the California end of the line.

Chapter Problems

P 1.1 To begin, we calculate the number of pixels that make up the display:

npixels = (1280)(1024) = 1,310,720 pixels

Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixelrequires 3 bytes of information. We can calculate the number of bytes ofinformation required for the display by multiplying the number of pixels in thedisplay by 3 bytes per pixel:

nbytes =1,310,720 pixels

1 display· 3 bytes

1 pixel= 3,932,160 bytes/display

Finally, we use the fact that there are 106 bytes per MB:

3,932,160 bytes1 display

· 1 MB106bytes

= 3.93 MB/display

Problems 1–5

P 1.2 c = 3 × 108 m/s so12c = 1.5 × 108 m/s

1.5 × 108 m1 s

=5 × 106 m

x sso x =

5 × 106

1.5 × 108 = 33.3 ms

P 1.3 We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First,recall that 1 mm = 103µm. Let’s also express the rate of growth of bamboo using theunits mm/s instead of mm/day. Use a product of ratios to perform this conversion:

250 mm1 day

· 1 day24 hours

· 1 hour60 min

· 1 min60 sec

=250

(24)(60)(60)=

103456

mm/s

Use a ratio to determine the time it takes for the bamboo to grow 10 µm:

10/3456 × 10−3 m1 s

=10 × 10−6 m

x sso x =

10 × 10−6

10/3456 × 10−3 = 3.456 s

P 1.4 Volume = area × thickness

106 = (10 × 106)(thickness)

⇒ thickness =106

10 × 106 = 0.10 mm

P 1.5300 × 109 dollars

1 year· 100 pennies

1 dollar· 1 year365.25 days

· 1 day24 hr

· 1 hr3600 s

· 1.5 mm1 penny

· 1 m1000 mm

= 1426 m/s

P 1.6 Our approach is as follows: To determine the area of a bit on a track, we need toknow the height and width of the space needed to store the bit. The height of thespace used to store the bit can be determined from the width of each track on thedisk. The width of the space used to store the bit can be determined by calculatingthe number of bits per track, calculating the circumference of the inner track, anddividing the number of bits per track by the circumference of the track. Thecalculations are shown below.

Width of track =1 in

77 tracks25,400µm

in= 329.87µm/track

Bits on a track =1.4 MB2 sides

8 bitsbyte

1 side77 tracks

= 72,727.273 bits/track

Circumference of inner track = 2π(1/2′′)(25,400µm/in) = 79,796.453µm

Width of bit on inner track =79,796.453µm72,727.273 bits

= 1.0972µm/bit

Area of bit on inner track = (1.0972)(329.87) = 361.934µm2


P 1.7 C/m3 = 1.6022 × 10−19 × 1029 = 1.6022 × 1010 C/m3

C/m = (1.6022 × 1010)(5.4 × 10−4) = 8.652 × 106 C/m

Therefore, (8.652 × 106)Cm

× ave vel(

m

s

)= i

Thus, average velocity =14008.652

× 10−6 = 161.81 µm/s

P 1.8 [a] n =20 × 10−6 C/s

1.6022 × 10−19 C/elec= 1.25 × 1014 elec/s

[b] m = 3303 mi · 5280 ft1 mi

· 12 in1 ft

· 2.54 cm1 in

· 104 µm1 cm

= 5.32 × 1012 µm

Therefore,n

m= 23.5

The number of electrons/second is approximately 23.5 times the number ofmicrometers between Sydney and San Francisco.

P 1.9 First we use Eq. (1.2) to relate current and charge:

i =dq

dt= 20 cos 5000t

Therefore, dq = 20 cos 5000t dt

To find the charge, we can integrate both sides of the last equation. Note that wesubstitute x for q on the left side of the integral, and y for t on the right side of theintegral:

∫ q(t)

q(0)dx = 20

∫ t

0cos 5000y dy

We solve the integral and make the substitutions for the limits of the integral,remembering that sin 0 = 0:

q(t) − q(0) = 20sin 5000y

5000

∣∣∣∣t

0=

205000

sin 5000t − 205000

sin 5000(0) =20

5000sin 5000t

But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value att = 0, so q(t) = 4 × 10−3 sin 5000t C = 4 sin 5000t mC

P 1.10 w = qV = (1.6022 × 10−19)(9) = 14.42 × 10−19 = 1.442 aJ

Problems 1–7

P 1.11 p = (6)(100 × 10−3) = 0.6 W; 3 hr · 3600 s1 hr

= 10,800 s

w(t) =∫ t

0p dt w(10,800) =

∫ 10,800

00.6 dt = 0.6(10,800) = 6480 J

P 1.12 Assume we are standing at box A looking toward box B. Then, using the passivesign convention p = vi, since the current i is flowing into the + terminal of thevoltage v. Now we just substitute the values for v and i into the equation for power.Remember that if the power is positive, B is absorbing power, so the power must beflowing from A to B. If the power is negative, B is generating power so the powermust be flowing from B to A.

[a] p = (20)(15) = 300 W 300 W from A to B

[b] p = (100)(−5) = −500 W 500 W from B to A

[c] p = (−50)(4) = −200 W 200 W from B to A

[d] p = (−25)(−16) = 400 W 400 W from A to B

P 1.13 [a]

p = vi = (−20)(5) = −100 WPower is being delivered by the box.

[b] Leaving

[c] Gaining

P 1.14 [a] p = vi = (−20)(−5) = 100 W, so power is being absorbed by the box.

[b] Entering

[c] Losing

P 1.15 [a] In Car A, the current i is in the direction of the voltage drop across the 12 Vbattery(the current i flows into the + terminal of the battery of Car A).Therefore using the passive sign convention, p = vi = (−40)(12) = −480W.Since the power is negative, the battery in Car A is generating power, so Car Bmust have the ”dead” battery.


[b] w(t) =∫ t

0p dx; 1.5 min = 1.5 · 60 s

1 min= 90 s

w(90) =∫ 90

0480 dx

w = 480(90 − 0) = 480(90) = 43,200 J = 43.2 kJ

P 1.16 p = vi; w =∫ t

0p dx

Since the energy is the area under the power vs. time plot, let us plot p vs. t.

Note that in constructing the plot above, we used the fact that 60 hr = 216,000 s =216 ks

p(0) = (6)(15 × 10−3) = 90 × 10−3 W

p(216 ks) = (4)(15 × 10−3) = 60 × 10−3 W

w = (60 × 10−3)(216 × 103) +12(90 × 10−3 − 60 × 10−3)(216 × 103) = 16.2 kJ

P 1.17 [a] To find the power at 625 µs, we substitute this value of time into both theequations for v(t) and i(t) and multiply the resulting numbers to get p(625 µs):

v(625 µs) = 50e−1600(625×10−6) − 50e−400(625×10−6) = 18.394 − 38.94 = −20.546 V

i(625 µs) = 5 × 10−3e−1600(625×10−6) − 5 × 10−3e−400(625×10−6)

= 0.0018394 − 0.003894 = −0.0020546 A

p(625 µs) = (−20.546)(−0.0020546) = 42.2 mW

[b] To find the energy at 625 µs, we need to integrate the equation for p(t) from 0 to625 µs. To start, we need an expression for p(t):

p(t) = v(t)i(t) = (50)(5 × 10−3)(e−1600t − e−400t)(e−1600t − e−400t)

Problems 1–9

=14(e−3200t − 2e−2000t + e−800t)

Now we integrate this expression for p(t) to get an expression for w(t). Notewe substitute x for t on the right side of the integral.

w(t) =14

∫ t

0(e−3200x − 2e−2000x + e−800x)dx

=14

[e−3200x

−3200+

e−2000x

1000− e−800x

800

]∣∣∣∣∣t

0

=14

[e−3200t

−3200+

e−2000t

1000− e−800t

800−( 1

−3200+

11000

− 1800

)]

=14

[e−3200t

−3200+

e−2000t

1000− e−800t

800+ 5.625 × 10−4

]

Finally, substitute t = 625 µs into the equation for w(t):

w(625 µs) =14[−4.2292 × 10−5 + 2.865 × 10−4 − 7.5816 × 10−4 + 5.625 × 10−4]

= 12.137 µJ

[c] To find the total energy, we let t → ∞ in the above equation for w(t). Note thatthis will cause all expressions of the form e−nt to go to zero, leaving only theconstant term 5.625 × 10−4. Thus,

wtotal =14[5.625 × 10−4] = 140.625 µJ

P 1.18 [a] v(20 ms) = 100e−1 sin 3 = 5.19 Vi(20 ms) = 20e−1 sin 3 = 1.04 A p(20 ms) = vi = 5.39 W

[b] p = vi = 2000e−100t sin.2 150t

= 2000e−100t[12

− 12

cos 300t]

= 1000e−100t − 1000e−100t cos 300t

w =∫ ∞

01000e−100t dt −

∫ ∞

01000e−100t cos 300t dt

= 1000e−100t

−100

∣∣∣∣∣∞

0

−1000

e−100t

(100)2 + (300)2 [−100 cos 300t + 300 sin 300t]∣∣∣∣∣

∞

0

= 10 − 1000[ 1001 × 104 + 9 × 104

]= 10 − 1

w = 9 J


P 1.19 [a] 0 s ≤ t < 1 s:

v = 5 V; i = 20t A; p = 100t W

1 s < t ≤ 2 s:

v = 0 V; i = 20 A; p = 0 W

2 s ≤ t < 3 s:

v = 0 V; i = 20 A; p = 0 W

3 s < t ≤ 4 s:

v = −5 V; i = 80 − 20t A; p = −400 + 100t W

4 s ≤ t < 5 s:

v = −5 V; i = 80 − 20t A; p = −400 + 100t W

5 s < t ≤ 6 s:

v = 5 V; i = −120 + 20t A; p = −600 + 100t W

6 s ≤ t < 7 s:

v = 5 V; i = −120 + 20t A; p = −600 + 100t W

t > 7 s:

v = 0 V; i = 20 A; p = 0 W

[b] Calculate the area under the curve from zero up to the desired time:

w(1) = 12(1)(100) = 50 J

w(6) = 12(1)(100) − 1

2(1)(100) + 12(1)(100) − 1

2(1)(100) = 0 J

w(10) = 12(1)(100) − 1

2(1)(100) + 12(1)(100) − 1

2(1)(100) + 12(1)(100) = 50 J

P 1.20 [a] p = vi = (100e−500t)(0.02 − 0.02e−500t) = (2e−500t − 2e−1000t) W

dp

dt= −1000e−500t + 2000e−1000t = 0 so 2e−1000t = e−500t

2 = e500t so ln 2 = 500t thus p is maximum at t = 1.4 ms

Problems 1–11

pmax = p(1.4 ms) = 0.5 W

[b] w =∫ ∞

0[2e−500t − 2e−1000t] dt =

[ 2−500

e−500t − 2−1000

e−1000t

∣∣∣∣∞

0

]

=4

1000− 2

1000= 2 mJ

P 1.21 [a] p = vi = 200 cos(500πt)4.5 sin(500πt) = 450 sin(1000πt) WTherefore, pmax = 450 W

[b] pmax(extracting) = 450 W

[c]

pavg =1

4 × 10−3

∫ 4×10−3

0450 sin(1000πx) dx =

4504 × 10−3

[− cos 1000πt

1000π

]4×10−3

0

=−4504π

[cos 4π − cos 0] =−4504π

[1 − 1] = 0 W

[d] pavg =−4504π

[cos 15π − cos 0] =−4504π

[−1 − 1] =9004π

= 71.62 W

P 1.22 [a] q = area under i vs. t plot

=6(5000)

2+ 6(5000) +

6(10,000)2

+ 8(15,000) +8(5000)

2= 15,000 + 30,000 + 30,000 + 120,000 + 20,000 = 215,000 C

[b] w =∫

pdt =∫

vi dt

v = 0.2 × 10−3t + 8 0 ≤ t ≤ 20 ks0 ≤ t ≤ 5000s

i = 20 − 1.2 × 10−3t

p = (8 + 0.2 × 10−3t)(20 − 1.2 × 10−3t)

= 160 − 5.6 × 10−3t − 2.4 × 10−7t2

w1 =∫ 5000

0(160 − 5.6 × 10−3t − 2.4 × 10−7t2) dt

=

(160t − 5.6 × 10−3

2t2 − 2.4 × 10−7

3t3)∣∣∣∣∣

5000

0

= 720 kJ

5000 ≤ t ≤ 15,000s

i = 17 − 0.6 × 10−3t

p = (8 + 0.2 × 10−3t)(17 − 0.6 × 10−3t)

= 136 − 1.4 × 10−3t − 1.2 × 10−7t2

w2 =∫ 15,000

5000(136 − 1.4 × 10−3t − 1.2 × 10−7t2) dt

=

(136t − 1.4 × 10−3

2t2 − 1.2 × 10−7

3t3)∣∣∣∣∣

15,000

5000

= 1090 kJ


15,000 ≤ t ≤ 20,000s

i = 32 − 1.6 × 10−3t

p = (8 + 0.2 × 10−3t)(32 − 1.6 × 10−3t)

= 256 − 6.4 × 10−3t − 3.2 × 10−7t2

w3 =∫ 20,000

15,000(256 − 6.4 × 10−3t − 3.2 × 10−7t2) dt

=

(256t − 6.4 × 10−3

2t2 − 3.2 × 10−7

3t3)∣∣∣∣∣

20,000

15,000

= 226,666.67 J

wT = w1 + w2 + w3 = 720,000 + 1,090,000 + 226,666.67 = 2036.67 kJ

P 1.23 [a] p = vi = [104t + 5)e−400t][(40t + 0.05)e−400t]

= 400 × 103t2e−800t + 700te−800t + 0.25e−800t

= e−800t[400,000t2 + 700t + 0.25]dp

dt= e−800t[800 × 103t + 700] − 800e−800t[400,000t2 + 700t + 0.25]= [−3,200,000t2 + 2400t + 5]100e−800t

Therefore,dp

dt= 0 when 3,200,000t2 − 2400t − 5 = 0

so pmax occurs at t = 1.68 ms.

[b] pmax = [400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168)

= 666.34 mW

[c] w =∫ t

0pdx

w =∫ t

0400,000x2e−800x dx +

∫ t

0700xe−800x dx +

∫ t

00.25e−800x dx

=400,000e−800x

−512 × 106 [64 × 104x2 + 1600x + 2]∣∣∣∣∣t

0

+

700e−800x

64 × 104 (−800x − 1)∣∣∣∣∣t

0

+ 0.25e−800x

−800

∣∣∣∣∣t

0When t → ∞ all the upper limits evaluate to zero, hence

w =(400,000)(2)512 × 106 +

70064 × 104 +

0.25800

= 2.97 mJ.

P 1.24 [a] We can find the time at which the power is a maximum by writing an expressionfor p(t) = v(t)i(t), taking the first derivativeof p(t) and setting it to zero, then solving for t. The calculations are shown below:

p = 0 t < 0, p = 0 t > 40 s

p = vi = (t − 0.025t2)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W 0 < t < 40 sdp

dt= 4 − 0.6t + 0.015t2 = 0

Problems 1–13

Use a calculator to find the two solutions to this quadratic equation:

t1 = 8.453 s; t2 = 31.547 s

Now we must find which of these two times gives the minimum power bysubstituting each of these values for t into the equation for p(t):

p(t1) = (8.453 − 0.025(8.453)2)(4 − 0.2 · 8.453) = 15.396 W

p(t2) = (31.547 − 0.025(31.547)2)(4 − 0.2 · 31.547) = −15.396 W

Therefore, maximum power is being delivered at t = 8.453 s.

[b] The maximum power was calculated in part (a) to determine the time at whichthe power is maximum: pmax = 15.396 W (delivered)

[c] As we saw in part (a), the other “maximum” power is actually a minimum, orthe maximum negative power. As we calculated in part (a), maximum power isbeing extracted at t = 31.547 s.

[d] This maximum extracted power was calculated in part (a) to determine the timeat which power is maximum: pmaxext = 15.396 W (extracted)

[e] w =∫ t

0pdx =

∫ t

0(4x − 0.3x2 + 0.005x3)dx = 2t2 − 0.1t3 + 0.00125t4

w(0) = 0 J w(30) = 112.50 J

w(10) = 112.50 J w(40) = 0 J

w(20) = 200 JTo give you a feel for the quantities of voltage, current, power, and energy andtheir relationships among one another, they are plotted below:


Problems 1–15

P 1.25 [a] p = vi = (8 × 104te−500t)(15te−500t) = 12 × 105t2e−1000t Wdp

dt= 12 × 105[t2(−1000)e−1000t + e−1000t(2t)]

= 12 × 105e−1000t[t(2 − 1000t)]dp

dt= 0 at t = 0, t = 2 ms

We know p is a minimum at t = 0 since v and i are zero at t = 0.

[b] pmax = 12 × 105(2 × 10−3)2e−2 = 649.61 mW

[c] w = 12 × 105∫ ∞

0t2e−1000t dt

= 12 × 105

e−1000t

(−1000)3 [106t2 + 2,000t + 2]∣∣∣∣∣∞

0

= 2400µJ

P 1.26 We use the passive sign convention to determine whether the power equation isp = vi or p = −vi and substitute into the power equation the values for v and i, asshown below:

pa = −vaia = −(−18)(−51) = −918 W

pb = vbib = (−18)(45) = −810 W

pc = vcic = (2)(−6) = −12 W

pd = −vdid = −(20)(−20) = 400 W

pe = −veie = −(16)(−14) = 224 W

pf = vfif = (36)(31) = 1116 WRemember that if the power is positive, the circuit element is absorbing power,whereas is the power is negative, the circuit element is developing power. We canadd the positive powers together and the negative powers together — if the powerbalances, these power sums should be equal:∑

Pdev = 918 + 810 + 12 = 1740 W;∑Pabs = 400 + 224 + 1116 = 1740 W

Thus, the power balances and the total power developed in the circuit is 1740 W.


P 1.27 [a] From the diagram and the table we have

pa = −vaia = −(900)(−22.5) = 20,250 W

pb = −vbib = −(105)(−52.5) = 5512.5 W

pc = −vcic = −(−600)(−30) = −18,000 W

pd = vdid = (585)(−52.5) = −30,712.5 W

pe = −veie = −(−120)(30) = 3600 W

pf = vfif = (300)(60) = 18,000 W

pg = −vgig = −(585)(82.5) = −48,262.5 W

ph = −vhih = −(−165)(82.5) = 13,612.5 W∑Pdel = 18,000 + 30,712.5 + 48,262.5 = 96,975 W∑Pabs = 20,250 + 5512.5 + 3600 + 18,000 + 13,612.5 = 60,975 W

Therefore,∑

Pdel = ∑Pabs and the subordinate engineer is correct.

[b] The difference between the power delivered to the circuit and the powerabsorbed by the circuit is

96,975 − 60,975 = 36,000

One-half of this difference is 18,000W, so it is likely that pc or pf is in error.Either the voltage or the current probably has the wrong sign. (In Chapter 2,we will discover that using KCL at the top node, the current ic should be 30 A,not −30 A!) If the sign of pc is changed from negative to positive, we canrecalculate the power delivered and the power absorbed as follows:∑

Pdel = 30,712.5 + 48,262.5 = 78,975 W∑Pabs = 20,250 + 5512.5 + 18,000 + 3600 + 18,000 + 13,612.5 = 78,975 W

Now the power delivered equals the power absorbed and the power balancesfor the circuit.

P 1.28 pa = vaia = (9)(1.8) = 16.2 W

pb = −vbib = −(−15)(1.5) = 22.5 W

pc = −vcic = −(45)(−0.3) = 13.5 W

pd = −vdid = −(54)(−2.7) = 145.8 W

pe = veie = (−30)(−1) = 30 W

pf = −vfif = −(−240)(4) = 960 W

pg = −vgig = −(294)(4.5) = −1323 W

ph = vhih = (−270)(−0.5) = 135 W

Problems 1–17

Therefore,

∑Pabs = 16.2 + 22.5 + 13.5 + 145.8 + 3 − +960 + 135 = 1323 W

∑Pdel = 1323 W

∑Pabs =

∑Pdel

Thus, the interconnection satisfies the power check

P 1.29 pa = vaia = (−160)(−10) = 1600 W

pb = vbib = (−100)(−20) = 2000 W

pc = −vcic = −(−60)(6) = 360 W

pd = vdid = (800)(−50) = −40,000 W

pe = −veie = −(800)(−20) = 16,000 W

pf = −vfif = −(−700)(14) = 9800 W

pg = −vgig = −(640)(−16) = 10,240 W∑Pdel = 40,000 W∑Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,000 = 40,000 W

Therefore,∑

Pdel =∑

Pabs = 40,000 W

P 1.30 [a] From an examination of reference polarities, the following elements employ thepassive convention: a, c, e, and f .

[b] pa = −56 W

pb = −14 W

pc = 150 W

pd = −50 W

pe = −18 W

pf = −12 W∑Pabs = 150 W;

∑Pdel = 56 + 14 + 50 + 18 + 12 = 150 W.

2Circuit Elements

Assessment Problems

AP 2.1

[a] To find vg write a KVL equation clockwise around the left loop, starting belowthe dependent source:

−ib

4+ vg = 0 so vg =

ib

4To find ib write a KCL equation at the upper right node. Sum the currentsleaving the node:

ib + 8 A = 0 so ib = −8 A

Thus,

vg =−84

= −2 V

[b] To find the power associated with the 8 A source, we need to find the voltagedrop across the source, vi. To do this, write a KVL equation clockwise aroundthe left loop, starting below the voltage source:

−vg + vi = 0 so vi = vg = −2 V

Using the passive sign convention,

ps = (8 A)(vi) = (8 A)(−2 V) = −16 W

Thus the current source generated 16 W of power.

2–1

2–2 CHAPTER 2. Circuit Elements

AP 2.2

[a] Note from the circuit that vx = −25 V. To find α write a KCL equation at thetop left node, summing the currents leaving:

15 A + αvx = 0

Substituting for vx,

15 A + α(−25 V) = 0 so α(25 V) = 15 A

Thus α =15 A25 V

= 0.6 A/V

[b] To find the power associated with the voltage source we need to know thecurrent, iv. To find this current, write a KCL equation at the top left node,summing the currents leaving the node:

−αvx + iv = 0 so iv = αvx = (0.6)(−25) = −15 A

Using the passive sign convention,

ps = −(iv)(25 V) = −(−15 A)(25 V) = 375 W.

Thus the voltage source dissipates 375 W.

AP 2.3

[a] A KVL equation gives

−vg + vR = 0 so vR = vg = 1 kV

Note from the circuit that the current through the resistor is ig = 5 mA. UseOhm’s law to calculate the value of the resistor:

R =vR

ig

=1 kV5 mA

= 200 kΩ

Using the passive sign convention to calculate the power in the resistor,

pR = (vR)(ig) = (1 kV)(5 mA) = 5 W

The resistor is dissipating 5 W of power.

Problems 2–3

[b] Note from part (a) the vR = vg and iR = ig. The power delivered by the sourceis thus

psource = −vgig so vg = −psource

ig

= −(−3 W)75 mA

= 40 V

Since we now have the value of both the voltage and the current for theresistor, we can use Ohm’s law to calculate the resistor value:

R =vg

ig

=40 V

75 mA= 533.33 Ω

The power absorbed by the resistor must equal the power generated by thesource. Thus,

pR = −psource = −(−3 W) = 3 W

[c] Again, note the iR = ig. The power dissipated by the resistor can be determinedfrom the resistor’s current:

pR = R(iR)2 = R(ig)2

Solving for ig,

i2g =pr

R=

480 mW300 Ω

= 0.0016 so ig =√

0.0016 = 0.04 A = 40 mA

Then, since vR = vg

vR = RiR = Rig = (300 Ω)(40 mA) = 12 V so vg = 12 V

AP 2.4

[a] Note from the circuit that the current throught the conductance G is ig, flowingfrom top to bottom (from KCL), and the voltage drop across the current sourceis vg, positive at the top (from KVL). From a version of Ohm’s law,

vg =ig

G=

0.5 A50 mS

= 10 V

Now that we know the voltage drop across the current source, we can find thepower delivered by this source:

psource = −vgig = −(10)(0.5) = −5 W

Thus the current source delivers 5 W to the circuit.


[b] We can find the value of the conductance using the power, and the value of thecurrent using Ohm’s law and the conductance value:

pg = Gv2g so G =

pg

v2g

=9

152 = 0.04 S = 40 mS

ig = Gvg = (40 mS)(15 V) = 0.6 A

[c] We can find the voltage from the power and the conductance, and then use thevoltage value in Ohm’s law to find the current:

pg = Gv2g so v2

g =pg

G=

8 W200 µS

= 40,000

Thus vg =√

40,000 = 200 V

ig = Gvg = (200 µS)(200 V) = 0.04 A = 40 mA

AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every circuitelement.

Write a KVL equation clockwise around the circuit, starting below the voltagesource:

−24 V + v2 + v5 − v1 = 0

Next, use Ohm’s law to calculate the three unknown voltages from the threecurrents:

v2 = 3i2; v5 = 7i5; v1 = 2i1

A KCL equation at the upper right node gives i2 = i5; a KCL equation at thebottom right node gives i5 = −i1; a KCL equation at the upper left node givesis = −i2. Now replace the currents i1 and i2 in the Ohm’s law equations withi5:

v2 = 3i2 = 3i5; v5 = 7i5; v1 = 2i1 = −2i5

Now substitute these expressions for the three voltages into the first equation:

24 = v2 + v5 − v1 = 3i5 + 7i5 − (−2i5) = 12i5

Therefore i5 = 24/12 = 2 A

Problems 2–5

[b] v1 = −2i5 = −2(2) = −4 V

[c] v2 = 3i5 = 3(2) = 6 V

[d] v5 = 7i5 = 7(2) = 14 V

[e] A KCL equation at the lower left node gives is = i1. Since i1 = −i5, is = −2 A.We can now compute the power associated with the voltage source:

p24 = (24)is = (24)(−2) = −48 W

Therefore 24 V source is delivering 48 W.

AP 2.6 Redraw the circuit labeling all voltages and currents:

We can find the value of the unknown resistor if we can find the value of its voltageand its current. To start, write a KVL equation clockwise around the right loop,starting below the 24 Ω resistor:

−120 V + v3 = 0

Use Ohm’s law to calculate the voltage across the 8 Ω resistor in terms of its current:

v3 = 8i3

Substitute the expression for v3 into the first equation:

−120 V + 8i3 = 0 so i3 =1208

= 15 A

Also use Ohm’s law to calculate the value of the current through the 24 Ω resistor:

i2 =120 V24 Ω

= 5 A

Now write a KCL equation at the top middle node, summing the currents leaving:

−i1 + i2 + i3 = 0 so i1 = i2 + i3 = 5 + 15 = 20 A

Write a KVL equation clockwise around the left loop, starting below the voltagesource:

−200 V + v1 + 120 V = 0 so v1 = 200 − 120 = 80 V


Now that we know the values of both the voltage and the current for the unknownresistor, we can use Ohm’s law to calculate the resistance:

R =v1

i1=

8020

= 4 Ω

AP 2.7 [a] Plotting a graph of vt versus it gives

Note that when it = 0, vt = 25 V; therefore the voltage source must be 25 V.Since the plot is a straight line, its slope can be used to calculate the value ofresistance:

R =∆v

∆i=

25 − 00.25 − 0

=25

0.25= 100 Ω

A circuit model having the same v − i characteristic is a 25 V source in serieswith a 100Ω resistor, as shown below:

[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:

To find the power delivered to the 25 Ω resistor we must calculate the currentthrough the 25 Ω resistor. Do this by first using KCL to recognize that thecurrent in each of the components is it, flowing in a clockwise direction. Writea KVL equation in the clockwise direction, starting below the voltage source,and using Ohm’s law to express the voltage drop across the resistors in thedirection of the current it flowing through the resistors:

−25 V + 100it + 25it = 0 so 125it = 25 so it =25125

= 0.2 A

Thus, the power delivered to the 25 Ω resistor is

p25 = (25)i2t = (25)(0.2)2 = 1 W.

Problems 2–7

AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0,it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot is astraight line, its slope can be used to calculate the value of resistance:

R =∆v

∆i=

25 − 00.25 − 0

=25

0.25= 100 Ω

A circuit model having the same v − i characteristic is a 0.25 A current sourcein parallel with a 100Ω resistor, as shown below:

[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:

Note that by writing a KVL equation around the right loop we see that thevoltage drop across both resistors is vt. Write a KCL equation at the top centernode, summing the currents leaving the node. Use Ohm’s law to specify thecurrents through the resistors in terms of the voltage drop across the resistorsand the value of the resistors.

−0.25 +vt

100+

vt

25= 0, so 5vt = 25, thus vt = 5 V

p25 =v2

t

25= 1 W.

AP 2.9 First note that we know the current through all elements in the circuit except the 6kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1; thecurrent in the three elements to the right of the 6 kΩ resistor is 30i1). To find thecurrent in the 6 kΩ resistor, write a KCL equation at the top node:

i1 + 30i1 = i6k = 31i1

We can then use Ohm’s law to find the voltages across each resistor in terms of i1.


The results are shown in the figure below:

[a] To find i1, write a KVL equation around the left-hand loop, summing voltages ina clockwise direction starting below the 5V source:

−5 V + 54,000i1 − 1 V + 186,000i1 = 0

Solving for i1

54,000i1 + 189,000i1 = 6 V so 240,000i1 = 6 V

Thus,

i1 =6

240,000= 25 µA

[b] Now that we have the value of i1, we can calculate the voltage for eachcomponent except the dependent source. Then we can write a KVL equationfor the right-hand loop to find the voltage v of the dependent source. Sum thevoltages in the clockwise direction, starting to the left of the dependent source:

+v − 54,000i1 + 8 V − 186,000i1 = 0

Thus,

v = 240,000i1 − 8 V = 240,000(25 × 10−6) − 8 V = 6 V − 8 V = −2 V

We now know the values of voltage and current for every circuit element. Let’sconstruct a power table:

Problems 2–9

Element Current Voltage Power Power

(µA) (V) Equation (µW)

5 V 25 5 p = −vi −125

54 kΩ 25 1.35 p = Ri2 33.75

1 V 25 1 p = −vi −25

6 kΩ 775 4.65 p = Ri2 3603.75

Dep. source 750 −2 p = −vi 1500

1.8 kΩ 750 1.35 p = Ri2 1012.5

8 V 750 8 p = −vi −6000

[c] The total power generated in the circuit is the sum of the negative power valuesin the power table:

−125 µW + −25 µW + −6000 µW = −6150 µW

Thus, the total power generated in the circuit is 6150 µW.

[d] The total power absorbed in the circuit is the sum of the positive power values inthe power table:

33.75 µW + 3603.75 µW + 1500µW + 1012.5 µW = 6150µW

Thus, the total power absorbed in the circuit is 6150 µW.

AP 2.10 Given that iφ = 2 A, we know the current in the dependent source is 2iφ = 4 A. Wecan write a KCL equation at the left node to find the current in the 10 Ω resistor.Summing the currents leaving the node,

−5 A + 2 A + 4 A + i10Ω = 0 so i10Ω = 5 A − 2 A − 4 A = −1 A

Thus, the current in the 10 Ω resistor is 1 A, flowing right to left, as seen in thecircuit below.


[a] To find vs, write a KVL equation, summing the voltages counter-clockwisearound the lower right loop. Start below the voltage source.

−vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0 so vs = 10 V + 60 V = 70 V

[b] The current in the voltage source can be found by writing a KCL equation at theright-hand node. Sum the currents leaving the node

−4 A + 1 A + iv = 0 so iv = 4 A − 1 A = 3 A

The current in the voltage source is 3 A, flowing top to bottom. The powerassociated with this source is

p = vi = (70 V)(3 A) = 210 W

Thus, 210 W are absorbed by the voltage source.

[c] The voltage drop across the independent current source can be found by writinga KVL equation around the left loop in a clockwise direction:

−v5A + (2 A)(30 Ω) = 0 so v5A = 60 V

The power associated with this source is

p = −v5Ai = −(60 V)(5 A) = −300 W

This source thus delivers 300 W of power to the circuit.

[d] The voltage across the controlled current source can be found by writing a KVLequation around the upper right loop in a clockwise direction:

+v4A + (10 Ω)(1 A) = 0 so v4A = −10 V

The power associated with this source is

p = v4Ai = (−10 V)(4 A) = −40 W

This source thus delivers 40 W of power to the circuit.

[e] The total power dissipated by the resistors is given by

(i30Ω)2(30 Ω) + (i10Ω)2(10 Ω) = (2)2(30 Ω) + (1)2(10 Ω) = 120 + 10 = 130 W

Problems 2–11

Problems

P 2.1

Vbb = no-load voltage of battery

Rbb = internal resistance of battery

Rx = resistance of wire between battery and switch

Ry = resistance of wire between switch and lamp A

Ra = resistance of lamp A

Rb = resistance of lamp B

Rw = resistance of wire between lamp A and lamp B

Rg1 = resistance of frame between battery and lamp A

Rg2 = resistance of frame between lamp A and lamp B

S = switch

P 2.2 Since we know the device is a resistor, we can use Ohm’s law to calculate theresistance. From Fig. P2.2(a),

v = Ri so R =v

i

Using the values in the table of Fig. P2.2(b),

R =−160−0.02

=−80

−0.01=

800.01

=1600.02

=2400.03

= 8kΩ

P 2.3 The resistor value is the ratio of the power to the square of the current:

50012 =

200022 =

450032 =

800042 =

12,50052 =

18,00062 = 500 Ω

P 2.4 Since we know the device is a resistor, we can use the power equation. From Fig.P2.4(a),

p = vi =v2

Rso R =

v2

p

Using the values in the table of Fig. P2.4(b)

R =(−8)2

3.2=

(−4)2

0.8=

(4)2

0.8=

(8)2

3.2=

(12)2

7.2=

(16)2

12.8= 20 Ω


P 2.5 [a] Yes, independent voltage sources can carry whatever current is required by theconnection; independent current source can support any voltage required bythe connection.

[b] 18 V source: absorbing

5 mA source: delivering

7 V source: absorbing

[c] P18V = (5 × 10−3)(18) = 90 mW (abs)

P5mA = −(5 × 10−3)(25) = −125 mW (del)

P7V = (5 × 10−3)(7) = 35 mW (abs)∑Pabs =

∑Pdel = 125 mW

[d] Yes; 18 V source is delivering, the 5 mA source is absorbing, and the 7 V sourceis absorbing

P18V = −(5 × 10−3)(18) = −90 mW (del)

P5mA = (5 × 10−3)(11) = 55 mW (abs)

P7V = (5 × 10−3)(7) = 35 mW (abs)∑Pabs =

∑Pdel = 90 mW

P 2.6

Write the two KCL equations, summing the currents leaving the node:

KCL, top node: 25A − 20A − 5A = 0A

KCL, bottom node: − 25A + 20A + 5A = 0A

Write the three KVL equations, summing the voltages in a clockwise direction:

KVL, left loop: − v25 + v20 = 0

KVL, right loop: 60V − 100V − v5 − v20 = 0

Problems 2–13

KVL, outer loop: 60V − 100V − v5 − v25 = 0

Note that since v5, v20, and v25 are not specified, we can choose values that satisfythe equations. For example, let v5 = −80V, v20 = 40V, and v25 = 40V. There aremany other voltage values that will satisfy the equations, too.Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. Wecan now calculate the power developed by the two voltage sources:

pv−sources = p60 + p100 = −(60)(5) + (100)(5) = 200 W.

Since the power is positive, the sources are absorbing 200 W of power, ordeveloping −200 W of power.

P 2.7


KCL, top node: − 30A − i8 + 10A = 0A

KCL, bottom node: 30A + i8 − 10A = 0A

Note that the value i8 = −20A satisfies these two equations.


KVL, left loop: − v30 − 16V + 8V = 0

KVL, right loop: − 10V + v10 − 8V = 0

KVL, outer loop: − 16V − 10V + v10 − v30 = 0

Note that v30 = −8V and v10 = 18V satisfy the three KVL equations.The interconnection is valid, since neither of Kirchhoff’s laws is violated. We usethe values of i8, v30 and v10 stated above to calculate the power associated with eachsource:

p30A = −(30)(−8) = 240 W p16V = −(30)(16) = −480 W


p8V = −(−20)(8) = 160 W p10V = −(10)(10) = −100 W

p10A = (10)(18) = 180 W∑

Pabs =∑

Pdel = 580 W

Power developed by the current sources:

pi−sources = p30A + p10A = 240 + 180 = 420 W

Since power is positive, the sources are absorbing 420 W of power, or developing−420 W of power.

P 2.8 The interconnect is valid since it does not violate Kirchhoff’s laws.

−10 + 40 + v5A − 50 = 0 so v5A = 20 V (KVL)

15 + 5 + i50V = 0 so i50V = −20 A (KCL)

p15A = −(15)(50) = −750 W p50V = (20)(50) = 1000 W

p5A = −(5)(20) = −100 W p10V = (5)(10) = 50 W

p40V = −(5)(40) = −200 W∑

Pdev =∑

Pabs = 1050 W

P 2.9 First there is no violation of Kirchhoff’s laws, hence the interconnection is valid.Kirchhoff’s voltage law requires

−20 + 60 + v1 − v2 = 0 so v1 − v2 = −40 V

The conservation of energy law requires

−(5 × 10−3)v2 − (15 × 10−3)v2 − (20 × 10−3)(20) + (20 × 10−3)(60) + (20 × 10−3)v1 = 0

or

v1 − v2 = −40 V

Hence any combination of v1 and v2 such that v1 − v2 = −40 V is a valid solution.

Problems 2–15

P 2.10

The interconnection is invalid because KCL is violated at the right-hand node.Summing the currents leaving,

−(−5A) − 3A + 8A = 10A = 0

Note that KCL is also violated at the left-hand node.

P 2.11


KCL, top node: 75A − 5v∆ − 25A = 0A

KCL, bottom node: − 75A + 5v∆ + 25A = 0A

To satisfy KCL, note that v∆ = 10 V.


KVL, left loop: − v75 − 50V + vdep − 20V = 0

KVL, right loop: 20V − vdep + v∆ = 0

KVL, outer loop: − v75 − 50V + v∆ = 0


Substitute the value v∆ = 10 V into the second KVL equation and find vdep = 30 V.Substitute the value v∆ = 10 V into the third equation and find v75 = −40 V. Thesevalues satisfy the first equation.Thus, the interconnection is valid because it does not violate Kirchhoff’s laws.Use the values for v∆, v75, and vdep above to calculate the total power developed inthe circuit:

p50V = (75)(50) = 3750 W p75A = (75)(−40) = −3000 W

p20V = [5(10)](20) = 1000 W pds = −(50)(30) = −1500 W

p25A = −(25)(10) = −250 W

∑Pdev = 3750 + 1000 = 4750 W =

∑Pabs

P 2.12 [a] Yes, Kirchhoff’s laws are not violated. (Note that i∆ = −8 A.)

[b] No, because the voltages across the independent and dependent current sourcesare indeterminate. For example, define v1, v2, and v3 as shown:

Kirchhoff’s voltage law requires

v1 + 20 = v3

v2 + 100 = v3

Conservation of energy requires

−8(20) − 8v1 − 16v2 − 16(100) + 24v3 = 0

or

v1 + 2v2 − 3v3 = −220

Now arbitrarily select a value of v3 and show the conservation of energy willbe satisfied. Examples:If v3 = 200 V then v1 = 180 V and v2 = 100 V. Then

180 + 200 − 600 = −220 (CHECKS)

If v3 = −100 V, then v1 = −120 V and v2 = −200 V. Then

−120 − 400 + 300 = −220 (CHECKS)

Problems 2–17

P 2.13 First, 10va = 5 V, so va = 0.5 VKVL for the outer loop: 5 − 20 + v9A = 0 so v9A = 15 VKVL for the right loop: 5 − 0.5 + vg = 0 so vg = −4.5 VKCL at the top node: 9 + 6 + ids = 9 so ids = −15 AThus,

p9A = −(9)(15) = −135 W p20V = (9)(20) = 180 W

pvg = −(6)(−4.5) = 27 W p6A = (6)(0.5) = 3 W

pds = −(15)(5) = −75 W

∑Pdev =

∑Pabs = 210 W

P 2.14

[a] Write a KVL equation clockwise aroud the right loop, starting below the 300 Ωresistor:

−va + vb = −0 so va = vb

Using Ohm’s law,

va = 300ia and vb = 75ib

Substituting,

300ia = 75ib so ib = 4ia

Write a KCL equation at the top middle node, summing the currents leaving:

−ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia

Write a KVL equation clockwise around the left loop, starting below thevoltage source:

−200 V + v40 + va = 0

From Ohm’s law,

v40 = 40ig and va = 300ia


Substituting,

−200 V + 40ig + 300ia = 0

Subsituting for ig:

−200 V + 40(5ia) + 300ia = −200 V + 200ia + 300ia = −200 V + 500ia = 0

Thus,

500ia = 200 V so ia =200 V500

= 0.4 A

[b] From part (a), ib = 4ia = 4(0.4 A) = 1.6 A.

[c] From the circuit, vo = 75 Ω(ib) = 75 Ω(1.6 A) = 120 V.

[d] Use the formula pR = Ri2R to calculate the power absorbed by each resistor:

p40Ω = i2g(40 Ω) = (5ia)2(40 Ω) = [5(0.4)]2(40 Ω) = (2)2(40 Ω) = 160 W

p300Ω = i2a(300 Ω) = (0.4)2(300 Ω) = 48 W

p75Ω = i2b(75 Ω) = (4ia)2(75 Ω) = [4(0.4)]2(75 Ω) = (1.6)2(75 Ω) = 192 W

[e] Using the passive sign convention,

psource = −(200 V)ig = −(200 V)(5ia) = −(200 V)[5(0.4 A)]

= −(200 V)(2 A) = −400 W

Thus the voltage source delivers 400 W of power to the circuit. Check:∑Pdis = 160 + 48 + 192 = 400 W

∑Pdel = 400 W

P 2.15 [a] vo = 8ia + 14ia + 18ia = 40(20) = 800 V

800 = 10io

io = 800/10 = 80 A

[b] ig = ia + io = 20 + 80 = 100 A

[c] pg(delivered) = (100)(800) = 80,000 W = 80 kW

Problems 2–19

P 2.16

[a] Write a KVL equation clockwise around the right loop:

−v60 + v30 + v90 = 0

From Ohm’s law,

v60 = (60 Ω)(4 A) = 240 V, v30 = 30io, v90 = 90io

Substituting,

−240 V + 30io + 90io = 0 so 120io = 240 V

Thus io =240 V120

= 2 A

Now write a KCL equatiohn at the top middle node, summing the currentsleaving:

−ig + 4 A + io = 0 so ig = 4 A + 2 A = 6 A

[b] Write a KVL equation clockwise around the left loop:

−vo + v60 = 0 so vo = v60 = 240 V

[c] Calculate power using p = vi for the source and p = Ri2 for the resistors:

psource = −voig = −(240 V)(6 A) = −1440 W

p60Ω = 42(60) = 960 W

p30Ω = 30i2o = (30)22 = 120 W

p90Ω = 90i2o = (90)22 = 360 W∑

Pdev = 1440 W∑

Pabs = 960 + 120 + 360 = 1440 W


P 2.17 [a]

v2 = 2(20) = 40 Vv8Ω = 80 − 40 = 40 Vi2 = 40 V/8 Ω = 5 Ai3 = io − i2 = 2 − 5 = −3 Av4Ω = (−3)(4) = −12 Vv1 = 4i3 + v2 = −12 + 40 = 28 Vi1 = 28 V/4 Ω = 7 A

[b] i4 = i1 + i3 = 7 − 3 = 4 A

p13Ω = 42(13) = 208 W

p8Ω = (5)2(8) = 200 W

p4Ω = 72(4) = 196 W

p4Ω = (−3)2(4) = 36 W

p20Ω = 22(20) = 80 W

[c]∑

Pdis = 208 + 200 + 196 + 36 + 80 = 720 W

ig = i4 + i2 = 4 + 5 = 9 A

Pdev = (9)(80) = 720 W

Problems 2–21

P 2.18 [a]

vo = 20(8) + 16(15) = 400 V

io = 400/80 = 5 A

ia = 25 A

P230 (supplied) = (230)(25) = 5750 W

ib = 5 + 15 = 20 A

P260 (supplied) = (260)(20) = 5200 W

[b]∑

Pdis = (25)2(2) + (20)2(8) + (5)2(4) + (15)216 + (20)22 + (5)2(80)

= 1250 + 3200 + 100 + 3600 + 800 + 2000 = 10,950 W∑Psup = 5750 + 5200 = 10,950 W

Therefore,∑

Pdis =∑

Psup = 10,950 W

P 2.19 [a]

v2 = 100 + 4(15) = 160 V; v1 = 160 − 30(2) = 100 V

i1 =v1

20=

10020

= 5 A; i3 = i1 − 2 = 5 − 2 = 3 A


vg = v1 + 30i3 = 100 + 30(3) = 190 V

vg − 5i4 = v2 so 5i4 = vg − v2 = 190 − 160 = 30 V

Thus i4 =305

= 6 A

ig = i3 + i4 = 3 + 6 = 9 A

[b] Calculate power using the formula p = Ri2:

p9 Ω = (9)(2)2 = 36 W; p11 Ω = (11)(2)2 = 44 W

p10 Ω = (10)(2)2 = 40 W; p30 Ω = (30)(3)2 = 270 W

p5 Ω = (5)(6)2 = 180 W; p4 Ω = (4)(5)2 = 100 W

p16 Ω = (16)(5)2 = 400 W; p15 Ω = (15)(4)2 = 240 W

[c] vg = 190 V

[d] Sum the power dissipated by the resistors:∑pdiss = 36 + 44 + 40 + 270 + 180 + 100 + 400 + 240 = 1310 W

The power associated with the sources is

pvolt−source = (100 V)(4 A) = 400 W

pcurr−source = −vgig = −(190 V)(9 A) = −1710 W

Thus the total power dissipated is 1310 + 400 = 1710 W and the total powerdeveloped is 1710 W, so the power balances.

P 2.20 [a] Plot the v − i characteristic

From the plot:

R =∆v

∆i=

(125 − 50)(15 − 0)

= 5 Ω

When it = 0, vt = 50 V; therefore the ideal current source has a current of 10 A

Problems 2–23

[b]

10 + it = i1 and 5i1 = −20it

Therefore, 10 + it = −4it so it = −2 A

P 2.21 [a] Plot the v—i characteristic:

From the plot:

R =∆v

∆i=

130 − (−30)8 − 0

= 20 Ω

When it = 0, vt = −30 V; therefore the ideal voltage source has a voltage of−30 V. Thus the device can be modeled as a −30 V source in series with a20 Ω resistor, as shown below:


[b] We attach a 40 Ω resistor to the device model developed in part (a):

Write a KVL equation clockwise around the circuit, using Ohm’s law toexpress the voltage drop across the resistors in terms of the current it throughthe resistors:

−(−30 V) − 20it − 40it = 0 so − 60it = −30 V

Thus it =−30 V−60

= +0.5 A

Now calculate the power dissipated by the resistor:

p40 Ω = 40i2t = (40)(0.5)2 = 10 W

P 2.22 [a]

[b] ∆v = 50 V; ∆i = 5 mA; R =∆v

∆i=

50 V5 mA

= 10 kΩ

Problems 2–25

[c]

10,000i1 = 2500is so i1 = 0.25is

0.02 = i1 + is = 0.25is + is = 1.25is

Thus, is =0.021.25

= 0.016 A = 16 mA

[d] Predicted open circuit voltage:

voc = vs = (0.02)(10,000) = 200 V

[e] From the table, the actual open circuit voltage is 140 V.

[f] This is a practical current source and is not a linear device.

P 2.23 [a] Begin by constructing a plot of voltage versus current:

[b] Since the plot is linear for 0 ≤ is ≤ 24 A amd since R = ∆v/∆i, we cancalculate R from the plotted values as follows:

R =∆v

∆i=

24 − 1824 − 0

=624

= 0.25 Ω

We can determine the value of the ideal voltage source by considering thevalue of vs when is = 0. When there is no current, there is no voltage dropacross the resistor, so all of the voltage drop at the output is due to the voltagesource. Thus the value of the voltage source must be 24 V. The model, validfor 0 ≤ is ≤ 24 A, is shown below:


[c] The circuit is shown below:

Write a KVL equation in the clockwise direction, starting below the voltagesource. Use Ohm’s law to express the voltage drop across the resistors in termsof the current i:

−24 V + 0.25i + 1i = 0 so 1.25i = 24 V

Thus, i =24 V

1.25 Ω= 19.2 A

[d] The circuit is shown below:

Write a KVL equation in the clockwise direction, starting below the voltagesource. Use Ohm’s law to express the voltage drop across the resistors in termsof the current i:

−24 V + 0.25i = 0 so 0.25i = 24 V

Thus, i =24 V

0.25 Ω= 96 A

[e] The short circuit current can be found in the table of values (or from the plot) asthe value of the current is when the voltage vs = 0. Thus,

isc = 48 A (from table)

[f] The plot of voltage versus current constructed in part (a) is not linear (it ispiecewise linear, but not linear for all values of is). Since the proposed circuitmodel is a linear model, it cannot be used to predict the nonlinear behaviorexhibited by the plotted data.

Problems 2–27

P 2.24

vab = 240 − 180 = 60 V; therefore, ie = 60/15 = 4 Aic = ie − 1 = 4 − 1 = 3 A; therefore, vbc = 10ic = 30 Vvcd = 180 − vbc = 180 − 30 = 150 V;

therefore, id = vcd/(12 + 18) = 150/30 = 5 Aib = id − ic = 5 − 3 = 2 Avac = vab + vbc = 60 + 30 = 90 VR = vac/ib = 90/2 = 45 Ω

CHECK: ig = ib + ie = 2 + 4 = 6 A

pdev = (240)(6) = 1440 W∑Pdis = 1(180) + 4(45) + 9(10) + 25(12) + 25(18) + 16(15) = 1440 W (CHECKS)

P 2.25 [a]

ib = 60 V/30 Ω = 2 Ava = (30 + 60)(2) = 180 V−500 + va + vb = 0 so vb = 500 − va = 500 − 180 = 320 Vie = vb/(60 + 36) = 320/96 = (10/3) Aid = ie − ib = (10/3) − 2 = (4/3) Avc = 30id + vb = 40 + 320 = 360 Vic = vc/180 = 360/180 = 2 Avd = 500 − vc = 500 − 360 = 140 Via = id + ic = 4/3 + 2 = (10/3) AR = vd/ia = 140/(10/3) = 42 Ω

[b] ig = ia + ib = (10/3) + 2 = (16/3) Apg (supplied) = (500)(16/3) = 2666.67 W


P 2.26 [a] Start with the 22.5 Ω resistor. Since the voltage drop across this resistor is 90 V,we can use Ohm’s law to calculate the current:

i22.5 Ω =90 V

22.5 Ω= 4 A

Next we can calculate the voltage drop across the 15 Ω resistor by writing aKVL equation around the outer loop of the circuit:

−240 V + 90 V + v15 Ω = 0 so v15 Ω = 240 − 90 = 150 V

Now that we know the voltage drop across the 15 Ω resistor, we can use Ohm’slaw to find the current in this resistor:

i15 Ω =150 V15 Ω

= 10 A

Write a KCL equation at the middle right node to find the current through the5 Ω resistor. Sum the currents entering:

4 A − 10 A + i5 Ω = 0 so i5 Ω = 10 A − 4 A = 6 A

Write a KVL equation clockwise around the upper right loop, starting belowthe 4 Ω resistor. Use Ohm’s law to express the voltage drop across the resistorsin terms of the current through the resistors:

−v4 Ω + 90 V + (5 Ω)(−6 A) = 0 so v4 Ω = 90 V − 30 V = 60 V

Using Ohm’s law we can find the current through the 4 Ω resistor:

i4 Ω =60 V4 Ω

= 15 A

Write a KCL equation at the middle node. Sum the currents entering:

15 A − 6 A − i20 Ω = 0 so i20 Ω = 15 A − 6 A = 9 A

Use Ohm’s law to calculate the voltage drop across the 20 Ω resistor:

v20 Ω = (20 Ω)(9 A) = 180 V

All of the voltages and currents calculated above are shown in the figure below:

Calculate the power dissipated by the resistors using the equation pR = Ri2R:

p4Ω = (4)(15)2 = 900 W p20Ω = (20)(9)2 = 1620 W

p5Ω = (5)(6)2 = 180 W p22.5Ω = (22.5)(4)2 = 360 W

p15Ω = (15)(10)2 = 1500 W

Problems 2–29

[b] We can calculate the current in the voltage source, ig by writing a KCL equationat the top middle node:

ig = 15 A + 4 A = 19 A

Now that we have both the voltage and the current for the source, we cancalculate the power supplied by the source:

pg = −240(19) = −4560 W thus pg (supplied) = 4560 W

[c]∑

Pdis = 900 + 1620 + 180 + 360 + 1500 = 4560 WTherefore,∑

Psupp =∑

Pdis

P 2.27 iE − iB − iC = 0

iC = βiB therefore iE = (1 + β)iB

i2 = −iB + i1

Vo + iERE − (i1 − iB)R2 = 0

−i1R1 + VCC − (i1 − iB)R2 = 0 or i1 =VCC + iBR2

R1 + R2

Vo + iERE + iBR2 − VCC + iBR2

R1 + R2R2 = 0

Now replace iE by (1 + β)iB and solve for iB. Thus

iB =[VCCR2/(R1 + R2)] − Vo

(1 + β)RE + R1R2/(R1 + R2)

P 2.28 [a] io = 0 because no current can exist in a single conductor connecting two partsof a circuit.

[b]

−200 + 8000ig + 12,000ig = 0 so ig = 200/20,000 = 10 mAv∆ = (12 × 103)(10 × 10−3) = 120 V5 × 10−3v∆ = 0.6 A9000i1 = 3000i2 so i2 = 3i10.6 + i1 + i2 = 0 so 0.6 + i1 + 3i1 = 0 thus i1 = −0.15 A


[c] i2 = 3i1 = −0.45 A

P 2.29 First note that we know the current through all elements in the circuit except the200 Ω resistor (the current in the three elements to the left of the 200 Ω resistor is iβ;the current in the three elements to the right of the 200 Ω resistor is 49iβ). To findthe current in the 200 Ω resistor, write a KCL equation at the top node:

iβ + 49iβ = i200 Ω = 50iβ

We can then use Ohm’s law to find the voltages across each resistor in terms of iβ .The results are shown in the figure below:

[a] To find iβ , write a KVL equation around the left-hand loop, summing voltagesin a clockwise direction starting below the 7.2V source:

−7.2 V + 55,000i1 + 0.7 V + 10,000iβ = 0

Solving for iβ

55,000iβ + 10,000iβ = 6.5 V so 65,000iβ = 6.5 V

Thus,

iβ =6.5

65,000= 100 µA

Now that we have the value of iβ , we can calculate the voltage for eachcomponent except the dependent source. Then we can write a KVL equationfor the right-hand loop to find the voltage vy of the dependent source. Sum thevoltages in the clockwise direction, starting to the left of the dependent source:

−vy − 24,500iβ + 9 V − 10,000iβ = 0

Thus,

vy = 9 V − 34,500iβ = 9 V − 34,500(100 × 10−6) = 9 V − 3.45 V = 5.55 V

Problems 2–31

[b] We now know the values of voltage and current for every circuit element. Let’sconstruct a power table:

Element Current Voltage Power Power

(µA) (V) Equation (µW)

7.2 V 100 7.2 p = −vi −720

55 kΩ 100 5.5 p = Ri2 550

0.7 V 100 0.7 p = vi 70

200 Ω 5000 1 p = Ri2 5000

Dep. source 4900 5.55 p = vi 27,195

500 Ω 4900 2.45 p = Ri2 12,005

9 V 4900 9 p = −vi −44,100

The total power generated in the circuit is the sum of the negative power valuesin the power table:

−720 µW + −44,100 µW = −44,820 µW

Thus, the total power generated in the circuit is 44,820 µW. The total powerabsorbed in the circuit is the sum of the positive power values in the powertable:

550 µW + 70 µW + 5000µW + 27,195 µW + 12,005 µW = 44,820 µW

Thus, the total power absorbed in the circuit is 44,820 µW and the power in thecircuit balances.

P 2.30 [a] 12 − 2iσ = 5i∆

5i∆ = 8iσ + 2iσ = 10iσ

Therefore, 12 − 2iσ = 10iσ, so iσ = 1 A

5i∆ = 10iσ = 10; so i∆ = 2 A

vo = 2iσ = 2 V

[b] ig = current out of the positive terminal of the 12 V sourcevd = voltage drop across the 8i∆ source

ig = i∆ + iσ + 8i∆ = 9i∆ + iσ = 19 A

vd = 2 + 8 = 10 V


∑Pgen = 12ig + 8i∆(8) = 12(19) + 8(2)(8) = 356 W∑Pdiss = 2iσig + 5i2∆ + 8iσ(iσ + 8i∆) + 2i2σ + 8i∆vd

= 2(1)(19) + 5(2)2 + 8(1)(17) + 2(1)2 + 8(2)(10)

= 356 W; Therefore,∑Pgen =

∑Pdiss = 356 W

P 2.31 40i2 +540

+510

= 0 so i2 = −15.625 mA

v1 = 80i2 = −1.25 V

25i1 +−1.25

20+ (−15.625 × 10−3) = 0 so i1 = 3.125 mA

vg = 60i1 + 260i1 = 320i1

Therefore, vg = 1 V

P 2.32VCCR2

R1 + R2=

(10)(60 × 103)100 × 103 = 6V

R1R2

R1 + R2=

(40 × 103)(60 × 103)100 × 103 = 24 kΩ

iB =6 − 0.6

24 × 103 + 50(120)=

5.4(24 + 6) × 103 = 0.18 mA

iC = βiB = (49)(0.18) = 8.82 mA

iE = iC + iB = 8.82 + 0.18 = 9 mA

v3d = (0.009)(120) = 1.08 V

vbd = Vo + v3d = 1.68 V

i2 =vbd

R2=

1.6860

× 10−3 = 28 µA

i1 = i2 + iB = 28 µA + 180 µA = 208 µA

vab = (40 × 103)(208 × 10−6) = 8.32V

iCC = iC + i1 = 8.82 mA + 208 µA = 9.028 mA

v13 + (8.82 × 10−3)(750) + 1.08 = 10

Thus, v13 = 2.305V

Problems 2–33

P 2.33 [a]

[b]

P 2.34 From the simplified circuit model, using Ohm’s law and KVL:

400i + 50i + 200i − 250 = 0 so i = 250/650 = 385 mA

This current is nearly enough to stop the heart, according to Table 2.1, so a warningsign should be posted at the 250 V source.

P 2.35

P 2.36 [a] p = i2R

parm =(250

650

)2

(400) = 59.17 W

pleg =(250

650

)2

(200) = 29.59 W

ptrunk =(250

650

)2

(50) = 7.40 W


[b]

(dT

dt

)arm

=2.39 × 10−4parm

4= 35.36 × 10−4 C/s

tarm =5

35.36× 104 = 1414.23 s or 23.57 min

(dT

dt

)leg

=2.39 × 10−4

10Pleg = 7.07 × 10−4 C/s

tleg =5 × 104

7.07= 7,071.13 s or 117.85 min

(dT

dt

)trunk

=2.39 × 10−4(7.4)

25= 0.71 × 10−4 C/s

ttrunk =5 × 104

0.71= 70,677.37 s or 1,177.96 min

[c] They are all much greater than a few minutes.

P 2.37 [a] Rarms = 400 + 400 = 800Ω

iletgo = 50 mA (minimum)

vmin = (800)(50) × 10−3 = 40 V

[b] No, 12/800 = 15 mA. Note this current is sufficient to give a perceptible shock.

P 2.38 Rspace = 1 MΩ

ispace = 3 mA

v = ispaceRspace = 3000 V.

3Simple Resistive Circuits

Assessment Problems

AP 3.1

Start from the right hand side of the circuit and make series and parallelcombinations of the resistors until one equivalent resistor remains. Begin bycombining the 6 Ω resistor and the 10 Ω resistor in series:

6 Ω + 10 Ω = 16 Ω

Now combine this 16 Ω resistor in parallel with the 64 Ω resistor:

16 Ω‖64 Ω =(16)(64)16 + 64

=102480

= 12.8 Ω

This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor:

12.8 Ω + 7.2 Ω = 20 Ω

Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor:

20 Ω‖30 Ω =(20)(30)20 + 30

=60050

= 12 Ω

Thus, the simplified circuit is as shown:

3–1

3–2 CHAPTER 3. Simple Resistive Circuits

[a] With the simplified circuit we can use Ohm’s law to find the voltage across boththe current source and the 12 Ω equivalent resistor:

v = (12 Ω)(5 A) = 60 V

[b] Now that we know the value of the voltage drop across the current source, wecan use the formula p = −vi to find the power associated with the source:

p = −(60 V)(5 A) = −300 W

Thus, the source delivers 300 W of power to the circuit.

[c] We now can return to the original circuit, shown in the first figure. In this circuit,v = 60 V, as calculated in part (a). This is also the voltage drop across the 30 Ωresistor, so we can use Ohm’s law to calculate the current through this resistor:

iA =60 V30 Ω

= 2 A

Now write a KCL equation at the upper left node to find the current iB:

−5 A + iA + iB = 0 so iB = 5 A − iA = 5 A − 2 A = 3 A

Next, write a KVL equation around the outer loop of the circuit, using Ohm’slaw to express the voltage drop across the resistors in terms of the currentthrough the resistors:

−v + 7.2iB + 6iC + 10iC = 0

So 16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V

Thus iC =38.416

= 2.4 A

Now that we have the current through the 10 Ω resistor we can use the formulap = Ri2 to find the power:

p10 Ω = (10)(2.4)2 = 57.6 W

AP 3.2

[a] We can use voltage division to calculate the voltage vo across the 75 kΩ resistor:

vo(no load) =75,000

75,000 + 25,000(200 V) = 150 V

Problems 3–3

[b] When we have a load resistance of 150 kΩ then the voltage vo is across theparallel combination of the 75 kΩ resistor and the 150 kΩ resistor. First,calculate the equivalent resistance of the parallel combination:

75 kΩ‖150 kΩ =(75,000)(150,000)75,000 + 150,000

= 50,000 Ω = 50 kΩ

Now use voltage division to find vo across this equivalent resistance:

vo =50,000

50,000 + 25,000(200 V) = 133.3 V

[c] If the load terminals are short-circuited, the 75 kΩ resistor is effectivelyremoved from the circuit, leaving only the voltage source and the 25 kΩresistor. We can calculate the current in the resistor using Ohm’s law:

i =200 V25 kΩ

= 8 mA

Now we can use the formula p = Ri2 to find the power dissipated in the 25 kΩresistor:

p25k = (25,000)(0.008)2 = 1.6 W

[d] The power dissipated in the 75 kΩ resistor will be maximum at no load since vo

is maximum. In part (a) we determined that the no-load voltage is 150 V, so becan use the formula p = v2/R to calculate the power:

p75k(max) =(150)2

75,000= 0.3 W

AP 3.3

[a] We will write a current division equation for the current throught the 80Ωresistor and use this equation to solve for R:

i80Ω =R

R + 40 Ω + 80 Ω(20 A) = 4 A so 20R = 4(R + 120)

Thus 16R = 480 and R =48016

= 30 Ω


[b] With R = 30 Ω we can calculate the current through R using current division,and then use this current to find the power dissipated by R, using the formulap = Ri2:

iR =40 + 80

40 + 80 + 30(20 A) = 16 A so pR = (30)(16)2 = 7680 W

[c] Write a KVL equation around the outer loop to solve for the voltage v, and thenuse the formula p = −vi to calculate the power delivered by the current source:

−v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0 so v = 1200 + 480 = 1680 V

Thus, psource = −(1680 V)(20 A) = −33,600 W

Thus, the current source generates 33,600 W of power.

AP 3.4

[a] First we need to determine the equivalent resistance to the right of the 40 Ω and70 Ω resistors:

Req = 20 Ω‖30 Ω‖(50 Ω + 10 Ω) so1

Req=

120 Ω

+1

30 Ω+

160 Ω

=1

10 Ω

Thus, Req = 10 Ω

Now we can use voltage division to find the voltage vo:

vo =40

40 + 10 + 70(60 V) = 20 V

[b] The current through the 40 Ω resistor can be found using Ohm’s law:

i40Ω =vo

40=

20 V40 Ω

= 0.5 A

This current flows from left to right through the 40 Ω resistor. To use currentdivision, we need to find the equivalent resistance of the two parallel branchescontaining the 20 Ω resistor and the 50 Ω and 10 Ω resistors:

20 Ω‖(50 Ω + 10 Ω) =(20)(60)20 + 60

= 15 Ω

Now we use current division to find the current in the 30 Ω branch:

i30Ω =15

15 + 30(0.5 A) = 0.16667 A = 166.67 mA

Problems 3–5

[c] We can find the power dissipated by the 50 Ω resistor if we can find the currentin this resistor. We can use current division to find this current from the currentin the 40 Ω resistor, but first we need to calculate the equivalent resistance ofthe 20 Ω branch and the 30 Ω branch:

20 Ω‖30 Ω =(20)(30)20 + 30

= 12 Ω

Current division gives:

i50Ω =12

12 + 50 + 10(0.5 A) = 0.08333 A

Thus, p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW

AP 3.5 [a]

We can find the current i using Ohm’s law:

i =1 V

100 Ω= 0.01 A = 10 mA

[b]

Rm = 50 Ω‖5.555 Ω = 5 Ω

We can use the meter resistance to find the current using Ohm’s law:

imeas =1 V

100 Ω + 5 Ω= 0.009524 = 9.524 mA

AP 3.6 [a]


Use voltage division to find the voltage v:

v =75,000

75,000 + 15,000(60 V) = 50 V

[b]

The meter resistance is a series combination of resistances:

Rm = 149,950 + 50 = 150,000 Ω

We can use voltage division to find v, but first we must calculate the equivalentresistance of the parallel combination of the 75 kΩ resistor and the voltmeter:

75,000 Ω‖150,000 Ω =(75,000)(150,000)75,000 + 150,000

= 50 kΩ

Thus, vmeas =50,000

50,000 + 15,000(60 V) = 46.15 V

AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite resistorsmust be equal. Therefore,

100Rx = (1000)(150) so Rx =(1000)(150)

100= 1500 Ω = 1.5 kΩ

[b] When the bridge is balanced, there is no current flowing through the meter, sothe meter acts like an open circuit. This places the following branches inparallel: The branch with the voltage source, the branch with the seriescombination R1 and R3 and the branch with the series combination of R2 andRx. We can find the current in the latter two branches using Ohm’s law:

iR1,R3 =5 V

100 Ω + 150 Ω= 20 mA; iR2,Rx =

5 V1000 + 1500

= 2 mA

We can calculate the power dissipated by each resistor using the formulap = Ri2:

p100Ω = (100 Ω)(0.02 A)2 = 40 mW

p150Ω = (150 Ω)(0.02 A)2 = 60 mW

p1000Ω = (1000 Ω)(0.002 A)2 = 4 mW

p1500Ω = (1500 Ω)(0.002 A)2 = 6 mW

Since none of the power dissipation values exceeds 250 mW, the bridge can beleft in the balanced state without exceeding the power-dissipating capacity ofthe resistors.

Problems 3–7

AP 3.8 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connectedresistors Ra, Rb, and Rc. To assist you the figure below has both the Y-connectedresistors and the ∆-connected resistors

Ra =(5)(10) + (5)(20) + (10)(20)

20= 17.5 Ω

Rb =(5)(10) + (5)(20) + (10)(20)

10= 35 Ω

Rc =(5)(10) + (5)(20) + (10)(20)

5= 70 Ω

The circuit with these new ∆-connected resistors is shown below:

From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor:

70 Ω‖28 Ω =(70)(28)70 + 28

= 20 Ω

Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor:

17.5 Ω‖105 Ω =(17.5)(105)17.5 + 105

= 15 Ω

Once the parallel combinations are made, we can see that the equivalent 20 Ωresistor is in series with the equivalent 15 Ω resistor, giving an equivalent resistance


of 20 Ω + 15 Ω = 35 Ω. Finally, this equivalent 35 Ω resistor is in parallel with theother 35 Ω resistor:

35 Ω‖35 Ω =(35)(35)35 + 35

= 17.5 Ω

Thus, the resistance seen by the 2 A source is 17.5 Ω, and the voltage can becalculated using Ohm’s law:

v = (17.5 Ω)(2 A) = 35 V

Problems 3–9

Problems

P 3.1 [a] The 6 Ω and 12 Ω resistors are in series, as are the 9 Ω and 7 Ω resistors. Thesimplified circuit is shown below:

[b] The 3 kΩ, 5 kΩ, and 7 kΩ resistors are in series. The simplified circuit is shownbelow:

[c] The 300 Ω, 400 Ω, and 500 Ω resistors are in series. The simplified circuit isshown below:

P 3.2 [a] The 10 Ω and 40 Ω resistors are in parallel, as are the 100 Ω and 25 Ω resistors.


The simplified circuit is shown below:

[b] The 9 kΩ, 18 kΩ, and 6 kΩ resistors are in parallel. The simplified circuit isshown below:

[c] The 600 Ω, 200 Ω, and 300 Ω resistors are in series. The simplified circuit isshown below:

P 3.3 [a] p4Ω = i2s4 = (12)24 = 576 W p18Ω = (4)218 = 288 W

p3Ω = (8)23 = 192 W p6Ω = (8)26 = 384 W

[b] p120V(delivered) = 120is = 120(12) = 1440 W

[c] pdiss = 576 + 288 + 192 + 384 = 1440 W

P 3.4 [a] From Ex. 3-1: i1 = 4 A, i2 = 8 A, is = 12 Aat node x: −12 + 4 + 8 = 0, at node y: 12 − 4 − 8 = 0

Problems 3–11

[b] v1 = 4is = 48 V v3 = 3i2 = 24 V

v2 = 18i1 = 72 V v4 = 6i2 = 48 Vloop abda: −120 + 48 + 72 = 0,loop bcdb: −72 + 24 + 48 = 0,loop abcda: −120 + 48 + 24 + 48 = 0

P 3.5 Always work from the side of the circuit furthest from the source. Remember thatthe current in all series-connected circuits is the same, and that the voltage dropacross all parallel-connected resistors is the same.

[a] Req = 6 + 12 + [4‖(9 + 7)] = 18 + (4‖16) = 18 + 3.2 = 21.2 Ω

[b] Req = 4 k + [10 k‖(3 k + 5 k + 7 k)] = 4 k + (10 k‖15 k) = 4 k + 6 k = 10 kΩ

[c] Req = (300 + 400 + 500) + (600‖1200) = 1200 + 400 = 1600 Ω

P 3.6 Always work from the side of the circuit furthest from the source. Remember thatthe current in all series-connected circuits is the same, and that the voltage dropacross all parallel-connected resistors is the same.

[a] Req = 18 + (100‖25‖(22 + (10‖40))) = 18 + (20‖(22 + 8) = 18 + 12 = 30 Ω

[b] Req = 10 k‖(5 k + 2 k + (9 k‖18 k‖6 k)) = 10 k‖(7 k + 3 k) = 10 k‖10 k = 5 kΩ

[c] Req = 600‖200‖300‖(250 + 150) = 600‖200‖300‖400 = 80 Ω

P 3.7 [a] Req = 12 + (24‖(30 + 18)) + 10 = 12 + (24‖48) + 10 = 12 + 16 + 10 = 38 Ω

[b] Req = 4 k‖30 k‖60 k‖(1.2 k + (7.2 k‖2.4 k) + 2 k) = 4 k‖30 k‖60 k‖(3.2 k + 1.8 k)

= 4 k‖30 k‖60 k‖5 k = 2 kΩ

P 3.8 [a] 5‖20 = 100/25 = 4 Ω 5‖20 + 9‖18 + 10 = 20 Ω

9‖18 = 162/27 = 6 Ω 20‖30 = 600/50 = 12 Ω

Rab = 5 + 12 + 3 = 20 Ω


[b] 5 + 15 = 20 Ω 30‖20 = 600/50 = 12 Ω

20‖60 = 1200/80 = 15 Ω 3‖6 = 18/9 = 2 Ω

15 + 10 = 25 Ω 3‖6 + 30‖20 = 2 + 12 = 14 Ω

25‖75 = 1875/100 = 18.75 Ω 26‖14 = 364/40 = 9.1 Ω

18.75 + 11.25 = 30 Ω Rab = 2.5 + 9.1 + 3.4 = 15 Ω

[c] 3 + 5 = 8 Ω 60‖40 = 2400/100 = 24 Ω

8‖12 = 96/20 = 4.8 Ω 24 + 6 = 30 Ω

4.8 + 5.2 = 10 Ω 30‖10 = 300/40 = 7.5 Ω

45 + 15 = 60 Ω Rab = 1.5 + 7.5 + 1.0 = 10 Ω

P 3.9 [a] For circuit (a)

Rab = 360‖(90 + 120‖(160 + 200)) = 360‖(90 + (120‖360)) = 360‖(90 + 90)

= 360‖180 = 120 Ω

For circuit (b)

1Re

=120

+115

+120

+14

+112

=3060

=12

Re = 2 Ω

Re + 16 = 18 Ω

18‖18 = 9 Ω

Rab = 10 + 8 + 9 = 27 Ω

For circuit (c)

15‖30 = 10 Ω

10 + 20 = 30 Ω

60‖30 = 20 Ω

20 + 10 = 30 Ω

30‖80‖(40 + 20) = 30‖80‖60 = 16 Ω

Rab = 16 + 24 + 10 = 50 Ω

Problems 3–13

[b] Pa = (0.032)(120) = 108 mW

Pb =1442

27= 768 W

Pc =0.082

50= 128 µ W

P 3.10 The equivalent resistance to the right of the 10 Ω resistor is

(6 + 5‖(8 + 12)) = 6 + 5‖20 = 6 + 4 = 10 Ω.

We can use current division to see that the current then splits equally between thetwo 10 Ω branches. Thus the current through the 6 Ω branch in the original circuit is5 A. This 5 A current splits between the branch with the 5 Ω resistor and the branchwith the 8 + 12 = 20 Ω resistor and we use current division to determine the currentin the 5 Ω resistor:

i5Ω =20

20 + 5(5) = 4 A

Thus the power in the 5 Ω resistor is

p5Ω = i25Ω(5) = 42(5) = 80 W

P 3.11 [a]

Req = 2 + 2 + (1/4 + 1/5 + 1/20)−1 = 6 Ω

ig = 120/6 = 20 A

v4Ω = 120 − (2 + 2)20 = 40 V

io = 40/4 = 10 A

i(15+5)Ω = 40/(15 + 5) = 2 A

vo = (5)(2) = 10 V

[b] i15Ω = 2 A; P15Ω = (2)2(15) = 60 W

[c] P120V = (120)(20) = 2.4 kW


P 3.12 [a] Req = R‖R =R2

2R=

R

2

[b] Req = R‖R‖R‖ · · · ‖R (n R’s)

= R‖ R

n − 1

=R2/(n − 1)

R + R/(n − 1)=

R2

nR=

R

n

[c] One solution:

400 =2000

nso n =

2000400

= 5

You can place 5 identical 2 kΩ resistors in parallel to get an equivalentresistance of 400 Ω.

[d] One solution:

12,500 =100,000

nso n =

100,00012,500

= 8

You can place 8 identical 100 kΩ resistors in parallel to get an equivalentresistance of 12.5 kΩ.

P 3.13 [a] We can calculate the no-load voltage using voltage division to determine thevoltage drop across the 500 Ω resistor:

vo =500

(2000 + 500)(75 V) = 15 V

[b] We can calculate the power if we know the current in each of the resistors.Under no-load conditions, the resistors are in series, so we can use Ohm’s lawto calculate the current they share:

i =75 V

2000 Ω + 500 Ω= 0.03 A = 30 mA

Now use the formula p = Ri2 to calculate the power dissipated by eachresistor:

PR1 = (2000)(0.03)2 = 1.8 W = 1800 mW

PR2 = (500)(0.03)2 = 0.45 W = 450 mW

[c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the no-loadvoltage requirement, first pick R1 to meet the 1 W specification

iR1 =75 − 15

R1, Therefore,

( 60R1

)2

R1 ≤ 1

Thus, R1 ≥ 602

1or R1 ≥ 3600 Ω

Problems 3–15

Now use the voltage specification:

R2

R2 + 3600(75) = 15

Thus, R2 = 900 Ω

R1 = 1600 Ω and R2 = 400 Ω are the smallest values of resistors that satisfythe 1 W specification.

P 3.14 Use voltage division to determine R2 from the no-load voltage specification:

6 V =R2

(R2 + 40)(18 V); so 18R2 = 6(R2 + 40)

Thus, 12R2 = 240 so R2 =24012

= 20 Ω

Now use voltage division again, this time to determine the value of Re, the parallelcombination of R2 and RL. We use the loaded voltage specification:

4 V =Re

(40 + Re)(18 V) so 18Re = 4(40 + Re)

Thus, 14Re = 160 so Re =16014

= 11.43 Ω

Now use the definition Re to calculate the value of RL given that R2 = 20 Ω:

Re =20RL

20 + RL

= 11.43 so 20RL = 11.43(RL + 20)

Therefore, 8.57RL = 228.6 and RL =226.88.57

= 26.67 Ω

P 3.15 [a] From the constraint on the no-load voltage,

R2

R1 + R2(40) = 8 so R1 = 4R2

From the constraint on the loaded voltage divider:

7.5 =

3600R2

3600 + R2

R1 +3600R2

3600 + R2

(40)

=

3600R2

3600 + R2

4R2 +3600R2

3600 + R2

(40)


=3600R2

4R2(3600 + R2) + 3600R2(40) =

144,000R2

4R22 + 18,000R2

So,144,000

4R2 + 18,000= 7.5 ·. . R2 = 300 Ω and R1 = 4R2 = 1200 Ω

[b] Power dissipated in R1 will be maximum when the voltage across R1 ismaximum. This will occur under load conditions.

vR1 = 40 − 7.5 = 32.5 V; PR1 =(32.5)2

1200= 880.2 mW

So specify a 1 W power rating for the resistor R1.The power dissipated in R2 will be maximum when the voltage drop across R2

is maximum. This occurs under no-load conditions with vo = 8 V.

PR2 =(8)2

300= 213.3 m W

So specify a 1/4 W power rating for the resistor R2.

P 3.16 Refer to the solution of Problem 3.15. The divider will reach its dissipation limitwhen the power dissipated in R1 equals 1 W

So (v2R1

/1200) = 1; vR1 = 34.641 V vo = 40 − 34.641 = 5.359 V

Therefore,Re

1200 + Re(40) = 5.359, and Re = 185.641 Ω

1200RL

1200 + RL= 185.641 ·. . RL = 219.62 Ω

P 3.17 [a]

120 kΩ + 30 kΩ = 150 kΩ

75 kΩ‖150 kΩ = 50 kΩ

vo1 =240

(25,000 + 50,000)(50,000) = 160 V

vo =120,000

(150,000)(vo1) = 128 V, vo = 128 V

Problems 3–17

[b]

i =240

100,000= 2.4 mA

75,000i = 180 V

vo =120,000150,000

(180) = 144 V; vo = 144 V

[c] It removes loading effect of second voltage divider on the first voltage divider.Observe that the open circuit voltage of the first divider is

v′o1 =

75,000(100,000)

(240) = 180 V

Now note this is the input voltage to the second voltage divider when thecurrent controlled voltage source is used.

P 3.18(24)2

R1 + R2 + R3= 36, Therefore, R1 + R2 + R3 = 16 Ω

(R1 + R2)24(R1 + R2 + R3)

= 12

Therefore, 2(R1 + R2) = R1 + R2 + R3

Thus, R1 + R2 = R3; 2R3 = 16; R3 = 8 Ω

R2(24)R1 + R2 + R3

= 6

4R2 = R1 + R2 + R3 so R2 = R3/2 = 4 Ω

R2 = 4 Ω; R1 = 16 − 8 − 4 = 4 Ω

P 3.19 Note – in the problem description, the first equation defines R1 not RL.

[a] At no load: vo = kvs =R2

R1 + R2vs.

At full load: vo = αvs =Re

R1 + Revs, where Re =

RoR2

Ro + R2

Therefore k =R2

R1 + R2and R1 =

(1 − k)k

R2

α =Re

R1 + Reand R1 =

(1 − α)α

Re


Thus(1 − α

α

) [R2Ro

Ro + R2

]=

(1 − k)k

R2

Solving for R2 yields R2 =(k − α)α(1 − k)

Ro

Also, R1 =(1 − k)

kR2

·. . R1 =(k − α)

αkRo

[b] R1 =(0.05

0.68

)Ro = 2.5 kΩ

R2 =(0.05

0.12

)Ro = 14.167 kΩ

[c]

Maximum dissipation in R2 occurs at no load, therefore,

PR2(max) =[(60)(0.85)]2

14,167= 183.6 mW

Maximum dissipation in R1 occurs at full load.

PR1(max) =[60 − 0.80(60)]2

2500= 57.60 mW

[d ]

PR1 =(60)2

2500= 1.44 W = 1440 mW

PR2 =(0)2

14,167= 0 W

P 3.20 [a] Let vo be the voltage across the parallel branches, positive at the upper terminal,then

ig = voG1 + voG2 + · · · + voGN = vo(G1 + G2 + · · · + GN)

It follows that vo =ig

(G1 + G2 + · · · + GN)

Problems 3–19

The current in the kth branch is ik = voGk; Thus,

ik =igGk

[G1 + G2 + · + GN ]

[b] io =120(0.00125)

[0.0025 + 0.0004167 + 0.00125 + 0.000625 + 0.0002083]= 30 mA

P 3.21 Begin by using the relationships among the branch currents to express all branchcurrents in terms of i4:

i1 = 2i2 = 2(10i3) = 20i4

i2 = 10i3 = 10i4

i3 = i4

Now use KCL at the top node to relate the branch currents to the current supplied bythe source.

i1 + i2 + i3 + i4 = 8 mA

Express the branch currents in terms of i4 and solve for i4:

8 mA = 20i4 + 10i4 + i4 + i4 = 32i4 so i4 =0.00832

= 0.00025 = 0.25 mA

Since the resistors are in parallel, the same voltage, 4 V appears across each of them.We know the current and the voltage for R4 so we can use Ohm’s law to calculate R4:

R4 =vg

i4=

4 V0.25 mA

= 16 kΩ

Calculate i3 from i4 and use Ohm’s law as above to find R3:

i3 = i4 = 0.25 mA ·. . R3 =vg

i3=

4 V0.25 mA

= 16 kΩ


i2 = 10i4 = 10(0.25 mA) = 2.5 mA ·. . R2 =vg

i2=

4 V2.5 mA

= 1.6 kΩ


i1 = 20i4 = 20(0.25 mA) = 5 mA ·. . R1 =vg

i1=

4 V5 mA

= 800 Ω

The resulting circuit is shown below:


P 3.22 [a]

Using voltage division,

v18Ω =18

18 + 30(40) = 15 V positive at the top

[b]

Using current division,

i30Ω =24

24 + 30 + 18(60 × 10−3) = 20 mA flowing from right to left

[c]

The 9 mA current in the 1.2 kΩ resistor is also the current in the 2 kΩ resistor.It then divides among the 4 kΩ, 30 kΩ, and 60 kΩ resistors.

4 kΩ‖60 kΩ = 3.75 kΩ


i30 kΩ =3.75 k

30 k + 3.75 k(9 × 10−3) = 1 m A, flowing bottom to top

[d]

Problems 3–21

The voltage drop across the 4 kΩ resistor is the same as the voltage drop acrossthe series combination of the 1.2 kΩ, the (7.2 k‖2.4 k)Ω combined resistor,and the 2 kΩ resistor. Note that

7.2 k‖2.4 k =(7200)(2400)

9600= 1.8 kΩ


vo =1800

1200 + 1800 + 2000(50) = 18 V positive at the top

P 3.23 [a]

First, note the following: 18‖9 = 6 Ω; 20‖5 = 4 Ω; and the voltage drop acrossthe 18 Ω resistor is the same as the voltage drop across the parallel combinationof the 18 Ω and 9 Ω resistors. Using voltage division,

vo =6

6 + 4 + 10(0.1 V) = 30 mV positive at the left

[b]

The equivalent resistance of the 5 Ω, 15 Ω, and 60 Ω resistors is

Re = (5 + 15)‖60 = 15 Ω

Using voltage division to find the voltage across the equivalent resistance,

vRe =15

15 + 10(10) = 6 V

Using voltage division again,

vo =15

5 + 15(6) = 4.5 V positive at the top


[c]

Find equivalent resistance on the right side

Rr = 5.2 +(12)(5 + 3)(12 + 3 + 5)

= 10 Ω

Find voltage bottom to top across Rr

(10)(3) = 30 V

Find the equivalent resistance on the left side

Rl = 6 +(40)(45 + 15)(40 + 45 + 15)

= 30 Ω

The current in the 6 Ω is

i6 Ω =3030

= 1 A left to right

Use current division to find io

io = (1)( 40

40 + 15 + 45

)= 0.4 A bottom to top

P 3.24 [a] v20k =20

20 + 5(45) = 36 V

v90k =90

90 + 60(45) = 27 V

vx = v20k − v90k = 36 − 27 = 9 V

[b] v20k =2025

(Vs) = 0.8Vs

v90k =90150

(Vs) = 0.6Vs

vx = 0.8Vs − 0.6Vs = 0.2Vs

P 3.25 150‖75 = 50 Ω

The equivalent resistance to the right of the 90 Ω resistor is

(50 + 40)‖(60 + 30) = 45 Ω

Problems 3–23

The voltage drop across this equivalent resistance is

4590 + 45

(3) = 1 V

Use voltage division to find v1, which is the voltage drop across the parallelcombination whose equivalent resistance is 50 Ω:

v1 =50

50 + 40(1) = 5/9 V

Use voltage division to find v2:

v2 =30

30 + 60(1) = 1/3 V

P 3.26 i300Ω =1000 + 200

1000 + 200 + 300 + 300(15 × 10−3) = 10 mA

v300Ω = (300)(10 × 10−3) = 3 V

i200Ω = i1 kΩ = 15 × 10−3 − i300Ω = 5 mA

v1k = (1000)(5 × 10−3) = 5 V

vo = 3 − 5 = −2 V

P 3.27 5 Ω‖20 Ω = 4 Ω; 4 Ω + 6 Ω = 10 Ω; 10‖40 = 8 Ω;

Therefore, ig =125

8 + 2= 12.5 A

i6Ω =(40)(12.5)

50= 10 A; io =

(5)(10)25

= 2 A

P 3.28 [a] Combine resistors in series and parallel to find the equivalent resistance seen bythe source. Use this equivalent resistance to find the current through thesource, and use current division to find io:

80 + 70 = 150 Ω 100‖150‖90 = 36 Ω 36 + 24 = 60 Ω

i24Ω =60 V60Ω

= 1 A

io =100‖90‖150

150(1) =

36150

= 0.24 A


[b] Use current division to find the current through the 90 Ω resistor from the sourcecurrent found in part (a), and use the calculated current to find the power in the90 Ω resistor:

i90Ω =100‖90‖150

90(1) =

3690

= 0.4 A

p90Ω = i290Ω(90) = (0.4)2(90) = 14.4 W

P 3.29 [a] v9Ω = (1)(9) = 9 V

i2Ω = 9/(2 + 1) = 3 A

i4Ω = 1 + 3 = 4 A;

v25Ω = (4)(4) + 9 = 25 V

i25Ω = 25/25 = 1 A;

i3Ω = i25Ω + i9Ω + i2Ω = 1 + 1 + 3 = 5 A;

v40Ω = v25Ω + v3Ω = 25 + (5)(3) = 40 V

i40Ω = 40/40 = 1 A

i5‖20Ω = i40Ω + i25Ω + i4Ω = 1 + 1 + 4 = 6 A

v5‖20Ω = (4)(6) = 24 V

v32Ω = v40Ω + v5‖20Ω = 40 + 24 = 64 V

i32Ω = 64/32 = 2 A;

i10Ω = i32Ω + i5‖20Ω = 2 + 6 = 8 A

vg = 10(8) + v32Ω = 80 + 64 = 144 V.

[b] P20Ω =(v5‖20Ω)2

20=

242

20= 28.8 W

P 3.30

40‖10 = 8 Ω 15‖60 = 12 Ω

Problems 3–25

i1 =(3)(40)(60)

= 2 A; vx = 8i1 = 16 V

vg = 20i1 = 40 V

v60 = vg − vx = 24 V

Pdevice =242

60+

162

10+

402

40= 75.2 W

P 3.31 [a] The model of the ammeter is an ideal ammeter in parallel with a resistor whoseresistance is given by

Rs =100 µV10 µA

= 10 Ω.

We can calculate the current through the real meter using current division:

im =(10/99)

10 + (10/99)(imeas) =

10990 + 10

(imeas) =1

100imeas

[b] Rs =100 µV10 µA

= 10 Ω.

im =(100/999,990)

10 + (100/999,990)(imeas) =

1100,000

(imeas)

[c] Yes

P 3.32 Measured value: 60‖20.1 = 15.056 Ω

ig =50

(15.056 + 10)= 1.9955 A; imeas = (1.9955)

6080.1

= 1.495 A

True value: 60‖20 = 15 Ω

ig =50

(15 + 10)=

5025

= 2.0 A; itrue = (2)(60

80

)= 1.5 A

% error =[1.495

1.5− 1

]× 100 = −0.3488%


P 3.33 Begin by using current division to find the actual value of the current io:

itrue =15

15 + 45(50 mA) = 12.5 mA

imeas =15

15 + 45 + 0.1(50 mA) = 12.48 mA

% error =[12.48

12.5− 1

]100 = −0.1664%

P 3.34 For all full-scale readings the total resistance is

RV + Rmovement =full-scale reading

10−3

We can calculate the resistance of the movement as follows:

Rmovement =20 mV1 mA

= 20 Ω

Therefore, RV = 1000 (full-scale reading) − 20

[a] RV = 1000(50) − 20 = 49, 980 Ω[b] RV = 1000(5) − 20 = 4980 Ω[c] RV = 1000(0.25) − 20 = 230 Ω[d] RV = 1000(0.025) − 20 = 5 Ω

P 3.35 [a] vmeas = (50 × 10−3)[15‖45‖(4980 + 20)] = 0.5612 V

[b] vtrue = (50 × 10−3)(15‖45) = 0.5625 V

% error =(0.5612

0.5625− 1

)× 100 = −0.23%

P 3.36

Original meter: Re =50 × 10−3

5= 0.01 Ω

Modified meter: Re =(0.02)(0.01)

0.03= 0.00667 Ω

·. . (Ifs)(0.00667) = 50 × 10−3

·. . Ifs = 7.5 A

Problems 3–27

P 3.37 At full scale the voltage across the shunt resistor will be 100 mV; therefore thepower dissipated will be

PA =(100 × 10−3)2

RA

Therefore RA ≥ (100 × 10−3)2

0.25= 40 mΩ

Otherwise the power dissipated in RA will exceed its power rating of 0.25 WWhen RA = 40 mΩ, the shunt current will be

iA =100 × 10−3

40 × 10−3 = 2.5 A

The measured current will be imeas = 2.5 + 0.001 = 2.501 A·. . Full-scale reading is for practical purposes is 2.5 A

P 3.38 The current in the shunt resistor at full-scale deflection is

iA = ifullscale − 20 × 10−6

The voltage across RA at full-scale deflection is always 10 mV, therefore

RA =10 × 10−3

ifullscale − 2 × 10−3 =10

1000ifs − 0.02

[a] RA =10

10,000 − 0.02= 1 mΩ

[b] RA =10

1000 − 0.02= 10 mΩ

[c] RA =10

100 − 0.02= 1 Ω

[d] RA =10

0.1 − 0.02= 125 Ω

P 3.39 [a]


10 × 103i1 + 50 × 103(i1 − iB) = 12

50 × 103(i1 − iB) = 0.4 + 30iB(0.3 × 103)

·. . 60i1 − 50iB = 12 × 10−3

50i1 − 59iB = 0.4 × 10−3

Calculator solution yields iB = 553.85 µA

[b] With the insertion of the ammeter the equations become

60i1 − 50iB = 12 × 10−3 (no change)

50 × 103(i1 − iB) = 2 × 103iB + 0.4 + 30iB(300)

50i1 − 61iB = 0.4 × 10−3

Calculator solution yields iB = 496.6 µA

[c] % error =( 496.6

553.85− 1

)100 = −10.34%

P 3.40 [a] vmeter = 100 V

[b] Rmeter = (100 Ω/V)(100 V) = 10 kΩ

10 k‖60 k = 8.57 kΩ

vmeter =8.57 k23.57 k

(100) = 36.36 V

[c] 10 k‖1 k = 6 kΩ

vmeter =666

(100) = 9.09 V

Problems 3–29

[d] vmeter a = 100 V

vmeter b + vmeter c = 45.45 V

No, because of the loading effect of the meter.

P 3.41 [a] Since the unknown voltage is greater than either voltmeter’s maximum reading,the only possible way to use the voltmeters would be to connect them in series.

[b]

Rm1 = (300)(1000) = 300 kΩ; Rm2 = (150)(800) = 120 kΩ

·. . Rm1 + Rm2 = 420 kΩ

i1 max =300300

× 10−3 = 1 mA; i2 max =150120

× 10−3 = 1.25 mA

·. . imax = 1 mA since meters are in series

vmax = 10−3(300 + 120)103 = 420 V

Thus the meters can be used to measure the voltage

[c] im =399

420 × 103 = 0.95 mA

vm1 = (0.95)(300) = 285 V vm2 = (0.95)(120) = 114 V

P 3.42 The current in the series-connected voltmeters is

im =288300

= 0.96 mA

v80 kΩ = (0.96)(80) = 76.8 V

Vpower supply = 288 + 115.2 + 76.8 = 480 V

P 3.43 Rmeter = Rm + Rmovement =750 V1.5 mA

= 500 kΩ

vmeas = (25 kΩ‖125 kΩ‖500 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V

vtrue = (25 kΩ‖125 kΩ)(30 mA) = (20.833 kΩ)(30 mA) = 625 V

% error =(600

625− 1

)100 = −4%


P 3.44 Note – the upper terminal of the voltmeter should be labeled 820 V, not 300 V.

[a] Rmeter = 360 kΩ + 200 kΩ‖50 kΩ = 400 kΩ

400‖600 = 240 kΩ

Vmeter =240300

(300) = 240 V

[b] What is the percent error in the measured voltage?

True value =600660

(300) = 272.73 V

% error =( 240

272.73− 1

)100 = −12%

P 3.45 [a] R1 =100 V2 mA

= 50 kΩ

R2 =10 V2 mA

= 5 kΩ

R3 =1 V

2 mA= 500 Ω

[b] Let ia = actual current in the movement

id = design current in the movement

Then % error =(

iaid

− 1)

100

For the 100 V scale:

ia =100

50,000 + 25=

10050,025

, id =100

50,000

iaid

=50,00050,025

= 0.9995 % error = (0.9995 − 1)100 = −0.05%

For the 10 V scale:iaid

=50005025

= 0.995 % error = (0.995 − 1.0)100 = −0.5%

For the 1 V scale:iaid

=500525

= 0.9524 % error = (0.9524 − 1.0)100 = −4.76%

P 3.46 [a] Rmovement = 50 Ω

R1 + Rmovement =30

1 × 10−3 = 30 kΩ ·. . R1 = 29,950 Ω

R2 + R1 + Rmovement =150

1 × 10−3 = 150 kΩ ·. . R2 = 120 kΩ

Problems 3–31

R3 + R2 + R1 + Rmovement =300

1 × 10−3 = 300 kΩ

·. . R3 = 150 kΩ


[b]

imove =288300

(1) = 0.96 mA

v1 = (0.96 m)(150 k) = 144 V

i1 =144

750 k= 0.192 mA

i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA

vmeas = vx = 144 + 150i2 = 316.8 V

[c] v1 = 150 V; i2 = 1 m + 0.20 m = 1.20 mA

i1 = 150/750,000 = 0.20 mA

·. . vmeas = vx = 150 + (150 k)(1.20 m) = 330 V

P 3.47 From the problem statement we have

50 =Vs(10)10 + Rs

(1) Vs in mV; Rs in MΩ

48.75 =Vs(6)6 + Rs

(2)

[a] From Eq (1) 10 + Rs = 0.2Vs

·. . Rs = 0.2Vs − 10

Substituting into Eq (2) yields

48.75 =6Vs

0.2Vs − 6or Vs = 52 mV

[b] From Eq (1)

50 =520

10 + Rs

or 50Rs = 20

So Rs = 400 kΩ

Problems 3–33

P 3.48 Since the bridge is balanced, we can remove the detector without disturbing thevoltages and currents in the circuit.

It follows that

i1 =ig(R2 + Rx)

R1 + R2 + R3 + Rx

=ig(R2 + Rx)∑

R

i2 =ig(R1 + R3)

R1 + R2 + R3 + Rx

=ig(R1 + R3)∑

R

v3 = R3i1 = vx = i2Rx

·. .R3ig(R2 + Rx)∑

R=

Rxig(R1 + R3)∑R

·. . R3(R2 + Rx) = Rx(R1 + R3)

From which Rx =R2R3

R1

P 3.49 [a]

The condition for a balanced bridge is that the product of the opposite resistorsmust be equal:

(200)(Rx) = (500)(800) so Rx =(500)(800)

200= 2000 Ω


[b] The source current is the sum of the two branch currents. Each branch currentcan be determined using Ohm’s law, since the resistors in each branch are inseries and the voltage drop across each branch is 6 V:

is =6 V

200 Ω + 800 Ω+

6 V500 Ω + 2000 Ω

= 8.4 mA

[c] We can use current division to find the current in each branch:

ileft =500 + 2000

500 + 2000 + 200 + 800(8.4 mA) = 6 mA

iright = 8.4 mA − 6 mA = 2.4 mA

Now we can use the formula p = Ri2 to find the power dissipated by eachresistor:

p200 = (200)(0.006)2 = 7.2 mW p800 = (800)(0.006)2 = 28.8 mW

p500 = (500)(0.0024)2 = 2.88 mW p2000 = (2000)(0.0024)2 = 11.52 mW

Thus, the 800 Ω resistor absorbs the most power; it absorbs 28.8 mW of power.

[d] From the analysis in part (c), the 500 Ω resistor absorbs the least power; itabsorbs 2.88 mW of power.

P 3.50 Redraw the circuit, replacing the detector branch with a short circuit.

6 kΩ‖30 kΩ = 5 kΩ

12 kΩ‖20 kΩ = 7.5 kΩ

ig =75

5000 + 7500= 6 mA

v1 = 6 mA(5000) = 30 V

v2 = 6 mA(7500) = 45 V

Problems 3–35

i1 =30 V

6000 Ω= 5 mA

i2 =45 V

12,000 Ω= 3.75 mA

id = i1 − i2 = 5 mA − 3.75 mA = 1.25 mA

P 3.51 Note the bridge structure is balanced, that is 15 × 5 = 25 × 3, hence there is nocurrent in the 5 kΩ resistor. It follows that the equivalent resistance of the circuit is

Req = 0.750 + 11.25 = 12 kΩ

The source current is 192/12,000 = 16 mA.The current down through the 3 kΩ resistor is

i3k = 163048

= 10 mA

·. . p3k = (10 × 10−3)2(3 × 103) = 300 mW

P 3.52 In order that all four decades (1, 10, 100, 1000) that are used to set R3 contribute tothe balance of the bridge, the ratio R2/R1 should be set to 0.001.

P 3.53 Begin by transforming the Y-connected resistors (10 Ω, 40 Ω, 50 Ω) to ∆-connectedresistors. Both the Y-connected and ∆-connected resistors are shown below to assistin using Eqs. 3.44 – 3.46:

Now use Eqs. 3.44 – 3.46 to calculate the values of the ∆-connected resistors:

R1 =(40)(10)

10 + 40 + 50= 4 Ω; R2 =

(50)(10)10 + 40 + 50

= 5 Ω; R3 =(40)(50)

10 + 40 + 50= 20 Ω

The transformed circuit is shown below:


The equivalent resistance seen by the 24 V source can be calculated by makingseries and parallel combinations of the resistors to the right of the 24 V source:

Req = (15 + 5)‖(4 + 1) + 20 = 20‖5 + 20 = 4 + 20 = 24 Ω

Therefore, the current i in the 24 V source is given by

i =24 V24 Ω

= 1 A

Use current division to calculate the currents i1 and i2. Note that the current i1 flowsin the branch containing the 15 Ω and 5 Ω series connected resistors, while thecurrent i2 flows in the parallel branch that contains the series connection of the 1 Ωand 4 Ω resistors:

i1 =1 + 4

1 + 4 + 15 + 5(i) =

525

(1 A) = 0.2 A, and i2 = 1 A − 0.2 A = 0.8 A

Now use KVL and Ohm’s law to calculate v1. Note that v1 is the sum of the voltagedrop across the 4 Ω resistor, 4i2, and the voltage drop across the 20 Ω resistor, 20i:

v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V

Finally, use KVL and Ohm’s law to calculate v2. Note that v2 is the sum of thevoltage drop across the 5 Ω resistor, 5i1, and the voltage drop across the 20 Ωresistor, 20i:

v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 V

P 3.54 [a] Calculate the values of the Y-connected resistors that are equivalent to the10 Ω, 40 Ω, and 50Ω ∆-connected resistors:

RX =(10)(50)

10 + 40 + 50= 5 Ω; RY =

(40)(50)10 + 40 + 50

= 20 Ω;

RZ =(10)(40)

10 + 40 + 50= 4 Ω

Replacing the R2—R3—R4 delta with its equivalent Y gives

Problems 3–37

Now calculate the equivalent resistance Rab by making series and parallelcombinations of the resistors:

Rab = 13 + 5 + [(4 + 8)‖(20 + 4)] + 7 = 33 Ω

[b] Calculate the values of the ∆-connected resistors that are equivalent to the8 Ω, 10 Ω, and 40 Ω Y-connected resistors:

RX =(10)(40) + (40)(8) + (8)(10)

8=

8008

= 100 Ω

RY =(10)(40) + (40)(8) + (8)(10)

10=

80010

= 80 Ω

RZ =(10)(40) + (40)(8) + (8)(10)

40=

80040

= 20 Ω

Replacing the R2, R4, R5 wye with its equivalent ∆ gives

Make series and parallel combinations of the resistors to find the equivalentresistance Rab:

100 Ω‖50 Ω = 33.33 Ω; 80 Ω‖4 Ω = 3.81 Ω

·. . 100‖50 + 80‖4 = 33.33 + 3.81 = 37.14 Ω

·. . 37.14‖20 =(37.14)(20)

57.14= 13 Ω

·. . Rab = 13 + 13 + 7 = 33 Ω


[c] Convert the delta connection R4—R5—R6 to its equivalent wye.Convert the wye connection R3—R4—R6 to its equivalent delta.

P 3.55 Replace the upper and lower deltas with the equivalent wyes:

R1U =(50)(10)

100= 5 Ω; R2U =

(50)(40)100

= 20 Ω; R3U =(40)(10)

100= 4 Ω

R1L =(60)(10)

100= 6 Ω; R2L =

(60)(30)100

= 18 Ω; R3L =(30)(10)

100= 3 Ω


Now make series and parallel combinations of the resistors:

(4 + 6)‖(20 + 32 + 20 + 18) = 10‖90 = 9 Ω

Rab = 33 + 5 + 9 + 3 + 40 = 90 Ω

P 3.56 18 + 2 = 20 Ω

20‖80 = 16 Ω

16 + 4 = 20 Ω

20‖30 = 12 Ω

12 + 8 = 20 Ω

20‖60 = 15 Ω

15 + 5 = 20 Ω

ig =240 V20 Ω

= 12 A

Problems 3–39

io =60

60 + 20(12 A) = 9 A

i30Ω =20

20 + 30(9 A) = 3.6 A

p30Ω = (30)(3.6)2 = 388.8 W

P 3.57 The top of the pyramid can be replaced by a resistor equal to

R1 =(18)(9)

27= 6 Ω

The lower left and right deltas can be replaced by wyes. Each resistance in the wyeequals 3 Ω. Thus our circuit can be reduced to

Now the 12 Ω in parallel with 6 Ω reduces to 4 Ω.

·. . Rab = 3 + 4 = 3 = 10 Ω

P 3.58 Note – the top resistor to the right of the 1.5 Ω resistor is 20 Ω.

[a] Convert the upper delta to a wye.

R1 =(50)(50)

200= 12.5 Ω

R2 =(50)(100)

200= 25 Ω

R3 =(50)(100)

200= 25 Ω

Convert the lower delta to a wye.

R4 =(60)(80)

200= 24 Ω

R5 =(60)(60)

200= 18 Ω


R6 =(60)(80)

200= 24 Ω

Now redraw the circuit using the wye equivalents.

Rab = 1.5 + 12.5 + (25 + 71 + 24)‖(25 + 31 + 24) + 18

= 1.5 + 12.5 + (120‖85) + 18 = 1.5 + 12.5 + 48 + 18 = 80 Ω

[b] When vab = 400 V

ig =40080

= 5 A

io =120

120 + 80(5) = 3 A

p31Ω = (31)(3)2 = 279 W

P 3.59 [a] After the 20 Ω—80 Ω—40 Ω wye is replaced by its equivalent delta, the circuitreduces to

Problems 3–41

Now the circuit can be reduced to

Req = 44 + 280‖92.5 = 113.53 Ω

ig = 5/113.53 = 44.04 mA

i = (280/372.5)(44) = 33.11 mA

v52.5Ω = (52.5)(33.11 m) = 1.74 V

io = 1.74/210 = 8.28 mA

[b] v40Ω = (40)(33.11 m) = 1.32 V

i1 = 1.32/56 = 23.65 mA

[c] Now that io and i1 are known return to the original circuit

i80Ω = 44.04 m − 23.65 m = 20.39 mA

i20Ω = 23.65 m − 8.28 m = 15.37 mA

i2 = i80Ω + i20Ω = 35.76 mA

[d] pdel = (5)(44.04 m) = 220.2 mW


P 3.60 [a] After the 30 Ω—60 Ω—10 Ω delta is replaced by its equivalent wye, the circuitreduces to

Use current division to calculate i1:

i1 =22 + 18

22 + 18 + 4 + 6(5 A) =

4050

(5 A) = 4 A

[b] Return to the original circuit and write a KVL equation around the upper leftloop:

(22 Ω)i22Ω + v − (4 Ω)(i1) = 0

so v = (4 Ω)(4 A) − (22 Ω)(5 A − 4 A) = −6 V

[c] Write a KCL equation at the lower center node of the original circuit:

i2 = i1 +v

60= 4 +

−660

= 3.9 A

[d] Write a KVL equation around the bottom loop of the original circuit:

−v5A + (4 Ω)(4 A) + (10 Ω)(3.9 A) + (1 Ω)(5 A) = 0

So, v5A = (4)(4) + (10)(3.9) + (1)(5) = 60 V

Thus, p5A = (5 A)(60 V) = 300 W

P 3.61 Subtracting Eq. 3.42 from Eq. 3.43 gives

R1 − R2 = (RcRb − RcRa)/(Ra + Rb + Rc).

Adding this expression to Eq. 3.41 and solving for R1 gives

R1 = RcRb/(Ra + Rb + Rc).

To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. To findR3, subtract Eq. 3.41 from Eq. 3.42 and add this result to Eq. 3.43. Using the hint,Eq. 3.43 becomes

R1 + R3 =Rb[(R2/R3)Rb + (R2/R1)Rb]

(R2/R1)Rb + Rb + (R2/R3)Rb=

Rb(R1 + R3)R2

(R1R2 + R2R3 + R3R1)

Problems 3–43

Solving for Rb gives Rb = (R1R2 + R2R3 + R3R1)/R2. To find Ra: First useEqs. 3.44–3.46 to obtain the ratios (R1/R3) = (Rc/Ra) or Rc = (R1/R3)Ra and(R1/R2) = (Rb/Ra) or Rb = (R1/R2)Ra. Now use these relationships to eliminateRb and Rc from Eq. 3.42. To find Rc, use Eqs. 3.44–3.46 to obtain the ratiosRb = (R3/R2)Rc and Ra = (R3/R1)Rc. Now use the relationships to eliminate Rb

and Ra from Eq. 3.41.

P 3.62 Ga =1Ra

=R1

R1R2 + R2R3 + R3R1

=1/G1

(1/G1)(1/G2) + (1/G2)(1/G3) + (1/G3)(1/G1)

=(1/G1)(G1G2G3)G1 + G2 + G3

=G2G3

G1 + G2 + G3Similar manipulations generate the expressions for Gb and Gc.

P 3.63 [a] Rab = 2R1 +R2(2R1 + RL)2R1 + R2 + RL

= RL

Therefore 2R1 − RL +R2(2R1 + RL)2R1 + R2 + RL

= 0

Thus R2L = 4R2

1 + 4R1R2 = 4R1(R1 + R2)

When Rab = RL, the current into terminal a of the attenuator will be vi/RL

Using current division, the current in the RL branch will be

vi

RL· R2

2R1 + R2 + RL

Therefore vo =vi

RL· R2

2R1 + R2 + RLRL

andvo

vi

=R2

2R1 + R2 + RL

[b] (600)2 = 4(R1 + R2)R1

9 × 104 = R21 + R1R2

vo

vi

= 0.6 =R2

2R1 + R2 + 600

·. . 1.2R1 + 0.6R2 + 360 = R2

0.4R2 = 1.2R1 + 360

R2 = 3R1 + 900

·. . 9 × 104 = R21 + R1(3R1 + 900) = 4R2

1 + 900R1

·. . R21 + 225R1 − 22,500 = 0


R1 = −112.5 ±√

(112.5)2 + 22,500 = −112.5 ± 187.5

·. . R1 = 75 Ω

·. . R2 = 3(75) + 900 = 1125 Ω

P 3.64 [a] After making the Y-to-∆ transformation, the circuit reduces to

Combining the parallel resistors reduces the circuit to

Now note: 0.75R +3RRL

3R + RL=

2.25R2L + 3.75RRL

3R + RL

Therefore Rab =3R

(2.25R2

L + 3.75RRL

3R + RL

)

3R +(

2.25R2L + 3.75RRL

3R + RL

) =3R(3R + 5RL)

15R + 9RL

When Rab = RL, we have 15RRL + 9R2L = 9R2 + 15RRL

Therefore R2L = R2 or RL = R

Problems 3–45

[b] When R = RL, the circuit reduces to

io =ii(3RL)4.5RL

=1

1.5ii =

11.5

vi

RL, vo = 0.75RLio =

12vi,

Thereforevo

vi

= 0.5

P 3.65 [a] 3.5(3R − RL) = 3R + RL

10.5R − 1050 = 3R + 300

7.5R = 1350, R = 180 Ω

R2 =2(180)(300)2

3(180)2 − (300)2 = 4500 Ω

[b]

vo =vi

3.5=

423.5

= 12 V

io =12300

= 40 mA

i1 =42 − 124500

=30

4500= 6.67 mA

ig =42300

= 140 mA

i2 = 140 m − 6.67 m = 133.33 mA

i3 = 40 m − 6.67 m = 33.33 mA

i4 = 133.33 m − 33.33 m = 100 mA


p4500 top = (6.67 × 10−3)2(4500) = 0.2 W

p180 left = (133.33 × 10−3)2(180) = 3.2 W

p180 right = (33.33 × 10−3)2(180) = 0.2 W

p180 vertical = (100 × 10−3)2(180) = 1.8 W

p300 load = (40 × 10−3)2(300) = 0.48 W

The 180 Ω resistor carrying i2 dissipates the most power.

[c] p180 left = 3.2 W

[d] Two resistors dissipate minimum power – the 4500 Ω and the 180 Ω carrying i3.

[e] Both resistors dissipate 0.2 W or 200 mW.

P 3.66 [a]

va =vinR4

Ro + R4 + ∆R

vb =R3

R2 + R3vin

vo = va − vb =R4vin

Ro + R4 + ∆R− R3

R2 + R3vin

When the bridge is balanced,

R4

Ro + R4vin =

R3

R2 + R3vin

·. .R4

Ro + R4=

R3

R2 + R3

Thus, vo =R4vin

Ro + R4 + ∆R− R4vin

Ro + R4

= R4vin

[ 1Ro + R4 + ∆R

− 1Ro + R4

]

=R4vin(−∆R)

(Ro + R4 + ∆R)(Ro + R4)

≈ −(∆R)R4vin

(Ro + R4)2

Problems 3–47

[b] ∆R = 0.03Ro

Ro =R2R4

R3=

(1000)(5000)500

= 10,000 Ω

∆R = (0.03)(104) = 300 Ω

·. . vo ≈ −300(5000)(6)(15,000)2 = −40 mV

[c] vo =−(∆R)R4vin

(Ro + R4 + ∆R)(Ro + R4)

=−300(5000)(6)

(15,300)(15,000)

= −39.2157 mV

P 3.67 [a] approx value =−(∆R)R4vin

(Ro + R4)2

true value =−(∆R)R4vin

(Ro + R4 + ∆R)(Ro + R4)

·. .approx value

true value=

(Ro + R4 + ∆R)(Ro + R4)

·. . % error =[Ro + R4 + ∆R

Ro + R4− 1

]× 100 =

∆R

Ro + R4× 100

But Ro =R2R4

R3

·. . % error =R3∆R

R4(R2 + R3)

[b] % error =(500)(300)

(5000)(1500)× 100 = 2%

P 3.68∆R(R3)(100)(R2 + R3)R4

= 0.5

∆R(500)(100)(1500)(5000)

= 0.5

·. . ∆R = 75 Ω

% change =75

10,000× 100 = 0.75%


P 3.69 [a] From Eq 3.64 we have(i1i2

)2

=R2

2

R21(1 + 2σ)2

Substituting into Eq 3.63 yields

R2 =R2

2

R21(1 + 2σ)2R1

Solving for R2 yields

R2 = (1 + 2σ)2R1

[b] From Eq 3.63 we have

i1ib

=R2

R1 + R2 + 2Ra

But R2 = (1 + 2σ)2R1 and Ra = σR1 therefore

i1lb

=(1 + 2σ)2R1

R1 + (1 + 2σ)2R1 + 2σR1=

(1 + 2σ)2

(1 + 2σ) + (1 + 2σ)2

=1 + 2σ

2(1 + σ)

It follows that(i1ib

)2

=(1 + 2σ)2

4(1 + σ)2

Substituting into Eq 3.66 gives

Rb =(1 + 2σ)2Ra

4(1 + σ)2 =(1 + 2σ)2σR1

4(1 + σ)2

P 3.70 From Eq 3.69

i1i3

=R2R3

D

But D = (R1 + 2Ra)(R2 + 2Rb) + 2RbR2

where Ra = σR1; R2 = (1 + 2σ)2R1 and Rb =(1 + 2σ)2σR1

4(1 + σ)2

Therefore D can be written as

Problems 3–49

D = (R1 + 2σR1)[(1 + 2σ)2R1 +

2(1 + 2σ)2σR1

4(1 + σ)2

]+

2(1 + 2σ)2R1

[(1 + 2σ)2σR1

4(1 + σ)2

]

= (1 + 2σ)3R21

[1 +

σ

2(1 + σ)2 +(1 + 2σ)σ2(1 + σ)2

]

=(1 + 2σ)3R2

1

2(1 + σ)2 2(1 + σ)2 + σ + (1 + 2σ)σ

=(1 + 2σ)3R2

1

(1 + σ)2 1 + 3σ + 2σ2

D =(1 + 2σ)4R2

1

(1 + σ)

·. .i1i3

=R2R3(1 + σ)(1 + 2σ)4R2

1

=(1 + 2σ)2R1R3(1 + σ)

(1 + 2σ)4R21

=(1 + σ)R3

(1 + 2σ)2R1When this result is substituted into Eq 3.69 we get

R3 =(1 + σ)2R2

3R1

(1 + 2σ)4R21

Solving for R3 gives

R3 =(1 + 2σ)4R1

(1 + σ)2

P 3.71 From the dimensional specifications, calculate σ and R3:

σ =y

x=

0.0251

= 0.025; R3 =V 2

dc

p=

122

120= 1.2 Ω

Calculate R1 from R3 and σ:

R1 =(1 + σ)2

(1 + 2σ)4R3 = 1.0372 Ω

Calculate Ra, Rb, and R2:

Ra = σR1 = 0.0259 Ω Rb =(1 + 2σ)2σR1

4(1 + σ)2 = 0.0068 Ω


R2 = (1 + 2σ)2R1 = 1.1435 Ω

Using symmetry,

R4 = R2 = 1.1435 Ω R5 = R1 = 1.0372 Ω

Rc = Rb = 0.0068 Ω Rd = Ra = 0.0259 Ω

Test the calculations by checking the power dissipated, which should be 120 W/m.Calculate D, then use Eqs. (3.58)-(3.60) to calculate ib, i1, and i2:

D = (R1 + 2Ra)(R2 + 2Rb) + 2R2Rb = 1.2758

ib =Vdc(R1 + R2 + 2Ra)

D= 21 A

i1 =VdcR2

D= 10.7561 A i2 =

Vdc(R1 + 2Ra)D

= 10.2439 A

It follows that i2bRb = 3 W and the power dissipation per meter is 3/0.025 = 120W/m. The value of i21R1 = 120 W/m. The value of i22R2 = 120 W/m. Finally,i21Ra = 3 W/m.

P 3.72 From the solution to Problem 3.71 we have ib = 21 A and i3 = 10 A. By symmetryic = 21 A thus the total current supplied by the 12 V source is 21 + 21 + 10 or 52 A.Therefore the total power delivered by the source is p12 V (del) = (12)(52) = 624 W.We also have from the solution that pa = pb = pc = pd = 3 W. Therefore the totalpower delivered to the vertical resistors is pV = (8)(3) = 24 W. The total powerdelivered to the five horizontal resistors is pH = 5(120) = 600 W.

·. .∑

pdiss = pH + pV = 624 W =∑

pdel

P 3.73 [a] σ = 0.03/1.5 = 0.02Since the power dissipation is 150 W/m the power dissipated in R3 must be200(1.5) or 300 W. Therefore

R3 =122

300= 0.48 Ω

From Table 3.1 we have

R1 =(1 + σ)2R3

(1 + 2σ)4 = 0.4269 Ω

Ra = σR1 = 0.0085 Ω

R2 = (1 + 2σ)2R1 = 0.4617 Ω

Problems 3–51

Rb =(1 + 2σ)2σR1

4(1 + σ)2 = 0.0022 Ω

Therefore

R4 = R2 = 0.4617 Ω R5 = R1 = 0.4269 Ω

Rc = Rb = 0.0022 Ω Rd = Ra = 0.0085 Ω

[b] D = [0.4269 + 2(0.0085)][0.4617 + 2(0.0022)] + 2(0.4617)(0.0022) = 0.2090

i1 =VdcR2

D= 26.51 A

i21R1 = 300 W or 200 W/m

i2 =R1 + 2Ra

DVdc = 25.49 A

i22R2 = 300 W or 200 W/m

i21Ra = 6 W or 200 W/m

ib =R1 + R2 + 2Ra

DVdc = 52 A

i2bRb = 6 W or 200 W/m

isource = 52 + 52 +12

0.48= 129 A

pdel = 12(129) = 1548 W

pH = 5(300) = 1500 W

pV = 8(6) = 48 W∑

pdel =∑

pdiss = 1548 W

4Techniques of Circuit Analysis

Assessment Problems

AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:

The two node voltage equations are

−15 +v1

60+

v1

15+

v1 − v2

5= 0

5 +v2

2+

v2 − v1

5= 0

Place these equations in standard form:

v1

( 160

+115

+15

)+ v2

(−1

5

)= 15

v1

(−1

5

)+ v2

(12

+15

)= −5

Solving, v1 = 60 V and v2 = 10 V;Therefore, i1 = (v1 − v2)/5 = 10 A

[b] p15A = −(15 A)v1 = −(15 A)(60 V) = −900 W = 900 W(delivered)

[c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W= −50 W(delivered)

4–1

4–2 CHAPTER 4. Techniques of Circuit Analysis

AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:

The two node voltage equations are:

−4.5 +v1

1+

v1 − v2

6 + 2= 0

v2

12+

v2 − v1

6 + 2+

v2 − 304

= 0


v1

(1 +

18

)+ v2

(−1

8

)= 4.5

v1

(−1

8

)+ v2

( 112

+18

+14

)= 7.5

Solving, v1 = 6 V v2 = 18 VTo find the voltage v, first find the current i through the series-connected 6 Ω and 2 Ωresistors:

i =v1 − v2

6 + 2=

6 − 188

= −1.5 A

Using a KVL equation, calculate v:

v = 2i + v2 = 2(−1.5) + 18 = 15 V

AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as shown:

The node voltage equations are:v1 − 50

6+

v1

8+

v1 − v2

2− 3i1 = 0

−5 +v2

4+

v2 − v1

2+ 3i1 = 0

Problems 4–3

The dependent source requires the following constraint equation:

i1 =50 − v1

6Place these equations in standard form:

v1

(16

+18

+12

)+ v2

(−1

2

)+ i1(−3) =

506

v1

(−1

2

)+ v2

(14

+12

)+ i1(3) = 5

v1

(16

)+ v2(0) + i1(1) =

506

Solving, v1 = 32 V; v2 = 16 V; i1 = 3 AUsing these values to calculate the power associated with each source:

p50V = −50i1 = −150 W

p5A = −5(v2) = −80 W

p3i1 = 3i1(v2 − v1) = −144 W

[b] All three sources are delivering power to the circuit because the powercomputed in (a) for each of the sources is negative.

AP 4.4 Redraw the circuit and label the reference node and the node at which the nodevoltage equation will be written:

The node voltage equation is

vo

40+

vo − 1010

+vo + 20i∆

20= 0

The constraint equation required by the dependent source is

i∆ = i10 Ω + i30 Ω =10 − vo

10+

10 + 20i∆30



vo

( 140

+110

+120

)+ i∆(1) = 1

vo

( 110

)+ i∆

(1 − 20

30

)= 1 +

1030

Solving, vo = 24 V i∆ = −3.2 A

AP 4.5 Redraw the circuit identifying the three node voltages and the reference node:

Note that the dependent voltage source and the node voltages v and v2 form asupernode. The v1 node voltage equation is

v1

7.5+

v1 − v

2.5− 4.8 = 0

The supernode equation is

v − v1

2.5+

v

10+

v2

2.5+

v2 − 121

= 0

The constraint equation due to the dependent source is

ix =v1

7.5

The constraint equation due to the supernode is

v + ix = v2

Place this set of equations in standard form:

v1

( 17.5

+1

2.5

)+ v

(− 1

2.5

)+ v2(0) + ix(0) = 4.8

v1

(− 1

2.5

)+ v

( 12.5

+110

)+ v2

( 12.5

+ 1)

+ ix(0) = 12

v1

(− 1

7.5

)+ v(0) + v2(0) + ix(1) = 0

v1(0) + v(1) + v2(−1) + ix(1) = 0

Solving this set of equations for v gives v = 8 V

v1 = 15 V, v2 = 10 V, ix = 2 A

Problems 4–5

AP 4.6 Redraw the circuit identifying the reference node and the two unknown nodevoltages. Note that the right-most node voltage is the sum of the 60 V source and thedependent source voltage.

The node voltage equation at v1 is

v1 − 602

+v1

24+

v1 − (60 + 6iφ)3

= 0


iφ =60 + 6iφ − v1

3

Place these two equations in standard form:

v1

(12

+124

+13

)+ iφ(−2) = 30 + 20

v1

(13

)+ iφ(1 − 2) = 20

Solving, v1 = 48 V iφ = −4 A

AP 4.7 [a] Redraw the circuit identifying the three mesh currents:

The mesh current equations are:

−80 + 5(i1 − i2) + 26(i1 − i3) = 0

30i2 + 90(i2 − i3) + 5(i2 − i1) = 0

8i3 + 26(i3 − i1) + 90(i3 − i2) = 0



31i1 − 5i2 − 26i3 = 80

−5i1 + 125i2 − 90i3 = 0

−26i1 − 90i2 + 124i3 = 0Solving,

i1 = 5 A; i2 = 2 A; i3 = 2.5 A

p80V = −(80)i1 = −(80)(5) = −400 W

Therefore the 80 V source is delivering 400 W to the circuit.

[b] p8Ω = (8)i23 = 8(2.5)2 = 50 W, so the 8 Ω resistor dissipates 50 W.

AP 4.8 [a] b = 8, n = 6, b − n + 1 = 3

[b] Redraw the circuit identifying the three mesh currents:

The three mesh-current equations are

−25 + 2(i1 − i2) + 5(i1 − i3) + 10 = 0

−(−3vφ) + 14i2 + 3(i2 − i3) + 2(i2 − i1) = 0

1i3 − 10 + 5(i3 − i1) + 3(i3 − i2) = 0The dependent source constraint equation is

vφ = 3(i3 − i2)

Place these four equations in standard form:

7i1 − 2i2 − 5i3 + 0vφ = 15

−2i1 + 19i2 − 3i3 + 3vφ = 0

−5i1 − 3i2 + 9i3 + 0vφ = 10

0i1 + 3i2 − 3i3 + 1vφ = 0Solving

i1 = 4 A; i2 = −1 A; i3 = 3 A; vφ = 12 V

Problems 4–7

pds = −(−3vφ)i2 = 3(12)(−1) = −36 W

Thus, the dependent source is delivering 36 W, or absorbing −36 W.

AP 4.9 Redraw the circuit identifying the three mesh currents:


−25 + 6(ia − ib) + 8(ia − ic) = 0

2ib + 8(ib − ic) + 6(ib − ia) = 0

5iφ + 8(ic − ia) + 8(ic − ib) = 0

The dependent source constraint equation is iφ = ia. We can substitute this simpleexpression for iφ into the third mesh equation and place the equations in standardform:

14ia − 6ib − 8ic = 25

−6ia + 16ib − 8ic = 0

−3ia − 8ib + 16ic = 0

Solving,

ia = 4 A; ib = 2.5 A; ic = 2 A

Thus,

vo = 8(ia − ic) = 8(4 − 2) = 16 V

AP 4.10 Redraw the circuit identifying the mesh currents:


Since there is a current source on the perimeter of the i3 mesh, we know thati3 = −16 A. The remaining two mesh equations are

−30 + 3i1 + 2(i1 − i2) + 6i1 = 0

8i2 + 5(i2 + 16) + 4i2 + 2(i2 − i1) = 0


11i1 − 2i2 = 30

−2i1 + 19i2 = −80

Solving: i1 = 2 A, i2 = −4 A, i3 = −16 AThe current in the 2 Ω resistor is i1 − i2 = 6 A ·. . p2 Ω = (6)2(2) = 72 WThus, the 2 Ω resistors dissipates 72 W.

AP 4.11 Redraw the circuit and identify the mesh currents:

There are current sources on the perimeters of both the ib mesh and the ic mesh, sowe know that

ib = −10 A; ic =2vφ

5

The remaining mesh current equation is

−75 + 2(ia + 10) + 5(ia − 0.4vφ) = 0

The dependent source requires the following constraint equation:

vφ = 5(ia − ic) = 5(ia − 0.4vφ)

Place the mesh current equation and the dependent source equation is standard form:

7ia − 2vφ = 55

5ia − 3vφ = 0

Solving: ia = 15 A; ib = −10 A; ic = 10 A; vφ = 25 VThus, ia = 15 A.

Problems 4–9


The 2 A current source is shared by the meshes ia and ib. Thus we combine thesemeshes to form a supermesh and write the following equation:

−10 + 2ib + 2(ib − ic) + 2(ia − ic) = 0

The other mesh current equation is

−6 + 1ic + 2(ic − ia) + 2(ic − ib) = 0

The supermesh constraint equation is

ia − ib = 2

Place these three equations in standard form:

2ia + 4ib − 4ic = 10

−2ia − 2ib + 5ic = 6

ia − ib + 0ic = 2

Solving, ia = 7 A; ib = 5 A; ic = 6 AThus, p1 Ω = i2c(1) = (6)2(1) = 36 W

AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1:


v1 − 2015

− 2 +v1 − 25

10= 0


Rearranging and solving,

v1

( 115

+110

)= 2 +

2015

+2510

·. . v1 = 35 V

p2A = −35(2) = −70 W

Thus the 2 A current source delivers 70 W.


There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. The other twomesh current equations are

−128 + 4(i1 − 4) + 6(i1 − i2) + 2i1 = 0

30ix + 5i2 + 6(i2 − i1) + 3(i2 − 4) = 0


ix = i1 − i3 = i1 − 4

Substitute the constraint equation into the second mesh equation and place theresulting two mesh equations in standard form:

12i1 − 6i2 = 144

24i1 + 14i2 = 132

Solving,

i1 = 9 A; i2 = −6 A; i3 = 4 A; ix = 9 − 4 = 5 A

·. . v4A = 3(i3 − i2) − 4ix = 10 V

p4A = −v4A(4) = −(10)(4) = −40 W

Thus, the 2 A current source delivers 40 W.

Problems 4–11

AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled:

Transform the 120 V source in series with the 20 Ω resistor into a 6 A source inparallel with the 20 Ω resistor. Also transform the −60 V source in series withthe 5 Ω resistor into a −12 A source in parallel with the 5 Ω resistor. The resultis the following circuit:

Combine the three current sources into a single current source, using KCL, andcombine the 20 Ω, 5 Ω, and 6 Ω resistors in parallel. The resulting circuit isshown on the left. To simplify the circuit further, transform the resulting 30 Asource in parallel with the 2.4 Ω resistor into a 72 V source in series with the2.4 Ω resistor. Combine the 2.4 Ω resistor in series with the 1.6 Ω resisor to geta very simple circuit that still maintains the voltage v. The resulting circuit ison the right.

Use voltage division in the circuit on the right to calculate v as follows:

v =812

(72) = 48 V

[b] Calculate i in the circuit on the right using Ohm’s law:

i =v

8=

488

= 6 A


Now use i to calculate va in the circuit on the left:

va = 6(1.6 + 8) = 57.6 V

Returning back to the original circuit, note that the voltage va is also thevoltage drop across the series combination of the 120 V source and 20 Ωresistor. Use this fact to calculate the current in the 120 V source, ia:

ia =120 − va

20=

120 − 57.620

= 3.12 A

p120V = −(120)ia = −(120)(3.12) = −374.40 W

Thus, the 120 V source delivers 374.4 W.

AP 4.16 To find RTh, replace the 72 V source with a short circuit:

Note that the 5 Ω and 20 Ω resistors are in parallel, with an equivalent resistance of5‖20 = 4 Ω. The equivalent 4 Ω resistance is in series with the 8 Ω resistor for anequivalent resistance of 4 + 8 = 12 Ω. Finally, the 12 Ω equivalent resistance is inparallel with the 12 Ω resistor, so RTh = 12‖12 = 6 Ω.

Use node voltage analysis to find vTh. Begin by redrawing the circuit and labelingthe node voltages:

The node voltage equations arev1 − 72

5+

v1

20+

v1 − vTh

8= 0

vTh − v1

8+

vTh − 7212

= 0

Problems 4–13


v1

(15

+120

+18

)+ vTh

(−1

8

)=

725

v1

(−1

8

)+ vTh

(18

+112

)= 6

Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent circuit isa 64.8 V source in series with a 6 Ω resistor.

AP 4.17 We begin by performing a source transformation, turning the parallel combination ofthe 15 A source and 8 Ω resistor into a series combination of a 120 V source and an8 Ω resistor, as shown in the figure on the left. Next, combine the 2 Ω, 8 Ω and 10 Ωresistors in series to give an equivalent 20 Ω resistance. Then transform the seriescombination of the 120 V source and the 20 Ω equivalent resistance into a parallelcombination of a 6 A source and a 20 Ω resistor, as shown in the figure on the right.

Finally, combine the 20 Ω and 12 Ω parallel resistors to give RN = 20‖12 = 7.5 Ω.Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a7.5 Ω resistor.

AP 4.18 Find the Thévenin equivalent with respect to A, B using source transformations. Tobegin, convert the series combination of the −36 V source and 12 kΩ resistor into aparallel combination of a −3 mA source and 12 kΩ resistor. The resulting circuit isshown below:

Now combine the two parallel current sources and the two parallel resistors to give a−3 + 18 = 15 mA source in parallel with a 12 k‖60 k= 10 kΩ resistor. Thentransform the 15 mA source in parallel with the 10 kΩ resistor into a 150 V source inseries with a 10 kΩ resistor, and combine this 10 kΩ resistor in series with the 15 kΩresistor. The Thévenin equivalent is thus a 150 V source in series with a 25 kΩ


resistor, as seen to the left of the terminals A,B in the circuit below.

Now attach the voltmeter, modeled as a 100 kΩ resistor, to the Thévenin equivalentand use voltage division to calculate the meter reading vAB:

vAB =100,000125,000

(150) = 120 V

AP 4.19 Begin by calculating the open circuit voltage, which is also vTh, from the circuitbelow:

Summing the currents away from the node labeled vTh We have

vTh

8+ 4 + 3ix +

vTh − 242

= 0

Also, using Ohm’s law for the 8 Ω resistor,

ix =vTh

8Substituting the second equation into the first and solving for vTh yields vTh = 8 V.

Now calculate RTh. To do this, we use the test source method. Replace the voltagesource with a short circuit, the current source with an open circuit, and apply the testvoltage vT, as shown in the circuit below:

Problems 4–15

Write a KCL equation at the middle node:

iT = ix + 3ix + vT/2 = 4ix + vT/2

Use Ohm’s law to determine ix as a function of vT:

ix = vT/8

Substitute the second equation into the first equation:

iT = 4(vT/8) + vT/2 = vT

Thus,

RTh = vT/iT = 1 Ω

The Thévenin equivalent is an 8 V source in series with a 1 Ω resistor.

AP 4.20 Begin by calculating the open circuit voltage, which is also vTh, using the nodevoltage method in the circuit below:

The node voltage equations are

v

60+

v − (vTh + 160i∆)20

− 4 = 0,

vTh

40+

vTh

80+

vTh + 160i∆ − v

20= 0

The dependent source constraint equation is

i∆ =vTh

40


Substitute the constraint equation into the node voltage equations and put the twoequations in standard form:

v( 1

60+

120

)+ vTh

(− 5

20

)= 4

v(− 1

20

)+ vTh

( 140

+180

+520

)= 0

Solving, v = 172.5 V and vTh = 30 V.

Now use the test source method to calculate the test current and thus RTh. Replacethe current source with a short circuit and apply the test source to get the followingcircuit:

Write a KCL equation at the rightmost node:

iT =vT

80+

vT

40+

vT + 160i∆80


i∆ =vT

40

Substitute the constraint equation into the KCL equation and simplify the right-handside:

iT =vT

10

Therefore,

RTh =vT

iT= 10 Ω

Thus, the Thévenin equivalent is a 30 V source in series with a 10 Ω resistor.

AP 4.21 First find the Thévenin equivalent circuit. To find vTh, create an open circuitbetween nodes a and b and use the node voltage method with the circuit

Problems 4–17

below:

The node voltage equations are:

vTh − (100 + vφ)4

+vTh − v1

4= 0

v1 − 1004

+v1 − 20

4+

v1 − vTh

4= 0


vφ = v1 − 20

Place these three equations in standard form:

vTh

(14

+14

)+ v1

(−1

4

)+ vφ

(−1

4

)= 25

vTh

(−1

4

)+ v1

(14

+14

+14

)+ vφ (0) = 30

vTh (0) + v1 (1) + vφ (−1) = 20

Solving, vTh = 120 V, v1 = 80 V, and vφ = 60 V.

Now create a short circuit between nodes a and b and use the mesh current methodwith the circuit below:

The mesh current equations are

−100 + 4(i1 − i2) + vφ + 20 = 0

−vφ + 4i2 + 4(i2 − isc) + 4(i2 − i1) = 0

−20 − vφ + 4(isc − i2) = 0



vφ = 4(i1 − isc)


4i1 − 4i2 + 0isc + vφ = 80

−4i1 + 12i2 − 4isc − vφ = 0

0i1 − 4i2 + 4isc − vφ = 20

4i1 + 0i2 − 4isc − vφ = 0

Solving, i1 = 45 A, i2 = 30 A, isc = 40 A, and vφ = 20 V. Thus,

RTh =vTh

isc=

12040

= 3 Ω

[a] For maximum power transfer, R = RTh = 3 Ω

[b] The Thévenin voltage, vTh = 120 V, splits equally between the Théveninresistance and the load resistance, so

vload =1202

= 60 V

Therefore,

pmax =v2

load

Rload=

602

3= 1200 W

AP 4.22 Sustituting the value R = 3 Ω into the circuit and identifying three mesh currents wehave the circuit below:


−100 + 4(i1 − i2) + vφ + 20 = 0

−vφ + 4i2 + 4(i2 − i3) + 4(i2 − i1) = 0

−20 − vφ + 4(i3 − i2) + 3i3 = 0

Problems 4–19


vφ = 4(i1 − i3)


4i1 − 4i2 + 0i3 + vφ = 80

−4i1 + 12i2 − 4i3 − vφ = 0

0i1 − 4i2 + 7i3 − vφ = 20

4i1 + 0i2 − 4i3 − vφ = 0

Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and vφ = 40 V.

[a] p100V = −(100)i1 = −(100)(30) = −3000 W. Thus, the 100 V source isdelivering 3000 W.

[b] pdepsource = −vφi2 = −(40)(20) = −800 W. Thus, the dependent source isdelivering 800 W.

[c] From Assessment Problem 4.21(b), the power delivered to the load resistor is1200 W, so the load power is (1200/3800)100 = 31.58% of the combinedpower generated by the 100 V source and the dependent source.


Problems

P 4.1

[a] 11 branches, 8 branches with resistors, 2 branches with independent sources, 1branch with a dependent source

[b] The current is unknown in every branch except the one containing the 8 Acurrent source, so the current is unknown in 10 branches.

[c] 9 essential branches – R4 − R5 forms an essential branch as does R8 − 10 V. Theremaining seven branches are essential branches that contain a single element.

[d] The current is known only in the essential branch containing the current source,and is unknown in the remaining 8 essential branches

[e] From the figure there are 6 nodes – three identified by rectangular boxes, twoidentified with single black dots, and one identified by a triangle.

[f] There are 4 essential nodes, three identified with rectangular boxes and oneidentified with a triangle

[g] A mesh is like a window pane, and as can be seen from the figure there are 6window panes or meshes.

P 4.2

Problems 4–21

[a] As can be seen from the figure, the circuit has 2 separate parts.

[b] There are 5 nodes – the four black dots and the node betweem the voltagesource and the resistor R1.

[c] There are 7 branches, each containing one of the seven circuit components.

[d] When a conductor joins the lower nodes of the two separate parts, there is nowonly a single part in the circuit. There would now be 4 nodes, because the twolower nodes are now joined as a single node. The number of branches remainsat 7, where each branch contains one of the seven individual circuitcomponents.

P 4.3 [a] From Problem 4.1(d) there are 8 essential branches were the current isunknown, so we need 8 simultaneous equations to describe the circuit.

[b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at(4 − 1) = 3 of these essential nodes. These would also be a dependent sourceconstraint equation.

[c] The remaining 4 equations needed to describe the circuit will be derived fromKVL equations.

[d] We must avoid using the topmost mesh and the leftmost mesh. Each of thesemeshes contains a current source, and we have no way of determining thevoltage drop across a current source.

P 4.4 [a] There are six circuit components, five resistors and the current source. Since thecurrent is known only in the current source, it is unknown in the five resistors.Therefore there are five unknown currents.

[b] There are four essential nodes in this circuit, identified by the dark black dots inFig. P4.4. At three of these nodes you can write KCL equations that will beindependent of one another. A KCL equation at the fourth node would bedependent on the first three. Therefore there are three independent KCLequations.

[c]

Sum the currents at any three of the four essential nodes a, b, c, and d. Usingnodes a, b, and c we get

−ig + i1 + i2 = 0


−i1 + i4 + i3 = 0

i5 − i2 − i3 = 0

[d] There are three meshes in this circuit: one on the left with the components ig,R1, and R4; one on the top right with components R1, R2, and R3; and one onthe bottom right with components R3, R4, and R5. We cannot write a KVLequation for the left mesh because we don’t know the voltage drop across thecurrent source. Therefore, we can write KVL equations for the two meshes onthe right, giving a total of two independent KVL equations.

[e] Sum the voltages around two independent closed paths, avoiding a path thatcontains the independent current source since the voltage across the currentsource is not known. Using the upper and lower meshes formed by the fiveresistors gives

R1i1 + R3i3 − R2i2 = 0

R3i3 + R5i5 − R4i4 = 0

P 4.5

[a] At node 1: − ig + i1 + i2 = 0

At node 2: − i2 + i3 + i4 = 0

At node 3: ig − i1 − i3 − i4 = 0

[b] There are many possible solutions. For example, solve the equation at node 1for ig:

ig = i1 + i2

Substitute this expression for ig into the equation at node 3:

(i1 + i2) − i1 − i3 − i4 = 0 so i2 − i3 − i4 = 0

Multiply this last equation by -1 to get the equation at node 2:

−(i2 − i3 − i4) = −0 so − i2 + i3 + i4 = 0

Problems 4–23

P 4.6

Note that we have chosen the lower node as the reference node, and that the voltageat the upper node with respect to the reference node is vo. Write a KCL equation(node voltage equation)by summing the currents leaving the upper node:vo + 25120 + 5

+vo

25+ 0.04 = 0

Solve by multiplying both sides of the KCL equation by 125 and collecting theterms involving vo on one side of the equation and the constants on the other side ofthe equation:vo + 25 + 5vo + 5 = 0 ·. . 6vo = −30 so vo = −30/6 = −5 V

P 4.7 [a] From the solution to Problem 4.6 we know vo = −5 V; thereforep40mA = (−5)(0.04) = −0.2 WThe power developed by the 40 mA source is 200 mW

[b] The current into the negative terminal of the 25 V source in the figure ofProblem 4.6 isig = (−5 + 25)/125 = 160 mAThe power in the 25 V source isp25V = −(25)(0.16) = −4 WThe power developed by the 25 V source is 4 W

[c] p5Ω = (0.16)2(5) = 128 mW

p120Ω = (0.16)2(120) = 3.072 W

p25Ω = (−5)2/25 = 1 W∑pdis = 0.128 + 3.072 + 1 = 4.2 W∑pdev = 0.2 + 4 = 4.2 W (checks!)

P 4.8

[a] The node voltage equation is:vo + 25

125+

vo

25+ 0.04 = 0


Solving,vo + 25 + 5vo + 5 = 0 ·. . 6vo = −30 so vo = −5 V

[b] Let vx = voltage drop across 40 mA source:vx = vo − (100)(0.04) = −5 − 4 = −9 Vp40mA = (−9)(0.04) = −360 mWThe power developed by the 40 mA source is 360 mW

[c] Let ig = current into negative terminal of 25 V source:ig = (−5 + 25)/125 = 160 mAp25V = −(25)(0.16) = −4 WThe power developed by the 25 V source is 4 W

[d] p5Ω = (0.16)2(5) = 128 mW

p120Ω = (0.16)2(120) = 3.072 W

p25Ω = (−5)2/25 = 1 W

p100Ω = (0.04)2(100) = 160 mW∑pdis = 0.128 + 3.072 + 1 + 0.160 = 4.36 W∑pdev = 0.360 + 4 = 4.36 W (checks!)

[e] vo is independent of any finite resistance connected in series with the 40 mAcurrent source

P 4.9


−6 +v1

40+

v1 − v2

8= 0

v2 − v1

8+

v2

80+

v2

120+ 1 = 0


v1

( 140

+18

)+ v2

(−1

8

)= 6

v1

(−1

8

)+ v2

(18

+180

+1

120

)= −1

Solving, v1 = 120 V and v2 = 96 V.

Check this result by calculating the power associated with each component:

Problems 4–25

Component Power Delivered (W) Power Absorbed (W)

6A −(6 A)(120 V) = −720

40 Ω1202

40= 360

8 Ω(120 − 96)2

8= 72

80 Ω962

80= 115.2

120 Ω962

120= 76.8

1 A (96 V)(1 A) = 96

Total −720 720

P 4.10 [a]

The two node voltage equations are:v1

6+

v1 − 444

+v1 − v2

1= 0

v2

3+

v2 − v1

1+

v2 + 22

= 0


v1

(16

+14

+ 1)

+ v2(−1) =444

v1(−1) + v2

(13

+ 1 +12

)= −2

2Solving, v1 = 12 V; v2 = 6 VNow calculate the branch currents from the node voltage values:

ia =44 − 12

4= 8 A

ib =126

= 2 A

ic =12 − 6

1= 6 A

id =63

= 2 A

ie =6 + 2

2= 4 A


[b] psources = p44V + p2V = −(44)ia − (2)ie = −(44)(8) − (2)(4) = −352 − 8 = −360 WThus, the power developed in the circuit is 360 W. Note that the resistorscannot develop power!

P 4.11 [a]

v1 − 1102

+v1 − v2

8+

v1 − v3

16= 0 so 11v1 − 2v2 − v3 = 880

v2 − v1

8+

v2

3+

v2 − v3

24= 0 so −3v1 + 12v2 − v3 = 0

v3 + 1102

+v3 − v2

24+

v3 − v1

16= 0 so −3v1 − 2v2 + 29v3 = −2640

Solving, v1 = 74.64 V; v2 = 11.79 V; v3 = −82.5 V

Thus, i1 =110 − v1

2= 17.68 A i4 =

v1 − v2

8= 7.86 A

i2 =v2

3= 3.93 A i5 =

v2 − v3

24= 3.93 A

i3 =v3 + 110

2= 13.75 A i6 =

v1 − v3

16= 9.82 A

[b]∑

Pdev = 110i1 + 110i3 = 3457.14 W∑Pdis = i21(2) + i22(3) + i23(2) + i24(8) + i25(24) + i26(16) = 3457.14 W

P 4.12

The two node voltage equations are:v1 − 150

20+

v1

80+

v1 − v2

40= 0

v2 − v1

40− 11.25 +

v2

4= 0

Problems 4–27


v1

( 120

+180

+140

)+ v2(− 1

40) =

15020

v1

(− 1

40

)+ v2

( 140

+14

)= 11.25

Solving, v1 = 100 V; v2 = 50 V

P 4.13

At vo : −2 +vo

50+

vo − 454 + 1

= 0

Solving, vo = 50 V

p2A = −(50)(2) = −100 W

Thus, the 2 A current source delivers 100 W, or the current source extracts −100 Wfrom the circuit.

P 4.14

The three node voltage equations are:v1 − 40

4+

v1

40+

v1 − v2

2= 0

v2 − v1

2+

v2 − v3

4− 28 = 0

v3

2+

v3 − v2

4+ 28 = 0


v1

(14

+140

+12

)+ v2

(−1

2

)+ v3(0) =

404

v1

(−1

2

)+ v2

(12

+14

)+ v3

(−1

4

)= 28

v1(0) + v2

(−1

4

)+ v3

(12

+14

)= −28


Solving, v1 = 60 V; v2 = 73 V; v3 = −13 V.

p28A = −va(28 A) = −(v2 − v3)(28 A) = −(73 + 13)(28) = −2408 W

The 28 A source delivers 2408 W.

P 4.15

The node voltage equations are:v1 + 40

12+

v1

25+ 5 +

v1 − v2

20= 0

v2 − v1

20+

v2 − v3

40− 7.5 − 5 = 0

v3

40+

v3 − v2

40+ 7.5 = 0


v1

( 112

+125

+120

)+ v2

(− 1

20

)+ v3(0) = −40

12− 5

v1

(− 1

20

)+ v2

( 120

+140

)+ v3

(− 1

40

)= 12.5

v1(0) + v2

(− 1

40

)+ v3

( 140

+140

)= −7.5

Solving, v1 = −10 V; v2 = 132 V; v3 = −84 V.Find the power:

i40V = (−10 + 40)/12 = 2.5 A

p40V = −(2.5)(40) = −100 W (del)

p5A = (5)(−10 − 132) = −710 W (del)

p7.5A = (7.5)(−84 − 132) = −1620 W (del)

p12Ω = (−10 + 40)2/12 = 75 W (abs)

p25Ω = (−10)2/25 = 4 W (abs)

Problems 4–29

p20Ω = (132 + 10)2/20 = 1008.2 W (abs)

p40Ω = (132 + 84)2/40 = 1166.4 W (abs)

p40Ω = (−84)2/40 = 176.4 W (abs)

∑pdiss = 75 + 4 + 1008.2 + 1166.4 + 176.4 = 2430 W

∑pdev = 100 + 710 + 1620 W = 2430 W (CHECKS)

P 4.16 [a]vo − v1

R+

vo − v2

R+

vo − v3

R+ · · · +

vo − vn

R= 0

·. . nvo = v1 + v2 + v3 + · · · + vn

·. . vo =1n

[v1 + v2 + v3 + · · · + vn] =1n

∑n

k=1vk

[b] vo =13(120 + 60 − 30) = 50 V

P 4.17 [a]

The node voltage equation is:

−0.45 +vo

100+

vo − 6.25i∆5

+vo − 45

25= 0

The dependent source constraint equation is:

i∆ =45 − vo


vo

( 1100

+15

+125

)+ i∆

(−6.25

5

)=

4525

+ 0.45

vo

( 125

)+ i∆(1) =

4525

Solving, vo = 15 V; i∆ = 1.2 A

[b] ids =vo − 6.25i∆

5=

15 − 7.55

= 1.5 A

pds = [6.25(1.2)](1.5) = 11.25 WThus, the dependent source absorbs 11.25 W


[c] p450mA = −(0.45)(15) = −6.75 Wp45V = −(1.2)(45) = −54 W∑

pdev = 6.75 + 54 = 60.75 WThus the independent sources develop 60.75 WAlso,∑

pdis = pds + p100Ω + p5Ω + p25Ω

= 11.25 + (15)2/100 + (1.5)2(5) + (1.2)2(25)= 11.25 + 2.25 + 11.25 + 36 = 60.75 W (checks!)

P 4.18 [a]


−5io +v1

20+

v1 − v2

5= 0

v2 − v1

5+

v2

40+

v2 − v3

10= 0

v3 − v2

10+

v3 − 11.5io

5+

v3 − 964

= 0

The dependent source constraint equation is:io = v2/40Place these equations in standard form:

v1

( 120

+15

)+ v2

(−1

5

)+ v3(0) + io(−5) = 0

v1

(−1

5

)+ v2

(15

+140

+110

)+ v3

(− 1

10

)+ io(0) = 0

v1(0) + v2

(− 1

10

)+ v3

( 110

+15

+14

)+ io

(−11.5

5

)=

964

v1(0) + v2

(− 1

40

)+ v3(0) + io(1) = 0

Solving, v1 = 156 V; v2 = 120 V; v3 = 78 V; io = 3 A

Problems 4–31

[b] Calculate the power:

pcccs = −[5(3)](156) = −2340 W

p20Ω = (156)2/20 = 1216.8 W

p5Ω = (156 − 120)2/5 = 259.2 W

p40Ω = (120)2/40 = 360 W

p10Ω = (120 − 78)2/10 = 176.4 W

p5Ω = (78 − 11.5 · 3)2/5 = 378.45 W

p4Ω = (78 − 96)2/4 = 81 W

p96V = [(78 − 96)/4](96) = −432 W

pccvs = [(78 − 3 · 11.5)/5](11.5 · 3) = 300.15 W∑pdev = 2340 + 432 = 2772 W∑pdis = 1216.8 + 259.2 + 360 + 176.4 + 378.45 + 81 + 300.15 = 2772 W

(checks)Thus, the circuit dissipates 2772 W

P 4.19


vo − 16010

+vo

100+

vo − 150iσ

30 + 20= 0


iσ = − vo


vo

( 110

+1

100+

150

)+ iσ

(−150

50

)=

16010

vo

( 1100

)+ iσ(1) = 0

Solving, vo = 100 V; iσ = −1 ANow find the power:

io =160 − 100

10− 1 = 5 A

pds = [150(−1)](5) = −750 W.Thus, the dependent source delivers 750 W


P 4.20 [a]


−25 +v1

40+

v1

160+

v1 − v2

10= 0

v2 − v1

10+

v2

20+

v2 − 84i∆8

= 0

The dependent source constraint equation is:i∆ = v1/160Place these three equations in standard form:

v1

( 140

+1

160+

110

)+ v2

(− 1

10

)+ i∆(0) = 25

v1

(− 1

10

)+ v2

( 110

+120

+18

)+ i∆

(−84

8

)= 0

v1

(− 1

160

)+ v2(0) + i∆(1) = 0

Solving, v1 = 352 V; v2 = 212 V; i∆ = 2.2 ANow calculate the power. Only the two sources can develop power, so focus onthe sources:p25A = −(352)(25) = −8800 W

idep source = (v2 − 84i∆)/8 = (212 − 84 · 2.2)/8 = 3.4 A

pdep source = (84 · 2.2)(3.4) = 628.32 W

Thus, only the current source develops power, so the total power developed inthe circuit is 8800 W

[b] The dependent source and all of the resistors dissipate the power developed bythe current source. Check that the power developed equals the powerdissipated:

p40Ω = (352)2/40 = 3097.6 W

p160Ω = (352)2/160 = 774.4 W

p10Ω = (352 − 212)2/10 = 1960 W

p20Ω = (212)2/20 = 2247.2 W

p8Ω = (212 − 84 · 2.2)2/8 = 92.48 W∑pdiss = 628.32 + 3097.6 + 774.4 + 1960 + 2247.2 + 92.48 = 8800 W so the

power balances.

Problems 4–33

P 4.21

The two node voltage equations are:v1 − 40

5+

v1

50+

v1 − v2

10= 0

v2 − v1

10− 10 +

v2

40+

v2 − 408

= 0


v1

(15

+150

+110

)+ v2

(− 1

10

)=

405

v1

(− 1

10

)+ v2

( 110

+140

+18

)= 10 +

408

Solving, v1 = 50 V; v2 = 80 V.Thus, vo = v1 − 40 = 50 − 40 = 10 V.POWER CHECK:ig = (50 − 40)/5 + (80 − 40)/8 = 7 A

p40V = (40)(7) = 280 W (abs)

p5Ω = (50 − 40)2/5 = 20 W (abs)

p8Ω = (80 − 40)2/8 = 200 W (abs)

p10Ω = (80 − 50)2/10 = 90 W (abs)

p50Ω = 502/50 = 50 W (abs)

p40Ω = 802/40 = 160 W (abs)

p10A = −(80)(10) = −800 W (del)∑pabs = 280 + 20 + 200 + 90 + 50 + 160 = 800 W =

∑pdel


P 4.22

The node voltage equations are:v1 − 2.26

20+

v1 − v2

50+

v1

25= 0

v2 − 2.2640

+v2 − v1

50+

v2

100= 0


v1

( 120

+150

+125

)+ v2

(− 1

50

)=

2.2620

v1

(− 1

50

)+ v2

( 140

+150

+1

100

)=

2.2640

Solving, v1 = 1.3 V; v2 = 1.5 V.

Thus, io =v1 − v2

50=

1.3 − 1.550

= −4 mA

P 4.23 [a]


Problems 4–35

v2 − 2301

+v2 − v4

1+

v2 − v3

1= 0

v3 − v2

1+

v3 − v5

1+

v3

1= 0

v4 − 2305 + 1

+v4 − v2

1+

v4 − v5

2= 0

v5 − v4

2+

v5 − v3

1+

v5

5 + 1= 0


v2(1 + 1 + 1) + v3(−1) + v4(−1) + v5(0) = 230

v2(−1) + v3(1 + 1 + 1) + v4(0) + v5(−1) = 0

v2(−1) + v3(0) + v4

(16

+ 1 +12

)+ v5

(−1

2

)=

2306

v2(0) + v3(−1) + v4

(−1

2

)+ v5

(12

+ 1 +16

)= 0

Solving, v2 = 150 V; v3 = 80 V; v4 = 140 V; v5 = 90 VFind the power dissipated by the 2 Ω resistor:

i2Ω =v4 − v5

2=

140 − 902

= 25 A

p2Ω = (25)2(2) = 1250 W

[b] Find the power developed by the 230 V source:

i230V =v2 − 230

1+

v4 − 2306

= −80 − 15 = −95 A

p230V = (230)(−95) = −21,850 W, so the source supplies 21,850 WCheck:

∑Pdis = (80)2(1) + (15)2(1) + (15)2(5) + (70)2(1) + (10)2(1)

+(25)2(2) + (10)2(1) + (80)2(1) + (15)2(5) + (15)2(1)

= 21,850 W(checks)


P 4.24 [a]

There is only one node voltage equation:va + 305000

+va

500+

va − 801000

+ 0.01 = 0

Solving,va + 30 + 10va + 5va − 400 + 50 = 0 so 16va = 320·. . va = 20 V

Calculate the currents:i1 = (−30 − 20)/5000 = −10 mA

i2 = 20/500 = 40 mA

i4 = 80/4000 = 20 mA

i5 = (80 − 20)/1000 = 60 mA

i3 + i4 + i5 − 10 mA = 0 so i3 = 0.01 − 0.02 − 0.06 = −0.07 = −70 mA

[b] p30V = (30)(−0.01) = −0.3 W

p10mA = (20 − 80)(0.01) = −0.6 W

p80V = (80)(−0.07) = −5.6 W

p5k = (−0.01)2(5000) = 0.5 W

p500Ω = (0.04)2(500) = 0.8 W

p1k = (80 − 20)2/(1000) = 3.6 W

p4k = (80)2/(4000) = 1.6 W∑pabs = 0.5 + 0.8 + 3.6 + 1.6 = 6.5 W∑pdel = 0.3 + 0.6 + 5.6 = 6.5 W (checks!)

Problems 4–37

P 4.25


7 +vb

3+

vb − vc

1= 0

−2vx +vc − vb

1+

vc − 42

= 0

The constraint equation for the dependent source is:vx = vc − 4


vb

(13

+ 1)

+ vc(−1) + vx(0) = −7

vb(−1) + vc

(1 +

12

)+ vx(−2) =

42

vb(0) + vc(1) + vx(−1) = 4

Solving, vo = vb = 1.5 V Also, vc = 9 V and vx = 5 V.

P 4.26

This circuit has a supernode includes the nodes v1, v2 and the 25 V source. Thesupernode equation is

2 +v1

50+

v2

150+

v2

20 + 55= 0

The supernode constraint equation is

v2 + 25 = v1

Place these two equations in standard form:


v1

( 150

)+ v2

( 1150

+175

)= −2

v1(1) + v2(−1) = 25

Solving, v1 = −37.5 V and v2 = −62.5 V.

p25V = (25)i25

i25 = −2 A − i50 = −2 A − v1

50= 2 A − −37.5

50= −2 A + 0.75 A = −1.25 A

Thus, p25V = (25)(−1.25) = −31.25 W

The 25 V source delivers 31.25 W.

P 4.27

The supernode equation is:

v1 − 10010

+v1

60+

v1 − 4v∆

20+

v1 − 4v∆

30= 0

The constraint equation for the dependent source is:4v∆ = v1 − v∆


v1

( 110

+160

+120

+130

)+ v∆

(− 4

20− 4

30

)=

10010

v1(1) + v∆(−5) = 0

Solving, v1 = 75 V; v∆ = 15 VThus, vo = 100 − v1 = 25 V

P 4.28 Calculate currents and voltages needed to calculate the power for the variouscomponents:

iφ =v4 − v3

8=

81.6 − 1088

= −3.3 A

403

iφ =403

(−3.3) = −44 V

v1 = v4 +403

iφ = 81.6 − 44 = 37.6 V

Problems 4–39

v3 + v∆ = 120 ·. . v∆ = 120 − 108 = 12 V

1.75v∆ = (1.75)(12) = 21 A

i120V =v1 − 120

4+

v3 − 1202

=37.6 − 120

4+

108 − 1202

= −26.6 A

iccvs =0 − v1

20+

v2 − v1

4=

−37.620

+120 − 37.6

4= 18.72 A

Now calculate the power associated with each circuit element:

p20Ω = (37.6)2/20 = 70.688 W

p4Ω = (37.6 − 120)2/4 = 1697.44 W

p120V = (120)(−26.6) = −3192 W

p2Ω = (12)2/2 = 72 W

p40Ω = (108)2/40 = 291.6 W

p8Ω = (108 − 81.6)2/8 = 87.12 W

p80Ω = (81.6)2/80 = 83.232 W

pvccs = (81.6)[1.75(12)] = 1713.6 W∑

pabs =∑

pdel = 4015.6 W

pccvs = (18.72)(−44) = −823.68 W

Now sum the powers:∑ptotal = 70.688 + 1697.44 − 3192 + 72 + 291.6 + 87.12

+83.232 + 1712.6 − 823.68 = 0 WThus, the power balances and the staff analyst has correctly calculated the voltagevalues

P 4.29

The supernode equation is:v1

100+

v1 − 6010

+v3 − v2

20+

v3

400− 0.625v∆ = 0

The node voltage equation at v2 is:


v2 − 605

+v2

200+

v2 − v3

20= 0

The supernode constraint equation is:v3 − v1 = 175iφ

The two dependent source constraint equations are:v∆ = v2 − 60iφ = −v2/200Place the four equations above in standard form:

v1

( 1100

+110

)+ v2

(− 1

20

)+ v3

( 1400

+120

)+ iφ(0) + v∆(−0.625) =

6010

v1(0) + v2

(15

+1

200+

120

)+ v3

(− 1

20

)+ iφ(0) + v∆(0) =

605

v1(1) + v2(0) + v3(−1) + iφ(175) + v∆(0) = 0

v1(0) + v2(1) + v3(0) + iφ(0) + v∆(−1) = 60

v1(0) + v2

( 1200

)+ v3(0) + iφ(1) + v∆(0) = 0

Solving,v1 = −60.75 V v2 = 30 V; v3 = −87 V; iφ = −0.15 A; v∆ = −30 VCalculate the power for the 60 V source:

i60V =v1 − 60

10+

v2 − 605

=−60.75 − 60

10+

30 − 605

= −18.075 A

p60V = (60)(−18.075) = −1084.5 W

Thus, the 60 V source delivers 1084.5 W

P 4.30 From Eq. 4.16, iB = vc/(1 + β)RE

From Eq. 4.17, iB = (vb − Vo)/(1 + β)RE

From Eq. 4.19,

iB =1

(1 + β)RE

[VCC(1 + β)RER2 + VoR1R2

R1R2 + (1 + β)RE(R1 + R2)− Vo

]

=VCCR2 − Vo(R1 + R2)

R1R2 + (1 + β)RE(R1 + R2)=

[VCCR2/(R1 + R2)] − Vo

[R1R2/(R1 + R2)] + (1 + β)RE

Problems 4–41

P 4.31 [a]


−60 + 4i1 + 10(i1 − i2) + 1i1 = 0

20 + 3i2 + 10(i2 − i1) + 2i2 = 0Place the equations in standard form:

i1(4 + 10 + 1) + i2(−10) = 60

i1(−10) + i2(3 + 10 + 2) = −20Solving, i1 = 5.6 A; i2 = 2.4 A

Now solve for the requested currents:ia = i1 = 5.6 A; ib = i1 − i2 = 3.2 A; ic = −i2 = −2.4 A

[b] If the polarity of the 60 V source is reversed, we have the following meshcurrent equations in standard form:

i1(4 + 10 + 1) + i2(−10) = −60

i1(−10) + i2(3 + 10 + 2) = −20Solving, i1 = −8.8 A; i2 = −7.2 A

Now solve for the requested currents:ia = i1 = −8.8 A; ib = i1 − i2 = −1.6 A; ic = −i2 = 7.2 A

P 4.32 [a]



−230 + 1(i1 − i2) + 2(i1 − i3) + 115 + 4i1 = 0

6i2 + 3(i2 − i3) + 1(i2 − i1) = 0

460 + 5i3 − 115 + 2(i3 − i1) + 3(i3 − i2) = 0Place these equations in standard form:

i1(1 + 2 + 4) + i2(−1) + i3(−2) = 115

i1(−1) + i2(6 + 3 + 1) + i3(−3) = 0

i1(−2) + i2(−3) + i3(5 + 2 + 3) = −345Solving, i1 = 4.4 A; i2 = −10.6 A; i3 = −36.8 AThe only components that can develop power in the circuit are the sources:

p230V = −(230)(4.4) = −1012 W

p115V = −(115)(−36.8 − 4.4) = 4738 W

p460V = (460)(−36.8) = −16,928 W

·. .∑

pdev = 1012 + 16,928 = 17940 W

[b] From part (a) we know that the 115 V source is dissipating power; compute thepower dissipated by the resistors:

p1Ω = (1)(4.4 + 10.6)2 = 225 W

p4Ω = (4)(4.4)2 = 77.44 W

p6Ω = (6)(−10.6)2 = 674.16 W

p2Ω = (2)(4.4 + 36.8)2 = 3394.88 W

p3Ω = (3)(−10.6 + 36.8)2 = 2059.32 W

p5Ω = (5)(−36.8)2 = 6771.2 W

·. .∑

pdis = 4738 + 225 + 77.44 + 674.16 + 3394.88 + 2059.32 + 6771.2= 17940 W (checks!)

P 4.33


Problems 4–43

−135 + 3(i1 − i2) + 20(i1 − i3) + 2i1 = 0

5i2 + 4(i2 − i3) + 3(i2 − i1) = 0

10iσ + 1i3 + 20(i3 − i1) + 4(i3 − i2) = 0

The dependent source constraint equation is:iσ = i2 − i1Place these equations in standard form:

i1(3 + 20 + 2) + i2(−3) + i3(−20) + iσ(0) = 135

i1(−3) + i2(5 + 4 + 3) + i3(−4) + iσ(0) = 0

i1(−20) + i2(−4) + i3(1 + 20 + 4) + iσ(10) = 0

i1(1) + i2(−1) + i3(0) + iσ(1) = 0

Solving, i1 = 64.8 A, i2 = 39 A; i3 = 68.4 A; iσ = −25.8 ACalculate the power:p20Ω = 20(68.4 − 64.8)2 = 259.2 WThus the 20 Ω resistor dissipates 259.2 W.

P 4.34

The mesh current equations:

−132 + 1i1 + 3(i1 − i3) + 2(i1 − i2) = 0

−7iφ + 2(i2 − i1) + 10(i2 − i3) = 0

5i3 + 10(i3 − i2) + 3(i3 − i1) = 0

The dependent source constraint equation:iφ = i2 − i3


i1(1 + 3 + 2) + i2(−2) + i3(−3) + iφ(0) = 132

i1(−2) + i2(10 + 2) + i3(−10) + iφ(−7) = 0

i1(−3) + i2(−10) + i3(5 + 10 + 3) + iφ(0) = 0

i1(0) + i2(−1) + i3(1) + iφ(1) = 0


Solving, i1 = 48 A; i2 = 36 A; i3 = 28 A; iφ = 8 A

Solve for the power:pdep source = −7(iφ)i2 = −7(8)(36) = −2016 WThus, the dependent source is developing 2016 W.

P 4.35


53(i2 − i3) + 5(i1 − i3) + 3(i1 − i2) = 0

30 + 3(i2 − i1) + 20(i2 − i3) + 7i2 = 0

−30 + 2i3 + 20(i3 − i2) + 5(i3 − i1) = 0


i1(5 + 3) + i2(53 − 3) + i3(−53 − 5) = 0

i1(−3) + i2(3 + 20 + 7) + i3(−20) = −30

i1(−5) + i2(−20) + i3(2 + 20 + 5) = 30

Solving, i1 = 186 A; i2 = 81.6 A; i3 = 96 A

Calculate the power:

p30V(left) = (30)(81.6) = 2448 W

p30V(right) = −(30)(96) = −2880 W

pdep source = 53(81.6 − 96)(186) = −141,955.2 W

p3Ω = (3)(186 − 81.6)2 = 32,698.08 W

p5Ω = (5)(186 − 96)2 = 40,500 W

p20Ω = (20)(81.6 − 96)2 = 4147.2 W

p7Ω = (7)(81.6)2 = 46,609.92 W

p2Ω = (2)(96)2 = 18,432 W

∑pdev = 2880 + 141,955.2 = 144,835.2 W

Problems 4–45

∑pdis = 2448 + 32,698.08 + 40,500 + 4147.2 + 46,609.92 + 18,432

= 144,835.2 W(checks)

Thus the dependent source develops 141,955.2 W.

P 4.36 [a]

10 = 18i1 − 16i2

0 = −16i1 + 28i2 + 4i∆

4 = 8i∆

Solving, i1 = 1 A; i2 = 0.5 A; i∆ = 0.5 A

v0 = 16(i1 − i2) = 16(0.5) = 8 V

[b] p4i∆ = 4i∆i2 = (4)(0.5)(0.5) = 1 W (abs)

·. . p4i∆ (deliver) = −1 W

P 4.37

600 = 25.6i1 − 16i2 − 5.6i3

−424 = −16i1 + 20i2 − 0.8i3

30 = i3

Solving, i1 = 35 A; i2 = 8 A; i3 = 30 A


[a] v30A = 0.8(i2 − i3) + 5.6(i1 − i3)

= 0.8(8 − 30) + 5.6(35 − 30) = 10.4 V

p30A = 30v30A = 30(10.4) = 312 W (abs)

Therefore, the 30 A source delivers −312 W.

[b] p600V = −600(35) = −21,000 W(del)

p424V = 424(8) = 3392 W(abs)

Therefore, the total power delivered is 21,000 W

[c] p4Ω = (35)2(4) = 4900 W

p3.2Ω = (8)2(3.2) = 204.8 W

p16Ω = (35 − 8)2(16) = 11,664 W

p5.6Ω = (35 − 30)2(5.6) = 140 W

p0.8Ω = (−30 + 8)2(0.8) = 387.2 W

∑presistors = 17,296 W

∑pabs = 17,296 + 312 + 3392 = 21,000 W (CHECKS)

P 4.38 [a]

The mesh current equation for the right mesh is:5400(i1 − 0.005) + 3700i1 − 150(0.005 − i1) = 0Solving, 9250i1 = 27.75 ·. . i1 = 3 mAThen, i∆ = 0.005 − i1 = 0.005 − 0.003 = 0.002 = 2 mA

[b] vo = (0.005)(10,000) + (0.002)(5400) = 60.8 Vp5mA = −(60.8)(0.005) = −304 mWThus, the 5 mA source delivers 304 mW

Problems 4–47

[c] 150i∆ = 150(0.002) = 0.3 Vpdep source = 150i∆i1 = −(0.3)(0.003) = −0.9 mWThe dependent source delivers 0.9 mW.

P 4.39

Mesh equations:

7i1 + 1(i1 − i3) + 2(i1 − i2) = 0

−125 + 2(i2 − i1) + 3(i2 − i3) + 75 = 0

Constraint equations:i3 = −0.5v∆; v∆ = 2(i1 − i2)Place these equations in standard form:

i1(7 + 1 + 2) + i2(−2) + i3(−1) + v∆(0) = 0

i1(−2) + i2(2 + 3) + i3(−3) + v∆(0) = 50

i1(0) + i2(0) + i3(1) + v∆(0.5) = 0

i1(2) + i2(−2) + i3(0) + v∆(−1) = 0

Solving, i1 = 6 A; i2 = 22 A; i3 = 16 A; v∆ = −32 VSolve the outer loop KVL equation to find vcs:−125 + 7i1 + vcs = 0; ·. . vcs = 125 − 7(6) = 83 VCalculate the power:

p125V = −(125)(22) = −2750 W

p75V = (75)(22 − 16) = 450 W

pdep source = −(83)[0.5(−32)] = 1328 W

Thus, the total power developed is 2750 W.CHECK:p7Ω = (6)2(7) = 252 W

p2Ω = (22 − 6)2(2) = 512 W

p3Ω = (22 − 16)2(3) = 108 W

p1Ω = (16 − 6)2(1) = 100 W


·. .∑

pabs = 450 + 1328 + 252 + 512 + 108 + 100 = 2750 W (checks!)

P 4.40

Since the bottom left mesh current value is known, we need only two mesh currentequations:

1i1 + 4(i1 − i2) + 5(i1 − 20) = 0

6.5i1 + 20(i2 − 20) + 4(i2 − i1) = 0


i1(1 + 4 + 5) + i2(−4) = 100

i1(6.5 − 4) + i2(20 + 4) = 400

Solving, i1 = 16 A; i2 = 15 AFind v:−v + 5(20 − i1) + 20(20 − i2) = 0 ·. . v = 5(4) + 20(5) = 120 VCalculate the power:

p20A = −(120)(20) = −2400 W

pdep source = [6.5(16)](15) = 1560 W

p1Ω = 1(16)2 = 256 W

p5Ω = 5(20 − 16)2 = 80 W

p4Ω = 4(16 − 15)2 = 4 W

p20Ω = 20(20 − 15)2 = 500 W

∑pdev = 2400 W

∑pdis = 1560 + 256 + 80 + 4 + 500 = 2400 W (checks)

The power developed by the 20 A source is 2400 W

Problems 4–49

P 4.41 [a]


−20 + 1(i1 − i3) + 25(i1 − i2) + 4i1 = 0

80 + 3i2 + 25(i2 − i1) + 2(i2 − i3) = 0The constraint equation is:i3 = 45i∆ = 45(i1 − i2)Place these equations in standard form:

i1(1 + 25 + 4) + i2(−25) + i3(−1) = 20

i1(−25) + i2(3 + 25 + 2) + i3(−2) = −80

i1(−45) + i2(45) + i3(1) = 0Solving, i1 = 8 A; i2 = 7 A; i3 = 45 A

Find the power in the 2 Ω resistor:p2Ω = 2(i2 − i3)2 = 2(−38)2 = 2888 WThe 2 Ω resistor dissipates 2888 W.

[b] Find the power developed by the sources:vo +80+3(7)+4(8)−20 = 0 ·. . vo = 20−80−21−32 = −113 V

pdep source = (−113)[45(8 − 7)] = −5085 W

p80V = (80)(7) = 560 W

p20V = −(20)(8) = −160 W

∑pdev = 5085 + 160 = 5245 W

The percent of the power developed that is deliverd to the 2 Ω resistor is:

28885245

× 100 = 55.06%


P 4.42 [a]


75 + 6i1 + 12(i1 − i2) − 7i∆ = 0

15i2 + 60(i2 − i3) + 7i∆ + 12(i2 − i1) = 0The two constraint equations are:

i∆ = −i2

i3 = 1.6v∆ = 1.6(6i1) = 9.6i1Place these equations in standard form:

i1(6 + 12) + i2(−12) + i3(0) + i∆(−7) = −75

i1(−12) + i2(15 + 60 + 12) + i3(−60) + i∆(7) = 0

i1(0) + i2(1) + i3(0) + i∆(1) = 0

i1(9.6) + i2(0) + i3(−1) + i∆(0) = 0Solving, i1 = 4 A; i2 = 29.4 A; i3 = 38.4 A; i∆ = −29.4 ACalculate the power associated with the three sources:

v = 60(i2 − i3) = −540 V

v∆ = 6i1 = 6(4) = 24 V

p75V = (75)(4) = 300 W

pCCVS = −7(−29.4)(4 − 29.4) = −5227.32 W

pVCCS = (−540)[1.6(24)] = −20,736 W

The two dependent sources are generating a total of5227.32 + 20,736 = 25,963.32 W.

[b] Find the power dissipated. Remember that the 75 V source is generating 300 W,as calculated in part (a):

p6Ω = (6)(4)2 = 96 W

p12Ω = (12)(4 − 29.4)2 = 7741.92 W

p15Ω = (15)(29.4)2 = 12,965.4 W

p60Ω = (60)(29.4 − 38.4)2 = 4860 W∑

pdis = 300 + 96 + 7741.92 + 12,965.4 + 4860 = 25,963.32 W(checks)

Problems 4–51

Thus the power dissipated in the circuit is 25,963.32 W.

P 4.43

The supermesh equation is:−20 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0The supermesh constraint equation is :i1 − i2 = 6Place these equations in standard form:

i1(4 + 1) + i2(9 + 6) = 20 + 90

i1(1) + i2(−1) = 6

Solving, i1 = 10 A; i2 = 4 ANow find the power:

p4Ω = 102(4) = 400 W

p1Ω = 102(1) = 100 W

p9Ω = 42(9) = 144 W

p6Ω = 42(6) = 96 W

p20V = −(20)(10) = −200 W

v6A = 9i2 − 90 + 6i2 = (9)(4) − 90 + (6)(4) = −30 V

p6A = (−30)(6) = −180 W

p90V = −(90)(4) = −360 W

In summary:∑pdev = 200 + 180 + 360 = 740 W∑pdiss = 400 + 100 + 144 + 96 = 740 W

Thus the power dissipated in the circuit is 740 W


P 4.44


i1(4 + 1) + i2(9 + 6) = 120 + 90

i1(1) + i2(−1) = 6

Solving, i1 = 15 A; i2 = 9 ANow find the power:

p4Ω = 152(4) = 900 W

p1Ω = 152(1) = 225 W

p9Ω = 92(9) = 729 W

p6Ω = 92(6) = 486 W

p120V = −(120)(15) = −1800 W

vo = 9i2 − 90 + 6i2 = 9(9) − 90 + 6(9) = 45 V

p6A = (45)(6) = 270 W

p90V = −(90)(9) = −810 W

In summary:∑pdev = 900 + 225 + 729 + 486 + 270 = 2610 W (note that the 6 A source is

now dissipating power!)∑pdiss = 1800 + 810 = 2610 W


P 4.45 [a]

Problems 4–53


i1(4 + 1) + i2(9 + 6) = 60 + 90

i1(1) + i2(−1) = 6Solving, i1 = 12 A; i2 = 6 ANow find the power:

p4Ω = 122(4) = 576 W

p1Ω = 122(1) = 144 W

p9Ω = 62(9) = 324 W

p6Ω = 62(6) = 216 W

p60V = −(60)(20) = −720 W

vo = 9i2 − 90 + 6i2 = 9(6) − 90 + 6(6) = 0 V

(the 6 A source acts like a short circuit carrying 6 A of current)

p6A = (0)(6) = 0 W

p90V = −(90)(6) = −540 W

In summary:∑pdev = 576 + 144 + 324 + 216 = 1260 W (note that the power of the 6

A source is zero)∑pdiss = 720 + 540 = 1260 W


[b]

Now there is no longer a supermesh. The two simple mesh current equationsare:−60 + 4i1 + 1i1 = 0

−90 + 6i2 + 9i2 = 0Since these equations are uncoupled, each can be solved separately:


5i1 = 60 ·. . i1 = 60/5 = 12 A

15i2 = 90 ·. . i2 = 90/15 = 6 A

Since the currents are the same as in part (a), the power will be the same ascalculated in part (a). Thus, the power dissipated in the circuit is again 1260 W.

[c] As noted in part (a), the 6 A source has zero voltage drop, so is equivalent to ashort circuit (which has no voltage drop by definition) carrying 6 A of current,as in the circuit of part (b).

P 4.46 [a]

The i1 mesh current equation:−100 + 5(i1 − i2) + 10(i1 − i3) + 2i1 = 0The i2 — i3 supermesh equationa:2i2 + 20i3 + 10(i3 − i1) + 5(i2 − i1) = 0The supermesh constraint:i3 − i2 = 1.2ib = 1.2i1Place these equations in standard form:

i1(5 + 10 + 2) + i2(−5) + i3(−10) = 100

i1(−10 − 5) + i2(2 + 5) + i3(20 + 10) = 0

i1(1.2) + i2(1) + i3(−1) = 0Solving, i1 = 7.4 A; i2 = −4.2 A; i3 = 4.68 ASolve for the requested currents:

ia = i2 = −4.2 A

ib = i1 = 7.4 A

ic = i3 = 4.68 A

id = i1 − i2 = 11.6 A

ie = i1 − i3 = 2.72 A

[b] Find vcs:2i2 + vcs + 5(i2 − i1) = 0 ·. . vcs = −2(−4.2) − 5(−4.2 − 7.4) = 66.4 V

Problems 4–55

Calculate the power:

p100V = −(100)(7.4) = −740 W

pdep source = −(66.4)[1.2(7.4)] = −589.632 W

p2Ω = 2(−4.2)2 = 35.28 W

p5Ω = 5(7.4 + 4.2)2 = 672.8 W

p2Ω = 2(7.4)2 = 109.52 W

p10Ω = 10(7.4 − 4.68)2 = 73.984 W

p20Ω = 20(4.68)2 = 438.048 W∑

pdev = 740 + 589.632 = 1329.632 W

∑pdis = 35.28 + 672.8 + 109.52 + 73.984 + 438.048 = 1329.632 W

P 4.47 [a]

The i2 mesh current equation:−4id + 10(i2 − i4) + 5(i2 − i3) = 0The i3 — i4 supermesh equation:40(i3 − 19) + 5(i3 − i2) + 10(i4 − i2) − 240 = 0The supermesh constraint equation:i4 − i3 = 2ib = 2(i2 − i3)Place the equations in standard form:

i2(10 + 5) + i3(−5) + i4(−10 − 4) = 0

i2(−5 − 10) + i3(40 + 5) + i4(10) = 240 + (40)(19)

i2(2) + i3(−1) + i4(−1) = 0Solving, i2 = 18 A; i3 = 26 A; i4 = 10 ASolve for the requested currents:


ia = 19 − i3 = 19 − 26 = −7 A

ib = i2 − i3 = 18 − 26 = −8 A

ic = i2 − i4 = 18 − 10 = 8 A

id = i4 = 10 A

ie = i2 = 18 A

[b] Find the power in the circuit:

va = 40ia = 40(−7) = −280 V

vb = −10ic − 240 = −10(8) − 240 = −320 V

p19A = −(−280)(19) = 5320 W

pCCCS = −(−320)(2)(−8) = −5120 W

pCCVS = −(4)(10)(18) = −720 W

p240V = −(240)(10) = −2400 W

p40Ω = (40)(−7)2 = 1960 W

p5Ω = (5)(−8)2 = 320 W

p10Ω = (10)(8)2 = 640 W∑

pdev = 5120 + 720 + 2400 = 8240 W

∑pdis = 5320 + 1960 + 320 + 640 = 8240 W(checks)

P 4.48 [a]

125 = 10i1 − 0.4i2 − 9.4i3

125 = −0.4i1 + 20i2 − 19.4i3

0 = −9.4i1 − 19.4i2 + 50i3Solving, i1 = 23.93 A; i2 = 17.79 A; i3 = 11.40 A

Problems 4–57

v1 = 9.4(i1 − i3) = 117.76 V

v2 = 19.4(i2 − i3) = 123.90 V

v3 = 21.2i3 = 241.66 V

[b] pR1 = (i1 − i3)2(9.4) = 1475.22 W

pR2 = (i2 − i3)2(19.4) = 791.29 W

pR3 = i23(21.2) = 2754.64 W

[c]∑

pdev = 125(i1 + i2) = 5213.99 W

∑pload = 5021.15 W

% delivered =5021.155213.99

× 100 = 96.3%

[d]

250 = 29.2i1 − 28.8i2

0 = −28.8i1 + 50i2

Solving, i1 = 19.82 A; i2 = 11.42 A

i1 − i2 = 8.41 A

v1 = (8.41)(9.4) = 79.01 V

v2 = 8.41(19.4) = 163.06 V

Note v1 is low and v2 is high. Therefore, loads designed for 125 V would notfunction properly, and could be damaged.


P 4.49

125 = (R1 + 0.6)ia − 0.4ib − R1ic

125 = −0.4ia + (R2 + 0.6)ib − R2ic

0 = −R1ia − R2ib + (R1 + R2 + 21.2)ic

∆ =

∣∣∣∣∣∣∣∣∣∣∣

(R1 + 0.6) −0.4 −R1

−0.4 (R2 + 0.6) −R2

−R1 −R2 (R1 + R2 + 21.2)

∣∣∣∣∣∣∣∣∣∣∣When R1 = R2, ∆ reduces to

∆ = 21.6R21 + 25.84R1 + 4.24.

Na =

∣∣∣∣∣∣∣125 −0.4 −R1

125 (R2 + 0.6) −R2

0 −R2 (R1 + R2 + 21.2)

∣∣∣∣∣∣∣= 125 [2R1R2 + R1 + 22.2R2 + 21.2]

Nb =

∣∣∣∣∣∣∣(R1 + 0.6) 125 −R1

−0.4 125 −R2

−R1 0 (R1 + R2 + 21.2)

∣∣∣∣∣∣∣= 125 [2R1R2 + 22.2R1 + R2 + 21.2]

ia =Na

∆, ib =

Nb

∆

ineutral = ia − ib =Na − Nb

∆=

125[(R1 − R2) + 22.2(R2 − R1)]∆

Problems 4–59

Now note that when R1 = R2, ineutral reduces to

ineutral =0∆

= 0

P 4.50


−240 + 12i1 + 20(i1 − i2) = 0

20(i2 − i1) + 15(i2 + 4) + 50(i2 + idc) + 40i2 = 0


i1(12 + 20) + i2(−20) + idc(0) = 240

i1(−20) + i2(20 + 15 + 50 + 40) + idc(50) = −60

But if the power associated with the 4 A source is zero, the voltage drop across thesource must be zero. This means that the voltage drop across the 15 Ω resistor is alsozero, so the 15 Ω resistor is effectively removed from the circuit. Once this happens,i2 = −4 A. Substitute this value into the first equation and solve for i1:32i1 − 20(−4) = 240 ·. . 32i1 = 160 so i1 = 5 ANow substitute this value for i1 into the second equation and solve for idc:−20(5) + 125(−4) + 50idc = −60 so 50idc = −60 + 100 + 500 = 540

·. . idc = 540/50 = 10.8 A


P 4.51 [a]

Write the mesh current equations. Note that if io = 0, then i1 = 0:

−23 + 5(−i2) + 10(−i3) + 46 = 0

30i2 + 15(i2 − i3) + 5i2 = 0

Vdc + 25i3 − 46 + 10i3 + 15(i3 − i2) = 0Place the equations in standard form:

i2(−5) + i3(−10) + Vdc(0) = −23

i2(30 + 15 + 5) + i3(−15) + Vdc(0) = 0

i2(−15) + i3(25 + 10 + 15) + Vdc(1) = 46Solving, i2 = 0.6 A; i3 = 2 A; Vdc = −45 VThus, the value of Vdc required to make io = 0 is −45 V.

[b] Calculate the power:

p23V = −(23)(0) = 0 W

p46V = −(46)(2) = −92 W

pVdc = (−45)(2) = −90 W

p30Ω = (30)(0.6)2 = 10.8 W

p5Ω = (5)(0.6)2 = 1.8 W

p15Ω = (15)(2 − 0.6)2 = 29.4 W

p10Ω = (10)(2)2 = 40 W

p20Ω = (20)(0)2 = 0 W

p25Ω = (25)(2)2 = 100 W

∑pdev = 92 + 90 = 182 W

∑pdis = 10.8 + 1.8 + 29.4 + 40 + 0 + 100 = 182 W(checks)

Problems 4–61

P 4.52 [a] There are three unknown node voltages and only two unknown mesh currents.Use the mesh current method to minimize the number of simultaneousequations.

[b]


2i1 + 10(i1 − i2) + 8(i1 − 4) = 0

4i2 + 1(i2 − 4) + 10(i2 − i1) = 0Place the equations in standard form:

i1(2 + 10 + 8) + i2(−10) = 32

i1(−10) + i2(4 + 1 + 10) = 4Solving, i1 = 2.6 A; i2 = 2 AFind the power in the 10 Ω resistor:i10Ω = i1 − i2 = 0.6 Ap10Ω = (0.6)2(10) = 3.6 W

[c] No, the voltage across the 4 A current source is readily available from the meshcurrents, and solving two simultaneous mesh-current equations is less workthan solving three node voltage equations.

[d] vg = 2i1 + 4i2 = 2(2.6) + 4(2) = 13.2 Vp4A = −(13.2)(4) = −52.8 WThus the 4 A source develops 52.8 W.

P 4.53 [a] There are three unknown node voltages and three unknown mesh currents, sothe number of simultaneous equations required is the same for both methods.The node voltage method has the advantage of having to solve the threesimultaneous equations for one unknown voltage provided the connection ateither the top or bottom of the circuit is used as the reference node. Thereforerecommend the node voltage method.

[b]


The node voltage equations are:v1

1+

v1 − v2

8+

v1 − v3

10= 0

−4 +v2

20+

v2 − v1

8+

v2 − v3

2= 0

v3 − v1

10+

v3 − v2

2+

v3

4= 0

Put the equations in standard form:

v1

(1 +

18

+110

)+ v2

(−1

8

)+ v3

(− 1

10

)= 0

v1

(−1

8

)+ v2

( 120

+18

+12

)+ v3

(−1

2

)= 4

v1

(− 1

10

)+ v2

(−1

2

)+ v3

(12

+110

+14

)= 0

Solving, v1 = 1.72 V; v2 = 11.33 V; v3 = 6.87 Vp4A = −(11.33)(4) = −45.32 WTherefore, the 4 A source is developing 45.32 W

P 4.54 [a] The node voltage method requires summing the currents at two supernodes interms of four node voltages and using two constraint equations to reduce thesystem of equations to two unknowns. If the connection at the bottom of thecircuit is used as the reference node, then the voltages controlling thedependent sources are node voltages. This makes it easy to formulate theconstraint equations. The current in the 20 V source is obtained by summingthe currents at either terminal of the source.

The mesh current method requires summing the voltages around the twomeshes not containing current sources in terms of four mesh currents. Inaddition the voltages controlling the dependent sources must be expressed interms of the mesh currents. Thus the constraint equations are morecomplicated, and the reduction to two equations and two unknowns involvesmore algebraic manipulation. The current in the 20 V source is found bysubtracting two mesh currents.

Because the constraint equations are easier to formulate in the node voltagemethod, it is the preferred approach.

Problems 4–63

[b]

Node voltage equations:v1

100+

v2

250− 0.2 + 3 × 10−3v3 = 0

v3

500+

v4

200− 3 × 10−3v3 + 0.2 = 0

Constraints:

v2 − v1 = 20; v4 − v3 = 0.4vα; vα = v2

Solving, v2 = 44 V

io = 0.2 − 44/250 = 24 mA

p20V = 20io = 480 mW (abs)

P 4.55 [a] Apply source transformations to both current sources to get

io =−(5.4 + 0.6)

2700 + 2300 + 1000= −1 mA

[b]

The node voltage equations:


−2 × 10−3 +v1

2700+

v1 − v2

2300= 0

v2

1000+

v2 − v1

2300+ 0.6 × 10−3 = 0


v1

( 12700

+1

2300

)+ v2

(− 1

2300

)= 2 × 10−3

v1

(− 1

2300

)+ v2

( 11000

+1

2300

)= −0.6 × 10−3

Solving, v1 = 2.7 V; v2 = 0.4 V

·. . io =v2 − v1

2300= −1 mA

P 4.56 [a]

Problems 4–65

io = −135/30,000 = −4.5 mA

[b]

va = (7500)(−0.0045) = −33.75 V

ia =va

90,000=

−33.7590,000

= −0.375 mA

ib = −8.4 × 10−3 + 0.375 × 10−3 + 4.5 × 10−3 = −3.525 mA

vb = (6000)(3.525 × 10−3) − 33.75 = −12.6 V

ig =−12.6 − 120

40,000= −3.315 mA

p120V = (120)(−3.315 × 10−3) = −397.8 mW

Check:p8.4mA = (−33.75)(8.4 × 10−3) = −283.5 mW∑

Pdev = 397.8 + 283.5 = 681.3 mW∑Pdis = (40,000)(−3.315 × 10−3)2 +

(−12.6)2

60,000+

(−33.75)2

90,000+(6000)(−3.525 × 10−3)2 + (7500)(−4.5 × 10−3)2

= 681.3 mW

P 4.57 [a]


·. . vo =250300

(480) = 400 V; io =400250

= 1.6 A

[b]

p520V = −(520)(3.6) = −1872 WTherefore, the 520 V source is developing 1872 kW.

[c] v = −(16)(1) − 40(2.6) = −120 Vp1A = (−120)(1) = −120 WTherefore the 1 A source is developing 120 W.

[d] Calculate the power dissipated by the resistors:

p16Ω = (16)(1)2 = 16 W

p260Ω = (260)(2)2 = 1040 W

p40Ω = (40)(2.6)2 = 270.4 W

p4Ω = (4)(1.6)2 = 10.24 W

p250Ω = (250)(1.6)2 = 640 W

p6Ω = (6)(1.6)2 = 15.36 W

∑pdev = 120 + 1872 = 1992 W

∑pdev = 16 + 1040 + 270.4 + 10.24 + 640 + 15.36 = 1992 W (CHECKS)

Problems 4–67

P 4.58 [a] Applying a source transformation to each current source yields

Now combine the 12 V and 5 V sources into a single voltage source and the6 Ω, 6 Ω and 5 Ω resistors into a single resistor to get

Now use a source transformation on each voltage source, thus

which can be reduced to

·. . io =8.510

(−1) = −0.85 A

[b]



6(ia − 2) + 6ia + 5(ia − 1) + 17(ia − io) − 34 = 0

1.5io + 34 + 17(io − ia) = 0Put these equations in standard form:

ia(6 + 6 + 5 + 17) + io(−17) = 12 + 5 + 34

ia(−17) + io(1.5 + 17) = −34Solving, ia = 1.075 A; io = −0.85 A

P 4.59 VTh =30

30 + 10(80) = 60 V

RTh = 10‖30 + 2.5 = 10 Ω

P 4.60

Write and solve the node voltage equation at v1:v1 − 60

10+

v1

40− 4 = 0

4v1 − 240 + v1 − 160 = 0 ·. . v1 = 400/5 = 80 VCalculate VTh:VTh = v1 + (8)(4) = 80 + 32 = 112 VCalculate RTh by removing the independent sources and making series and parallelcombinations of the resistors:RTh = 8 + 40‖10 = 8 + 8 = 16 Ω

Problems 4–69

P 4.61 After making a source transformation the circuit becomes


−500 + 8(i1 − i2) + 12i1 = 0

−300 + 30i2 + 5.2i2 + 8(i2 − i1) = 0

Put the equations in standard form:

i1(8 + 12) + i2(−8) = 500

i1(−8) + i2(30 + 5.2 + 8) = 300

Solving, i1 = 30 A; i2 = 12.5 AVTh = 5.2i2 + 12i1 = 425 VRTh = (8‖12 + 5.2)‖30 = 7.5 Ω

P 4.62 First we make the observation that the 10 mA current source and the 10 kΩ resistorwill have no influence on the behavior of the circuit with respect to the terminals a,b.This follows because they are in parallel with an ideal voltage source. Hence ourcircuit can be simplified to


or

Therefore the Norton equivalent is determined by adding the current sources andcombining the resistors in parallel:

P 4.63 [a] First, find the Thévenin equivalent with respect to a,b using a succession ofsource transformations.

Problems 4–71

·. . VTh = 54 V RTh = 4.5 kΩ

vmeas =85.590

(54) = 51.3 V

[b] %error =(51.3 − 54

54

)× 100 = −5%

P 4.64 [a] Open circuit:

The node voltage equations are:v1 − 17.4

40+

v1

15+ 0.1 = 0

−0.1 +v2

14+

v2 − 17.426

= 0

The above equations are decoupled, so just solve the second equation for v2

and use v2 to solve for voc:−36.4 + 26v2 + 14v2 − 243.6 = 0 ·. . v2 = 280/40 = 7 V

voc =10

10 + 4(7) = 5 V

Short circuit:


Write a node voltage equation at v2:

−0.1 +v2 − 17.4

26+

v2

4= 0

Solving,−5.2 + 2v2 − 34.8 + 13v2 = 0 ·. . v2 = 40/15 VCalculate the short circuit current:isc = (40/15)/4 = 2/3 ATherefore, RTh = 5/(2/3) = 7.5 Ω

[b]

RTh = 10‖(26 + 4) = 7.5 Ω (CHECKS)

Problems 4–73

P 4.65

OPEN CIRCUITUse Ohm’s law to solve for v2 on the right hand side of the circuit:v2 = −80ib(50,000) = −40 × 105ibUse this value of v2 to express the value of the dependent voltage source in terms ofib:4 × 10−5v2 = 4 × 10−5(−40 × 105ib) = −160ibWrite the mesh current equation for the ib mesh:1310ib − 160ib + 100(ib − 500 × 10−6) = 0Solving,1250ib = 0.05 ·. . ib = 0.05/1250 = 40µ AThus,VTh = v2 = −40 × 105ib = −40 × 105(40 × 10−6) = −160 VSHORT CIRCUIT

v2 = 0; isc = −80ib

Calculate ib using current division on the left hand side of the circuit:

ib =100

100 + 1310500 × 10−6 = 35.461 µ A

Calculate the short circuit current from the right hand side of the circuit:

isc = −80(35.461 × 10−6) = −2.8369 × 10−3 mA

Calculate RTh from the short circuit current and open circuit voltage:

RTh =−160

−2.8369 × 10−3 = 56.4 kΩ


P 4.66

12.72 = VTh − 2RTh

12 = VTh − 20RTh

Solving the above equations for VTh and RTh yields

VTh = 12.8 V, RTh = 40 mΩ

·. . IN = 320 A, RN = 40 mΩ

P 4.67 First, find the Thévenin equivalent with respect to Ro.

Problems 4–75

Ro io vo Ro io vo

0 12 0 20 4 80

2 10 20 30 3 90

6 7.5 45 40 2.4 96

10 6 60 50 2 100

15 4.8 72 70 1.5 105

P 4.68

The node voltage equations are:v1 − 9015,000

+v1

10,000+

v1 − v2

4000= 0

v2 − v1

4000+

v2

40,000+

v2 − v3

5000− 19i∆ = 0

v3 − v2

5000+

v3

89,000+ 19i∆ = 0


i∆ =v1 − v2

4000Substitute the constraint equation into the node voltage equations and put the threeremaining equations in standard form:

v1

(1

15,000+

110,000

+1

4000

)+ v2

(− 1

4000

)+ v3(0) =

9015,000

v1

(− 1

4000− 19

4000

)+ v2

(1

4000+

140,000

+1

5000+

194000

)+ v3

(− 1

5000

)= 0

v1

( 194000

)+ v2

(− 1

5000− 19

4000

)+ v3

(1

5000+

189,000

)= 0

Solving, v1 = 32.75 V; v2 = 30.58 V; v3 = −19.8 VVTh = v3 = −19.8 V



−90 + 15,000i1 + 10,000(i1 − i∆) = 0

4000i∆ + 40,000(i∆ − isc) + 10,000(i∆ − i1) = 0

40,000(isc − i∆) + 5000(isc + 19i∆) = 0

Put these equations in standard form:

i1(25,000) + i∆(−10,000) + isc(0) = +90

i1(−10,000) + i∆(54,000) + isc(−40,000) = 0

i1(0) + i∆(55,000) + isc(45,000) = 0

Solving, i1 = 3745.62 µA; i∆ = 364.04 µ A; isc = −444.94 µ Aisc = −444.94 µ ARTh = −19.8/ − 444.94 × 10−6 = 44.5 kΩ

P 4.69 [a] Use source transformations to simplify the left side of the circuit.

ib =7.7 − 5.522,000

= 0.1 mA

Problems 4–77

Let Ro = Rmeter‖1.3 kΩ = 5.5/4.4 × 10−3 = 1250 Ω

·. .(Rmeter)(1300)Rmeter + 1300

= 1250; Rmeter =(1250)(1300)

50= 32.5 kΩ

[b] Actual value of ve:

ib =7.7

22,000 + 44(1300)= 97.22 µ A

ve = 44ib(1300) = 5.56 V

% error =(5.5 − 5.56

5.56

)× 100 = −1.10%

P 4.70 [a] Find the Thévenin equivalent with respect to the terminals of the ammeter. Thisis most easily done by first finding the Thévenin with respect to the terminalsof the 4.8 Ω resistor.Thévenin voltage: note iφ is zero.

VTh − 242

+VTh

100+

VTh

25+

VTh

20= 0

50VTh + VTh + 4VTh + 5VTh = 50(24) ·. . VTh = 50(24)/60 = 20 V

Short-circuit current:

isc = 12 + 2isc, ·. . isc = −12 A

RTh =20

−12= −1.67 Ω


Rtotal =206

= 3.333 Ω

Rmeter = 3.333 − 3.133 = 0.20 Ω

[b] Actual current:

iactual =20

3.133= 6.383 A

% error =6 − 6.383

6.383× 100 = −6%

P 4.71

i1 = 100/20,000 = 5 mA

100 = VTh − 0.005RTh, VTh = 100 + 0.005RTh

Problems 4–79

i2 = 200/50,000 = 4 mA

200 = VTh − 0.004RTh, VTh = 200 + 0.004RTh

·. . 100 + 0.005RTh = 200 + 0.004RTh so RTh = 100 kΩ

VTh = 100 + 500 = 600 V

P 4.72

Use voltage division to calculate v1 and v2:

v1 =501

501 + 100(5) = 4.168053 V

v2 =5000

5000 + 1000(5) = 4.1666667 V

Now calculate VTh:VTh = v1 − v2 = 4.168053 − 4.1666667 = 1.3866 mVCalculate RTh by removing the voltage source and creating series and parallelcombinations of the resisitors:


RTh = 100‖501 + 1000‖5000 =(100)(501)

601+

(1000)(5000)6000

= 916.69 ΩThe resulting Thévenin equivalent circuit is shown below:

Use KVL to calculate igal:

igal =1.3866 × 10−3

916.69 + 50= 1.43 µA

P 4.73 VTh = 0, since circuit contains no independent sources.

iT =vT − v1

20+

vT − 40i∆60

v1 − 40i∆16

+v1

80+

v1 − vT

20= 0

·. . 10v1 − 200i∆ = 4vT i∆ =−v1

80, 200i∆ = −2.5v1

·. . 12.5v1 = 4vT; v1 = 0.32vT

60iT = 4vT − 2.5v1 = 3.2vT

·. .vT

iT=

603.2

= 18.75 Ω

RTh = 18.75 Ω

Problems 4–81

P 4.74 VTh = 0 since there are no independent sources in the circuit. To find RTh, apply a 1A test source and calculate the voltage drop across the test source. Use the meshcurrent method.

The mesh current equations for the two meshes on the left:

−10ix + 5(ix − iy) + 40ix = 0

10ix + 20(iy − 1) + 10iy + 5(iy − ix) = 0


ix(−10 + 5 + 40) + iy(−5) = 0

ix(10 − 5) + iy(20 + 10 + 5) = 20

Solving, ix = 80 mA; iy = 560 mAFind the voltage drop across the 1 A source:vT = 20(1 − 0.56) = 8.8 V·. . RTh = vT/1 A = 8.8/1 = 8.8 Ω

P 4.75 We begin by finding the Thévenin equivalent with respect to Ro. After making acouple of source transformations the circuit simplifies to

i∆ =160 − 30i∆

50; i∆ = 2 A


VTh = 20i∆ + 30i∆ = 50i∆ = 100 V

Using the test-source method to find the Thévenin resistance gives

iT =vT

30+

vT − 30(−vT/30)20

iTvT

=130

+110

=430

=215

RTh =vT

iT=

152

= 7.5 Ω

Thus our problem is reduced to analyzing the circuit shown below.

p =( 100

7.5 + Ro

)2

Ro = 250

104

R2o + 15Ro + 56.25

Ro = 250

104Ro

250= R2

o + 15Ro + 56.25

40Ro = R2o + 15Ro + 56.25

R2o − 25Ro + 56.25 = 0

Ro = 12.5 ± √156.25 − 56.25 = 12.5 ± 10

Ro = 22.5 Ω

Ro = 2.5 Ω

Problems 4–83

P 4.76 [a] Find the Thévenin equivalent with respect to the terminals of RL.Open circuit voltage:


−240 + 3(i1 − i2) + 20(i1 − i3) + 2i1 = 0

2i2 + 4(i2 − i3) + 3(i2 − i1) = 0

10iβ + 1i3 + 20(i3 − i1) + 4(i3 − i2) = 0The dependent source constraint equation is:iβ = i2 − i1


i1(3 + 20 + 2) + i2(−3) + i3(−20) + iβ(0) = 240

i1(−3) + i2(2 + 4 + 3) + i3(−4) + iβ(0) = 0

i1(−20) + i2(−4) + i3(4 + 1 + 20) + iβ(10) = 0

i1(1) + i2(−1) + i3(0) + iβ(1) = 0Solving, i1 = 99.6 A; i2 = 78 A; i3 = 100.8 A; iβ = −21.6 AVTh = 20(i1 − i3) = −24 VShort-circuit current:



−240 + 3(i1 − i2) + 2i1 = 0

2i2 + 4(i2 − i3) + 3(i2 − i1) = 0

10iβ + 1i3 + 4(i3 − i2) = 0The dependent source constraint equation is:iβ = i2 − i1


i1(3 + 2) + i2(−3) + i3(0) + iβ(0) = 240

i1(−3) + i2(2 + 4 + 3) + i3(−4) + iβ(0) = 0

i1(0) + i2(−4) + i3(4 + 1) + iβ(10) = 0

i1(1) + i2(−1) + i3(0) + iβ(1) = 0Solving, i1 = 92 A; i2 = 73.33 A; i3 = 96 A; iβ = −18.67 A

isc = i1 − i3 = −4 A; RTh =VTh

isc=

−24−4

= 6 Ω

RL = RTh = 6 Ω

[b] pmax =122

6= 24 W

P 4.77 [a]

v − 1212,000

+v − 1020,000

+v

12,500= 0

Solving, v = 7.03125 V

v10k =10,00012,500

(7.03125) = 5.625 V

·. . VTh = v − 10 = −4.375 V

Problems 4–85

RTh = [(12,000‖20,000) + 2500] = 5 kΩ

Ro = RTh = 5 kΩ

[b]

pmax = (−437.5 × 10−6)2(5000) = 957.03 µ W

P 4.78 Write KCL equations at each of the labeled nodes, place them in standard form, andsolve:

At v1: − 3 × 10−3 +v1

4000+

v1 − v2

8000= 0

At v2:v2 − v1

8000+

v2 − 1020,000

+v2 − v3

2500= 0

At v3:v3 − v2

2500+

v3

10,000+

v3 − 105000

= 0


Standard form:

v1

( 14000

+1

8000

)+ v2

(− 1

8000

)+ v3(0) = 0.003

v1

(− 1

8000

)+ v2

(1

8000+

120,000

+1

2500

)+ v3

(− 1

2500

)=

1020,000

v1(0) + v2

(− 1

2500

)+ v3

(1

2500+

110,000

+1

5000

)=

105000

Calculator solution:

v1 = 10.890625 V v2 = 8.671875 V v3 = 7.8125 V

Calculate currents:

i2 =10 − v2

20,000= 66.40625µ A i3 =

10 − v3

5000= 437.5 µ A

Calculate power delivered by the sources:

p3mA = (3 × 10−3)v1 = (3 × 10−3)(10.890625) = 32.671875 mW

p10Vmiddle = i2(10) = (66.40625 × 10−6)(10) = 0.6640625 mW

p10Vtop = i3(10) = (437.5 × 10−6)(10) = 4.375 mW

pdeliveredtotal = 32.671875 + 0.6640625 + 4.375 = 37.7109375 mW

Calculate power absorbed by the 5 kΩ resistor and the percentage power:

p5k = i23(5000) = (437.5 × 10−6)2(5000) = 0.95703125 mW

% delivered to Ro:0.9579312537.7109375

(100) = 2.54%

P 4.79 [a] From the solution of Problem 4.68 we have RTh = 44.5 kΩ and VTh = −19.8 V.Therefore

Ro = RTh = 44.5 kΩ

[b] p =(−9.9)2

44,500= 2.2 mW

Problems 4–87

[c]

The node voltage equations are:v1 − 9015,000

+v1

10,000+

v1 − v2

4000= 0

v2 − v1

4000+

v2

40,000+

v2 − v3

5000− 19i∆ = 0

v3 − v2

5000+

v3

89,000+ 19i∆ +

v3

44,500= 0


i∆ =v1 − v2


v1

(1

15,000+

110,000

+1

4000

)+ v2

(− 1

4000

)+ v3(0) + i∆(0) =

9015,000

v1

(− 1

4000

)+ v2

(1

4000+

140,000

+1

5000

)+ v3

(− 1

5000

)+ i∆(−19) = 0

v1(0) + v2

(− 1

5000

)+ v3

(1

5000+

189,000

+1

44,500

)+ i∆(19) = 0

v1

( 14000

)+ v2

(− 1

4000

)+ v3(0) + i∆(−1) = 0

Solving,v1 = 33.2818 V; v2 = 31.4697 V; v3 = −9.9 V; i∆ = 453 µACalculate the power:

ig =90 + 33.2818

15,000= 3.78 mA

p90V = −(90)(3.78 × 10−3) = −340.31 mWpdep source = (v3 − v2)(19i∆) = −356.07 mW∑

pdev = 340.31 + 356.07 = 696.38 mW

% delivered =2.2 × 10−3

696.38 × 10−3 × 100 = 0.316%

P 4.80 [a] From the solution to Problem 4.67 we have


Ro(Ω) Po(W) Ro(Ω) Po(W)

0 0 20 320.00

2 200.00 30 270.00

6 337.50 40 230.40

10 360.00 50 200.00

15 345.60 70 157.50

[b]

[c] Ro = 10 Ω, Po (max) = 360 W

P 4.81 Find the Thévenin equivalent with respect to the terminals of Ro.Open circuit voltage:

(440 − 220) = 5ia − 2ib − 3ic

0 = −2ia + 10ib − ic

ic = 0.5v∆; v∆ = 2(ia − ib); ic = ia − ib

Problems 4–89

Solving, ia = 96.8 A; ib = 26.4 A; ic = 70.4 A; v∆ = 140.8 V

·. . VTh = 7ib = 184.8 V

Short circuit current:

440 − 220 = 5ia − 2isc − 3ic

0 = −2ia + 3isc − 1ic

ic = 0.5v∆; v∆ = 2(ia − isc) ·. . ic = ia − isc

Solving, isc = 60 A; ia = 80 A; ic = 20 A; v∆ = 40 V

RTh = VTh/isc = 184.8/60 = 3.08 Ω

Ro = 3.08 Ω

pRo =(92.4)2

3.08= 2772 W

With Ro equal to 3.08 Ω the circuit becomes


220 = 5i1 − 3(0.5)(2)(i1 − i3) − 2i3 = 2i1 + i3

·. . 2i1 = 220 − i3 = 220 − 43.2 = 176.8 ·. . i1 = 88.4 A

v∆ = 2(i1 − i3) = 90.4 V

i2 = 0.5v∆ = 45.2 A

Thus we have

vc = 220 + 3(43.2) − 2 = 347.6 V

Problems 4–91

Therefore, the only source developing power is the 440 V source.

p440V = −(440)(88.4) = −38,896 W Power delivered is 38,896 W

% delivered =2772

38,896(100) = 7.13%

P 4.82 [a] We begin by finding the Thévenin equivalent with respect to the terminals of Ro.Open circuit voltage


−100 + 4(i1 − i2) + 80(i1 − i3) + 16i1 = 0

124i∆ + 8(i2 − i3) + 4(i2 − i1) = 0

50 + 12i3 + 80(i3 − i1) + 8(i3 − i2) = 0The constraint equation is:i∆ = i3 − i1Place these equations in standard form:

i1(4 + 80 + 16) + i2(−4) + i3(−80) + i∆(0) = 100

i1(−4) + i2(8 + 4) + i3(−8) + i∆(124) = 0

i1(−80) + i2(−8) + i3(12 + 80 + 8) + i∆(0) = −50

i1(1) + i2(0) + i3(−1) + i∆(1) = 0Solving, i1 = 4.7 A; i2 = 10.5 A; i3 = 4.1 A; i∆ = −0.6 AAlso, VTh = vab = −80i∆ = 48 VNow find the short-circuit current.


Note with the short circuit from a to b that i∆ is zero, hence 124i∆ is also zero.

The mesh currents are:−100 + 4(i1 − i2) + 16i1 = 0

8(i2 − i3) + 4(i2 − i1) = 0

50 + 12i3 + 8(i3 − i2) = 0Place these equations in standard form:

i1(4 + 16) + i2(−4) + i3(0) = 100

i1(−4) + i2(8 + 4) + i3(−8) = 0

i1(0) + i2(−8) + i3(12 + 8) = −50Solving, i1 = 5 A; i2 = 0 A; i3 = −2.5 AThen, isc = i1 − i3 = 7.5 ARTh = 48/7.5 = 6.4 Ω

For maximum power transfer Ro = RTh = 6.4 Ω

[b] pmax =242

6.4= 90 W

P 4.83 From the solution of Problem 4.82 we know that when Ro is 6.4 Ω, the voltageacross Ro is 24 V, positive at the upper terminal. Therefore our problem reduces tothe analysis of the following circuit. In constructing the circuit we have used the factthat i∆ is −0.3 A, and hence 124i∆ is −37.2 V.

Problems 4–93

Using the node voltage method to find v1 and v2 yields

4.05 +24 − v1

4+

24 − v2

8= 0

2v1 + v2 = 104.4; v1 + 37.2 = v2

Solving, v1 = 22.4 V; v2 = 59.6 V.It follows that

ig1 =22.4 − 100

16= −4.85 A

ig2 =59.6 − 50

12= 0.8 A

i2 =59.6 − 24

8= 4.45 A

ids = −4.45 − 0.8 = −5.25 A

p100V = 100ig1 = −485 W

p50V = 50ig2 = 40 W

pds = 37.2ids = −195.3 W

·. .∑

pdev = 485 + 195.3 = 680.3 W

·. . % delivered =90

680.3(100) = 13.23%

·. . 13.23% of developed power is delivered to load


P 4.84 [a] Open circuit voltage

Node voltage equations:v1 − 60

2+

v1 − 4i∆5

+v1 − v2

4= 0

v2 − v1

4+ 2v∆ = 0

Constraint equations:

v∆ = 60 − v1

i∆ =v1 − v2

4Place the equations in standard form:

v1

(12

+15

+14

)+ v2

(−1

4

)+ i∆

(−4

5

)+ v∆(0) = 30

v1

(−1

4

)+ v2

(14

)+ i∆(0) + v∆(2) = 0

v1(1) + v2(0) + i∆(0) + v∆(1) = 60

v1(1) + v2(−1) + i∆(−4) + v∆(0) = 0

Solving, v1 = 20 V; v2 = −300 V; i∆ = 80 A; v∆ = 40 VShort circuit current:

The node voltage equation:v1 − 60

2+

v1 − 4i∆5

+v1

4= 0

The constraint equation:

Problems 4–95

i∆ = v1/4Place these equations in standard form:

v1

(12

+15

+14

)+ i∆

(−4

5

)= 30

v1

(14

)+ i∆(−1) = 0

Solving, v1 = 40 V; i∆ = 10 AThen, v∆ = 60 − 40 = 20 Vand isc = i∆ − 2v∆ = 10 − 40 = −30 AThus, RTh = −300/ − 30 = 10 Ω

[b]

pmax =(150)2

10= 2250 W

[c]

The node voltage equation:va − 60

2+

va − 4i∆5

+va + 150

4= 0

The constraint equation is:

i∆ =va + 150

4Place the equations in standard form:

va

(12

+15

+14

)+ i∆

(−4

5

)= 30 − 150

4

va

(−1

4

)+ i∆(1) =

1504

Solving, va = 30 V; i∆ = 45 ACalculate the power:


i60V =va − 60

2= −15 A

p60V = (60)(−15) = −900 W

iccvs =va − 4i∆

5= −30 A

pccvs = 4(45)(−30) = −5400 W

pvccs = (−150)[2(30)] = −9000 W

∑pdev = 900 + 5400 + 9000 = 15,300 W

% delivered =2250

15,3000× 100 = 14.7%

P 4.85 [a] First find the Thévenin equivalent with respect to Ro.Open circuit voltage: iφ = 0; 50iφ = 0

v1

100+

v1 − 28010

+v1 − 280

25+

v1

400+ 0.5125v∆ = 0

v∆ =(280 − v1)

255 = 56 − 0.2v1

v1 = 210 V; v∆ = 14 V

VTh = 280 − v∆ = 280 − 14 = 266 V

Short circuit current

Problems 4–97

v1

100+

v1 − 28010

+v2

20+

v2

400+ 0.5125(280) = 0

v∆ = 280 V

v2 + 50iφ = v1

iφ =2805

+v2

20= 56 + 0.05v2

v2 = −968 V; v1 = −588 V

iφ = isc = 56 + 0.05(−968) = 7.6 A

RTh = VTh/isc = 266/7.6 = 35 Ω

·. . Ro = 35 Ω

[b]

pmax = (133)2/35 = 505.4 W


[c]

v1

100+

v1 − 28010

+v2 − 133

20+

v2

400+ 0.5125(280 − 133) = 0

v2 + 50iφ = v1; iφ = 133/35 = 3.8 A

Therefore, v1 = −189 V and v2 = −379 V; thus,

ig =280 − 133

5+

280 + 18910

= 76.30 A

p280V (dev) = (280)(76.3) = 21,364 W

P 4.86 [a] Since 0 ≤ Ro < ∞ maximum power will be delivered to the 8 Ω resistor whenRo = 0.

[b] P =242

8= 72 W

P 4.87 [a] 110 V source acting alone:

Re =10(14)

24=

356

Ω

i′ =110

5 + 35/6=

13213

A

v′ =(35

6

)(13213

)=

77013

V

4 A source acting alone:

Problems 4–99

5 Ω‖10 Ω = 50/15 = 10/3 Ω

10/3 + 2 = 16/3 Ω

16/3‖12 = 48/13 Ω

Hence our circuit reduces to:

It follows that

v′′a = 4(48/13) = (192/13) V

and

v′′ =−v′′

a

(16/3)(10/3) = −5

8v′′

a = −(120/13) V

·. . v = v′ + v′′ =77013

− 12013

= 50 V

[b] p =v2

o

10= 250 W


P 4.88 70-V source acting alone:

v′ = 70 − 4i′b

i′s =v′

b

2+

v′

10= i′a + i′b

70 = 20i′a + v′b

i′a =70 − v′

b

20

·. . i′b =v′

b

2+

v′

10− 70 − v′

b

20=

1120

v′b +

v′

10− 3.5

v′ = v′b + 2i′b

·. . v′b = v′ − 2i′b

·. . i′b =1120

(v′ − 2i′b) +v′

10− 3.5 or i′b =

1342

v′ − 7042

·. . v′ = 70 − 4(13

42v′ − 70

42

)or v′ =

322094

=161047

V

50-V source acting alone:

v′′ = −4i′′b

Problems 4–101

v′′ = v′′b + 2i′′b

v′′ = −50 + 10i′′d

·. . i′′d =v′′ + 50

10

i′′s =v′′

b

2+

v′′ + 5010

i′′b =v′′

b

20+ i′′s =

v′′b

20+

v′′b

2+

v′′ + 5010

=1120

v′′b +

v′′ + 5010

v′′b = v′′ − 2i′′b

·. . i′′b =1120

(v′′ − 2i′′b ) +v′′ + 50

10or i′′b =

1342

v′′ +10042

Thus, v′′ = −4(13

42v′′ +

10042

)or v′′ = −200

47V

Hence, v = v′ + v′′ =161047

− 20047

=141047

= 30 V

P 4.89 10 V source acting alone:

vo1 =20

20 + 5 + 15(10) = 5 V

20 V source acting alone:

vo2 =13.333

13.333 + 10 + 30(20) = 5 V


6 A current source acting alone:

Node voltage equations:v1

15+

v1 − v2

5− 6 = 0

v2 − v1

5+

v2

40+

v2 − v3

10= 0

v3 − v2

10+

v3

30+ 6 = 0

In standard form:

v1

( 115

+15

)+ v2

(−1

5

)+ v3(0) = 6

v1

(−1

5

)+ v2

(15

+140

+110

)+ v3

(− 1

10

)= 0

v1(0) + v2

(− 1

10

)+ v3

( 110

+130

)= −6

Solving, v1 = 22.5 V; v2 = 0 V; v3 = −45 VNote that vo3 = v2 = 0 VFinally, vo = vo1 + vo2 + vo3 = 5 + 5 + 0 = 10 V

P 4.90 Voltage source acting alone:

vo1 − 254000

+vo1

20,000− 2.2

(vo1 − 25

4000

)= 0

Problems 4–103

Simplifying 5vo1 − 125 + vo1 − 11vo1 + 275 = 0

·. . vo1 = 30 V

Current source acting alone:

vo2

4000+

vo2

20,000+ 0.005 − 2.2

(vo2

4000

)= 0

Simplifying 5vo2 + vo2 + 100 − 11vo2 = 0

·. . vo2 = 20 V

vo = vo1 + vo2 = 30 + 20 = 50 V

P 4.91 Voltage source acting alone:

io1 =−135

40 + 100‖25= −2.25 A

vo1 =6090

(−135) = −90 V



v1

30+

v1

60+ 18 = 0 ·. . v1 = −360 V; vo2 = 360 V

−18 +v2

80+

v2 − v3

20= 0

v3 − v2

20+

v3

25+

v3

40= 0

·. . v2 = 441.6 V; v3 = 192 V; io2 = 192/40 = 4.8 A

·. . vo = vo1 + vo2 = −90 + 360 = 270 V

io = io1 + io2 = −2.25 + 4.8 = 2.55 A

P 4.92 6 A source:

30 Ω‖5 Ω‖60 Ω = 4 Ω

·. . io1 =20

20 + 5(6) = 4.8 A

10 A source:

Problems 4–105

io2 =425

(10) = 1.6 A

75 V source:

io3 = − 425

(15) = −2.4 A

io = io1 + io2 + io3 = 4.8 + 1.6 − 2.4 = 4 A

P 4.93 [a] By hypothesis i′o + i′′o = 3.5 mA.

i′′′o =20008000

(−0.005) = −1.25 mA; ·. . io = 3.5 − 1.25 = 2.25 mA

[b] With all three sources in the circuit write a single node voltage equation.

vb − 82000

+vb

6000+ 0.005 − 0.010 = 0


·. . vb = 13.5 V

io =vb

6000=

13.56000

= 2.25 mA

P 4.94 [a]

voc = VTh = 75 V; iL =6020

= 3 A; iL =75 − 60

RTh=

15RTh

Therefore RTh =153

= 5 Ω

[b] iL =vo

RL

=VTh − vo

RTh

Therefore RTh =VTh − vo

vo/RL

=(

VTh

vo

− 1)

RL

P 4.95 [a]

v − v1

2xr+

v

R+

v − v2

2r(L − x)= 0

v

[1

2xr+

1R

+1

2r(L − x)

]=

v1

2xr+

v2

2r(L − x)

v =v1RL + xR(v2 − v1)RL + 2rLx − 2rx2

[b] Let D = RL + 2rLx − 2rx2

dv

dx=

(RL + 2rLx − 2rx2)R(v2 − v1) − [v1RL + xR(v2 − v1)]2r(L − 2x)D2

dv

dx= 0 when numerator is zero.

The numerator simplifies to

x2 +2L − v1

(v2 − v1)x +

RL(v2 − v1) − 2rv1L2

2r(v2 − v1)= 0

Problems 4–107

Solving for the roots of the quadratic yields

x =L

v2 − v1

−v1 ±

√v1v2 − R

2rL(v2 − v1)2

[c] x =L

v2 − v1

−v1 ±

√v1v2 − R

2rL(v1 − v2)2

v2 = 1200 V, v1 = 1000 V, L = 16 km

r = 5 × 10−5 Ω/m; R = 3.9 Ω

L

v2 − v1=

16,0001200 − 1000

= 80; v1v2 = 1.2 × 106

R

2rL(v1 − v2)2 =

3.9(−200)2

(10 × 10−5)(16 × 103)= 0.975 × 105

x = 80−1000 ±√

1.2 × 106 − 0.0975 × 106= 80−1000 ± 1050 = 80(50) = 4000 m

[d]

vmin =v1RL + R(v2 − v1)xRL + 2rLx − 2rx2

=(1000)(3.9)(16 × 103) + 3.9(200)(4000)

(3.9)(16,000) + 10 × 10−5(16,000)(4000) − 10 × 10−5(16 × 106)= 975 V

P 4.96 [a] In studying the circuit in Fig. P4.96 we note it contains six meshes and sixessential nodes. Further study shows that by replacing the parallel resistorswith their equivalent values the circuit reduces to four meshes and fouressential nodes as shown in the following diagram.

The node Voltage approach will require solving three node Voltage equationsalong with equations involving v∆ and iβ .

The mesh-current approach will require writing one supermesh equation plusthree constraint equations involving the three current sources. Thus at theoutset we know the supermesh equation can be reduced to a single unknowncurrent. Since we are interested in the power developed by the 1 V source, wewill retain the mesh current ib and eliminate the mesh currents ia, ic And id.

The supermesh is denoted by the dashed line in the following figure.


[b] Summing the voltages around the supermesh yields

−9iβ +43ia + 0.75ib + 1 + 5ib + 7(ic − id) + 8ic = 0

Note that iβ = ib And multiply the equation by 12:

−108ib + 16ia + 9ib + 12 + 60ib + 84(ic − id) + 96ic = 0

or

16ia − 39ib + 180ic − 84id = −12

Now note:

ib − ic = 3iβ = 3ib; ·. . ic = −2ib

whence

16ia − 39ib − 360ib − 84id = −12

Now use the constraint that

ia − ic = −2

ia = −2 + ic = −2 − 2ib

Therefore

−32 − 32ib − 399ib − 84id = −12

−431ib − 84id = 20

Now use the constraint

id = −6v∆ = −6(−4

3ia

)= 8ia = −16 − 16ib

Problems 4–109

Therefore

−431ib − 84(−16 − 16ib) = 20

or

913ib = −1324

·. . ib ≈ −1.45 A

p1V = 1ib ∼= −1.45 W; ·. . p1V (developed) ∼= 1.45 W

P 4.97

B–C supernode:vB − 3vx

4+

vB − vE

7− 0.1 = 0

At node E:vE

6+

vE − 3vx

5+

vE − vB

7+ 5 = 0

At node D:vD + 13v∆

3− 5 + 0.1 +

vD

2= 0

Constraint: v∆ = vB − vE

Constraint: vx = −v∆ + 5i∆ − 0.9Constraint: i∆ = (3vx − vB)/4In standard form:

vB

(14

+17

)+ vD(0) + vE

(−1

7

)+ v∆(0) + vx

(−3

4

)+ i∆(0) = 0.1

vB(0) + vD

(12

+13

)+ vE(0) + v∆

(133

)+ vx(0) + i∆(0) = 4.9

vB

(−1

7

)+ vD(0) + vE

(16

+15

+17

)+ v∆(0) + vx

(−3

5

)+ i∆(0) = −5

vB(−1) + vD(0) + vE(1) + v∆(1) + vx(0) + i∆(0) = 0

vB(0) + vD(0) + vE(0) + v∆(1) + vx(1) + i∆(−5) = −0.9

vB(1) + vD(0) + vE(0) + v∆(0) + vx(−3) + i∆(4) = 0


Solving, vB = −11.17 V; vD = −20.95 V; vE = −16.33 V;v∆ = 5.16 V; vx = −2.87 V; i∆ = 0.64 A

p5A = (vE − vD)(5) = 23.1 WThe 5 A source absorbs 23.1 W

P 4.98

The mesh equations are:

−125 + 0.15ia + 18.4(ia − ic) + 0.25(ia − ib) = 0

−125 + 0.25(ib − ia) + 38.4(ib − id) + 0.15ib = 0

0.15ic + 18.4(ic − ie) + 0.25(ic − id) + 18.4(ic − ia) = 0

0.15id + 38.4(id − ib) + 0.25(id − ic) + 38.4(id − ie) = 0

11.6ie + 38.4(ie − id) + 18.4(ie − ic) = 0


ia(18.8) + ib(−0.25) + ic(−18.4) + id(0) + ie(0) = 125

ia(−0.25) + ib(38.8) + ic(0) + id(−38.4) + ie(0) = 125

ia(−18.4) + ib(0) + ic(37.2) + id(−0.25) + ie(−18.4) = 0

ia(0) + ib(−38.4) + ic(−0.25) + id(77.2) + ie(−38.4) = 0

ia(0) + ib(0) + ic(−18.4) + id(−38.4) + ie(68.4) = 0

Solving,ia = 32.77 A; ib = 26.46 A; ic = 26.33 A; id = 23.27 A; ie = 20.14 AFind the requested voltages:v1 = 18.4(ic − ie) = 113.90 Vv2 = 38.4(id − ie) = 120.19 Vv3 = 11.6ie = 233.62 V

Problems 4–111

P 4.99

100 = 6ia − 1ib + 0ic − 2id − 2ie + 0if − 1ig

0 = −1ia + 4ib − 2ic + 0id + 0ie + 0if + 0ig

0 = 0ia − 2ib + 13ic − 3id + 0ie + 0if + 0ig

0 = −2ia + 0ib − 3ic + 9id − 4ie + 0if + 0ig

0 = −2ia + 0ib + 0ic − 4id + 9ie − 3if + 0ig

0 = 0ia + 0ib + 0ic + 0id − 3ie + 13if − 2ig

0 = −1ia + 0ib + 0ic + 0id + 0ie − 2if + 4ig

A computer solution yields

ia = 30 A; ie = 15 A;

ib = 10 A; if = 5 A;

ic = 5 A; ig = 10 A;

id = 15 A

·. . i = id − ie = 0 A

CHECK: p1T = p1B = (ib)2 = (ig)2 = 100 Wp1L = (ia − ib)2 = (ia − ig)2 = 400 Wp2C = 2(ib − ic)2 = (ig − if )2 = 50 Wp3 = 3(ic − id)2 = 3(ie − if )2 = 300 Wp4 = 4(id − ie)2 = 0 Wp8 = 8(ic)2 = 8(if )2 = 200 Wp2L = 2(ia − id)2 = 2(ia − ie)2 = 450 W


∑pabs = 100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300+

200 + 50 + 400 + 100 = 3000 W∑pgen = 100ia = 100(30) = 3000 W (CHECKS)

P 4.100dv1

dIg1=

−R1[R2(R3 + R4) + R3R4](R1 + R2)(R3 + R4) + R3R4

dv1

dIg2=

R1R3R4

(R1 + R2)(R3 + R4) + R3R4

dv2

dIg1+

−R1R3R4

(R1 + R2)(R3 + R4) + R3R4

dv2

dIg2=

R3R4(R1 + R2)(R1 + R2)(R3 + R4) + R3R4

P 4.101 From the solution to Problem 4.100 we have

dv1

dIg1=

−25[5(125) + 3750]30(125) + 3750

= −17512

V/A

and

dv2

dIg1=

−(25)(50)(75)30(125) + 3750

= −12.5 V/A

By hypothesis, ∆Ig1 = 11 − 12 = −1 A

·. . ∆v1 = (−17512

)(−1) =17512

= 14.5833 V

Thus, v1 = 25 + 14.5833 = 39.5833 VAlso,

∆v2 = (−12.5)(−1) = 12.5 V

Thus, v2 = 90 + 12.5 = 102.5 VThe PSpice solution is

v1 = 39.5830 V

and

v2 = 102.5000 V

These values are in agreement with our predicted values.

Problems 4–113

P 4.102 From the solution to Problem 4.100 we have

dv1

dIg2=

(25)(50)(75)30(125) + 3750

= 12.5 V/A

and

dv2

dIg2=

(50)(75)(30)30(125) + 3750

= 15 V/A

By hypothesis, ∆Ig2 = 17 − 16 = 1 A

·. . ∆v1 = (12.5)(1) = 12.5 V

Thus, v1 = 25 + 12.5 = 37.5 VAlso,

∆v2 = (15)(1) = 15 V

Thus, v2 = 90 + 15 = 105 VThe PSpice solution is

v1 = 37.5 V

and

v2 = 105 V


P 4.103 From the solutions to Problems 4.100 — 4.102 we have

dv1

dIg1= −175

12V/A;

dv1

dIg2= 12.5 V/A

dv2

dIg1= −12.5 V/A;

dv2

dIg2= 15 V/A

By hypothesis,

∆Ig1 = 11 − 12 = −1 A

∆Ig2 = 17 − 16 = 1 A

Therefore,

∆v1 =17512

+ 12.5 = 27.0833 V


∆v2 = 12.5 + 15 = 27.5 V

Hence

v1 = 25 + 27.0833 = 52.0833 V

v2 = 90 + 27.5 = 117.5 V

The PSpice solution is

v1 = 52.0830 V

and

v2 = 117.5 V


P 4.104 By hypothesis,

∆R1 = 27.5 − 25 = 2.5 Ω

∆R2 = 4.5 − 5 = −0.5 Ω

∆R3 = 55 − 50 = 5 Ω

∆R4 = 67.5 − 75 = −7.5 Ω

So

∆v1 = 0.5833(2.5) − 5.417(−0.5) + 0.45(5) + 0.2(−7.5) = 4.9168 V

·. . v1 = 25 + 4.9168 = 29.9168 V

∆v2 = 0.5(2.5) + 6.5(−0.5) + 0.54(5) + 0.24(−7.5) = −1.1 V

·. . v2 = 90 − 1.1 = 88.9 V

The PSpice solution is

v1 = 29.6710 V

and

v2 = 88.5260 V

Note our predicted values are within a fraction of a volt of the actual values.

5The Operational Amplifier

Assessment Problems

AP 5.1 [a] This is an inverting amplifier, so

vo = (−Rf/Ri)vs = (−80/16)vs, so vo = −5vs

vs( V) 0.4 2.0 3.5 −0.6 −1.6 −2.4

vo( V) −2.0 −10.0 −15.0 3.0 8.0 10.0Two of the vs values, 3.5 V and −2.4 V, cause the op amp to saturate.

[b] Use the negative power supply value to determine the largest input voltage:

−15 = −5vs, vs = 3 V

Use the positive power supply value to determine the smallest input voltage:

10 = −5vs, vs = −2 V

Therefore − 2 ≤ vs ≤ 3 V

AP 5.2 From Assessment Problem 5.1

vo = (−Rf/Ri)vs = (−Rx/16,000)vs

= (−Rx/16,000)(−0.640) = 0.64Rx/16,000 = 4×10−5Rx

Use the negative power supply value to determine one limit on the value of Rx:

4×10−5Rx = −15 so Rx = −15/4×10−5 = −375 kΩ

Since we cannot have negative resistor values, the lower limit for Rx is 0. Now usethe positive power supply value to determine the upper limit on the value of Rx:

4×10−5Rx = 10 so Rx = 10/4×10−5 = 250 kΩ

Therefore,

0 ≤ Rx ≤ 250 kΩ

5–1

5–2 CHAPTER 5. The Operational Amplifier

AP 5.3 [a] This is an inverting summing amplifier so

vo = (−Rf/Ra)va + (−Rf/Rb)vb = −(250/5)va − (250/25)vb = −50va − 10vb

Substituting the values for va and vb:

vo = −50(0.1) − 10(0.25) = −5 − 2.5 = −7.5 V

[b] Substitute the value for vb into the equation for vo from part (a) and use thenegative power supply value:

vo = −50va − 10(0.25) = −50va − 2.5 = −10 V

Therefore 50va = 7.5, so va = 0.15 V

[c] Substitute the value for va into the equation for vo from part (a) and use thenegative power supply value:

vo = −50(0.10) − 10vb = −5 − 10vb = −10 V;

Therefore 10vb = 5, so vb = 0.5 V

[d] The effect of reversing polarity is to change the sign on the vb term in eachequation from negative to positive.Repeat part (a):

vo = −50va + 10vb = −5 + 2.5 = −2.5 V

Repeat part (b):

vo = −50va + 2.5 = −10 V; 50va = 12.5, va = 0.25 V

Repeat part (c):

vo = −5 + 10vb = 15 V; 10vb = 20; vb = 2.0 V

AP 5.4 [a] Write a node voltage equation at vn; remember that for an ideal op amp, thecurrent into the op amp at the inputs is zero:

vn

4500+

vn − vo

63,000= 0

Solve for vo in terms of vn by multiplying both sides by 63,000 and collectingterms:

14vn + vn − vo = 0 so vo = 15vn

Now use voltage division to calculate vp. We can use voltage division becausethe op amp is ideal, so no current flows into the non-inverting input terminaland the 400 mV divides between the 15 kΩ resistor and the Rx resistor:

vp =Rx

15,000 + Rx

(0.400)

Problems 5–3

Now substitute the value Rx = 60 kΩ:

vp =60,000

15,000 + 60,000(0.400) = 0.32 V

Finally, remember that for an ideal op amp, vn = vp, so substitute the value ofvp into the equation for v0

vo = 15vn = 15vp = 15(0.32) = 4.8 V

[b] Substitute the expression for vp into the equation for vo and set the resultingequation equal to the positive power supply value:

vo = 15(

0.4Rx

15,000 + Rx

)= 5

15(0.4Rx) = 5(15,000 + Rx) so Rx = 75 kΩ

AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the outputvoltage in terms of the input voltages and the resistor values given in Eq. 5.22:

vo =20(60)10(24)

vb − 5010

va

Simplify this expression and substitute in the value for vb:

vo = 5(vb − va) = 20 − 5va

Set this expression for vo to the positive power supply value:

20 − 5va = 10 V so va = 2 V

Now set the expression for vo to the negative power supply value:

20 − 5va = −10 V so va = 6 V

Therefore 2 ≤ va ≤ 6 V

[b] Begin as before by substituting the appropriate values into Eq. 5.22:

vo =8(60)10(12)

vb − 5va = 4vb − 5va

Now substitute the value for vb:

vo = 4(4) − 5va = 16 − 5va

Set this expression for vo to the positive power supply value:

16 − 5va = 10 V so va = 1.2 V

Now set the expression for vo to the negative power supply value:

16 − 5va = −10 V so va = 5.2 V

Therefore 1.2 ≤ va ≤ 5.2 V


AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig. 5.15:

Write the node voltage equation at the left hand node:

vn

500,000+

vn − vg

5000+

vn − vo

100,000= 0

Multiply both sides by 500,000 and simplify:

vn + 100vn − 100vg + 5vn − 5v0 = 0 so 21.2vn − vo = 20vg

Write the node voltage equation at the right hand node:

vo − 300,000(−vn)5000

+vo − vn

100,000= 0

Multiply through by 100,000 and simplify:

20vo + 6 × 106vn + vo − vn = 0 so 6 × 106vn + 21vo = 0

Use Cramer’s method to solve for vo:

∆ =

∣∣∣∣∣∣∣21.2 −1

6 × 106 21

∣∣∣∣∣∣∣ = 6,000,445.2

No =

∣∣∣∣∣∣∣21.2 20vg

6 × 106 0

∣∣∣∣∣∣∣ = −120 × 106vg

vo =No

∆= −19.9985vg; so

vo

vg

= −19.9985

[b] Use Cramer’s method again to solve for vn:

N1 =

∣∣∣∣∣∣∣20vg −1

0 21

∣∣∣∣∣∣∣ = 420vg

vn =N1

∆= 6.9995 × 10−5vg

vg = 1 V, vn = 69.995 µ V

Problems 5–5

[c] The resistance seen at the input to the op amp is the ratio of the input voltage tothe input current, so calculate the input current as a function of the inputvoltage:

ig =vg − vn

5000=

vg − 6.9995 × 10−5vg

5000Solve for the ratio of vg to ig to get the input resistance:

Rg =vg

ig

=5000

1 − 6.9995 × 10−5 = 5000.35 Ω

[d] This is a simple inverting amplifier configuration, so the voltage gain is the ratioof the feedback resistance to the input resistance:

vo

vg

= −100,0005000

= −20

Since this is now an ideal op amp, the voltage difference between the two inputterminals is zero; since vp = 0, vn = 0Since there is no current into the inputs of an ideal op amp, the resistance seenby the input voltage source is the input resistance:

Rg = 5000 Ω


Problems

P 5.1 [a] The five terminals of the op amp are identified as follows:

[b] The input resistance of an ideal op amp is infinite, which constrains the value ofthe input currents to 0. Thus, in = 0 A.

[c] The open-loop voltage gain of an ideal op amp is infinite, which constrains thedifference between the voltage at the two input terminals to 0. Thus,(vp − vn) = 0.

[d] Write a node voltage equation at vn:

vn − 2.510,000

+vn − vo

40,000= 0

But vp = 0 and vn = vp = 0. Thus,

−2.510,000

− vo

40,000= 0 so vo = −10 V

P 5.2vb − va

20+

vb − vo

100= 0, therefore vo = 6vb − 5va

[a] va = 4 V, vb = 0 V, vo = −15 V (sat)

[b] va = 2 V, vb = 0 V, vo = −10 V

[c] va = 2 V, vb = 1 V, vo = −4 V

[d] va = 1 V, vb = 2 V, vo = 7 V

[e] If vb = 1.6 V, vo = 9.6 − 5va = ±15

·. . −1.08 ≤ va ≤ 4.92 V

P 5.3 vo = −(0.5 × 10−3)(10 × 103) = −5 V

·. . io =−5

5000= −1 mA

P 5.4 Since the current into the inverting input terminal of an ideal op-amp is zero, thevoltage across the 2.2 MΩ resistor is (2.2 × 106)(3.5 × 10−6) or 7.7 V. Thereforethe voltmeter reads 7.7 V.

Problems 5–7

P 5.5 [a] ia =25 × 10−3

5000= 5 µA

va = −50 × 103ia = −250 mV

[b]va

50,000+

va

10,000+

va − vo

40,000= 0

·. . 4va + 20va + 5va − 5vo = 0

·. . vo = 29va/5 = −1.45 V

[c] ia = 5 µA

[d] io =−vo

30,000+

va − vo

40,000= 78.33 µ A

P 5.6 [a] The gain of an inverting amplifier is the ratio of the feedback resistor to theinput resistor. If the gain of the inverting amplifier is to be 6, the feedbackresistor must be 6 times as large as the input resistor. There are many possibledesigns that use only 20 kΩ resistors. We present two here. Use a single 20 kΩresistor as the input resistor, and use six 20 kΩ resistors in series as thefeedback resistor to give a total of 120 kΩ.

Alternately, Use a single 20 kΩ resistor as the feedback resistor and use six 20kΩ resistors in parallel as the input resistor to give a total of 3.33 kΩ.

[b] To amplify a 3 V signal without saturating the op amp, the power supplyvoltages must be greater than or equal to the product of the input voltage andthe amplifier gain. Thus, the power supplies should have a magnitude of(3)(6) = 18 V.

P 5.7 [a] The circuit shown is a non-inverting amplifier.


[b] We assume the op amp to be ideal, so vn = vp = 3V. Write a KCL equation atvn:

340,000

+3 − vo

80,000= 0

Solving,

vo = 9 V.

P 5.8 vp =1824

(12) = 9 V = vn

vn − 2430

+vn − vo

20= 0

vo = (45 − 48)/3 = −1.0 V

iL =vo

5× 10−3 = − 1

5× 10−3 = −200 × 10−6

iL = −200 µA

P 5.9 [a] First, note that vn = vp = 2.5 VLet vo1 equal the voltage output of the op-amp. Then

2.5 − vg

5000+

2.5 − vo1

10,000= 0, ·. . vo1 = 7.5 − 2vg

Also note that vo1 − 2.5 = vo, ·. . vo = 5 − 2vg

[b] Yes, the circuit designer is correct!

Problems 5–9

P 5.10 [a] Let v∆ be the voltage from the potentiometer contact to ground. Then

0 − vg

2000+

0 − v∆

50,000= 0

−25vg − v∆ = 0, ·. . v∆ = −25(40 × 10−3) = −1 V

v∆

αR∆+

v∆ − 050,000

+v∆ − vo

(1 − α)R∆= 0

v∆

α+ 2v∆ +

v∆ − vo

1 − α= 0

v∆

( 1α

+ 2 +1

1 − α

)=

vo

1 − α

·. . vo = −1[1 + 2(1 − α) +

(1 − α)α

]

When α = 0.2, vo = −1(1 + 1.6 + 4) = −6.6 V

When α = 1, vo = −1(1 + 0 + 0) = −1 V

·. . −6.6 V ≤ vo ≤ −1 V

[b] −1[1 + 2(1 − α) +

(1 − α)α

]= −7

α + 2α(1 − α) + (1 − α) = 7α

α + 2α − 2α2 + 1 − α = 7α

·. . 2α2 + 5α − 1 = 0 so α ∼= 0.186

P 5.11 [a] Replace the combination of vg, 1.6 kΩ, and the 6.4 kΩ resistors with itsThévenin equivalent.

Then vo =−[12 + σ50]

1.28(0.2)

At saturation vo = −5 V; therefore

−(12 + σ50

1.28

)(0.2) = −5, or σ = 0.4

Thus for 0 ≤ σ < 0.40 the operational amplifier will not saturate.


[b] When σ = 0.272, vo =−(12 + 13.6)

1.28(0.2) = −4 V

Alsovo

10+

vo

25.6+ io = 0

·. . io = − vo

10− vo

25.6=

410

+4

25.6mA = 556.25 µA

P 5.12 [a]

vn − va

R+

vn − vo

R= 0

2vn − va = vo

va

Ra+

va − vn

R+

va − vo

R= 0

va

[ 1Ra

+2R

]− vn

R=

vo

R

va

(2 +

R

Ra

)− vn = vo

vn = vp = va + vg

·. . 2vn − va = 2va + 2vg − va = va + 2vg

·. . va − vo = −2vg (1)

2va + va

(R

Ra

)− va − vg = vo

·. . va

(1 +

R

Ra

)− vo = vg (2)

Now combining equations (1) and (2) yields

−vaR

Ra= −3vg

Problems 5–11

or va = 3vgRa

R

Hence ia =va

Ra=

3vg

RQ.E.D.

[b] At saturation Vo = ± Vcc

·. . va = ± Vcc − 2vg (3)

and

·. . va

(1 +

R

Ra

)= ± Vcc + vg (4)

Dividing Eq (4) by Eq (3) gives

1 +R

Ra=

± Vcc + vg

± Vcc − 2vg

·. .R

Ra=

± Vcc + vg

± Vcc − 2vg

− 1 =3vg

± Vcc − 2vg

or Ra =(± Vcc − 2vg)

3vg

R Q.E.D.

P 5.13 [a] Assume the op-amp is operating within its linear range, then

iL =8

4000= 2 mA

For RL = 4 kΩ vo = (4 + 4)(2) = 16 V

Now since vo < 20 V our assumption of linear operation is correct, therefore

iL = 2 mA

[b] 20 = 2(4 + RL); RL = 6 kΩ

[c] As long as the op-amp is operating in its linear region iL is independent of RL.From (b) we found the op-amp is operating in its linear region as long asRL ≤ 6 kΩ. Therefore when RL = 16 kΩ the op-amp is saturated. We canestimate the value of iL by assuming ip = in iL. TheniL = 20/(4,0000 + 16,000) = 1 mA. To justify neglecting the current into theop-amp assume the drop across the 50 kΩ resistor is negligible, and the inputresistance to the op-amp is at least 500 kΩ. Thenip = in = (8 − 4)/(500 × 103) = 8 µA. But 8 µA 1 mA, hence ourassumption is reasonable.


[d]

P 5.14 [a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output voltage ofthe amplifier on the right. Then

vo1 =−4710

(1) = −4.7 V; vo2 =−22033

(−0.15) = 1.0 V

ia =vo2 − vo1

1000= 5.7 mA

[b] ia = 0 when vo1 = v02 so from (a) vo2 = 1 V

Thus−4710

(vL) = 1

vL = −1047

= −212.77 mV

P 5.15 [a] p600Ω =(60 × 10−3)2

(600)= 6 µW

[b] v600Ω =600

30,000(60 × 10−3) = 1.2 mV

p600Ω =(1.2 × 10−3)2

(600)= 2.4 nW

[c]pa

pb=

6 × 10−6

2.4 × 10−9 = 2500

[d] Yes, the operational amplifier serves several useful purposes:

Problems 5–13

• First, it enables the source to control 2, 500 times as much power deliveredto the load resistor. When a small amount of power controls a largeramount of power, we refer to it as power amplification.

• Second, it allows the full source voltage to appear across the load resistor,no matter what the source resistance. This is the voltage follower functionof the operational amplifier.

• Third, it allows the load resistor voltage (and thus its current) to be setwithout drawing any current from the input voltage source. This is thecurrent amplification function of the circuit.

P 5.16 [a] This circuit is an example of an inverting summing amplifier.

[b] vo = −22033

va − 22022

vb − 22080

vc = −8 + 15 − 11 = −4 V

[c] vo = −19 − 10vb = ±6

·. . vb = −1.3 V when vo = −6 V;

vb = −2.5 V when vo = 6 V

·. . −2.5 V ≤ vb ≤ −1.3 V

P 5.17 [a] Write a KCL equation at the inverting input to the op amp:

vd − va

40,000+

vd − vb

22,000+

vd − vc

100,000+

vd

352,000+

vd − vo

220,000= 0

Multiply through by 220,000, plug in the values of input voltages, andrearrange to solve for vo:

vo = 220,000(

440,000

+−1

22,000+

−5100,000

+8

352,000+

8220,000

)= 14 V

[b] Write a KCL equation at the inverting input to the op amp. Use the given valuesof input voltages in the equation:

8 − va

40,000+

8 − 922,000

+8 − 13100,000

+8

352,000+

8 − vo

220,000= 0

Simplify and solve for vo:

44 − 5.5va − 10 − 11 + 5 + 8 − vo = 0 so vo = 36 − 5.5va

Set vo to the positive power supply voltage and solve for va:

36 − 5.5va = 15 ·. . va = 3.818 V

Set vo to the negative power supply voltage and solve for va:

36 − 5.5va = −15 ·. . va = 9.273 V


Therefore,

3.818 V ≤ va ≤ 9.273 V

P 5.18 [a]8 − 440,000

+8 − 922,000

+8 − 13100,000

+8

352,000+

8 − v0

Rf

= 0

8 − vo

Rf

= −2.7272 × 10−5 so Rf =8 − vo

−2.727 × 10−5

For vo = 15 V, Rf = 256.7 kΩ

For vo = −15 V, Rf < 0 so this solution is not possible.

[b] io = −(if + i10k) = −[

15 − 8256.7 × 103 +

1510,000

]= −1.527 mA

P 5.19 We want the following expression for the output voltage:

vo = −(2va + 4vb + 6vc + 8vd)

This is an inverting summing amplifier, so each input voltage is amplified by a gainthat is the ratio of the feedback resistance to the resistance in the forward path forthe input voltage:

vo = −[ 48Ra

va +48Rb

vb +48Rc

vc +48Rd

vd

]

Solve for each input resistance value to yield the desired gain:·. . Ra = 48,000/2 = 24 kΩ Rc = 48,000/6 = 8 kΩ

Rb = 48,000/4 = 12 kΩ Rd = 48,000/8 = 6 kΩ

The final circuit is shown here:

P 5.20 [a] vp = vs, vn =R1vo

R1 + R2, vn = vp

Therefore vo =(

R1 + R2

R1

)vs =

(1 +

R2

R1

)vs

Problems 5–15

[b] vo = vs

[c] Because vo = vs, thus the output voltage follows the signal voltage.

P 5.21 vo = −[

Rf

3000(0.15) +

Rf

5000(0.1) +

Rf

25,000(0.25)

]

−6 = −8 × 10−5Rf ; Rf = 75 kΩ; ·. . 0 ≤ Rf ≤ 75 kΩ

P 5.22 [a] This circuit is an example of the non-inverting amplifier.

[b] Use voltage division to calculate vp:

vp =10,000

10,000 + 30,000vs =

vs

4

Write a KCL equation at vn = vp = vs/4:

vs/44000

+vs/4 − vo

28,000= 0

Solving,

vo = 7vs/4 + vs/4 = 2vs

[c] 2vs = 8 so vs = 4 V

2vs = −12 so vs = −6 V

Thus, −6 V ≤ vs ≤ 4 V.

P 5.23 [a] vp = vn =68,00080,000

vg = 0.85vg

·. .0.85vg

30,000+

0.85vg − vo

63,000= 0;

·. . vo = 2.635vg = 2.635(4), vo = 10.54 V

[b] vo = 2.635vg = ±12

vg = ±4.554 V, −4.554 ≤ vg ≤ 4.554 V

[c]0.85vg

30,000+

0.85vg − vo

Rf= 0

(0.85Rf

30,000+ 0.85

)vg = vo = ±12

·. . 1.7×10−3Rf + 51 = ±360; 1.7×10−3Rf = 360 − 51; Rf = 181.76 kΩ

P 5.24 [a] This circuit is an example of a non-inverting summing amplifier.


[b] Write a KCL equation at vp and solve for vp in terms of vs:

vp − vs

15,000+

vp − 630,000

= 0

2vp − 2vs + vp − 6 = 0 so vp = 2vs/3 + 2

Now write a KCL equation at vn and solve for vo:

vn

20,000+

vn − vo

60,000= 0 so vo = 4vn

Since we assume the op amp is ideal, vn = vp. Thus,

vo = 4(2vs/3 + 2) = 8vs/3 + 8

[c] 8vs/3 + 8 = 16 so vs = 3 V

8vs/3 + 8 = −12 so vs = −7.5 V

Thus, −7.5 V ≤ vs ≤ 3 V.

P 5.25 [a] The circuit is a non-inverting summing amplifier.

[b]vp − va

3.3 × 103 +vp − vb

4.7 × 103 = 0

·. . vp = 0.5875va + 0.4125vb

vn

10,000+

vn − vo

100,000= 0

·. . vo = 11vn = 11vp = 6.4625va + 4.5375vb = 8.03 V

[c] vp = vn =vo

11= 730 mV

ia =va − vp

3.3 × 103 = −100 µA

ib =vb − vp

4.7 × 103 = 100 µA

[d] 6.4625 for va

4.5375 for vb

P 5.26 [a]vp − va

Ra+

vp − vb

Rb+

vp − vc

Rc+

vp

Rg

= 0

·. . vp =RbRcRg

Dva +

RaRcRg

Dvb +

RaRbRg

Dvc

where D = RbRcRg + RaRcRg + RaRbRg + RaRbRc

Problems 5–17

vn

Rs

+vn − vo

Rf= 0

vn

( 1Rs

+1Rf

)=

vo

Rf

·. . vo =(1 +

Rf

Rs

)vn = kvn

where k =(1 +

Rf

Rs

)

vp = vn

·. . vo = kvp

or

vo =kRgRbRc

Dva +

kRgRaRc

Dvb +

kRgRaRb

Dvc

kRgRbRc

D= 3 ·. .

Rb

Ra= 1.5

kRgRaRc

D= 2 ·. .

Rc

Rb= 2

kRgRaRb

D= 1 ·. .

Rc

Ra= 3

Since Ra = 2 kΩ Rb = 3 kΩ Rc = 6 kΩ

·. . D = [(3)(6)(4) + (2)(6)(4) + (2)(3)(4) + (2)(3)(6)] × 109 = 180 × 109

k(4)(3)(6) × 109

180 × 109 = 3

k =540 × 109

72 × 109 = 7.5

·. . 7.5 = 1 +Rf

Rs

Rf

Rs

= 6.5

Rf = (6.5)(12,000) = 78 kΩ

[b] vo = 3(0.8) + 2(1.5) + 2.10 = 7.5 V

vn = vp =7.57.5

= 1.0 V

ia =0.8 − 12000

=−0.22000

= −0.1 mA = −100 µA


ib =1.5 − 1.0

3000=

0.53000

= 166.67 µA

ic =2.10 − 1.0

6000=

1.16000

= 183.33 µA

ig =1

4000= 250 µA

is =vn

12,000=

112,000

= 83.33 µA

P 5.27 [a]vp − va

Ra+

vp − vb

Rb+

vp − vc

Rc= 0

·. . vp =RbRc

Dva +

RaRc

Dvb +

RaRb

Dvc

where D = RbRc + RaRc + RaRb

vn

10,000+

vn − vo

Rf= 0

(Rf

10,000+ 1

)vn = vo

LetRf

10,000+ 1 = k

vo = kvn = kvp

·. . vo =kRbRc

Dva +

kRaRc

Dvb +

kRaRb

Dvc

·. .kRbRc

D= 5 ·. .

Rc

Ra= 5

kRaRc

D= 4

kRaRb

D= 1 ·. .

Rc

Rb= 4

·. . Rc = 5Ra = 5 kΩ

Rb = Rc/4 = 1.25 kΩ

·. . D = (1.25)(5) + (1)(5) + (1.25)(1) = 12.5 × 106

·. . k =5D

RbRc=

(5)(12.5) × 106

(1.25)(5) × 106 = 10

·. .Rf

10,000+ 1 = 10, Rf = 90 kΩ

Problems 5–19

[b] vo = 5(0.5) + 4(1) + 1.5 = 8 V

vn = vo/10 = 0.8 V = vp

ia =va − vp

1000=

0.5 − 0.81000

= −300 µA

ib =vb − vp

1250=

1 − 0.81250

= 160 µA

ic =vc − vp

5000=

1.5 − 0.85000

= 140 µA

P 5.28 [a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0,therefore vn = 0 and we have

0 − va

Ra+

0 − v′o

Rb+ in = 0, in = 0

Therefore

v′o

Rb= − va

Ra, v′

o = −Rb

Rava

Assume vb is acting alone. Replace va with a short circuit. Now

vp = vn =vbRd

Rc + Rd

vn

Ra+

vn − v′′o

Rb+ in = 0, in = 0

( 1Ra

+1

Rb

)(Rd

Rc + Rd

)vb − v′′

o

Rb= 0

v′′o =

(Rb

Ra+ 1

)(Rd

Rc + Rd

)vb =

Rd

Ra

(Ra + Rb

Rc + Rd

)vb

vo = v′o + v′′

o =Rd

Ra

(Ra + Rb

Rc + Rd

)vb − Rb

Rava

[b]Rd

Ra

(Ra + Rb

Rc + Rd

)=

Rb

Ra, therefore Rd(Ra + Rb) = Rb(Rc + Rd)

RdRa = RbRc, thereforeRa

Rb=

Rc

Rd

WhenRd

Ra

(Ra + Rb

Rc + Rd

)=

Rb

Ra

Eq. (5.22) reduces to vo =Rb

Ravb − Rb

Rava =

Rb

Ra(vb − va).


P 5.29 Use voltage division to find vp:

vp =2000

2000 + 8000(5) = 1 V

Write a KCL equation at vn and solve it for vo:

vn − va

5000+

vn − vo

Rf

= 0 so(

Rf

5000+ 1

)vn − Rf

5000va = vo

Since the op amp is ideal, vn = vp = 1V, so

vo =(

Rf

5000+ 1

)− Rf

5000va

To satisfy the equation,

(Rf

5000+ 1

)= 5 and

Rf

5000= 4

Thus, Rf = 20 kΩ.

P 5.30 [a]

vp

72,000+

vp − vc

9,000+

vp − vd

24,000= 0

·. . vp = (2/3)vc + 0.25vd = vn

vn − va

12,000+

vn − vb

18,000+

vn − vo

144,000= 0

·. . vo = 21vn − 12va − 8vb

= 21[(2/3)vc + 0.25vd] − 12va − 8vb

= 21(0.4 + 0.2) − 12(0.5) − 8(0.3) = 4.2 V

Problems 5–21

[b] vo = 14vc + 4.2 − 6 − 2.4

±15 = 14vc − 4.2

·. . 14vc = ±15 + 4.2

·. . vc = 1.371 V and vc = −0.771 V

·. . −771 ≤ vc ≤ 1371 mV

P 5.31 [a] vo =Rd(Ra + Rb)Ra(Rc + Rd)

vb − Rb

Rava =

47(110)10(80)

(0.80) − 10(0.67)

vo = 5.17 − 6.70 = −1.53 V

[b] vn = vp =(800)(47)

80= 470 mV

ia =(670 − 470)10−3

10 × 103 = 20 µA

Ra =va

ia=

670 × 10−3

20 × 10−6 = 33.5 kΩ

[c] Rin b = Rc + Rd = 80 kΩ

P 5.32 vp =vbRb

Ra + Rb= vn

vn − va

4700+

vn − vo

Rf= 0

vn

(Rf

4700+ 1

)− vaRf

4700= vo

·. .(

Rf

4700+ 1

)Rb

Ra + Rbvb − Rf

4700va = vo

·. .Rf

4700= 10; Rf = 47 kΩ

·. .Rf

4700+ 1 = 11

·. . 11(

Rb

Ra + Rb

)= 10

11Rb = 10Rb + 10Ra Rb = 10Ra


Ra + Rb = 220 kΩ

11Ra = 220 kΩ

Ra = 20 kΩ

Rb = 220 − 20 = 200 kΩ

P 5.33 vp = vn = Rbib

Rbib − 3000ia3000

+Rbib − vo

Rf= 0

(Rb

3000+

Rb

Rf

)ib − ia =

vo

Rf

vo =[RbRf

3000+ Rb

]ib − Rfia

·. . Rf = 2000 Ω

(2/3)Rb + Rb = 2000

·. . Rb = 1200 Ω

P 5.34 vo =Rd(Ra + Rb)Ra(Rc + Rd)

vb − Rb

Rava

By hypothesis: Rb/Ra = 4; Rc + Rd = 470 kΩ;Rd(Ra + Rb)Ra(Rc + Rd)

= 3

·. .Rd

Ra

(Ra + 4Ra)470,000

= 3 so Rd = 282 kΩ; Rc = 188 kΩ

Also, when vo = 0 we have

vn − va

Ra+

vn

Rb= 0

·. . vn

(1 +

Ra

Rb

)= va; vn = 0.8va

ia =va − 0.8va

Ra= 0.2

va

Ra; Rin =

va

ia= 5Ra = 22 kΩ

·. . Ra = 4.4 kΩ; Rb = 17.6 kΩ

Problems 5–23

P 5.35 [a] vn = vp = αvg vo = (αvg − vg)4 + αvg

vn − vg

R1+

vn − vo

Rf= 0 = [(α − 1)4 + α]vg

(vn − vg)Rf

R1+ vn − vo = 0 = (5α − 4)vg

= (5α − 4)(2) = 10α − 8

α vo α vo α vo

0.0 −8 V 0.4 −4 V 0.8 0 V

0.1 −7 V 0.5 −3 V 0.9 1 V

0.2 −6 V 0.6 −2 V 1.0 2 V

0.3 −5 V 0.7 −1 V

[b] Rearranging the equation for vo from (a) gives

vo =(

Rf

R1+ 1

)vgα + −

(Rf

R1

)vg

Therefore,

slope =(

Rf

R1+ 1

)vg; intercept = −

(Rf

R1

)vg

[c] Using the equations from (b),

−6 =(

Rf

R1+ 1

)vg; 4 = −

(Rf

R1

)vg

Solving,

vg = −2 V;Rf

R1= 2

P 5.36 vp =15009000

(−18) = −3 V = vn

18 − 31600

+−3 − vo

Rf= 0


·. . vo =15

1600Rf − 3

vo = 9 V; Rf = 1280 Ω

vo = −9 V; Rf = −640 Ω

But Rf ≥ 0, ·. . Rf = 1280 Ω

P 5.37 [a] Adm =(24)(26) + (25)(25)

(2)(1)(25)= 24.98

[b] Acm =(1)(24) − 25(1)

1(25)= −0.04

[c] CMRR =∣∣∣∣24.980.04

∣∣∣∣ = 624.50

P 5.38 Acm =(20)(50) − (50)Rx

20(50 + Rx)

Adm =50(20 + 50) + 50(50 + Rx)

2(20)(50 + Rx)

Adm

Acm=

Rx + 1202(20 − Rx)

·. .Rx + 120

2(20 − Rx)= ±1000 for the limits on the value of Rx

If we use +1000 Rx = 19.93 kΩ

If we use −1000 Rx = 20.07 kΩ

19.93 kΩ ≤ Rx ≤ 20.07 kΩ

P 5.39 [a] Replace the op amp with the model from Fig. 5.15:

Problems 5–25

Write two node voltage equations, one at the left node, the other at the rightnode:vn − vg

5000+

vn − vo

100,000+

vn

500,000= 0

vo + 3 × 105vn

5000+

vo − vn

100,000+

vo

1000= 0

Simplify and place in standard form:

106vn − 5vo = 100vg

(6×106 − 1)vn + 121vo = 0

Let vg = 1 V and solve the two simultaneous equations:

vo = −19.9915 V; vn = 403.2 µV

[b] From the solution in part (a), vn = 403.2 µV.

[c] ig =vg − vn

5000=

vg − 403.2 × 10−6vg

5000

Rg =vg

ig

=5000

1 − 403.2 × 10−6 = 5002.02 Ω

[d] For an ideal op amp, the voltage gain is the ratio between the feedback resistorand the input resistor:

vo

vg

= −100,0005000

= −20

For an ideal op amp, the difference between the voltages at the input terminalsis zero, and the input resistance of the op amp is infinite. Therefore,

vn = vp = 0 V; Rg = 5000 Ω

P 5.40 Note – the load resistor should have the value 4 kΩ.

[a] Replace the op amp with the model shown in Fig. 5.15. The node voltageequation at the inverting input:

vn

40,000+

vn − vg

500,000+

vn − vo

80,000= 0

Simplify:

12.5vn + vn − vg + 6.25vn − 6.25vo = 0

The node voltage equation at the op amp output:

vo

4000+

vo − 20,000(vp − vn)5000

+vo − vn

80,000= 0


Simplify:

20vo + 16vo − 320,000(vp − vn) + vo − vn = 0

From the input,

vp − vn = 0.8(vg − vn)

Substituting into the equation written at the output,

20vo + 16vo − 256,000(vg − vn) + vo − vn = 0

Now let vg = 1 V; plug this value into both the input and output equations andsimplify into two simultaneous equations:

19.75vn − 6.25vo = 1

255,999vn + 37vo = 256,000

These equations are in standard form, so solve them to yieldvo = 2.9986 V; vn = 999.571 mVThus,

vo

vg

=2.9986

1= 2.9986

[b] From part (a), vn = 999.571 mV. Use this value to solve for vp:

vp = 0.8(1 − vn) + vn = 999.914 mV

[c] vp − vn = 343.6 µ V

[d] ig =vg − vp

100,000=

1 − 999.914 × 10−3

100,000= 859 pA

[e] For an ideal op amp, vn = vp = vg, so the KVL equation at the inverting node is

vo

40,000+

vg − vo

80,000= 0

Then,

vo = 3vg sovo

vg

= 3

Also,

vn = vp = 1 V; vp − vn = 0 V; ig = 0 A

Problems 5–27

P 5.41 [a]

vn − 0.881600

+vn

500,000+

vn − vTh

24,000= 0

vTh + 105vn

2000+

vTh − vn

24,000= 0

Solving, vTh = −13.198 V

Short-circuit current calculation:

vn

500,000+

vn − 0.881600

+vn − 024,000

= 0

·. . vn = 0.823 V

isc =vn

24,000− 105

2000vn = −41.13 A

RTh =vTh

isc= 320.90 mΩ


[b] The output resistance of the inverting amplifier is the same as the Théveninresistance, i.e.,

Ro = RTh = 320.90 mΩ

[c]

vo =( 330

330.32

)(−13.198) = −13.185 V

vn − 0.881600

+vn

500,000+

vn + 13.18524,000

= 0

·. . vn = 941.92 µV

ig =0.88 − 941.92 × 10−6

1600= 549.41 µA

Rg =0.88

0.88 − 941.92 × 10−6 (1600) = 1601.7 Ω

P 5.42 [a] vTh =−241.6

(0.88) = −13.2 V

RTh = 0, since op-amp is ideal

Problems 5–29

[b] Ro = RTh = 0 Ω

[c] Rg = 1.6 kΩ since vn = 0

P 5.43 [a]

vn − vg

15,000+

vn − vo

135,000= 0

·. . vo = 10vn − 9vg

Also vo = A(vp − vn) = −Avn

·. . vn =−vo

A

·. . vo

(1 +

10A

)= −9vg

vo =−9A

(10 + A)vg

[b] vo =−9(90)(0.4)(10 + 90)

= −3.24 V

[c] vo = −9(0.4) = −3.60 V

[d] −3.42 =−9(0.4)A10 + A

·. . A = 190

P 5.44 From Eq. 5.57,

vref

R + ∆R= vn

(1

R + ∆R+

1R − ∆R

+1

Rf

)− vo

Rf

Substituting Eq. 5.59 for vp = vn:

vref

R + ∆R=

vref

(1

R+∆R+ 1

R−∆R+ 1

Rf

)(R − ∆R)

(1

R+∆R+ 1

R−∆R+ 1

Rf

) − vo

Rf


Rearranging,

vo

Rf

= vref

( 1R − ∆R

− 1R + ∆R

)

Thus,

vo = vref

( 2∆R

R2 − ∆R2

)Rf

P 5.45

i1 =15 − 105000

= 1 mA

i2 + i1 + 0 = 10 mA; i2 = 9 mA

vo2 = 10 + (400)(9) × 10−3 = 13.6 V

i3 =15 − 13.6

2000= 0.7 mA

i4 = i3 + i1 = 1.7 mA

vo1 = 15 + 1.7(0.5) = 15.85 V

Problems 5–31

P 5.46 vp =5.68.0

vg = 0.7vg = 7 sin(π/3)t V

vn

15,000+

vn − vo

75,000= 0

6vn = vo; vn = vp

·. . vo = 42 sin(π/3)t V 0 ≤ t ≤ ∞

vo = 0 t ≤ 0

At saturation

42 sin(

π

3

)t = ±21; sin

π

3t = ±0.5

·. .π

3t =

π

6,

5π6

,7π6

,11π6

, etc.

t = 0.50 s, 2.50 s, 3.50 s, 5.50 s, etc.

P 5.47 It follows directly from the circuit that vo = −16vg

From the plot of vg we have vg = 0, t < 0

vg = t 0 ≤ t ≤ 0.5

vg = −t + 1 0.5 ≤ t ≤ 1.5

vg = t − 2 1.5 ≤ t ≤ 2.5

vg = −t + 3 2.5 ≤ t ≤ 3.5

vg = t − 4 3.5 ≤ t ≤ 4.5, etc.Therefore


vo = −16t 0 ≤ t ≤ 0.5

vo = 16t − 16 0.5 ≤ t ≤ 1.5

vo = −16t + 32 1.5 ≤ t ≤ 2.5

vo = 16t − 48 2.5 ≤ t ≤ 3.5

vo = −16t + 64 3.5 ≤ t ≤ 4.5, etc.These expressions for vo are valid as long as the op amp is not saturated. Since thepeak values of vo are ±5, the output is clipped at ±5. The plot is shown below.

P 5.48 [a] Use Eq. 5.61 to solve for Rf ; note that since we are using 1% strain gages,∆ = 0.01:

Rf =voR

2∆vref=

(5)(120)(2)(0.01)(15)

= 2 kΩ

[b] Now solve for ∆ given vo = 50 mV:

∆ =voR

2Rfvref=

(0.05)(120)2(2000)(15)

= 100 × 10−6

The change in strain gage resistance that corresponds to a 50 mV change inoutput voltage is thus

∆R = ∆R = (100 × 10−6)(120) = 12 mΩ

Problems 5–33

P 5.49 [a]

Let R1 = R + ∆R

vp

Rf

+vp

R+

vp − vin

R1= 0

·. . vp

[1

Rf

+1R

+1R1

]=

vin

R1

·. . vp =RRfvin

RR1 + RfR1 + RfR= vn

vn

R+

vn − vin

R+

vn − vo

Rf

= 0

vn

[1R

+1R

+1

Rf

]− vo

Rf

=vin

R

·. . vn

[R + 2Rf

RRf

]− vin

R=

vo

Rf

·. .vo

Rf

=[R + 2Rf

RRf

]RRfvin

[RR1 + RfR1 + RfR]− vin

R

·. .vo

Rf

=[

R + 2Rf

RR1 + RfR1 + RfR− 1

R

]vin

·. . vo =[R2 + 2RRf − R1(R + Rf ) − RRf ]Rf

R[R1(R + Rf ) + RRf ]vin

Now substitute R1 = R + ∆R and get

vo =−∆R(R + Rf )Rfvin

R[(R + ∆R)(R + Rf ) + RRf ]

If ∆R R

vo ≈ (R + Rf )Rf (−∆R)vin

R2(R + 2Rf )

[b] vo ≈ 47 × 104(48 × 104)(−95)15108(95 × 104)

≈ −3.384 V


[c] vo =−95(48 × 104)(47 × 104)15

104[(1.0095)104(48 × 104) + 47 × 108]= −3.368 V

P 5.50 [a] vo ≈ (R + Rf )Rf (−∆R)vin

R2(R + 2Rf )

vo =(R + Rf )(−∆R)Rfvin

R[(R + ∆R)(R + Rf ) + RRf ]

·. .approx value

true value=

R[(R + ∆R)(R + Rf ) + RRf ]R2(R + 2Rf )

·. . Error =R[(R + ∆R)(R + Rf ) + RRf ] − R2(R + 2Rf )

R2(R + 2Rf )

=∆R

R

(R + Rf )(R + 2Rf )

·. . % error =∆R(R + Rf )R(R + 2Rf )

× 100

[b] % error =95(48 × 104) × 100

104(95 × 104)= 0.48%

P 5.51 1 =∆R(48 × 104)104(95 × 104)

× 100

·. . ∆R =950048

= 197.91667 Ω

·. . % change in R =197.19667

104 × 100 ≈ 1.98%

P 5.52 [a] It follows directly from the solution to Problem 5.49 that

vo =[R2 + 2RRf − R1(R + Rf ) − RRf ]Rfvin

R[R1(R + Rf ) + RRf ]

Now R1 = R − ∆R. Substituting into the expression gives

vo =(R + Rf )Rf (∆R)vin

R[(R − ∆R)(R + Rf ) + RRf ]

Now let ∆R R and get

vo ≈ (R + Rf )Rf∆Rvin

R2(R + 2Rf )

Problems 5–35

[b] It follows directly from the solution to Problem 5.49 that

·. .approx value

true value=

R[(R − ∆R)(R + Rf ) + RRf ]R2(R + 2Rf )

·. . Error =(R − ∆R)(R + Rf ) + RRf − R(R + 2Rf )

R(R + 2Rf )

=−∆R(R + Rf )R(R + 2Rf )

·. . % error =−∆R(R + Rf )R(R + 2Rf )

× 100

[c] R − ∆R = 9810 Ω ·. . ∆R = 10,000 − 9810 = 190 Ω

·. . vo ≈ (48 × 104)(47 × 104)(190)(15)108(95 × 104)

≈ 6.768 V

[d] % error =−190(48 × 104)(100)

104(95 × 104)= −0.96%

6Inductance, Capacitance, and Mutual

Inductance

Assessment Problems

AP 6.1 [a] ig = 8e−300t − 8e−1200tA

v = Ldig

dt= −9.6e−300t + 38.4e−1200tV, t > 0+

v(0+) = −9.6 + 38.4 = 28.8 V

[b] v = 0 when 38.4e−1200t = 9.6e−300t or t = (ln 4)/900 = 1.54 ms

[c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W

[d]dp

dt= 0 when e1800t − 12.5e900t + 16 = 0

Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0

x = 1.45, t =ln 1.45900

= 411.05 µs

x = 11.05, t =ln 11.05

900= 2.67 ms

p is maximum at t = 411.05 µs

[e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W

[f] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78 A

wmax = (1/2)(4 × 10−3)(3.78)2 = 28.6 mJ

[g] W is max when i is max, i is max when di/dt is zero.

When di/dt = 0, v = 0, therefore t = 1.54 ms.

6–1

6–2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance

AP 6.2 [a] i = Cdv

dt= 24 × 10−6 d

dt[e−15,000t sin 30,000t]

= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, i(0+) = 0.72 A

[b] i(

π

80ms)

= −31.66 mA, v(

π

80ms)

= 20.505 V,

p = vi = −649.23 mW

[c] w =(1

2

)Cv2 = 126.13 µJ

AP 6.3 [a] v =( 1

C

) ∫ t

0−i dx + v(0−)

=1

0.6 × 10−6

∫ t

0−3 cos 50,000x dx = 100 sin 50,000t V

[b] p(t) = vi = [300 cos 50,000t] sin 50,000t

= 150 sin 100,000t W, p(max) = 150 W

[c] w(max) =(1

2

)Cv2

max = 0.30(100)2 = 3000µJ = 3 mJ

AP 6.4 [a] Leq =60(240)

300= 48 mH

[b] i(0+) = 3 + −5 = −2 A

[c] i =1256

∫ t

0+(−0.03e−5x) dx − 2 = 0.125e−5t − 2.125 A

[d] i1 =503

∫ t

0+(−0.03e−5x) dx + 3 = 0.1e−5t + 2.9 A

i2 =256

∫ t

0+(−0.03e−5x) dx − 5 = 0.025e−5t − 5.025 A

i1 + i2 = i

AP 6.5 v1 = 0.5 × 106∫ t

0+240 × 10−6e−10x dx − 10 = −12e−10t + 2 V

v2 = 0.125 × 106∫ t

0+240 × 10−6e−10x dx − 5 = −3e−10t − 2 V

v1(∞) = 2 V, v2(∞) = −2 V

W =[12(2)(4) +

12(8)(4)

]× 10−6 = 20 µJ

Problems 6–3

AP 6.6 [a] Summing the voltages around mesh 1 yields

4di1dt

+ 8d(i2 + ig)

dt+ 20(i1 − i2) + 5(i1 + ig) = 0

or

4di1dt

+ 25i1 + 8di2dt

− 20i2 = −(

5ig + 8dig

dt

)

Summing the voltages around mesh 2 yields

16d(i2 + ig)

dt+ 8

di1dt

+ 20(i2 − i1) + 780i2 = 0

or

8di1dt

− 20i1 + 16di2dt

+ 800i2 = −16dig

dt

[b] From the solutions given in part (b)

i1(0) = −0.4 − 11.6 + 12 = 0; i2(0) = −0.01 − 0.99 + 1 = 0

These values agree with zero initial energy in the circuit. At infinity,

i1(∞) = −0.4A; i2(∞) = −0.01A

When t = ∞ the circuit reduces to

·. . i1(∞) = −(7.8

20+

7.8780

)= −0.4A; i2(∞) = − 7.8

780= −0.01A

From the solutions for i1 and i2 we have

di1dt

= 46.40e−4t − 60e−5t

di2dt

= 3.96e−4t − 5e−5t

Also,dig

dt= 7.84e−4t

Thus

4di1dt

= 185.60e−4t − 240e−5t


25i1 = −10 − 290e−4t + 300e−5t

8di2dt

= 31.68e−4t − 40e−5t

20i2 = −0.20 − 19.80e−4t + 20e−5t

5ig = 9.8 − 9.8e−4t

8dig

dt= 62.72e−4t

Test:

185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t

+0.20 + 19.80e−4t − 20e−5t ?= −[9.8 − 9.8e−4t + 62.72e−4t]

−9.8 + (300 − 240 − 40 − 20)e−5t

+(185.60 − 290 + 31.68 + 19.80)e−4t ?= −(9.8 + 52.92e−4t)

−9.8 + 0e−5t + (237.08 − 290)e−4t ?= −9.8 − 52.92e−4t

−9.8 − 52.92e−4t = −9.8 − 52.92e−4t (OK)

Also,

8di1dt

= 371.20e−4t − 480e−5t

20i1 = −8 − 232e−4t + 240e−5t

16di2dt

= 63.36e−4t − 80e−5t

800i2 = −8 − 792e−4t + 800e−5t

16dig

dt= 125.44e−4t

Test:

371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t

−8 − 792e−4t + 800e−5t ?= −125.44e−4t

(8 − 8) + (800 − 480 − 240 − 80)e−5t

+(371.20 + 232 + 63.36 − 792)e−4t ?= −125.44e−4t

(800 − 800)e−5t + (666.56 − 792)e−4t ?= −125.44e−4t

−125.44e−4t = −125.44e−4t (OK)

Problems 6–5

Problems

P 6.1 [a] i = 0 t < 0

i = 50t A 0 ≤ t ≤ 5 ms

i = 0.5 − 50t A 5 ≤ t ≤ 10 ms

i = 0 10 ms < t

[b] v = Ldi

dt= 20 × 10−3(50) = 1 V 0 ≤ t ≤ 5 ms

v = 20 × 10−3(−50) = −1 V 5 ≤ t ≤ 10 ms

v = 0 t < 0

v = 1 V 0 < t < 5 ms

v = −1 V 5 < t < 10 ms

v = 0 10 ms < t

p = vi

p = 0 t < 0

p = (50t)(1) = 50t W 0 < t < 5 ms

p = (0.5 − 50t)(−1) = 50t − 0.5 W 5 < t < 10 ms

p = 0 10 ms < t

w = 0 t < 0

w =∫ t

0(50x) dx = 50

x2

2

∣∣∣∣t0= 25t2 J 0 < t < 5 ms

w =∫ t

0.005(50x − 0.5) dx + 0.625 × 10−3

= 25x2 − 0.5x∣∣∣∣t0.005

+0.625 × 10−3

= 25t2 − 0.5t + 2.5 × 10−3 J 5 < t < 10 ms

w = 0 10 ms < t

P 6.2 [a] 0 ≤ t ≤ 2 ms :

i =1L

∫ t

0vs dx + i(0) =

1200 × 10−6

∫ t

05 × 10−3 dx + 0


= 25x∣∣∣∣t0= 25t A

2 ms ≤ t < ∞ :

i =1

200 × 10−6

∫ t

2×10−3(0) dx + 25(2 × 10−3) = 50 mA

[b]

P 6.3 Note – the initial current should be 1 A.

0 ≤ t ≤ 2 s

iL =1

2.5 × 10−4

∫ t

03 × 10−3e−4x dx + 0 = 1.2

e−4x

−4

∣∣∣∣t0

+ 0

= 0.3 − 0.3e−4t A, 0 ≤ t ≤ 2 s

iL(2) = 0.3A

2 s < t < ∞

iL = −1.2(

e−4(x−2)

−4

∣∣∣∣t2

+ 0.3)

= 0.3e−4(t−2) A, 2 s ≤ t < ∞

Problems 6–7

P 6.4 [a] v = Ldi

dt

di

dt= 18[t(−10e−10t) + e−10t] = 18e−10t(1 − 10t)

v = (50 × 10−6)(18)e−10t(1 − 10t)

= 0.9e−10t(1 − 10t) mV, t > 0

[b] p = vi

v(200 ms) = 0.9e−2(1 − 2) = −121.8 µV

i(200 ms) = 18(0.2)e−2 = 487.2 mA

p(200 ms) = (−121.8 × 10−6)(487.2 × 10−3) = −59.34 µW

[c] delivering

[d] w =12Li2 =

12(50 × 10−6)(487.2 × 10−3)2 = 5.93 µJ

[e] The energy is a maximum where the current is a maximum:

diL

dt= 18[t(−10)e−10t + e−10t) = 18e−10t(1 − 10t)

diL

dt= 0 when t = 0.1 s


imax = 18(0.1)e−1 = 662.2 mA

wmax =12(50 × 10−6)(662.2 × 10−3)2 = 10.96 µJ

P 6.5 [a] 0 ≤ t ≤ 1 s :

v = −100t

i =15

∫ t

0−100x dx + 0 = −20

x2

2

∣∣∣∣t0

i = −10t2 A

1 s ≤ t ≤ 3 s :

v = −200 + 100t

i(1) = −10 A

·. . i =15

∫ t

1(100x − 200) dx − 10

= 20∫ t

1x dx − 40

∫ t

1dx − 10

= 10(t2 − 1) − 40(t − 1) − 10

= 10t2 − 40t + 20 A3 s ≤ t ≤ 5 s :

v = 100

i(3) = 10(9) − 120 + 20 = −10 A

i =15

∫ t

3100 dx − 10

= 20t − 60 − 10 = 20t − 70 A5 s ≤ t ≤ 6 s :

v = −100t + 600

i(5) = 100 − 70 = 30

i =15

∫ t

5(−100x + 600) dx + 30

= −20∫ t

5x dx + 120

∫ t

5dx + 30

= −10(t2 − 25) + 120(t − 5) + 30

= −10t2 + 120t − 320 A

Problems 6–9

[b] i(6) = −10(36) + 120(6) − 320 = 720 − 680 = 40 A, 6 ≤ t < ∞[c]

P 6.6 [a] vL = Ldi

dt= [125 sin 400t]e−200t V

·. .dvL

dt= 25,000(2 cos 400t − sin 400t)e−200t V/s

dvL

dt= 0 when tan 400t = 2

·. . t = 2.77 ms

Also 400t = 1.107 + π etc.

Because of the decaying exponential vL will be maximum the first time thederivative is zero.

[b] vL(max) = [125 sin 1.107]e−0.554 = 64.27 V

vL max = 64.27 V

Note: When t = (1.107 + π)/400; vL = −13.36 V

P 6.7 [a] i =1

15 × 10−3

∫ t

030 sin 500x dx − 4

= 2000∫ t

0sin 500x dx − 4

= 2000[− cos 500x

500

∣∣∣∣t

0− 4

= 4(1 − cos 500t) − 4

i = −4 cos 500t A


[b] p = vi = (30 sin 500t)(−4 cos 500t)

= −120 sin 500t cos 500t

p = −60 sin 1000t W

w =12Li2

=12(15 × 10−3)16 cos2 500t

= 120 cos2 500t mJ

w = [60 + 60 cos 1000t] mJ.

Problems 6–11

[c] Absorbing power: Delivering power:

π ≤ t ≤ 2π ms 0 ≤ t ≤ π ms

3π ≤ t ≤ 4π ms 2π ≤ t ≤ 3π ms

P 6.8 [a] i(0) = A1 + A2 = 0.04

di

dt= −10,000A1e

−10,000t − 40,000A2e−40,000t

v = −200A1e−10,000t − 800A2e

−40,000t V

v(0) = −200A1 − 800A2 = 28

Solving, A1 = 0.1 and A2 = −0.06

Thus,

i = 0.1e−10,000t − 0.06e−40,000t A, t ≥ 0

v = −20e−10,000t + 48e−40,000t V, t ≥ 0

[b] If p = 0 then either i = 0 or v = o. Suppose i = 0:

i = 0.1e−10,000t − 0.06e−40,000t = 0

·. . 0.1e30,000t = 0.06 so t = −17.03 µs

This answer is impossible! So assume that v = 0:

v = −20e−10,000t + 48e−40,000t = 0

Then, − 20e30,000t = −48 ·. . t = 29.18 µs

This answer makes sense; therefore, the power is 0 at t = 29.18 µs.


P 6.9 [a] From Problem 6.8 we have

A1 + A2 = 0.04

Now, we add the second equation for the coefficients:

−200A1 − 800A2 = −68

Solving, A1 = −0.06; A2 = 0.1Thus,

i = −0.06e−10,000t + 0.1e−40,000tA t ≥ 0

v = 12e−10,000t − 80e−40,000tA t ≥ 0

[b] i = 0 when 0.06e−10,000t = 0.1e−40,000t

·. . e30,0000t = 5/3 so t = 17.03 µs

Thus,

i > 0 for 0 ≤ t ≤ 17.03 µs and i < 0 for 17.03 µs ≤ t < ∞v = 0 when 12e−10,000t = 80e−40,000t

·. . e30,0000t = 20/3 so t = 63.24 µs

Thus,

v < 0 for 0 ≤ t ≤ 63.24 µs and v > 0 for 63.24 µs ≤ t < ∞Therefore,

p < 0 for 0 ≤ t ≤ 17.03 µs and 63.24 µs ≤ t < ∞(inductor delivers energy)

p > 0 for 17.03 µs ≤ t ≤ 63.24 µs (inductor stores energy)

[c] The energy stored at t = 0 is

w(0) =12L[i(0)]2 =

12(20 × 10−3)(40 × 10−3)2 = 16 µJ

The power for t > 0 is

p = vi = 6e−50,000t − 8e−80,000t − 0.72e−20,000t

The energy for t > 0 is

w =∫ ∞

0p dt =

∫ ∞

06e−50,000x dx −

∫ ∞

08e−80,000x dx −

∫ ∞

00.72e−20,000x dx

=6

50,000− 8

80,000− 0.72

20,000= −16 µJ

Thus, the energy stored at t = 0 equals the energy extracted for t > 0.

Problems 6–13

P 6.10 i = (B1 cos 1.6t + B2 sin 1.6t)e−0.4t

i(0) = B1 = 5 A

di

dt= (B1 cos 1.6t + B2 sin 1.6t)(−0.4e−0.4t) + e−0.4t(−1.6B1 sin 1.6t + 1.6B2 cos 1.6t)

= [(1.6B2 − 0.4B1) cos 1.6t − (1.6B1 + 0.4B2) sin 1.6t]e−0.4t

v = 2di

dt= [(3.2B2 − 0.8B1) cos 1.6t − (3.2B1 + 0.8B2) sin 1.6t]e−0.4t

v(0) = 28 = 3.2B2 − 0.8B1 = 3.2B2 − 4 ·. . B2 = 32/3.2 = 10 A

Thus,

i = (5 cos 1.6t + 10 sin 1.6t)e−0.4t A, t ≥ 0

v = (28 cos 1.6t − 24 sin 1.6t)e−0.4t V, t ≥ 0

i(5) = 1.24 A; v(5) = −3.76 V

p(5) = (1.24)(−3.76) = −4.67 W

The power delivered is 4.67 W.

P 6.11 For 0 ≤ t ≤ 1.6 s:

iL =15

∫ t

03 × 10−3 dx + 0 = 0.6 × 10−3t

iL(1.6 s) = (0.6 × 10−3)(1.6) = 0.96 mA

Rm = (20)(1000) = 20 kΩ

vm(1.6 s) = (0.96 × 10−3)(20 × 103) = 19.2 V

P 6.12 p = vi = 40t[e−10t − 10te−20t − e−20t]

W =∫ ∞

0p dx =

∫ ∞

040x[e−10x − 10xe−20x − e−20x] dx = 0.2 J

This is energy stored in the inductor at t = ∞.


P 6.13 [a] v(20 µs) = 12.5 × 109(20 × 10−6)2 = 5 V (end of first interval)

v(20 µs) = 106(20 × 10−6) − (12.5)(400) × 10−3 − 10

= 5 V (start of second interval)

v(40 µs) = 106(40 × 10−6) − (12.5)(1600) × 10−3 − 10

= 10 V (end of second interval)

[b] p(10µs) = 62.5 × 1012(10−5)3 = 62.5 mW, v(10 µs) = 1.25 V,

i(10µs) = 50 mA, p(10 µs) = vi = 62.5 mW,

p(30 µs) = 437.50 mW, v(30 µs) = 8.75 V, i(30 µs) = 0.05 A

[c] w(10 µs) = 15.625 × 1012(10 × 10−6)4 = 0.15625µJ

w = 0.5Cv2 = 0.5(0.2 × 10−6)(1.25)2 = 0.15625µJ

w(30 µs) = 7.65625µJ

w(30 µs) = 0.5(0.2 × 10−6)(8.75)2 = 7.65625µJ

P 6.14 iC = C(dv/dt)

0 < t < 0.5 :

vc = 30t2 V

iC = 20 × 10−6(60)t = 1.2t mA

0.5 < t < 1 :

vc = 30(t − 1)2 V

iC = 20 × 10−6(60)(t − 1) = 1.2(t − 1) mA

Problems 6–15

P 6.15 [a] 0 ≤ t ≤ 5 µs

C = 5 µF1C

= 2 × 105

v = 2 × 105∫ t

04 dx + 12

v = 8 × 105t + 12 V 0 ≤ t ≤ 5 µs

v(5 µs) = 4 + 12 = 16 V

[b] 5 µs ≤ t ≤ 20 µs

v = 2 × 105∫ t

5×10−6−2 dx + 16 = −4 × 105t + 2 + 16

v = −4 × 105t + 18V 5 ≤ t ≤ 20 µs

v(20 µs) = −4 × 105(20 × 10−6) + 18 = 10 V

[c] 20 µs ≤ t ≤ 25 µs

v = 2 × 105∫ t

20×10−66 dx + 10 = 12 × 105t − 24 + 10

v = 12 × 105t − 14 V, 20 µs ≤ t ≤ 25 µs

v(25 µs) = 12 × 105(25 × 10−6) − 14 = 16 V

[d] 25 µs ≤ t ≤ 35 µs

v = 2 × 105∫ t

25×10−64 dx + 16 = 8 × 105t − 20 + 16

v = 8 × 105t − 4 V, 25 µs ≤ t ≤ 35 µs

v(35 µs) = 8 × 105(35 × 10−6) − 4 = 24 V

[e] 35 µs ≤ t < ∞

v = 2 × 105∫ t

35×10−60 dx + 24 = 24

v = 24 V, 35 µs ≤ t < ∞[f]


P 6.16 v = −10 V, t ≤ 0; C = 0.8 µF

v = 40 − e−1000t(50 cos 500t + 20 sin 500t)V, t ≥ 0

[a] i = 0, t < 0

[b]dv

dt= 1000e−1000t(50 cos 500t + 20 sin 500t)

−e−1000t(−25,000 sin 500t + 10,000 cos 500t)

= e−1000t(50,000 cos 500t + 20,000 sin 500t

+25,000 sin 500t − 10,000 cos 500t)

= (40,000 cos 500t + 45,000 sin 500t)e−1000t

i = Cdv

dt= (32 cos 500t + 36 sin 500t)e−1000t mA

[c] no

[d] yes, from 0 to 32 mA

[e] v(∞) = 40 V

w =12Cv2 =

12(0.8 × 10−6)(40)2 = 640 µJ

P 6.17 [a] i =400 × 10−3

5 × 10−6 t = 8 × 104t 0 ≤ t ≤ 5 µs

i = 400 × 10−3 5 ≤ t ≤ 20 µs

q =∫ 5×10−6

08 × 104t dt +

∫ 15×10−6

5×10−6400 × 10−3 dt

= 8 × 104 t2

2

∣∣∣∣5×10−6

0+400 × 10−3(10 × 10−6)

= 8 × 104(12)(25 × 10−12) + 4 × 10−6

= 5 µC

[b] v =1

0.25 × 10−6

[∫ 5 µs

08 × 104x dx +

∫ 20 µs

5 µs0.4x dx +

∫ 30 µs

20 µs(104x − 0.5) dx

]

=1

0.25 × 10−6

[4 × 104t2

∣∣∣∣5 µs

0+0.4t

∣∣∣∣20 µs

5 µs+(5000t2 − 0.5t)

∣∣∣∣30 µs

20 µs

]

=1

0.25 × 10−6 [1 × 10−6 + 6 × 10−6 − 10.5 × 10−6 + 8 × 10−6] = 18V

Problems 6–17

[c] v(50 µs) = 18 +1

0.25 × 10−6 (5000t2 − 0.5t)∣∣∣∣50 µs

30 µs

= 18 +1

0.25 × 10−6 (−12.5 × 10−6 + 10.5 × 10−6) = 10V

w =12Cv2 =

12(0.25 × 10−6)(10)2 = 12.5 µJ

P 6.18 [a] v =1

0.5 × 10−6

∫ 500×10−6

050 × 10−3e−2000t dt − 20

= 100 × 103 e−2000t

−2000

∣∣∣∣500×10−6

0−20

= 50(1 − e−1) − 20 = 11.61 V

w = 12Cv2 = 1

2(0.5)(10−6)(11.61)2 = 33.7 µJ

[b] v(∞) = 50 − 20 = 30V

w(∞) =12(0.5 × 10−6)(30)2 = 225 µJ

P 6.19 [a] w(0) =12C[v(0)]2 =

12(0.25) × 10−6(50)2 = 312.5 µJ

[b] v = (A1t + A2)e−4000t

v(0) = A2 = 50 V

dv

dt= −4000e−4000t(A1t + A2) + e−4000t(A1)

= (−4000A1t − 4000A2 + A1)e−4000t

dv

dt(0) = A1 − 4000A2

i = Cdv

dt, i(0) = C

dv(0)dt

·. .dv(0)

dt=

i(0)C

=400 × 10−3

0.25 × 10−6 = 16 × 105

·. . 16 × 105 = A1 − 4000(50)

Thus, A1 = 16 × 105 + 2 × 105 = 18 × 105 Vs


[c] v = (18 × 105t + 50)e−4000t

i = Cdv

dt= 0.25 × 10−6 d

dt(18 × 105t + 50)e−4000t

i =d

dt[(0.45t + 12.5 × 10−6)e−4000t]

= (0.45t + 12.5 × 10−6)(−4000)e−4000t + e−4000t(0.45)

= (−1800t − 0.05 + 0.45)e−4000t

= (0.40 − 1800t)e−4000t A, t ≥ 0

P 6.20 5‖(12 + 8) = 4 H

4‖4 = 2 H

15‖(8 + 2) = 6 H

3‖6 = 2 H

6 + 2 = 8 H

P 6.21 30‖20 = 12 H

80‖(8 + 12) = 16 H

60‖(14 + 16) = 20 H

15‖(20 + 10) = 20 H

Lab = 5 + 10 = 15 H

P 6.22 [a]

i(t) = −12

∫ t

012e−x dx + 6

= 6e−x

∣∣∣∣t0

+6

= 6e−t − 6 + 6

i(t) = 6e−t A, t ≥ 0

Problems 6–19

[b] i1(t) = −13

∫ t

012e−x dx + 2

= 4e−x

∣∣∣∣t0

+2

= 4(e−t − 1) + 2

i1(t) = 4e−t − 2 A, t ≥ 0

[c] i2(t) = −16

∫ t

012e−x dx + 4

= 2e−x

∣∣∣∣t0

+4

= 2(e−t − 1) + 4

i2(t) = 2e−t + 2 A, t ≥ 0

[d] p = vi = (12e−t)(6e−t) = 72e−2t W

w =∫ ∞

0p dt =

∫ ∞

072e−2t dt

= 72e−2t

−2

∣∣∣∣∞0

= 36 J

[e] w =12(3)(2)2 +

12(6)(4)2 = 54 J

[f] wtrapped =12(3)(−2)2 +

12(6)(2)2 = 18 J

wtrapped = 54 − 36 = 18 J checks

[g] Yes, they agree.

P 6.23 [a] io(0) = i1(0) + i2(0) = 4 A

[b]

io = − 110

∫ t

0160e−4x dx + 4 = −16

[e−4x

−4

∣∣∣∣∣t

0

+ 4

= 4(e−4t − 1) + 4 = 4e−4t A, t ≥ 0


[c]

va = 8d

dt(4e−4t) = −128e−4t V

vc = va + vb = −128e−4t + 160e−4t

= 32e−4t V

i1 = − 13

∫ t

032e−4x dx + 1

= 2.67e−4t − 2.67 + 1

i1 = 2.67e−4t − 1.67 A, t ≥ 0

[d] i2 = −16

∫ t

032e−4x dx + 3

= 1.33e−4t − 1.33 + 3

i2 = 1.33e−4t + 1.67 A, t ≥ 0

[e] w(0) =12(3)(1)2 +

12(6)(3)2 +

12(8)(4)2 = 92.5 J

[f] wdel =12(10)(4)2 = 80 J

[g] wtrapped = 92.5 − 80 = 12.5 J

P 6.24 vb = 160e−4t V

io = 4e−4t A

p = 640e−8t W

w =∫ t

0640e−8x dx = 640

e−8x

−8

∣∣∣∣t0= 80(1 − e−8t) W

wtotal = 80 J

w(0.2) = 80(1 − e−1.6) = 63.85 J

Thus,

% delivered =63.8580

(100) = 79.8%

Problems 6–21

P 6.2514

+16

=512

·. . Ceq = 2.4 µF

14

+112

=412

·. . Ceq = 3 µF

124

+18

=424

·. . Ceq = 6 µF

P 6.26 Work from the right hand side of the circuit, simplifying step by step:

1. 48 µF in series with 16 µF : 1/C = 1/16 µ + 1/48 µ ·. . C = 12 µFThe voltages add in series, so the 12 µF capacitor has a voltage of 20 V,negative at the top.

2. Previous 12 µF in parallel with 3 µF : C = 12 µ + 3 µ = 15 µFThe voltage is 20 V, negative at the top.

3. Previous 15 µF in series with 30 µF :1/C = 1/15 µ + 1/30 µ ·. . C = 10 µF

The voltages add in series, so the 10 µF capacitor has a voltage of 10 V,positive at the right.


4. Previous 10 µF in parallel with 10 µF : C = 10 µ + 10 µ = 20 µFThe voltage is 10 V, negative at the top.

5. Previous 20 µF in series with 5 µF and 4 µF :1/C = 1/20 µ + 1/5 µ + 1/4 µ ·. . C = 2 µF

The voltages in series add: 5V − 10V + 30V = 25V positive at the top.

The equivalent capacitance is 2 µF with a voltage of 25 V, positive at the top.

P 6.27 [a]

vo = − 12 × 10−6

∫ t

020 × 10−6e−x dx + 10

= 10e−x

∣∣∣∣t0

+10

= 10e−t V, t ≥ 0

[b] v1 = − 13 × 10−6 (20 × 10−6)e−x

∣∣∣∣t0

+4

= 6.67e−t − 2.67 V, t ≥ 0

[c] v2 = − 16 × 10−6 (20 × 10−6)e−x

∣∣∣∣t0

+6

= 3.33e−t + 2.67 V, t ≥ 0

[d] p = vi = (10e−t)(20 × 10−6)e−t

= 200 × 10−6e−2t

w =∫ ∞

0200 × 10−6e−2t dt

= 200 × 10−6 e−2t

−2

∣∣∣∣∞0

= −100 × 10−6(0 − 1) = 100µJ

[e] w = 12(3 × 10−6)(4)2 + 1

2(6 × 10−9)(6)2

= 132 µJ

[f] wtrapped = 12(3 × 10−6)(8/3)2 + 1

2(6 × 10−6)(8/3)2

= 32 µJ

Problems 6–23

CHECK: 100 + 32 = 132µJ

[g] Yes, they agree.

P 6.28 C1 = 10 + 2 = 12µF

1C2

=1

12 µ+

18 µ

·. . C2 = 4.8 µF

vo(0) + v1(0) = −5 + 25 = 20 V

[a]

v2 = − 14.8 × 10−6

∫ t

01.92 × 10−3e−20x dx + 20

= −400e−20x

−20

∣∣∣∣t0

+20

= 20(e−20t − 1) + 20

= 20e−20t V, t ≥ 0

[b] vo = − 18 × 10−6

∫ t

01.92 × 10−3e−20x dx − 5

= −240e−20x

−20

∣∣∣∣t0

−5

= 12(e−20t − 1) − 5

= 12e−20t − 17 V, t ≥ 0

[c] v1 = − 112 × 10−6

∫ t

01.92 × 10−3e−20x dx + 25

= −160e−20x

−20

∣∣∣∣t0

+25

= 8(e−20t − 1) + 25

= 8e−20t + 17 V, t ≥ 0


[d] i1 = −10 × 10−6 d

dt[8e−20t + 17]

= −10 × 10−6(−20)8e−20t

= 1.6e−20t mA, t > 0

[e] i2 = −2 × 10−6 d

dt[8e−20t + 17]

= −2 × 10−6(−20)8e−20t

= 0.32e−20t mA, t > 0CHECK: i1 + i2 = 1.92e−20t mA = io

P 6.29 [a] w(0) = [12(8 × 10−6)(−5)2 + 12(10 × 10−6)(25)2 + 1

2(2 × 10−6)(25)2]

= 3850 µJ

[b] vo(∞) = −17 V

v1(∞) = 17 V

w(∞) = [12(8 × 10−6)(−17)2 + 12(12 × 10−6)(17)2]

= 2890 µJ

[c] w =∫ ∞

0(20e−20t)(1.92 × 10−3e−20t) dt = 960 µJ

CHECK: 3850 − 2890 = 960µJ

[d] % delivered =9603850

× 100 = 24.9%

[e] w(40 ms) =∫ 0.04

0(20e−20t)(1.92 × 10−3e−20t) dt

= 0.0384e−40t

−40

∣∣∣∣0.04

0

= 960 × 10−6(1 − e−1.6) = 766.2 µJ

% delivered =766.2960

(100) = 79.8%

P 6.30 From Figure 6.17(a) we have

v =1C1

∫ t

0i + v1(0) +

1C2

∫ t

0i dx + v2(0) + · · ·

v =[ 1C1

+1C2

+ · · ·] ∫ t

0i dx + v1(0) + v2(0) + · · ·

Problems 6–25

Therefore1

Ceq=[ 1C1

+1C2

+ · · ·], veq(0) = v1(0) + v2(0) + · · ·

P 6.31 From Fig. 6.18(a)

i = C1dv

dt+ C2

dv

dt+ · · · = [C1 + C2 + · · ·]dv

dt

Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the initialvoltage on every capacitor must be the same. This initial voltage would appear onCeq.

P 6.32dio

dt= 5e−2000t[−8000 sin 4000t + 4000 cos 4000t]

−2000e−2000t[2 cos 4000t + sin 4000t]dio

dt(0+) = 5[1(4000) + (−2000)(2)] = 0

v2(0+) = 10 × 10−3dio

dt(0+) = 0

v1(0+) = 40io(0+) + v2(0+) = 40(10) + 0 = 400V

P 6.33 vc = − 10.625 × 10−6

(∫ t

01.5e−16,000x dx −

∫ t

00.5e−4000x dx

)− 50

= 150(e−16,000t − 1) − 200(e−4000t − 1) − 50

= 150e−16,000t − 200e−4000t V

vL = 25 × 10−3dio

dt

= 25 × 10−3(−24,000e−16,000t + 2000e−4000t)

= −600e−16,000t + 50e−4000t V

vo = vc − vL

= (150e−16,000t − 200e−4000t) − (−600e−16,000t + 50e−4000t)

= 750e−16,000t − 250e−4000t V, t > 0

P 6.34 [a] −2dig

dt+ 16

di2dt

+ 32i2 = 0

16di2dt

+ 32i2 = 2dig

dt

[b] i2 = e−t − e−2t A

di2dt

= −e−t + 2e−2t A/s


ig = 8 − 8e−t A

dig

dt= 8e−t A/s

·. . −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t

[c] v1 = 4dig

dt− 2

di2dt

= 4(8e−t) − 2(−e−t + 2e−2t)

= 34e−t − 4e−2t V, t > 0

[d] v1(0) = 34 − 4 = 30 V; Also

v1(0) = 4dig

dt(0) − 2

di2dt

(0)

= 4(8) − 2(−1 + 2) = 32 − 2 = 30 VYes, the initial value of v1 is consistent with known circuit behavior.

P 6.35 [a] Yes, vo = 20(i2 − i1) + 60i2

[b] vo = 20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t)+

60(1 − 52e−5t + 51e−4t)

= 20(−3 − 116e−5t + 119e−4t) + 60 − 3120e−5t + 3060e−4t

vo = −5440e−5t + 5440e−4t V

[c] vo = L2d

dt(ig − i2) + M

di1dt

= 16d

dt(15 + 36e−5t − 51e−4t) + 8

d

dt(4 + 64e−5t − 68e−4t)

= −2880e−5t + 3264e−4t − 2560e−5t + 2176e−4t

vo = −5440e−5t + 5440e−4t V

P 6.36 [a] vg = 5(ig − i1) + 20(i2 − i1) + 60i2

= 5(16 − 16e−5t − 4 − 64e−5t + 68e−4t)+

20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t)+

60(1 − 52e−5t + 51e−4t)

= 60 + 5780e−4t − 5840e−5t V

[b] vg(0) = 60 + 5780 − 5840 = 0 V

Problems 6–27

[c] pdev = vgig

= 960 + 92,480e−4t − 94,400e−5t − 92,480e−9t+

93,440e−10tW

[d] pdev(∞) = 960 W

[e] i1(∞) = 4 A; i2(∞) = 1 A; ig(∞) = 16 A;

p5Ω = (16 − 4)2(5) = 720 W

p20Ω = 32(20) = 180 W

p60Ω = 12(60) = 60 W∑

pabs = 720 + 180 + 60 = 960 W

·. .∑

pdev =∑

pabs = 960 W

P 6.37 [a] Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2 and transfer theig terms to the right hand side of the equations. We get

4di1dt

+ 25i1 − 8di2dt

− 20i2 = 5ig − 8dig

dt

−8di1dt

− 20i1 + 16di2dt

+ 80i2 = 16dig

dt

[b] From the given solutions we have

di1dt

= −320e−5t + 272e−4t

di2dt

= 260e−5t − 204e−4t

Thus,

4di1dt

= −1280e−5t + 1088e−4t

25i1 = 100 + 1600e−5t − 1700e−4t

8di2dt

= 2080e−5t − 1632e−4t

20i2 = 20 − 1040e−5t + 1020e−4t

5ig = 80 − 80e−5t

8dig

dt= 640e−5t


Thus,

−1280e−5t + 1088e−4t + 100 + 1600e−5t − 1700e−4t − 2080e−5t

+1632e−4t − 20 + 1040e−5t − 1020e−4t ?= 80 − 80e−5t − 640e−5t

80 + (1088 − 1700 + 1632 − 1020)e−4t

+(1600 − 1280 − 2080 + 1040)e−5t ?= 80 − 720e−5t

80 + (2720 − 2720)e−4t + (2640 − 3360)e−5t = 80 − 720e−5t (OK)

8di1dt

= −2560e−5t + 2176e−4t

20i1 = 80 + 1280e−5t − 1360e−4t

16di2dt

= 4160e−5t − 3264e−4t

80i2 = 80 − 4160e−5t + 4080e−4t

16dig

dt= 1280e−5t

2560e−5t − 2176e−4t − 80 − 1280e−5t + 1360e−4t + 4160e−5t − 3264e−4t

+80 − 4160e−5t + 4080e−4t ?= 1280e−5t

(−80 + 80) + (2560 − 1280 + 4160 − 4160)e−5t

+(1360 − 2176 − 3264 + 4080)e−4t ?= 1280e−5t

0 + 1280e−5t + 0e−4t = 1280e−5t (OK)

P 6.38 [a] L2 =(

M2

k2L1

)=

(0.09)2

(0.75)2(0.288)= 50 mH

N1

N2=

√L1

L2=

√28850

= 2.4

[b] P1 =L1

N21

=0.288

(1200)2 = 0.2 × 10−6 Wb/A

P2 =L2

N22

=0.05

(500)2 = 0.2 × 10−6 Wb/A

P 6.39 P1 =L1

N21

= 2 nWb/A; P2 =L2

N22

= 2 nWb/A; M = k√

L1L2 = 180 µH

P12 = P21 =M

N1N2= 1.2 nWb/A

P11 = P1 − P21 = 0.8 nWb/A

Problems 6–29

P 6.40 [a] k =M√L1L2

=7.2√81

= 0.8

[b] M =√

81 = 9 mH

[c]L1

L2=

N21 P1

N22 P2

=(

N1

N2

)2

·. .(

N1

N2

)2

=273

= 9

N1

N2= 3

P 6.41 [a] M = k√

L1L2 = 0.8√

324 = 14.4 mH

P1 =L1

N21

=36 × 10−3

(200)2 = 900 nWb/A

dφ11

dφ21=

P11

P21= 0.1; P21 = 10P11

P1 = P11 + P21 = 11P11

P11 =111

P1 = 81.82 nWb/A

P21 = 10P11 = 818.18 nWb/A

N2 =M

N1P21=

14.4 × 10−3

(200)(818.18 × 10−9)= 88 turns

[b] P2 =L2

N22

=9 × 10−3

(88)2 = 1162.19 nWb/A

[c] P11 = 81.82 nWb/A [see part (a)]

[d]φ22

φ12=

P22

P12

P12 = P21 = 818.18 nWb/A

P22 = P2 − P12 = 1162.19 × 10−9 − 818.18 × 10−9 = 344.01 nWb/A

φ22

φ12=

344.01818.18

= 0.4205

P 6.42 [a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign thecurrent into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4.Hence, 1 and 4 or 2 and 3.

[b] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign thecurrent into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dotterminal 4. Hence, 2 and 4 or 1 and 3.


[c] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign thecurrent into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dotterminal 4. Hence, 2 and 4 or 1 and 3.

[d] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assign thecurrent into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4.Hence, 1 and 4 or 2 and 3.

P 6.43 [a]1k2 =

(1 +

P11

P12

)(1 +

P22

P12

)=(1 +

P11

P21

)(1 +

P22

P12

)Therefore

k2 =P12P21

(P21 + P11)(P12 + P22)

Now note that

φ1 = φ11 + φ21 = P11N1i1 + P21N1i1 = N1i1(P11 + P21)

and similarly

φ2 = N2i2(P22 + P12)

It follows that

(P11 + P21) =φ1

N1i1

and

(P22 + P12) =(

φ2

N2i2

)

Therefore

k2 =(φ12/N2i2)(φ21/N1i1)(φ1/N1i1)(φ2/N2i2)

=φ12φ21

φ1φ2

or

k =

√√√√(φ21

φ1

)(φ12

φ2

)

[b] The fractions (φ21/φ1) and (φ12/φ2) are by definition less than 1.0, thereforek < 1.

P 6.44 [a] vab = L1di

dt+ L2

di

dt+ M

di

dt+ M

di

dt= (L1 + L2 + 2M)

di

dt

It follows that Lab = (L1 + L2 + 2M)

[b] vab = L1di

dt− M

di

dt+ L2

di

dt− M

di

dt= (L1 + L2 − 2M)

di

dt

Therefore Lab = (L1 + L2 − 2M)

Problems 6–31

P 6.45 When the switch is opened the induced voltage is negative at the dotted terminal.Since the voltmeter kicks upscale, the induced voltage across the voltmeter must bepositive at its positive terminal. Therefore, the voltage is negative at the negativeterminal of the voltmeter.

Thus, the lower terminal of the unmarked coil has the same instantaneous polarity asthe dotted terminal. Therefore, place a dot on the lower terminal of the unmarkedcoil.

P 6.46 [a] vab = L1d(i1 − i2)

dt+ M

di2dt

0 = L1d(i2 − i1)

dt− M

di2dt

+ Md(i1 − i2)

dt+ L2

di2dt

Collecting coefficients of [di1/dt] and [di2/dt], the two mesh-current equationsbecome

vab = L1di1dt

+ (M − L1)di2dt

and

0 = (M − L1)di1dt

+ (L1 + L2 − 2M)di2dt

Solving for [di1/dt] gives

di1dt

=L1 + L2 − 2ML1L2 − M2 vab

from which we have

vab =(

L1L2 − M2

L1 + L2 − 2M

)(di1dt

)

·. . Lab =L1L2 − M2

L1 + L2 − 2M

[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, therefore

Lab =L1L2 − M2

L1 + L2 + 2M

P 6.47 [a] W = (0.5)L1i21 + (0.5)L2i

22 + Mi1i2

M = 0.85√

(18)(32) = 20.4 mH

W = [9(36) + 16(81) + 20.4(54)] = 2721.6 mJ

[b] W = [324 + 1296 + 1101.6] = 2721.6 mJ

[c] W = [324 + 1296 − 1101.6] = 518.4 mJ

[d] W = [324 + 1296 − 1101.6] = 518.4 mJ


P 6.48 [a] M = 1.0√

(18)(32) = 24 mH, i1 = 6 A

Therefore 16i22 + 144i2 + 324 = 0, i22 + 9i2 + 20.25 = 0

Therefore i2 = −(9

2

)±√(9

2

)2

− 20.25 = −4.5 ±√

0

Therefore i2 = −4.5 A

[b] No, setting W equal to a negative value will make the quantity under the squareroot sign negative.

P 6.49 When the button is not pressed we have

C2dv

dt= C1

d

dt(vs − v)

or

(C1 + C2)dv

dt= C1

dvs

dt

dv

dt=

C1

(C1 + C2)dvs

dt

Assuming C1 = C2 = C

dv

dt= 0.5

dvs

dt

or

v = 0.5vs(t) + v(0)

When the button is pressed we have

Problems 6–33

C1dv

dt+ C3

dv

dt+ C2

d(v − vs)dt

= 0

·. .dv

dt=

C2

C1 + C2 + C3

dvs

dt

Assuming C1 = C2 = C3 = C

dv

dt=

13

dvs

dt

v =13vs(t) + v(0)

Therefore interchanging the fixed capacitor and the button has no effect on thechange in v(t).

P 6.50 With no finger touching and equal 10 pF capacitors

v(t) =1020

(vs(t)) + 0 = 0.5vs(t)

With a finger touching

Let Ce = equivalent capacitance of person touching lamp

Ce =(10)(100)

110= 9.091 pF

Then C + Ce = 10 + 9.091 = 19.091 pF

·. . v(t) =10

29.091vs = 0.344vs

·. . ∆v(t) = (0.5 − 0.344)vs = 0.156vs


P 6.51 With no finger on the button the circuit is

C1dv

dt(v − vs) + C2

d

dt(v + vs) = 0

when C1 = C2 = C (2C)dv

dt= 0

With a finger on the button

C1d(v − vs)

dt+ C2

d(v + vs)dt

+ C3dv

dt= 0

(C1 + C2 + C3)dv

dt+ C2

dvs

dt− C1

dvs

dt= 0

when C1 = C2 = C3 = C (3C)dv

dt= 0

·. . there is no change in the output voltage of this circuit.

7Response of First-Order RL and RC

Circuits

Assessment Problems

AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a shortcircuit, effectively eliminating the 2 Ω resistor from the circuit.

First combine the 30 Ω and 6 Ω resistors in parallel:30‖6 = 5 ΩUse voltage division to find the voltage drop across the parallel resistors:

v =5

5 + 3(120) = 75 V

Now find the current using Ohm’s law:

i(0−) = −v

6= −75

6= −12.5 A

[b] w(0) =12Li2(0) =

12(8 × 10−3)(12.5)2 = 625 mJ

[c] To find the time constant, we need to find the equivalent resistance seen by theinductor for t > 0. When the switch opens, only the 2 Ω resistor remainsconnected to the inductor. Thus,

τ =L

R=

8 × 10−3

2= 4 ms

[d] i(t) = i(0−)et/τ = −12.5e−t/0.004 = −12.5e−250t A, t ≥ 0

[e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A

7–1

7–2 CHAPTER 7. Response of First-Order RL and RC Circuits

So w (5 ms) = 12Li2(5 ms) = 1

2(8) × 10−3(3.58)2 = 51.3 mJw (dis) = 625 − 51.3 = 573.7 mJ

% dissipated =(573.7

625

)100 = 91.8%

AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor:


i(0−) =10

10 + 6(6.4) = 4 A

Now use the circuit for t > 0 to find the equivalent resistance seen by theinductor, and use this value to find the time constant:

Req = 4‖(6 + 10) = 3.2 Ω, ·. . τ =L

Req=

0.323.2

= 0.1 s

Use the initial inductor current and the time constant to find the current in theinductor:i(t) = i(0−)e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0Use current division to find the current in the 10 Ω resistor:

io(t) =4

4 + 10 + 6(−i) =

420

(−4e−10t) = −0.8e−10t A, t ≥ 0+

Finally, use Ohm’s law to find the voltage drop across the 10 Ω resistor:vo(t) = 10io = 10(−0.8e−10t) = −8e−10t V, t ≥ 0+

[b] The initial energy stored in the inductor is

w(0) =12Li2(0−) =

12(0.32)(4)2 = 2.56 J

Find the energy dissipated in the 4 Ω resistor by integrating the power over alltime:

v4Ω(t) = Ldi

dt= 0.32(−10)(4e−10t) = −12.8e−10t V, t ≥ 0+

Problems 7–3

p4Ω(t) =v2

4Ω

4= 40.96e−20t W, t ≥ 0+

w4Ω(t) =∫ ∞

040.96e−20tdt = 2.048 J

Find the percentage of the initial energy in the inductor dissipated in the 4 Ωresistor:

% dissipated =(2.048

2.56

)100 = 80%

AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like anopen circuit.

Find the voltage drop across the open circuit by finding the voltage drop acrossthe 50 kΩ resistor. First use current division to find the current through the50 kΩ resistor:

i50k =80 × 103

80 × 103 + 20 × 103 + 50 × 103 (7.5 × 10−3) = 4 mA

Use Ohm’s law to find the voltage drop:v(0−) = (50 × 103)i50k = (50 × 103)(0.004) = 200 V

[b] To find the time constant, we need to find the equivalent resistance seen by thecapacitor for t > 0. When the switch opens, only the 50 kΩ resistor remainsconnected to the capacitor. Thus,τ = RC = (50 × 103)(0.4 × 10−6) = 20 ms

[c] v(t) = v(0−)e−t/τ = 200e−t/0.02 = 200e−50t V, t ≥ 0

[d] w(0) =12Cv2 =

12(0.4 × 10−6)(200)2 = 8 mJ

[e] w(t) =12Cv2(t) =

12(0.4 × 10−6)(200e−50t)2 = 8e−100t mJ

The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:

8 × 10−3e−100t = 2 × 10−3, e100t = 4, t = (ln 4)/100 = 13.86 ms

AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested voltage, vo,is the sum of the voltage drops for the two RC circuits. The circuit for t < 0 isshown below:


Find the current in the loop and use it to find the initial voltage drops across thetwo RC circuits:

i =15

75,000= 0.2 mA, v5(0−) = 4 V, v1(0−) = 8 V

There are two time constants in the circuit, one for each RC subcircuit. τ5 isthe time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant forthe 1 µF – 40 kΩ subcircuit:τ5 = (20 × 103)(5 × 10−6) = 100 ms; τ1 = (40 × 103)(1 × 10−6) = 40 msTherefore,v5(t) = v5(0−)e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0v1(t) = v1(0−)e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0Finally,vo(t) = v1(t) + v5(t) = [8e−25t + 4e−10t] V, t ≥ 0

[b] Find the value of the voltage at 60 ms for each subcircuit and use the voltage tofind the energy at 60 ms:v1(60 ms) = 8e−25(0.06) ∼= 1.79 V, v5(60 ms) = 4e−10(0.06) ∼= 2.20 Vw1(60 ms) = 1

2Cv21(60 ms) = 1

2(1 × 10−6)(1.79)2 ∼= 1.59 µJw5(60 ms) = 1

2Cv25(60 ms) = 1

2(5 × 10−6)(2.20)2 ∼= 12.05 µJw(60 ms) = 1.59 + 12.05 = 13.64 µJFind the initial energy from the initial voltage:w(0) = w1(0) + w2(0) = 1

2(1 × 10−6)(8)2 + 12(5 × 10−6)(4)2 = 72 µJ

Now calculate the energy dissipated at 60 ms and compare it to the initialenergy:wdiss = w(0) − w(60 ms) = 72 − 13.64 = 58.36 µJ

% dissipated = (58.36 × 10−6/72 × 10−6)(100) = 81.05 %

AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in theinductor:

Problems 7–5

i(0−) = 24/2 = 12 A = i(0+)Note that i(0−) = i(0+) because the current in an inductor is continuous.

[b] Use the circuit at t = 0+, shown below, to calculate the voltage drop across theinductor at 0+. Note that this is the same as the voltage drop across the 10 Ωresistor, which has current from two sources — 8 A from the current sourceand 12 A from the initial current through the inductor.

v(0+) = −10(8 + 12) = −200 V

[c] To calculate the time constant we need the equivalent resistance seen by theinductor for t > 0. Only the 10 Ω resistor is connected to the inductor fort > 0. Thus,τ = L/R = (200 × 10−3/10) = 20 ms

[d] To find i(t), we need to find the final value of the current in the inductor. Whenthe switch has been in position a for a long time, the circuit reduces to the onebelow:

Note that the inductor behaves as a short circuit and all of the current from the8 A source flows through the short circuit. Thus,if = −8 ANow,i(t) = if + [i(0+) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02

= −8 + 20e−50t A, t ≥ 0

[e] To find v(t), use the relationship between voltage and current for an inductor:

v(t) = Ldi(t)dt

= (200 × 10−3)(−50)(20e−50t) = −200e−50t V, t ≥ 0+


AP 7.6 [a]

From Example 7.6,

vo(t) = −60 + 90e−100t V

Write a KVL equation at the top node and use it to find the relationshipbetween vo and vA:vA − vo

8000+

vA

160,000+

vA + 7540,000

= 0

20vA − 20vo + vA + 4vA + 300 = 0

25vA = 20vo − 300

vA = 0.8vo − 12

Use the above equation for vA in terms of vo to find the expression for vA:

vA(t) = 0.8(−60 + 90e−100t) − 12 = −60 + 72e−100t V, t ≥ 0+

[b] t ≥ 0+, since there is no requirement that the voltage be continuous in a resistor.

AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage dropacross the capacitor:

i =(

40 × 103

125 × 103

)(10 × 10−3) = 3.2 mA

vc(0−) = (3.2 × 10−3)(25 × 103) = 80 V so vc(0+) = 80 V

Now use the next circuit, valid for 0 ≤ t ≤ 10 ms, to calculate vc(t) for thatinterval:

Problems 7–7

For 0 ≤ t ≤ 100 ms:

τ = RC = (25 × 103)(1 × 10−6) = 25 ms

vc(t) = vc(0−)et/τ = 80e−40t V, 0 ≤ t ≤ 10 ms

[b] Calculate the starting capacitor voltage in the interval t ≥ 10 ms, using thecapacitor voltage from the previous interval:vc(0.01) = 80e−40(0.01) = 53.63 VNow use the next circuit, valid for t ≥ 10 ms, to calculate vc(t) for that interval:

For t ≥ 10 ms :

Req = 25 kΩ‖100 kΩ = 20 kΩ

τ = ReqC = (20 × 103)(1 × 10−6) = 0.02 s

Therefore vc(t) = vc(0.01+)e−(t−0.01)/τ = 53.63e−50(t−0.01) V, t ≥ 0.01 s

[c] To calculate the energy dissipated in the 25 kΩ resistor, integrate the powerabsorbed by the resistor over all time. Use the expression p = v2/R tocalculate the power absorbed by the resistor.

w25k =∫ 0.01

0

[80e−40t]2

25,000dt +

∫ ∞

0.01

[53.63e−50(t−0.01)]2

25,000dt = 2.91 mJ

[d] Repeat the process in part (c), but recognize that the voltage across this resistoris non-zero only for the second interval:

w100kΩ =∫ ∞

0.01

[53.63e−50(t−0.01)]2

100,000dt = 0.29 mJ

We can check our answers by calculating the initial energy stored in thecapacitor. All of this energy must eventually be dissipated by the 25 kΩresistor and the 100 kΩ resistor.

Check: wstored = (1/2)(1 × 10−6)(80)2 = 3.2 mJ

wdiss = 2.91 + 0.29 = 3.2 mJ

AP 7.8 [a] Note – the 30 Ω resistor should be a 3 Ω resistor; the resistor in parallel with the8 A current source should be 9 Ω.Prior to switch a closing at t = 0, there are no sources connected to theinductor; thus, i(0−) = 0.At the instant A is closed, i(0+) = 0.


For 0 ≤ t ≤ 1 s,

The equivalent resistance seen by the 10 V source is 2 + (3‖0.8). The currentleaving the 10 V source is

102 + (3‖0.8)

= 3.8 A

The final current in the inductor, which is equal to the current in the 0.8 Ωresistor is

i(∞) =3

3 + 0.8(3.8) = 3 A

The resistance seen by the inductor is calculated to find the time constant:

0.8 + (2‖3) = 2 Ω τ =L

R=

22

= 1 s

Therefore,

i = i(∞) + [i(0+) − i(∞)]e−t/τ = 3 − 3e−t A, 0 ≤ t ≤ 1 s

For part (b) we need the value of i(t) at t = 1 s:

i(1) = 3 − 3e−1 = 1.896 A

.

[b] For t > 1 s

Use current division to find the final value of the current:

i =9

9 + 6(−8) = −4.8 A

The equivalent resistance seen by the inductor is used to calculate the timeconstant:

3‖(9 + 6) = 2.5 Ω τ =L

R=

22.5

= 0.8 s

Problems 7–9

Therefore,

i = i(∞) + [i(1+) − i(∞)]e−(t−1)/τ

= −4.8 + 6.696e−1.25(t−1) A, t ≥ 1 s

AP 7.9 0 ≤ t ≤ 32 ms:

vo = − 1RCf

∫ 32×10−3

0−10 dt + 0 = − 1

RCf

(−10t)∣∣∣∣32×10−3

0= − 1

RCf

(−320 × 10−3)

RCf = (200 × 103)(0.2 × 10−6) = 40 × 10−3 so1

RCf

= 25

vo = −25(−320 × 10−3) = 8 V

t ≥ 32 ms:

vo = − 1RCf

∫ t

32×10−35 dy + 8 = − 1

RCf

(5y)∣∣∣∣t32×10−3

+8 = − 1RCf

5(t − 32 × 10−3) + 8

RCf = (250 × 103)(0.2 × 10−6) = 50 × 10−3 so1

RCf

= 20

vo = −20(5)(t − 32 × 10−3) + 8 = −100t + 11.2

The output will saturate at the negative power supply value:

−15 = −100t + 11.2 ·. . t = 262 ms


AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at thenon-inverting input:

vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ

τ = (160 × 103)(10 × 10−9) = 10−3; 1/τ = 625

vp = −2 + 2e−625t V; vn = vp

Write a KVL equation at the inverting input, and use it to determine vo:

vn

10,000+

vn − vo

40,000= 0

·. . vo = 5vn = 5vp = −10 + 10e−625t V


−10 + 10e−625t = −5; e−625t = 1/2; t = ln 2/625 = 1.11 ms

[b] Use RC circuit analysis to determine the expression for the voltage at thenon-inverting input:

vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V

The analysis for vo is the same as in part (a):

vo = 5vp = −10 + 15e−625t V


−10 + 15e−625t = −5; e−625t = 1/3; t = ln 3/625 = 1.76 ms

Problems 7–11

Problems

P 7.1 [a] t < 0

2 kΩ‖6 kΩ = 1.5kΩ

Find the current from the voltage source by combining the resistors in seriesand parallel and using Ohm’s law:

ig(0−) =40

(1500 + 500)= 20 mA

Find the branch currents using current division:

i1(0−) =20008000

(0.02) = 5 mA

i2(0−) =60008000

(0.02) = 15 mA

[b] The current in an inductor is continuous. Therefore,

i1(0+) = i1(0−) = 5 mA

i2(0+) = −i1(0+) = −5 mA (when switch is open)

[c] τ =L

R=

0.4 × 10−3

8 × 103 = 5 × 10−5 s;1τ

= 20,000

i1(t) = i1(0+)e−t/τ = 5e−20,000t mA, t ≥ 0

[d] i2(t) = −i1(t) when t ≥ 0+

·. . i2(t) = −5e−20,000t mA, t ≥ 0+

[e] The current in a resistor can change instantaneously. The switching operationforces i2(0−) to equal 15 mA and i2(0+) = −5 mA.

P 7.2 [a] i(0) = 60 V/(10 Ω + 5 Ω) = 4 A

[b] τ =L

R=

445 + 5

= 80 ms


[c] i = 4e−t/0.08 = 4e−12.5t A, t ≥ 0

v1 = −45i = −180e−12.5t V t ≥ 0+

v2 = Ldi

dt= (4)(−12.5)(4e−12.5t) = −200e−12.5t V t ≥ 0+

[d] pdiss = i2(45) = 720e−25t W

wdiss =∫ t

0720e−25x dx = 720

e−25x

−25

∣∣∣∣t0= 28.8 − 28.8e−25t J

wdiss(40 ms) = 28.8 − 28.8e−1 = 18.205 J

w(0) =12(4)(4)2 = 32 J

% dissipated =18.205

32(100) = 56.89%

P 7.3 [a] io(0−) = 0 since the switch is open for t < 0.

[b] For t = 0− the circuit is:

120 Ω‖60 Ω = 40 Ω

·. . ig =12

10 + 40= 0.24 A = 240 mA

iL(0−) =(120

180

)ig = 160 mA

[c] For t = 0+ the circuit is:

120 Ω‖40 Ω = 30 Ω

Problems 7–13

·. . ig =12

10 + 30= 0.30 A = 300 mA

ia =(120

160

)300 = 225 mA

·. . io(0+) = 225 − 160 = 65 mA

[d] iL(0+) = iL(0−) = 160 mA

[e] io(∞) = ia = 225 mA

[f] iL(∞) = 0, since the switch short circuits the branch containing the 20 Ωresistor and the 100 mH inductor.

[g] τ =L

R=

100 × 10−3

20= 5 ms;

1τ

= 200

·. . iL = 0 + (160 − 0)e−200t = 160e−200t mA, t ≥ 0

[h] vL(0−) = 0 since for t < 0 the current in the inductor is constant

[i] Refer to the circuit at t = 0+ and note:

20(0.16) + vL(0+) = 0; ·. . vL(0+) = −3.2 V

[j] vL(∞) = 0, since the current in the inductor is a constant at t = ∞.

[k] vL(t) = 0 + (−3.2 − 0)e−200t = −3.2e−200t V, t ≥ 0+

[l] io = ia − iL = 225 − 160e−200t mA, t ≥ 0+

P 7.4 [a]v

i= R =

400e−5t

10e−5t= 40 Ω

[b] τ =15

= 200 ms

[c] τ =L

R= 200 × 10−3

L = (200 × 10−3)(40) = 8 H

[d] w(0) =12L[i(0)]2 =

12(8)(10)2 = 400 J

[e] wdiss =∫ t

04000e−10x dx = 400 − 400e−10t

0.8w(0) = (0.8)(400) = 320 J

400 − 400e−10t = 320 ·. . e10t = 5

Solving, t = 160.9 ms.


P 7.5 [a] iL(0) =126

= 2 A

io(0+) =122

− 2 = 6 − 2 = 4 A

io(∞) =122

= 6 A

[b] iL = 2e−t/τ ; τ =L

R=

14

s

iL = 2e−4t A

io = 6 − iL = 6 − 2e−4t A, t ≥ 0+

[c] 6 − 2e−4t = 5

1 = 2e−4t

e6t = 2 ·. . t = 173.3 ms

P 7.6 w(0) =12(30 × 10−3)(32) = 135 mJ

15w(0) = 27 mJ

iR = 3e−t/τ

pdiss = i2RR = 9Re−2t/τ

wdiss =∫ t

0R(9)e−2x/τ dx

wdiss = 9Re−2x/τ

−2/τ

∣∣∣∣to0= −4.5τR(e−2to/τ − 1) = 4.5L(1 − e−2to/τ )

4.5L = (4.5)(30) × 10−3 = 0.135; to = 15 µs

1 − e−2to/τ =15

e2to/τ = 1.25;2toτ

=2toRL

= ln 1.25

R =L ln 1.25

2to=

30 × 10−3 ln 1.2530 × 10−6 = 223.14 Ω

Problems 7–15

P 7.7 [a] w(0) =12LI2

g

wdiss =∫ to

0I2gRe−2t/τ dt = I2

gRe−2t/τ

(−2/τ)

∣∣∣∣to0

=12I2gRτ(1 − e−2to/τ ) =

12I2gL(1 − e−2to/τ )

wdiss = σw(0)

·. .12LI2

g (1 − e−2to/τ ) = τ(1

2LI2

g

)

1 − e−2to/τ = σ; e2to/τ =1

(1 − σ)

2toτ

= ln[

1(1 − σ)

];

R(2to)L

= ln[1/(1 − σ)]

R =L ln[1/(1 − σ)]

2to

[b] R =(30 × 10−3) ln[1/0.8]

30 × 10−6

R = 223.14 Ω

P 7.8 [a] t < 0

iL(0−) =150180

(12) = 10 A

t ≥ 0

τ =1.6 × 10−3

8= 200 × 10−6; 1/τ = 5000

io = −10e−5000t A t ≥ 0


[b] wdel =12(1.6 × 10−3)(10)2 = 80 mJ

[c] 0.95wdel = 76 mJ

·. . 76 × 10−3 =∫ to

08(100e−10,000t) dt

·. . 76 × 10−3 = −80 × 10−3e−10,000t

∣∣∣∣to0= 80 × 10−3(1 − e−10,000to)

·. . e−10,000to = 4 × 10−3 so to = 552.1 µs

·. .toτ

=552.1 × 10−6

200 × 10−6 = 2.76 so to ≈ 2.76τ

P 7.9 For t < 0+

ig =−48

6 + (18‖1.5)= −6.5 A

iL(0−) =18

18 + 1.5(−6.5) = −6 A = iL(0+)

For t > 0

iL(t) = iL(0+)e−t/τ A, t ≥ 0

τ =L

R=

0.510 + 12.45 + (54‖26)

= 0.0125 s;1τ

= 80

iL(t) = −6e−80t A, t ≥ 0

io(t) =5480

(−iL(t)) =5480

(6e−80t) = 4.05e−80t V, t ≥ 0+

Problems 7–17

P 7.10 From the solution to Problem 7.9,

i54Ω =2680

(−iL) = −1.95e−80t A

P54Ω = 54(i54Ω)2 = 205.335e−160tW

wdiss =∫ 0.0125

0205.335e−160t dt

=205.335−160

e−160t

∣∣∣∣0.0125

0

= 1.28(1 − e−2) = 1.11 J

wstored =12(0.5)(−6)2 = 9 mJ.

% diss =1.119

× 100 = 12.3%

P 7.11 [a] t < 0 :

iL(0−) = iL(0+) =70

70 + 4(11.84) = 11.2 A

i∆ =70160

iT = 0.4375iT

vT = 30i∆ + iT(90)(70)

160= 30(0.4375)iT +

(90)(70)160

iT = 52.5iT

vT

iT= RTh = 52.5 Ω


τ =L

R=

20 × 10−3

52.5= ·. .

1τ

= 2625

iL = 11.2e−2625t A, t ≥ 0

[b] vL = LdiLdt

= 20 × 10−3(−2625)(11.2e−2625t) = −588e−2625t V, t ≥ 0+

[c]

vL = 30i∆ + 90i∆ = 120i∆

i∆ =vL

120= −4.9e−2625t A t ≥ 0+

P 7.12 w(0) =12(20 × 10−3)(11.2)2 = 1254.4 mJ

p30i∆ = −30i∆iL = −30(−4.9e−2625t)(11.2e−2625t) = 1646.4e−5250t W

w30i∆ =∫ ∞

01646.4e−5250t dt = 1646.4

e−5250t

−5250

∣∣∣∣∞0

= 313.6mJ

% dissipated =313.61254.4

(100) = 25%

P 7.13 t < 0

iL(0−) = iL(0+) = 4 A

Problems 7–19

t > 0

Find Thévenin resistance seen by inductor

iT = 4vT ;vT

iT= RTh =

14

= 0.25 Ω

τ =L

R=

5 × 10−3

0.25= 20 ms; 1/τ = 50

io = 4e−50t A, t ≥ 0

vo = Ldiodt

= (5 × 10−3)(−200e−50t) = −e−50t V, t ≥ 0+

P 7.14 t < 0:

iL(0+) = 8 A


t > 0:

Re =(10)(40)

50+ 10 = 18 Ω

τ =L

Re

=0.07218

= 4 ms;1τ

= 250

·. . iL = 8e−250t A

·. . vo = −10iL − 0.072diLdt

= −80e−250t + 144e−250t

= 64e−250t A t ≥ 0+

P 7.15 w(0) =12(72 × 10−3)(8)2 = 2304 mJ

p40Ω =v2

o

40=

642

40e−500t = 102.4e−500t W

w40Ω =∫ ∞

0102.4e−500t dt = 204.8 mJ

%diss =204.82304

(100) = 8.89%

P 7.16 [a] vo(t) = vo(0+)e−t/τ

·. . vo(0+)e−1×10−3/τ = 0.5vo(0+)

·. . e1×10−3/τ = 2

·. . τ =L

R=

1 × 10−3

ln 2

·. . L =10 × 10−3

ln 2= 14.43 mH

Problems 7–21

[b] vo(0+) = −10iL(0+) = −10(1/10)30 × 10−3 = −30 mV

·. . vo = −0.03e−t/τ V, t ≥ 0+

p10Ω =v2

o

10= 9 × 10−5e−2t/τ

w10Ω(1 ms) =∫ 10−3

0+9 × 10−5e−2t/τ dt

= 4.5τ × 10−5(1 − e−2(0.001)/τ )

τ =1

1000 ln 2·. . w10Ω(1 ms) = 48.69 nJ

wL(0) =12Li2L(0) =

12(14.43 × 10−3)(3 × 10−3)2 = 64.92 nJ

%dissipated in 1 ms =48.6964.92

(100) = 75%

P 7.17 [a] t < 0 :

t = 0+:

33 = iab + 9 + 15, iab = 9 A, t = 0+

[b] At t = ∞:

iab = 165/5 = 33 A, t = ∞


[c] i1(0) = 9, τ1 =12.5 × 10−3

5= 2.5 ms

i2(0) = 15, τ2 =3.75 × 10−3

31.25 ms

i1(t) = 9e−400t A, t ≥ 0

i2(t) = 15e−800t A, t ≥ 0

iab = 33 − 9e−400t − 15e−800t A, t ≥ 0+

33 − 9e−400t − 15e−800t = 19

14 = 9e−400t + 15e−800t

Let x = e−400t ·. . x2 = e−800t

Substituting,

15x2 + 9x − 14 = 0 so x = 0.7116 = e−400t

·. . t =[ln(1/0.7116)]

400= 850.6 µs

P 7.18 [a] t < 0

1 kΩ‖4 kΩ = 0.8 kΩ

20 kΩ‖80 kΩ = 16 kΩ

(105 × 10−3)(0.8 × 103) = 84 V

Problems 7–23

iL(0−) =84

16,800= 5 mA

t > 0

τ =L

R=

624

× 10−3 = 250 µs;1τ

= 4000

iL(t) = 5e−4000t mA, t ≥ 0

p4k = 25 × 10−6e−8000t(4000) = 0.10e−8000t W

wdiss =∫ t

00.10e−8000x dx = 12.5 × 10−6[1 − e−8000t] J

w(0) =12(6)(25 × 10−6) = 75 µJ

0.10w(0) = 7.5 µJ

12.5(1 − e−8000t) = 7.5; ·. . e8000t = 2.5

t =ln 2.58000

= 114.54 µs

[b] wdiss(total) = 75(1 − e−8000t) µJ

wdiss(114.54 µs) = 45 µJ

% = (45/75)(100) = 60%


P 7.19 [a] t < 0:

t = 0+:

t > 0:

iR = −2e−t/τ A; τ =L

R=

57.5

= 666.67 ms ·. .1τ

= 1.5

iR = −2e−1.5t A

vR = (7.5)(−2e−1.5t) = −15e−1.5t V

Problems 7–25

v1 = 1.25[(−1.5)(−2e−1.5t)] = 3.75e−1.5t V,

vo = −v1 − vR = 11.25e−1.5t V t ≥ 0+

[b] io =16

∫ t

011.25e−1.5x dx + 0 = 1.25 − 1.25e−1.5t A t ≥ 0

P 7.20 [a] From the solution to Problem 7.19,

iR = −2e−1.5t A

pR = (−2e−1.5t)2(7.5) = 30e−3t W

wdiss =∫ ∞

030e−3t dt

= 30e−3t

−3

∣∣∣∣∞0

= 10 J

[b] wtrapped =12(10)(−1.25)2 +

12(6)(1.25)2 = 12.5 J

CHECK: w(0) = 12(1.25)(2)2 + 1

2(10)(2)2 = 22.5 J

·. . w(0) = wdiss + wtrapped

P 7.21 [a] v1(0−) = v1(0+) = 40 V v2(0+) = 0

Ceq = (1)(4)/5 = 0.8 µF

τ = (25 × 103)(0.8 × 10−6) = 20ms;1τ

= 50

i =40

25,000e−50t = 1.6e−50t mA, t ≥ 0+

v1 =−1

10−6

∫ t

01.6 × 10−3e−50x dx + 40 = 32e−50t + 8 V, t ≥ 0

v2 =1

4 × 10−6

∫ t

01.6 × 10−3e−50x dx + 0 = −8e−50t + 8 V, t ≥ 0


[b] w(0) =12(10−6)(40)2 = 800 µJ

[c] wtrapped =12(10−6)(8)2 +

12(4 × 10−6)(8)2 = 160 µJ.

The energy dissipated by the 25 kΩ resistor is equal to the energy dissipated bythe two capacitors; it is easier to calculate the energy dissipated by thecapacitors (final voltage on the equivalent capacitor is zero):

wdiss =12(0.8 × 10−6)(40)2 = 640 µJ.

Check: wtrapped + wdiss = 160 + 640 = 800µJ; w(0) = 800µJ.

P 7.22 [a] Calculate the initial voltage drop across the capacitor:

v(0) = (2.7 k‖3.3 k)(40 mA) = (1485)(40 × 10−3) = 59.4 V

The equivalent resistance seen by the capacitor is

Re = 3 k‖(2.4 k + 3.6 k) = 3 k‖6 k = 2 kΩ

τ = ReC = (2000)(0.5) × 10−6 = 1000µs;1τ

= 1000

v = v(0)e−t/τ = 59.4e−1000t V t ≥ 0

io =v

2.4 k + 3.6 k= 9.9e−1000t mA, t ≥ 0+

[b] w(0) =12(0.5 × 10−6)(59.4)2 = 882.09 µJ

i3k =59.4e−1000t

3000= 19.8e−1000t mA

p3k = [(19.8 × 10−3)e−1000t]2(3000) = 1.176e−2000t

w3k(500 µs) = 1.176e−2000x

−2000

∣∣∣∣500×10−6

0=

1.176−2000

(e−1 − 1) = 371.72 µJ

% =371.72882.09

× 100 = 42.14%

P 7.23 [a] R =v

i= 4 kΩ

[b]1τ

=1

RC= 25; C =

1(25)(4 × 103)

= 10 µF

[c] τ =125

= 40 ms

[d] w(0) =12(10 × 10−6)(48)2 = 11.52 mJ

Problems 7–27

[e] wdiss(60 ms) =∫ 0.06

0

v2

Rdt =

∫ 0.06

0

(48e−25t)2

(4 × 103)dt

= 0.576e−50t

−50

∣∣∣∣0.06

0= −5.74 × 10−4 + 0.01152 = 10.95 mJ

P 7.24 [a] t < 0:

i1(0−) = i2(0−) =3 V30 Ω

= 100 mA

[b] t > 0:

i1(0+) =0.22

= 100 mA

i2(0+) =−0.2

8= −25 mA

[c] Capacitor voltage cannot change instantaneously, therefore,

i1(0−) = i1(0+) = 100 mA

[d] Switching can cause an instantaneous change in the current in a resistivebranch. In this circuit

i2(0−) = 100 mA and i2(0+) = −25 mA

[e] vc = 0.2e−t/τ V, t ≥ 0 Re = 2||(5 + 3) = 1.6 Ω

τ = 1.6(2 × 10−6) = 3.2 × 10−6 s

vc = 0.2e−312,500t V, t ≥ 0

i1 =vc

2= 0.1e−312,500t A, t ≥ 0

[f] i2 =−vc

8= −25e−312,500t mA, t ≥ 0+


P 7.25 [a] t < 0:

Re = 12 k||68 k = 10.2 kΩ

vo(0) =10,200

10,200 + 1800(−120) = −102 V

t > 0:

τ = [(10/3) × 10−6)(12,000) = 40 ms;1τ

= 25

vo = −102e−25t V, t ≥ 0

p =v2

o

12,000= 867 × 10−3e−50t W

wdiss =∫ 12×10−3

0867 × 10−3e−50t dt

= 17.34 × 10−3(1 − e−50(12×10−3)) = 7.82 mJ

[b] w(0) =(1

2

)(103

)(102)2 × 10−6 = 17.34 mJ

0.75w(0) = 13 mJ∫ to

0867 × 10−3e−50x dx = 13 × 10−3

·. . 1 − e−50to = 0.75; e50to = 4; so to = 27.73 ms

Problems 7–29

P 7.26 [a]

vT = 20 × 103(iT + αv∆) + 5 × 103iT

v∆ = 5 × 103iT

vT = 25 × 103iT + 20 × 103α(5 × 103iT )

RTh = 25,000 + 100 × 106α

τ = RThC = 40 × 10−3 = RTh(0.8 × 10−6)

RTh = 50 kΩ = 25,000 + 100 × 106α

α =25,000

100 × 106 = 2.5 × 10−4 A/V

[b] vo(0) = (−5 × 10−3)(3600) = −18 V t < 0t > 0:

vo = −18e−25t V, t ≥ 0

v∆

5000+

v∆ − vo

20,000+ 2.5 × 10−4v∆ = 0


4v∆ + v∆ − vo + 5v∆ = 0

·. . v∆ =vo

10= −1.8e−25t V, t ≥ 0+

P 7.27 [a]

pds = (16.2e−25t)(−450 × 10−6e−25t) = −7290 × 10−6e−50t W

wds =∫ ∞

0pds dt = −145.8 µJ.

·. . dependent source is delivering 145.8 µJ

[b] w5k =∫ ∞

0(5000)(0.36 × 10−3e−25t)2 dt = 648 × 10−6

∫ ∞

0e−50t dt = 12.96 µJ

w20k =∫ ∞

0

(16.2e−25t)2

20,000dt = 13,122 × 10−6

∫ ∞

0e−50t dt = 262.44 µJ

wc(0) =12(0.8 × 10−6)(18)2 = 129.6 µJ

∑wdiss = 12.96 + 262.44 = 275.4 µJ

∑wdev = 145.8 + 129.6 = 275.4 µJ.

P 7.28 t < 0

Problems 7–31

t > 0

vT = −5io − 15io = −20io = 20iT ·. . RTh =vT

iT= 20 Ω

τ = RC = 40 µs;1τ

= 25,000

vo = 15e−25,000t V, t ≥ 0

io = − vo

20= −0.75e−25,000t A, t ≥ 0+

P 7.29 [a] The equivalent circuit for t > 0:

τ = 2 ms; 1/τ = 500

vo = 10e−500t V, t ≥ 0

io = e−500t mA, t ≥ 0+

i24kΩ = e−500t(16

40

)= 0.4e−500t mA, t ≥ 0+

p24kΩ = (0.16 × 10−6e−1000t)(24,000) = 3.84e−1000t mW

w24kΩ =∫ ∞

03.84 × 10−3e−1000t dt = −3.84 × 10−6(0 − 1) = 3.84 µJ


w(0) =12(0.25 × 10−6)(40)2 +

12(1 × 10−6)(50)2 = 1.45 mJ

% diss (24 kΩ) =3.84 × 10−6

1.45 × 10−3 × 100 = 0.26%

[b] p400Ω = 400(1 × 10−3e−500t)2 = 0.4 × 10−3e−1000t

w400Ω =∫ ∞

0p400 dt = 0.40 µJ

% diss (400Ω) =0.4 × 10−6

1.45 × 10−3 × 100 = 0.03%

i16kΩ = e−500t(24

40

)= 0.6e−500t mA, t ≥ 0+

p16kΩ = (0.6 × 10−3e−500t)2(16,000) = 5.76 × 10−3e−1000t W

w16kΩ =∫ ∞

05.76 × 10−3e−1000t dt = 5.76 µJ

% diss (16kΩ) = 0.4%

[c]∑

wdiss = 3.84 + 5.76 + 0.4 = 10 µJ

wtrapped = w(0) −∑wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ

% trapped =1.441.45

× 100 = 99.31%

Check: 0.26 + 0.03 + 0.4 + 99.31 = 100%

P 7.30 [a] Ce =(2 + 1)62 + 1 + 6

= 2 µF

vo(0) = −5 + 30 = 25 V

τ = (2 × 10−6)(250 × 103) = 0.5 s;1τ

= 2

vo = 25e−2t V, t > 0+

Problems 7–33

[b] wo =12(3 × 10−6)(30)2 +

12(6 × 10−6)(5)2 = 1425µJ

wdiss =12(2 × 10−6)(25)2 = 625 µJ

% diss =1425 − 625

1425× 100 = 56.14%

[c] io =vo

250 × 10−3 = 100e−2t µA

v1 = − 16 × 10−6

∫ t

0100 × 10−6e−2x dx − 5 = −16.67

∫ t

0e−2x dx − 5

= −16.67e−2x

−2

∣∣∣∣t0

−5 = 8.33e−2t − 13.33 V t ≥ 0

[d] v1 + v2 = vo

v2 = vo − v1 = 25e−2t − 8.33e−2t + 13.33 = 16.67e−2t + 13.33 V t ≥ 0

[e] wtrapped =12(6 × 10−6)(13.33)2 +

12(3 × 10−6)(13.33)2 = 800 µJ

wdiss + wtrapped = 625 + 800 = 1425µJ (check)

P 7.31 [a] At t = 0− the voltage on each capacitor will be 150 V(5 × 30), positive at theupper terminal. Hence at t ≥ 0+ we have

·. . isd(0+) = 5 +1500.2

+1500.5

= 1055 A

At t = ∞, both capacitors will have completely discharged.

·. . isd(∞) = 5 A

[b] isd(t) = 5 + i1(t) + i2(t)

τ1 = 0.2(10−6) = 0.2 µs

τ2 = 0.5(100 × 10−6) = 50 µs


·. . i1(t) = 750e−5×106t A, t ≥ 0+

i2(t) = 300e−20,000t A, t ≥ 0

·. . isd = 5 + 750e−5×106t + 300e−20,000t mA, t ≥ 0+

P 7.32 [a] t < 0:

io(0−) =6000

6000 + 4000(40 m) = 24 mA

vo(0−) = (3000)(24 m) = 72 V

i2(0−) = 40 − 24 = 16 mA

v2(0−) = (6000)(16 m) = 96 V

t > 0

τ = RC = (1000)(0.2 × 10−6) = 200 µs;1τ

= 5000

io(t) =24

1 × 103 e−t/τ = 24e−5000t mA, t ≥ 0+

[b]

Problems 7–35

vo =1

0.6 × 10−6

∫ t

024 × 10−3e−5000x dx + 72

= (40,000)e−5000x

−5000

∣∣∣∣t0

+72

= −8e−5000t + 8 + 72vo = [−8e−5000t + 80] V, t ≥ 0

[c] wtrapped = (1/2)(0.3 × 10−6)(80)2 + (1/2)(0.6 × 10−6)(80)2

wtrapped = 2880µJ.

Check:

wdiss =12(0.2 × 10−6)(24)2 = 57.6 µJ

w(0) =12(0.3 × 10−6)(96)2 +

12(0.6 × 10−6)(72)2 = 2937.6 µJ.

wtrapped + wdiss = w(0)

2880 + 57.6 = 2937.6 OK.

P 7.33 [a] t < 0

iL(0−) = −5 A

t > 0

iL(∞) =40 − 804 + 16

= −2 A

τ =L

R=

4 × 10−3

4 + 16= 200 µs;

1τ

= 5000


iL = iL(∞) + [iL(0+) − iL(∞)]e−t/τ

= −2 + (−5 + 2)e−5000t = −2 − 3e−5000t A, t ≥ 0

vo = 16iL + 80 = 16(−2 − 3e−5000t) + 80 = 48 − 48e−5000t V, t ≥ 0+

[b] vL = LdiLdt

= 4 × 10−3(−5000)[−3e−5000t] = 60e−5000t V, t ≥ 0+

vL(0+) = 60 V

From part (a) vo(0+) = 0 V

Check: at t = 0+ the circuit is:

vL(0+) = 40 + (5 A)(4 Ω) = 60 V, vo(0+) = 80 − (16 Ω)(5 A) = 0 V

P 7.34 [a] t < 0

KVL equation at the top node:

50 =vo

8+

vo

40+

vo

10Multiply by 40 and solve:

2000 = (5 + 1 + 4)vo; vo = 200 V

·. . io(0−) =vo

10= 200/10 = 20 A

t > 0

Problems 7–37

Use voltage division to find the Thévenin voltage:

VTh = vo =40

40 + 120(800) = 200 V

Remove the voltage source and make series and parallel combinations ofresistors to find the equivalent resistance:

RTh = 10 + 120‖40 = 10 + 30 = 40 Ω

The simplified circuit is:

τ =L

R=

40 × 10−3

40= 1 ms;

1τ

= 1000

io(∞) =20040

= 5 A

·. . io = io(∞) + [io(0+) − io(∞)]e−t/τ

= 5 + (20 − 5)e−1000t = 5 + 15e−1000t A, t ≥ 0

[b] vo = 10io + Ldiodt

= 10(5 + 15e−1000t) + 0.04(−1000)(15e−1000t)

= 50 + 150e−1000t − 600e−1000t

vo = 50 − 450e−1000t V, t ≥ 0+

P 7.35 After making a Thévenin equivalent we have

For t < 0, the 15 Ω resistor is bypassed:

io(0−) = io(0+) = 50/5 = 10 A


τ =L

R=

16 × 10−3

5 + 15= 8 × 10−4;

1τ

= 1250

i(∞) =V

Req=

505 + 15

= 2.5 A

io = io(∞) + [io(0+) − io(∞)]e−t/τ = 2.5 + (10 − 2.5)e−1250t = 2.5 + 7.5e−1250t A, t ≥ 0

vo = Ldiodt

= 16 × 10−3(−1250)(7.5e−1250t) = −150e−1250t V, t ≥ 0+

P 7.36 [a] vo(0+) = −IgR2; τ =L

R1 + R2

vo(∞) = 0

vo(t) = −IgR2e−[(R1+R2)/L]t V, t ≥ 0+

[b] vo = −(10)(15)e− (5+15)0.016 t = −150e−1250t V, t ≥ 0+

[c] vo(0+) → ∞, and the duration of vo(t) → zero

[d] vsw = R2io; τ =L

R1 + R2

io(0+) = Ig; io(∞) = IgR1

R1 + R2

Therefore io(t) = IgR1R1+R2

+[Ig − IgR1

R1+R2

]e−[(R1+R2)/L]t

io(t) = R1Ig

(R1+R2) + R2Ig

(R1+R2)e−[(R1+R2)/L]t

Therefore vsw = R1Ig

(1+R1/R2) + R2Ig

(1+R1/R2)e−[(R1+R2)/L]t, t ≥ 0+

[e] |vsw(0+)| → ∞; duration → 0

P 7.37 Opening the inductive circuit causes a very large voltage to be induced across theinductor L. This voltage also appears across the switch (part [e] of Problem 7.36)causing the switch to arc over. At the same time, the large voltage across L damagesthe meter movement.

P 7.38 [a] From Eqs. (7.35) and (7.42)

i =Vs

R+(Io − Vs

R

)e−(R/L)t

v = (Vs − IoR)e−(R/L)t

·. .Vs

R= 4; Io − Vs

R= 4

Problems 7–39

Vs − IoR = −80;R

L= 40

·. . Io = 4 +Vs

R= 8 A

Now since Vs = 4R we have

4R − 8R = −80; R = 20 Ω

Vs = 80 V; L =R

40= 0.5 H

[b] i = 4 + 4e−40t; i2 = 16 + 32e−40t + 16e−80t

w =12Li2 =

12(0.5)[16 + 32e−40t + 16e−80t] = 4 + 8e−40t + 4e−80t

·. . 4 + 8e−40t + 4e−80t = 9 or e−80t + 2e−40t − 1.25 = 0

Let x = e−40t:

x2 + 2x − 1.25 = 0; Solving, x = 0.5; x = −2.5

But x ≥ 0 for all t. Thus,

e−40t = 0.5; e40t = 2; t = 25 ln 2 = 17.33 ms

P 7.39 For t < 0

vx

15− 0.8vφ +

vx − 48021

= 0

vφ =vx − 480

21

vx

15− 0.8

(vx − 480

21

)+(

vx − 48021

)

=vx

15+ 0.2

(vs − 480

21

)= 21vx + 3(vx − 480) = 0


·. . 24vx = 1440 so vx = 60 V io(0−) =vx

15= 4 A

t > 0

Find Thévenin equivalent with respect to a, b

VTh − 3205

− 0.8(

VTh − 3205

)= 0 VTh = 320 V

vT = (iT + 0.8vφ)(5) =(iT + 0.8

vT

5

)(5)

Problems 7–41

vT = 5iT + 0.8vT·. . 0.2vT = 5iT

vT

iT= RTh = 25 Ω

io(∞) = 320/40 = 8 A

τ =80 × 10−3

40= 2 ms; 1/τ = 500

io = 8 + (4 − 8)e−500t = 8 − 4e−500t A, t ≥ 0

P 7.40 t > 0; calculate vo(0+)

va

15+

va − vo(0+)5

= 20 × 10−3

·. . va = 0.75vo(0+) + 75 × 10−3

15 × 10−3 +vo(0+) − va

5+

vo(0+)8

− 9i∆ + 50 × 10−3 = 0

13vo(0+) − 8va − 360i∆ = −2600 × 10−3

i∆ =vo(0+)

8− 9i∆ + 50 × 10−3

·. . i∆ =vo(0+)

80+ 5 × 10−3

·. . 360i∆ = 4.5vo(0+) + 1800 × 10−3


8va = 6vo(0+) + 600 × 10−3

·. . 13vo(0+) − 6vo(0+) − 600 × 10−3 − 4.5vo(0+)−

1800 × 10−3 = −2600 × 10−3

2.5vo(0+) = −200 × 10−3; vo(0+) = −80 mV

vo(∞) = 0

Find the Thévenin resistance seen by the 4 mH inductor:

iT =vT

20+

vT

8− 9i∆

i∆ =vT

8− 9i∆ ·. . 10i∆ =

vT

8; i∆ =

vT

80

iT =vT

20+

10vT

80− 9vT

80

iTvT

=120

+180

=580

=116

S

·. . RTh = 16Ω

τ =4 × 10−3

16= 0.25 ms; 1/τ = 4000

·. . vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV, t ≥ 0+

Problems 7–43

P 7.41 [a]

v

R+

1L

∫ t

0v dx =

Vs

R

1R

dv

dt+

v

L= 0

dv

dt+

R

Lv = 0

[b]dv

dt= −R

Lv

dv

dtdt = −R

Lv dt

·. .dv

v= −R

Ldt

∫ v(t)

v(0+)

dy

y= −R

L

∫ t

0+dx

ln y∣∣∣∣v(t)

v(0+)= −

(R

L

)t

ln[

v(t)v(0+)

]= −

(R

L

)t

v(t) = v(0+)e−(R/L)t; v(0+) =(

Vs

R− Io

)R = Vs − IoR

·. . v(t) = (Vs − IoR)e−(R/L)t

P 7.42 t > 0

τ =140


io = 5e−40t A, t ≥ 0

vo = 40io = 200e−40t V, t > 0+

200e−40t = 100; e40t = 2

·. . t =140

ln 2 = 17.33 ms

P 7.43 [a] wdiss =12Lei

2(0) =12(1)(5)2 = 12.5 J

[b] i3H =13

∫ t

0(200)e−40x dx − 5

= 1.67(1 − e−40t) − 5 = −1.67e−40t − 3.33 A

i1.5H =1

1.5

∫ t

0(200)e−40x dx + 0

= −3.33e−40t + 3.33 A

wtrapped =12(4.5)(3.33)2 = 25 J

[c] w(0) =12(3)(5)2 = 37.5 J

P 7.44 [a] t < 0

t > 0

iL(0−) = iL(0+) = 25 mA; τ =24 × 10−3

120= 0.2 ms;

1τ

= 5000

iL(∞) = −50 mA

iL = −50 + (25 + 50)e−5000t = −50 + 75e−5000t mA, t ≥ 0

vo = −120[75 × 10−3e−5000t] = −9e−5000t V, t ≥ 0+

Problems 7–45

[b] i1 =1

60 × 10−3

∫ t

0−9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA, t ≥ 0

[c] i2 =1

40 × 10−3

∫ t

0−9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA, t ≥ 0

P 7.45 [a] Let v be the voltage drop across the parallel branches, positive at the top node,then

−Ig +v

Rg

+1L1

∫ t

0v dx +

1L2

∫ t

0v dx = 0

v

Rg

+( 1

L1+

1L2

) ∫ t

0v dx = Ig

v

Rg

+1Le

∫ t

0v dx = Ig

1Rg

dv

dt+

v

Le

= 0

dv

dt+

Rg

Le

v = 0

Therefore v = IgRge−t/τ ; τ = Le/Rg

Thus

i1 =1L1

∫ t

0IgRge

−x/τ dx =IgRg

L1

e−x/τ

(−1/τ)

∣∣∣∣t0=

IgLe

L1(1 − e−t/τ )

i1 =IgL2

L1 + L2(1 − e−t/τ ) and i2 =

IgL1

L1 + L2(1 − e−t/τ )

[b] i1(∞) =L2

L1 + L2Ig; i2(∞) =

L1

L1 + L2Ig

P 7.46 For t < 0, i80mH(0) = 50 V/10 Ω = 5 AFor t > 0, after making a Thévenin equivalent we have

i =Vs

R+(Io − Vs

R

)e−t/τ


1τ

=R

L=

8100 × 10−3 = 80

Io = 5 A; If =Vs

R=

−808

= −10 A

i = −10 + (5 + 10)e−80t = −10 + 15e−80t A, t ≥ 0

vo = 0.08di

dt= 0.08(−1200e−80t) = −96e−80t V, t > 0+

P 7.47 For t < 0

Simplify the circuit:

80/10,000 = 8 mA, 10 kΩ‖40 kΩ‖24 kΩ = 6 kΩ

8 mA − 3 mA = 5 mA

5 mA × 6 kΩ = 30 V

Thus, for t < 0

·. . vo(0−) = vo(0+) = 30 V

t > 0

Problems 7–47

Simplify the circuit:

8 mA + 2 mA = 10 mA

10 k‖40 k‖24 k = 6 kΩ

(10 mA)(6 kΩ) = 60 V

Thus, for t > 0

vo(∞) = −60 V

τ = RC = (10 k)(0.05 µ) = 0.5 ms;1τ

= 2000

vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = −60 + [30 − (−60)]e−2000t

= −60 + 90e−2000t V t ≥ 0

P 7.48 [a] Simplify the circuit for t > 0 using source transformation:

Since there is no source connected to the capacitor for t < 0

vo(0−) = vo(0+) = 0 V

From the simplified circuit,

vo(∞) = 60 V

τ = RC = (20 × 103)(0.5 × 10−6) = 10 ms 1/τ = 100

vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = (60 − 60e−100t) V, t ≥ 0


[b] ic = Cdvo

dt

ic = 0.5 × 10−6(−100)(−60e−100t) = 3e−100t mA

v1 = 8000ic + vo = (8000)(3 × 10−3)e−100t + (60 − 60e−100t) = 60 − 36e−100t V

io =v1

60 × 103 = 1 − 0.6e−100t mA, t ≥ 0+

[c] i1(t) = io + ic = 1 + 2.4e−100t mA t ≥ 0+

[d] i2(t) =v1

15 × 103 = 4 − 2.4e−100t mA t ≥ 0+

[e] i1(0+) = 1 + 2.4 = 3.4 mA

At t = 0+:

Re = 15 k‖60 k‖8 k = 4800 Ω

v1(0+) = (5 × 10−3)(4800) = 24 V

i1(0+) =v1(0+)60,000

+v1(0+)8000

= 0.4 m + 3 m = 3.4mA (checks)

P 7.49 [a] v = IsR + (Vo − IsR)e−t/RC i =(Is − Vo

R

)e−t/RC

·. . IsR = 40, Vo − IsR = −24

·. . Vo = 16 V

Is − Vo

R= 3 × 10−3; Is − 16

R= 3 × 10−3; R =

40Is

·. . Is − 0.4Is = 3 × 10−3; Is = 5 mA

R =405

× 103 = 8 kΩ

1RC

= 2500; C =1

2500R=

10−3

20 × 103 = 50 nF; τ = RC =1

2500= 400 µs

[b] v(∞) = 40 V

w(∞) =12(50 × 10−9)(1600) = 40µJ

0.81w(∞) = 32.4 µJ

v2(to) =32.4 × 10−6

25 × 10−9 = 1296; v(to) = 36 V

40 − 24e−2500to = 36; e2500to = 6; ·. . to = 716.70 µs

Problems 7–49

P 7.50 [a] For t > 0:

τ = RC = 250 × 103 × 8 × 10−9 = 2 ms;1τ

= 500

vo = 50e−500t V, t ≥ 0+

[b] io =vo

250,000=

50e−500t

250,000= 200e−500t µA

v1 =−1

40 × 10−9 × 200 × 10−6∫ t

0e−500x dx + 50 = 10e−500t + 40 V, t ≥ 0

P 7.51 [a] w =12Ceqv

2o =

12(8 × 10−9)(502) = 10 µJ

[b] wtrapped =12(40)2(50 × 10−9) = 40 µJ

[c] w(0) =12(40 × 10−9)(502) = 50 µJ

P 7.52 For t > 0

VTh = (−25)(16,000)ib = −400 × 103ib

ib =33,00080,000

(120 × 10−6) = 49.5 µA

VTh = −400 × 103(49.5 × 10−6) = −19.8 V

RTh = 16 kΩ

vo(∞) = −19.8 V; vo(0+) = 0

τ = (16, 000)(0.25 × 10−6) = 4 ms; 1/τ = 250


vo = −19.8 + 19.8e−250t V, t ≥ 0

w(t) =12(0.25 × 10−6)v2

o = w(∞)(1 − e−250t)2 J

(1 − e−250t)2 =0.36w(∞)

w(∞)= 0.36

1 − e−250t = 0.6

e−250t = 0.4 ·. . t = 3.67 ms

P 7.53 [a]

io(0+) =−365000

= −7.2 mA

[b] io(∞) = 0

[c] τ = RC = (5000)(0.8 × 10−6) = 4 ms

[d] io = 0 + (−7.2)e−250t = −7.2e−250t mA, t ≥ 0+

[e] vo = −[36 + 1800(−7.2 × 10−3e−250t)] = −36 + 12.96e−250t V, t ≥ 0+

P 7.54 [a] vo(0−) = vo(0+) = 120 V

vo(∞) = −150 V; τ = 2 ms;1τ

= 500

vo = −150 + (120 − (−150))e−500t

vo = −150 + 270e−500t V, t ≥ 0

[b] io = −0.04 × 10−6(−500)[270e−500t] = 5.4e−500t mA, t ≥ 0+

Problems 7–51

[c] vg = vo − 12.5 × 103io = −150 + 202.5e−500t V

[d] vg(0+) = −150 + 202.5 = 52.5 VChecks:

vg(0+) = io(0+)[37.5 × 103] − 150 = 202.5 − 150 = 52.5 V

i50k =vg

50k= −3 + 4.05e−500t mA

i150k =vg

150k= −1 + 1.35e−500t mA

-io + i50k + i150k + 4 = 0 (ok)

P 7.55 For t < 0, vo(0) = (−3 m)(15 k) = −45 Vt > 0:

VTh = −20 × 103i∆ +1050

(75) = −20 × 103( −75

50 × 103

)+ 15 = 45 V

vT = −20 × 103i∆ + 8 × 103iT = −20 × 103(0.2)iT + 8 × 103iT = 4 × 103iT

RTh =vT

iT= 4 kΩ

t > 0

vo = 45 + (−45 − 45)e−t/τ


τ = RC = (4000)( 1

16× 10−6

)= 250 µs;

1τ

= 4000

vo = 45 − 90e−4000t V, t ≥ 0

P 7.56 vo(0) = 45 V; vo(∞) = −45 V

RTh = 20 kΩ

τ = (20 × 103)( 1

16× 10−6

)= 1.25 × 10−3;

1τ

= 800

v = −45 + (45 + 45)e−800t = −45 + 90e−800t V, t ≥ 0

P 7.57 t < 0;

io(0−) =20100

(10 × 10−3) = 2 mA; vo(0−) = (2 × 10−3)(50,000) = 100 V

t = ∞:

io(∞) = −5 × 10−3( 20

100

)= −1 mA; vo(∞) = io(∞)(50,000) = −50 V

RTh = 50 kΩ‖50 kΩ = 25 kΩ; C = 16 nF

τ = (25,000)(16 × 10−9) = 0.4 ms;1τ

= 2500

·. . vo(t) = −50 + 150e−2500t V, t ≥ 0

ic = Cdvo

dt= −6e−2500t mA, t ≥ 0+

Problems 7–53

i50k =vo

50,000= −1 + 3e−2500t mA, t ≥ 0+

io = ic + i50k = −(1 + 3e−2500t) mA, t ≥ 0+

P 7.58 [a] Let i be the current in the clockwise direction around the circuit. Then

Vg = iRg +1C1

∫ t

0i dx +

1C2

∫ t

0i dx

= iRg +( 1

C1+

1C2

) ∫ t

0i dx = iRg +

1Ce

∫ t

0i dx

Now differentiate the equation

0 = Rgdi

dt+

i

Ce

ordi

dt+

1RgCe

i = 0

Therefore i =Vg

Rg

e−t/RgCe =Vg

Rg

e−t/τ ; τ = RgCe

v1(t) =1C1

∫ t

0

Vg

Rg

e−x/τ dx =Vg

RgC1

e−x/τ

−1/τ

∣∣∣∣t0

= − VgCe

C1(e−t/τ − 1)

v1(t) =VgC2

C1 + C2(1 − e−t/τ ); τ = RgCe

v2(t) =VgC1

C1 + C2(1 − e−t/τ ); τ = RgCe

[b] v1(∞) =C2

C1 + C2Vg; v2(∞) =

C1

C1 + C2Vg

P 7.59 [a]

IsR = Ri +1C

∫ t

0+i dx + Vo

0 = Rdi

dt+

i

C+ 0

·. .di

dt+

i

RC= 0


[b]di

dt= − i

RC;

di

i= − dt

RC∫ i(t)

i(0+)

dy

y= − 1

RC

∫ t

0+dx

lni(t)

i(0+)=

−t

RC

i(t) = i(0+)e−t/RC ; i(0+) =IsR − Vo

R=(Is − Vo

R

)

·. . i(t) =(Is − Vo

R

)e−t/RC

P 7.60 [a] t < 0

t > 0

vo(0−) = vo(0+) = 40 V

vo(∞) = 80 V

τ = (0.16 × 10−6)(6.25 × 103) = 1 ms; 1/τ = 1000

vo = 80 − 40e−1000t V, t ≥ 0

[b] io = −Cdvo

dt= −0.16 × 10−6[40,000e−1000t]

= −6.4e−1000t mA; t ≥ 0+

Problems 7–55

[c] v1 =−1

0.2 × 10−6

∫ t

0−6.4 × 10−3e−1000x dx + 32

= 64 − 32e−1000t V, t ≥ 0

[d] v2 =−1

0.8 × 10−6

∫ t

0−6.4 × 10−3e−1000x dx + 8

= 16 − 8e−1000t V, t ≥ 0

[e] wtrapped =12(0.2 × 10−6)(64)2 +

12(0.8 × 10−6)(16)2 = 512 µJ.

P 7.61 [a] vc(0+) = 50 V

[b] Use voltage division to find the final value of voltage:

vc(∞) =20

20 + 5(−30) = −24 V

[c] Find the Thévenin equivalent with respect to the terminals of the capacitor:

VTh = −24 V, RTh = 20‖5 = 4 Ω,

Therefore τ = ReqC = 4(25 × 10−9) = 0.1 µs

The simplified circuit for t > 0 is:

[d] i(0+) =−24 − 50

4= −18.5 A

[e] vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ

= −24 + [50 − (−24)]e−t/τ = −24 + 74e−107t V, t ≥ 0

[f] i = Cdvc

dt= (25 × 10−9)(−107)(74e−107t) = −18.5e−107t A, t ≥ 0+

P 7.62 [a] Use voltage division to find the initial value of the voltage:

vc(0+) = v9k =9 k

9 k + 3 k(120) = 90 V

[b] Use Ohm’s law to find the final value of voltage:

vc(∞) = v40k = −(1.5 × 10−3)(40 × 103) = −60 V


[c] Find the Thévenin equivalent with respect to the terminals of the capacitor:

VTh = −60 V, RTh = 10 k + 40 k = 50 kΩ

τ = RThC = 1 ms = 1000µs

[d] vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ

= −60 + (90 + 60)e−1000t = −60 + 150e−1000t V, t ≥ 0

We want vc = −60 + 150e−1000t = 0:

Therefore t =ln(150/60)

1000= 916.3 µs

P 7.63 [a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and rightof the 400-mH inductor. We get

i(0−) =−60 − 20015 k + 5 k

= −13 mA

i(0−) = i(0+) = −13 mA

[b] For t > 0, the circuit reduces to

Therefore i(∞) = −60/5,000 = −12 mA

[c] τ =L

R=

400 × 10−3

5000= 80 µs

[d] i(t) = i(∞) + [i(0+) − i(∞)]e−t/τ

= −12 + [−13 + 12]e−12,500t = −12 − e−12,500t mA, t ≥ 0

P 7.64 [a] From Example 7.10,

Leq =L1L2 − M2

L1 + L2 − 2M=

36 − 1620 − 8

=53

H

τ =Leq

R=

(5/3)(50/3)

=110

Problems 7–57

io =100

(50/3)− 100

(50/3)e−10t = 6 − 6e−10t A t ≥ 0

[b] vo = 100 − 503

io = 100 − 503

(6 − 6e−10t) = 100e−10t V, t ≥ 0+

[c] vo = 2di1dt

+ 4di2dt

io = i1 + i2

diodt

=di1dt

+di2dt

di2dt

=diodt

− di1dt

= 60e−10t − di1dt

·. . 100e−10t = 2di1dt

+ 4(

60e−10t − di1dt

)

·. .di1dt

= 70e−10t

di1 = 70e−10t dt

∫ i1

0dx = 70

∫ t

0e−10y dy

·. . i1 = 70e−10y

−10

∣∣∣∣t0= 7 − 7e−10t A, t ≥ 0

[d] i2 = io − i1

= 6 − 6e−10t − 7 + 7e−10t

= −1 + e−10t A, t ≥ 0

[e] vo = L2di2dt

+ Mdi1dt

= 18(−10e−10t) + 4(70e−10t)

= 100e−10t V, t ≥ 0+ (checks)Also,

vo = L1di1dt

+ Mdi2dt

= 2(70e−10t) + 4(−10e−10t)

= 100e−10t V, t ≥ 0+ CHECKS


i1(0) = 7 − 7 = 0; agrees with initial conditions;i2(0) = −1 + 1 = 0; agrees with initial conditions;The final values of io, i1, and i2 can be checked via the conservation ofWb-turns:

io(∞)Leq = 6 × (5/3) = 10 Wb-turns

i1(∞)L1 + i2(∞)M = 7(2) − 1(4) = 10 Wb-turns

i2(∞)L2 + i1(∞)M = −1(18) + 7(4) = 10 Wb-turns

Thus our solutions make sense in terms of known circuit behavior.

P 7.65 [a] Leq =(3)(15)3 + 15

= 2.5 H

τ =Leq

R=

2.57.5

=13

s

io(0) = 0; io(∞) =1207.5

= 16 A

·. . io = 16 − 16e−3t A, t ≥ 0

vo = 120 − 7.5io = 120e−3t V, t ≥ 0+

i1 =13

∫ t

0120e−3x dx =

403

− 403

e−3t A, t ≥ 0

i2 = io − i1 =83

− 83e−3t A, t ≥ 0

[b] io(0) = i1(0) = i2(0) = 0, consistent with initial conditions.vo(0+) = 120 V, consistent with io(0) = 0.

vo = 3di1dt

= 120e−3t V, t ≥ 0+

or

vo = 15di2dt

= 120e−3t V, t ≥ 0+

The voltage solution is consistent with the current solutions.

λ1 = 3i1 = 40 − 40e−3t Wb-turns

λ2 = 15i2 = 40 − 40e−3t Wb-turns

·. . λ1 = λ2 as it must, since

vo =dλ1

dt=

dλ2

dt

λ1(∞) = λ2(∞) = 40 Wb-turns

Problems 7–59

λ1(∞) = 3i1(∞) = 3(40/3) = 40 Wb-turns

λ2(∞) = 15i2(∞) = 15(8/3) = 40 Wb-turns

·. . i1(∞) and i2(∞) are consistent with λ1(∞) and λ2(∞).

P 7.66 [a] From Example 7.10,

Leq =L1L2 − M2

L1 + L2 + 2M=

50 − 2515 + 10

= 1 H

τ =L

R=

120

;1τ

= 20

·. . io(t) = 4 − 4e−20t A, t ≥ 0

[b] vo = 80 − 20io = 80 − 80 + 80e−20t = 80e−20t V, t ≥ 0+

[c] vo = 5di1dt

− 5di2dt

= 80e−20t V

io = i1 + i2

diodt

=di1dt

+di2dt

= 80e−20t A/s

·. .di2dt

= 80e−20t − di1dt

·. . 80e−20t = 5di1dt

− 400e−20t + 5di1dt

·. . 10di1dt

= 480e−20t; di1 = 48e−20t dt

∫ t1

0dx =

∫ t

048e−20y dy

i1 =48

−20e−20y

∣∣∣∣t0= 2.4 − 2.4e−20t A, t ≥ 0

[d] i2 = io − i1 = 4 − 4e−20t − 2.4 + 2.4e−20t

= 1.6 − 1.6e−20t A, t ≥ 0

[e] io(0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.

vo = Leqdiodt

= 1(80)e−20t = 80e−20t V, t ≥ 0+ (checks)

Also,

vo = 5di1dt

− 5di2dt

= 80e−20t V, t ≥ 0+ (checks)


vo = 10di2dt

− 5di1dt

= 80e−20t V, t ≥ 0+ (checks)

vo(0+) = 80 V, which agrees with io(0+) = 0 A

io(∞) = 4 A; io(∞)Leq = (4)(1) = 4 Wb-turns

i1(∞)L1 + i2(∞)M = (2.4)(5) + (1.6)(−5) = 4 Wb-turns (ok)

i2(∞)L2 + i1(∞)M = (1.6)(10) + (2.4)(−5) = 4 Wb-turns (ok)

Therefore, the final values of io, i1, and i2 are consistent with conservation offlux linkage. Hence, the answers make sense in terms of known circuitbehavior.

P 7.67 [a] Leq = 5 + 10 − 2.5(2) = 10 H

τ =L

R=

1040

=14;

1τ

= 4

i = 2 − 2e−4t A, t ≥ 0

[b] v1(t) = 5di1dt

− 2.5di

dt= 2.5

di

dt= 2.5(8e−4t) = 20e−4t V, t ≥ 0+

[c] v2(t) = 10di1dt

− 2.5di

dt= 7.5

di

dt= 7.5(8e−4t) = 60e−4t V, t ≥ 0+

[d] i(0) = 2 − 2 = 0, which agrees with initial conditions.

80 = 40i1 + v1 + v2 = 40(2 − 2e−4t) + 20e−4t + 60e−4t = 80 V

Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, theanswers make sense in terms of known circuit behavior.

P 7.68 [a] Leq = 5 + 10 + 2.5(2) = 20 H

τ =L

R=

2040

=12;

1τ

= 2

i = 2 − 2e−2t A, t ≥ 0

[b] v1(t) = 5di1dt

+ 2.5di

dt= 7.5

di

dt= 7.5(4e−2t) = 30e−2t V, t ≥ 0+

[c] v2(t) = 10di1dt

+ 2.5di

dt= 12.5

di

dt= 12.5(4e−2t) = 50e−2t V, t ≥ 0+

[d] i(0) = 0, which agrees with initial conditions.

80 = 40i1 + v1 + v2 = 40(2 − 2e−2t) + 30e−2t + 50e−2t = 80 V

Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, theanswers make sense in terms of known circuit behavior.

Problems 7–61

P 7.69 Use voltage division to find the initial voltage:

vo(0) =60

40 + 60(50) = 30 V

Use Ohm’s law to find the final value of voltage:

vo(∞) = (−5 mA)(20 kΩ) = −100 V

τ = RC = (20 × 103)(250 × 10−9) = 5 ms;1τ

= 200

vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ

= −100 + (30 + 100)e−200t = −100 + 130e−200t V, t ≥ 0

P 7.70 [a] t < 0:

Using Ohm’s law,

ig =800

40 + 60‖40= 12.5 A


i(0−) =60

60 + 40(12.5) = 7.5 A = i(0+)

[b] 0 ≤ t ≤ 1 ms:

i = i(0+)e−t/τ = 7.5e−t/τ

1τ

=R

L=

40 + 120‖6080 × 10−3 = 1000

i = 7.5e−1000t

i(200µs) = 7.5e−103(200×10−6) = 7.5e−0.2 = 6.14 A


[c] i(1ms) = 7.5e−1 = 2.7591 A

1 ms ≤ t < ∞

1τ

=R

L=

4080 × 10−3 = 500

i = i(1 ms)e−(t−1 ms)/τ = 2.7591e−500(t−0.001) A

i(6ms) = 2.7591e−500(0.005) = 2.7591e−2.5 = 226.48 mA

[d] 0 ≤ t ≤ 1 ms:

i = 7.5e−1000t

v = Ldi

dt= (80 × 10−3)(−1000)(7.5e−1000t) = −600e−1000t V

v(1−ms) = −600e−1 = −220.73 V

[e] 1 ms ≤ t ≤ ∞:

i = 2.7591e−500(t−0.001)

v = Ldi

dt= (80 × 10−3)(−500)(2.591e−500(t−0.001))

= −110.4e−500(t−0.001) V

v(1+ms) = −110.4 V

P 7.71 Note that for t > 0, vo = (4/6)vc, where vc is the voltage across the 0.5 µFcapacitor. Thus we will find vc first.t < 0

vc(0) =315

(−75) = −15 V

Problems 7–63

0 ≤ t ≤ 800 µs:

τ = ReC, Re =(6000)(3000)

9000= 2 kΩ

τ = (2 × 103)(0.5 × 10−6) = 1 ms,1τ

= 1000

vc = −15e−1000t V, t ≥ 0

vc(800 µs) = −15e−0.8 = −6.74 V

800 µs ≤ t ≤ 1.1 ms:

τ = (6 × 103)(0.5 × 10−6) = 3 ms,1τ

= 333.33

vc = −6.74e−333.33(t−800×10−6) V

1.1 ms ≤ t < ∞:

τ = 1 ms,1τ

= 1000

vc(1.1ms) = −6.74e−333.33(1100−800)10−6= −6.74e−0.1 = −6.1 V


vc = −6.1e−1000(t−1.1×10−3) V

vc(1.5ms) = −6.1e−1000(1.5−1.1)10−3= −6.1e−0.4 = −4.09 V

vo = (4/6)(−4.09) = −2.73 V

P 7.72 w(0) =12(0.5 × 10−6)(−15)2 = 56.25 µJ

0 ≤ t ≤ 800 µs:

vc = −15e−1000t; v2c = 225e−2000t

p3k = 75e−2000t mW

w3k =∫ 800×10−6

075 × 10−3e−2000t dt

= 75 × 10−3 e−2000t

−2000

∣∣∣∣800×10−6

0

= −37.5 × 10−6(e−1.6 − 1) = 29.93 µJ

1.1 ms ≤ t ≤ ∞:

vc = −6.1e−1000(t−1.1×10−3) V; v2c = 37.19e−2000(t−1.1×10−3)

p3k = 12.4e−2000(t−1.1×10−3) mW

w3k =∫ ∞

1.1×10−312.4 × 10−3e−2000(t−1.1×10−3) dt

= 12.4 × 10−3 e−2000(t−1.1×10−3)

−2000

∣∣∣∣∞1.1×10−3

= −6.2 × 10−6(0 − 1) = 6.2 µJ

w3k = 29.93 + 6.2 = 36.13 µJ

% =36.1356.25

(100) = 64.23%

Problems 7–65

P 7.73 For t < 0:

i(0) =1015

(15) = 10 A

0 ≤ t ≤ 10 ms:

i = 10e−100t A

i(10ms) = 10e−1 = 3.68 A

10 ms ≤ t ≤ 20 ms:

Req =(5)(20)

25= 4 Ω

1τ

=R

L=

450 × 10−3 = 80

i = 3.68e−80(t−0.01) A

20 ms ≤ t ≤ ∞:

i(20ms) = 3.68e−80(0.02−0.01) = 1.65 A


i = 1.65e−100(t−0.02) A

vo = Ldi

dt; L = 50 mH

di

dt= 1.65(−100)e−100(t−0.02) = −165e−100(t−0.02)

vo = (50 × 10−3)(−165)e−100(t−0.02)

= −8.26e−100(t−0.02) V, t > 20+ ms

vo(25ms) = −8.26e−100(0.025−0.02) = −5 V

P 7.74 From the solution to Problem 7.73, the initial energy is

w(0) =12(50 mH)(10 A)2 = 2.5 J

0.04w(0) = 0.1 J

·. .12(50 × 10−3)i2L = 0.1 so iL = 2 A

Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms since

i(10 ms) = 3.68 A and i(20 ms) = 1.65 A

For 10 ms ≤ t ≤ 20 ms:

i = 3.68e−80(t−0.01) = 2

e80(t−0.01) =3.682

so t − 0.01 = 0.0076 ·. . t = 17.6 ms

P 7.75 0 ≤ t ≤ 10 µs:

τ = RC = (4 × 103)(20 × 10−9) = 80 µs; 1/τ = 12,500

Problems 7–67

vo(0) = 0 V; vo(∞) = −20 V

vo = −20 + 20e−12,500t V 0 ≤ t ≤ 10 µs

10 µs ≤ t ≤ ∞:

t = ∞:

i =−50 V20 kΩ

= −2.5 mA

vo(∞) = (−2.5 × 10−3)(16,000) + 30 = −10 V

vo(10 µs) = −20 + 20−0.125 = −2.35 V

vo = −10 + (−2.35 + 10)e−(t − 10×10−6)/τ

RTh = 4 kΩ‖16 kΩ = 3.2 kΩ

τ = (3200)(20 × 10−9) = 64 µs; 1/τ = 15,625

vo = −10 + 7.65e−15,625(t − 10×10−6) 10 µs ≤ t ≤ ∞

P 7.76 0 ≤ t ≤ 200 µs;

Re = 150‖100 = 60 kΩ; τ =(10

3× 10−9

)(60,000) = 200µs


vc = 300e−5000t V

vc(200 µs) = 300e−1 = 110.36 V

200 µs ≤ t ≤ ∞:

Re = 30‖60 + 120‖40 = 20 + 30 = 50 kΩ

τ =(10

3× 10−9

)(50,000) = 166.67 µs;

1τ

= 6000

vc = 110.36e−6000(t − 200 µs) V

vc(300 µs) = 110.36e−6000(100 µs) = 60.57 V

io(300 µs) =60.5750,000

= 1.21 mA

i1 =6090

io =23io; i2 =

40160

io =14io

isw = i1 − i2 =23io − 1

4io =

512

io =512

(1.21 × 10−3) = 0.50 mA

P 7.77 t < 0:

vc(0−) = (20 × 10−3)(500) = 10 V = vc(0+)

Problems 7–69

0 ≤ t ≤ 50 ms:

τ = ∞; 1/τ = 0; vo = 10e−0 = 10 V

50 ms ≤ t ≤ ∞:

τ = (6.25 k)(0.16 µ) = 1 ms; 1/τ = 1000; vo = 10e−1000(t − 0.05) V

Summary:

vo = 10 V, 0 ≤ t ≤ 50 ms

vo = 10e−1000(t − 0.05) V, 50 ms ≤ t ≤ ∞

P 7.78 t < 0:

iL(0−) = 10 V/5 Ω = 2 A = iL(0+)


0 ≤ t ≤ 5:

τ = 5/0 = ∞

iL(t) = 2e−t/∞ = 2e−0 = 2

iL(t) = 2 A, 0 ≤ t ≤ 5 s

5 ≤ t ≤ ∞:

τ =51

= 5 s; 1/τ = 0.2

iL(t) = 2e−0.2(t −5) A, t ≥ 5 s

P 7.79 [a] 0 ≤ t ≤ 2.5 ms

vo(0+) = 80 V; vo(∞) = 0

τ =L

R= 2 ms; 1/τ = 500

vo(t) = 80e−500t V, 0+ ≤ t ≤ 2.5 ms

vo(2.5− ms) = 80e−1.25 = 22.92 V

io(2.5− ms) =(80 − 22.92)

20= 2.85 A

vo(2.5+ ms) = −20(2.85) = −57.08 V

vo(∞) = 0; τ = 2 ms; 1/τ = 500

vo = −57.08e−500(t − 0.0025) V 2.5+ ms ≤ t ≤ ∞

Problems 7–71

[b]

[c] vo(5 ms) = −16.35 V

io =+16.35

20= 817.68 mA

P 7.80 [a] io(0) = 0; io(∞) = 25 mA

1τ

=R

L=

2000250

× 103 = 8000

io = (25 − 25e−8000t) mA, 0 ≤ t ≤ 75 µs

vo = 0.25diodt

= 50e−8000t V, 0+ ≤ t ≤ 75− µs

75+ µs ≤ t ≤ ∞:

io(75µs) = 25 − 25e−0.6 = 11.28 mA; io(∞) = 0

io = 11.28e−8000(t−75×10−6) mA

vo = (0.25)diodt

= −22.56e−8000(t−75µs)

·. . t < 0 : vo = 0

0+ ≤ t ≤ 75− µs : vo = 50e−8000t V

75+ µs ≤ t ≤ ∞ : vo = −22.56e−8000(t−75µs)

[b] vo(75−µs) = 50e−0.6 = 27.44 V

vo(75+µs) = −22.56 V

[c] io(75−µs) = io(75+µs) = 11.28 mA


P 7.81 [a] 0 ≤ t < 1 ms:

vc(0+) = 0; vc(∞) = 50 V;

RC = 400 × 103(0.01 × 10−6) = 4 ms; 1/RC = 250

vc = 50 − 50e−250t

vo = 50 − 50 + 50e−250t = 50e−250t V, 0 ≤ t ≤ 1 ms

1 ms < t ≤ ∞:

vc(1 ms) = 50 − 50e−0.25 = 11.06 V

vc(∞) = 0 V

τ = 4 ms; 1/τ = 250

vc = 11.06e−250(t − 0.001) V

vo = −vc = −11.06e−250(t − 0.001) V, 1 ms < t ≤ ∞[b]

P 7.82 [a] t < 0; vo = 00 ≤ t ≤ 4 ms:

τ = (200 × 103)(0.025 × 10−6) = 5 ms; 1/τ = 200

vo = 100 − 100e−200t V, 0 ≤ t ≤ 4 ms

vo(4 ms) = 100(1 − e−0.8) = 55.07 V

4 ms ≤ t ≤ 8 ms:

vo = −100 + 155.07e−200(t−0.004) V

vo(8 ms) = −100 + 155.07e−0.8 = −30.32 V

8 ms ≤ t ≤ ∞:

vo = −30.32e−200(t−0.008) V

Problems 7–73

[b]

[c] t ≤ 0 : vo = 00 ≤ t ≤ 4 ms:

τ = (50 × 103)(0.025 × 10−6) = 1.25 ms 1/τ = 800

vo = 100 − 100e−800t V, 0 ≤ t ≤ 4 ms

vo(4 ms) = 100 − 100e−3.2 = 95.92 V

4 ms ≤ t ≤ 8 ms:

vo = −100 + 195.92e−800(t−0.004) V, 4 ms ≤ t ≤ 8 ms

vo(8 ms) = −100 + 195.92e−3.2 = −92.01 V

8 ms ≤ t ≤ ∞:

vo = −92.01e−800(t−0.008) V, 8 ms ≤ t ≤ ∞


P 7.83 [a] τ = RC = (20,000)(0.2 × 10−6) = 4 ms; 1/τ = 250

io = vo = 0 t < 0

io(0+) = 20(16

20

)= 16 mA, io(∞) = 0

·. . io = 16e−250t mA 0+ ≤ t ≤ 2− ms

i16kΩ = 20 − 16e−250t mA

·. . vo = 320 − 256e−250t V 0+ ≤ t ≤ 2− ms

vc = vo − 4 × 103io = 320 − 320e−250t V 0 ≤ t ≤ 2 ms

vc(2 ms) = 320 − 320e−0.5 = 125.91 V

·. . io(2+ ms) = 16e−0.5 = 9.7 mA

io(∞) = 0

vc = 125.91e−250(t−0.002), 2+ ms ≤ t ≤ ∞

io = Cdvc

dt= (0.2 × 10−6)(−250)(125.91)e−250(t−0.002)

= −6.3e−250(t−0.002) mA, 2+ ms ≤ t ≤ ∞vo = 4000io + vc = 100.73e−250(t −0.002) V 2+ ms ≤ t ≤ ∞Summary part (a)

io = 0 t < 0

io = 16e−250t mA (0+ ≤ t ≤ 2− ms)

io = −6.3e−250(t −0.002) mA 2+ ms ≤ t ≤ ∞vo = 0 t < 0

vo = 320 − 256e−250t V, 0 ≤ t ≤ 2− ms

vo = 100.73e−250(t −0.002) V, 2+ ms ≤ t ≤ ∞[b] io(0−) = 0

io(0+) = 16 mA

io(2− ms) = 16e−0.5 = 9.7 mA

io(2+ ms) = −6.3 mA

Problems 7–75

[c] vo(0−) = 0

vo(0+) = 64 V

vo(2− ms) = 320 − 256e−0.5 = 164.73 V

vo(2+ ms) = 100.73

[d]

[e]

P 7.84 [a]

Using Ohm’s law,

vT = 5000iσ


iσ =20,000

20,000 + 5000(iT + βiσ) = 0.8iT + 0.8βiσ


Solve for iσ:

iσ(1 − 0.8β) = 0.8iT

iσ =0.8iT

1 − 0.8β; vT = 5000iσ =

4000iT(1 − 0.8β)

Find β such that RTh = −5 kΩ:

RTh =vT

iT=

40001 − 0.8β

= −5000

1 − 0.8β = −0.8 ·. . β = 2.25

[b] Find VTh;

Write a KCL equation at the top node:

VTh − 405000

+VTh

20,000− 2.25iσ = 0


iσ =(VTh − 40)

5000= 0

Solving,

VTh = 50 V

Write a KVL equation around the loop:

50 = −5000i + 0.2di

dt

Rearranging:

di

dt= 250 + 25,000i = 25,000(i + 0.01)

Problems 7–77

Separate the variables and integrate to find i;

di

i + 0.01= 25,000 dt

∫ i

0

dx

x + 0.01=∫ t

025,000 dx

·. . i = −10 + 10e25,000t mA

di

dt= (10 × 10−3)(25,000)e25,000t = 250e25,000t

Solve for the arc time:

v = 0.2di

dt= 50e25,000t = 45,000; e25,000t = 900

·. . t =ln 90025,000

= 272.1 µs

P 7.85 Find the Thévenin equivalent with respect to the terminals of the capacitor.RTh calculation:

iT =vT

2000+

vT

5000− 4

vT

5000

·. .iTvT

=5 + 2 − 810,000

= − 110,000

vT

iT= −10,000

1= −10 kΩ

Open circuit voltage calculation:


The node voltage equations:

voc

2000+

voc − v1

1000− 4i∆ = 0

v1 − voc

1000+

v1

4000− 5 × 10−3 = 0

The constraint equation:

i∆ =v1

4000

Solving, voc = −80 V, v1 = −60 V

vc(0) = 0; vc(∞) = −80 V

τ = RC = (−10,000)(1.6 × 10−6) = −16 ms;1τ

= −62.5

vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ = −80 + 80e62.5t = 14,400

Solve for the time of the maximum voltage rating:

e62.5t = 181; 62.5t = ln 181; t = 83.18 ms

P 7.86

vT = 2000iT + 4000(iT − 2 × 10−3vφ) = 6000iT − 8vφ

= 6000iT − 8(2000iT )

Problems 7–79

vT

iT= −10,000

τ =10

−10,000= −1 ms; 1/τ = −1000

i = 25e1000t mA

·. . 25e1000t × 10−3 = 5; t =ln 2001000

= 5.3 ms

P 7.87 t > 0:

vT = 12 × 104i∆ + 16 × 103iT

i∆ = − 20100

iT = −0.2iT

·. . vT = −24 × 103iT + 16 × 103iT

RTh =vT

iT= −8 kΩ

τ = RC = (−8 × 103)(2.5 × 10−6) = −0.02 1/τ = −50

vc = 20e50t V; 20e50t = 20,000

50t = ln 1000 ·. . t = 138.16 ms


P 7.88 [a]

τ = (25)(2) × 10−3 = 50 ms; 1/τ = 20

vc(0+) = 80 V; vc(∞) = 0

vc = 80e−20t V

·. . 80e−20t = 5; e20t = 16; t =ln 1620

= 138.63 ms

[b] 0+ < t < 138.63 ms:

i = (2 × 10−6)(−1600e−20t) = −3.2e−20t mA

138.63+ ms < t ≤ ∞:

τ = (2)(4) × 10−3 = 8 ms; 1/τ = 125

vc(138.63+ ms) = 5 V; vc(∞) = 80 V

vc = 80 − 75e−125(t−0.13863) V, 138.63+ ms ≤ t ≤ ∞

i = 2 × 10−6(9375)e−125(t−0.13863)

= 18.75e−125(t−0.13863) mA, 138.63+ ms ≤ t ≤ ∞

[c] 80 − 75e−125∆t = 0.85(80) = 68

80 − 68 = 75e−125∆t = 12

e125∆t = 6.25; ∆t =ln 6.2512.5

∼= 14.66 ms

P 7.89 Use voltage division to find the voltage at the non-inverting terminal:

vp =80100

(−45) = −36 V = vn

Problems 7–81

Write a KCL equation at the inverting terminal:

−36 − 1480,000

+ 2.5 × 10−6 d

dt(−36 − vo) = 0

·. . 2.5 × 10−6dvo

dt=

−5080,000

Separate the variables and integrate:

dvo

dt= −250 ·. . dvo = −250dt

∫ vo(t)

vo(0)dx = −250

∫ t

0dy ·. . vo(t) − vo(0) = −250t

vo(0) = −36 + 56 = 20 V

vo(t) = −250t + 20

Find the time when the voltage reaches 0:

0 = −250t + 20 ·. . t =20250

= 80 ms

P 7.90 The equation for an integrating amplifier:

vo =1

RC

∫ t

0(vb − va) dy + vo(0)

Find the values and substitute them into the equation:

RC = (100 × 103)(0.05 × 10−6) = 5 ms

1RC

= 200; vb − va = −15 − (−7) = −8 V

vo(0) = −4 + 12 = 8 V

vo = 200∫ t

0−8 dx + 8 = (−1600t + 8) V, 0 ≤ t ≤ tsat

RC circuit analysis for v2:

v2(0+) = −4 V; v2(∞) = −15 V; τ = RC = (100 k)(0.05 µ) = 5 ms


v2 = v2(∞) + [v2(0+) − v2(∞)]e−t/τ

= −15 + (−4 + 15)e−200t = −15 + 11e−200t V, 0 ≤ t ≤ tsat

vf + v2 = vo·. . vf = vo − v2 = 23 − 1600t − 11e−200t V, 0 ≤ t ≤ tsat

Note that

−1600tsat + 8 = −20 ·. . tsat =−28

−1600= 17.5 ms

so the op amp operates in its linear region until it saturates at 17.5 ms.

P 7.91 vo = − 1R(0.5 × 10−6)

∫ t

04 dx + 0 =

−4tR(0.5 × 10−6)

−4(15 × 10−3)R(0.5 × 10−6)

= −10

·. . R =−4(15 × 10−3)

−10(0.5 × 10−6)= 12 kΩ

P 7.92 vo =−4t

R(0.5 × 10−6)+ 6 =

−4(40 × 10−3)R(0.5 × 10−6)

+ 6 = −10

·. . R =−4(40 × 10−3)

−16(0.5 × 10−6)= 20 kΩ

P 7.93 [a]Cdvp

dt+

vp − vb

R= 0; therefore

dvp

dt+

1RC

vp =vb

RC

vn − va

R+ C

d(vn − vo)dt

= 0;

thereforedvo

dt=

dvn

dt+

vn

RC− va

RC

But vn = vp

Thereforedvn

dt+

vn

RC=

dvp

dt+

vp

RC=

vb

RC

Thereforedvo

dt=

1RC

(vb − va); vo =1

RC

∫ t

0(vb − va) dy

[b] The output is the integral of the difference between vb and va and then scaled bya factor of 1/RC.

Problems 7–83

[c] vo =1

RC

∫ t

0(vb − va) dx

RC = (50 × 103)(10 × 10−9) = 0.5 ms

vb − va = −25 mV

vo =1

0.0005

∫ t

0−25 × 10−3dx = −50t

−50tsat = −6; tsat = 120 ms

P 7.94 [a] RC = (25 × 103)(0.4 × 10−6) = 10 ms;1

RC= 100

vo = 0, t < 0

[b] 0 ≤ t ≤ 250 ms :

vo = −100∫ t

0−0.20 dx = 20t V

[c] 250 ms ≤ t ≤ 500 ms;

vo(0.25) = 20(0.25) = 5 V

vo(t) = −100∫ t

0.250.20 dx + 5 = −20(t − 0.25) + 5 = −20t + 10 V

[d] 500 ms ≤ t ≤ ∞ :

vo(0.5) = −10 + 10 = 0 V

vo(t) = 0 V


P 7.95 [a] vo = 0, t < 0

RC = (25 × 103)(0.4 × 10−6) = 10 ms1

RC= 100

[b] RfCf = (5 × 106)(0.4 × 10−6) = 2;1

RfCf

= 0.5

vo =−5 × 106

25 × 103 (−0.2)[1 − e−0.5t] = 40(1 − e−0.5t) V, 0 ≤ t ≤ 250 ms

[c] vo(0.25) = 40(1 − e−0.125) ∼= 4.70 V

vo =−VmRf

Rs

+VmRf

Rs

(2 − e−0.125)e−0.5(t−0.25)

= −40 + 40(2 − e−0.125)e−0.5(t−0.25)

= −40 + 44.70e−0.5(t−0.25) V, 250 ms ≤ t ≤ 500 ms

[d] vo(0.5) = −40 + 44.70e−0.125 ∼= −0.55 V

vo = −0.55e−0.5(t−0.5) V, 500 ms ≤ t ≤ ∞

P 7.96 [a] RC = (1000)(800 × 10−12) = 800 × 10−9;1

RC= 1,250,000

0 ≤ t ≤ 1 µs:

vg = 2 × 106t

vo = −1.25 × 106∫ t

02 × 106x dx + 0

= −2.5 × 1012x2

2

∣∣∣∣t0= −125 × 1010t2 V, 0 ≤ t ≤ 1 µs

Problems 7–85

vo(1 µs) = −125 × 1010(1 × 10−6)2 = −1.25 V

1 µs ≤ t ≤ 3 µs:

vg = 4 − 2 × 106t

vo = −125 × 104∫ t

1×10−6(4 − 2 × 106x) dx − 1.25

= −125 × 104

[4x

∣∣∣∣t1×10−6

−2 × 106x2

2

∣∣∣∣t1×10−6

]− 1.25

= −5 × 106t + 5 + 125 × 1010t2 − 1.25 − 1.25= 125 × 1010t2 − 5 × 106t + 2.5 V, 1 µs ≤ t ≤ 3 µs

vo(3 µs) = 125 × 1010(3 × 10−6)2 − 5 × 106(3 × 10−6) + 2.5

= −1.25

3 µs ≤ t ≤ 4 µs:

vg = −8 + 2 × 106t

vo = −125 × 104∫ t

3×10−6(−8 + 2 × 106x) dx − 1.25

= −125 × 104

[−8x

∣∣∣∣t3×10−6

+2 × 106x2

2

∣∣∣∣t3×10−6

]− 1.25

= 107t − 30 − 125 × 1010t2 + 11.25 − 1.25= −125 × 1010t2 + 107t − 20 V, 3 µs ≤ t ≤ 4 µs

vo(4 µs) = −125 × 1010(4 × 10−6)2 + 107(4 × 10−6) − 20 = 0

[b]

[c] The output voltage will also repeat. This follows from the observation that att = 4 µs the output voltage is zero, hence there is no energy stored in thecapacitor. This means the circuit is in the same state at t = 4 µs as it was att = 0, thus as vg repeats itself, so will vo.


P 7.97 [a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal. Atthe instant T1 turns ON the capacitor C2 is connected across b2 − e2, thusvbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity of thevoltage on C2 starts to reverse, that is, the right-hand terminal of C2 starts tocharge toward +VCC . At the same time, C1 is charging toward VCC , positiveon the right. At the instant the charge on C2 reaches zero, vbe2 is zero, T2 turnsON. This makes vbe1 = −VCC and T1 snaps OFF. Now the capacitors C1 andC2 start to charge with the polarities to turn T1 ON and T2 OFF. This switchingaction repeats itself over and over as long as the circuit is energized. At theinstant T1 turns ON, the voltage controlling the state of T2 is governed by thefollowing circuit:

It follows that vbe2 = VCC − 2VCCe−t/R2C2 .

[b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the voltageacross C1, thus

It follows that vce2 = VCC − VCCe−t/RLC1 .

[c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will becomepositive and turn T2 ON. vbe2 = 0 when VCC − 2VCCe−t/R2C2 = 0, or whent = R2C2 ln 2.

[d] When t = R2C2 ln 2, we have

vce2 = VCC − VCCe−[(R2C2 ln 2)/(RLC1)] = VCC − VCCe−10 ln 2 ∼= VCC

[e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have

Problems 7–87

ib1 =VCC

R1+

VCC

RLe−t/RLC1

[f] At the instant T2 turns back ON, t = R2C2 ln 2; therefore

ib1 =VCC

R1+

VCC

RLe−10 ln 2 ∼= VCC

R1

When T2 turns OFF, ib1 drops to zero instantaneously.

[g]

[h]

P 7.98 [a] tOFF2 = R2C2 ln 2 = 14.43 × 103(1 × 10−9) ln 2 ∼= 10 µs

[b] tON2 = R1C1 ln 2 ∼= 10 µs

[c] tOFF1 = R1C1 ln 2 ∼= 10 µs

[d] tON1 = R2C2 ln 2 ∼= 10 µs

[e] ib1 =10

1000+

1014,430

∼= 10.69 mA

[f] ib1 =10

14,430+

101000

e−10 ∼= 0.693 mA


[g] vce2 = 10 − 10e−10 ∼= 10 V

P 7.99 [a] tOFF2 = R2C2 ln 2 = (14.43 × 103)(0.8 × 10−9) ln 2 ∼= 8 µs

[b] tON2 = R1C1 ln 2 ∼= 10 µs

[c] tOFF1 = R1C1 ln 2 ∼= 10 µs

[d] tON1 = R2C2 ln 2 = 8 µs

[e] ib1 = 10.69 mA

[f] ib1 =10

14,430+

101000

e−8 ∼= 0.693 mA

[g] vce2 = 10 − 10e−8 ∼= 10 V

Note in this circuit T2 is OFF 8 µs and ON 10 µs of every cycle, whereas T1 isON 8 µs and OFF 10 µs every cycle.

P 7.100 If R1 = R2 = 50RL = 100 kΩ, then

C1 =48 × 10−6

100 × 103 ln 2= 692.49 pF; C2 =

36 × 10−6

100 × 103 ln 2= 519.37 pF

If R1 = R2 = 6RL = 12 kΩ, then

C1 =48 × 10−6

12 × 103 ln 2= 5.77 nF; C2 =

36 × 10−6

12 × 103 ln 2= 4.33 nF

Therefore 692.49 pF ≤ C1 ≤ 5.77 nF and 519.37 pF ≤ C2 ≤ 4.33 nF

P 7.101 [a] T2 is normally ON since its base current ib2 is greater than zero, i.e.,ib2 = VCC/R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0.When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C ischarged to VCC , positive at the left terminal. This is a stable state; there isnothing to disturb this condition if the circuit is left to itself.

[b] When S is closed momentarily, vbe2 is changed to −VCC and T2 snaps OFF. Theinstant T2 turns OFF, vce2 jumps to VCCR1/(R1 + RL) and ib1 jumps toVCC/(R1 + RL), which turns T1 ON.

[c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since vbe2 isthe same as the voltage across C, it starts to increase from −VCC toward+VCC . However, T2 turns ON as soon as vbe2 = 0. The equation for vbe2 isvbe2 = VCC − 2VCCe−t/RC . vbe2 = 0 when t = RC ln 2, therefore T2 staysOFF for RC ln 2 seconds.

P 7.102 [a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to

vce2 =(

VCC

R1 + RL

)R1 =

6(5)25

= 1.2 V

Problems 7–89

T2 remains open for (23,083)(250) × 10−12 ln 2 ∼= 4 µs.

[b] ib2 =VCC

R= 259.93 µA, −5 ≤ t ≤ 0 µs

ib2 = 0, 0 < t < RC ln 2

ib2 =VCC

R+

VCC

RLe−(t−RC ln 2)/RLC

= 259.93 + 300e−0.2×106(t−4×10−6) µA, RC ln 2 < t

P 7.103 [a] We want the lamp to be in its nonconducting state for no more than 10 s, thevalue of to:

10 = R(10 × 10−6) ln1 − 64 − 6

and R = 1.091 MΩ

[b] When the lamp is conducting

VTh =20 × 103

20 × 103 + 1.091 × 106 (6) = 0.108 V

RTh = 20 k||1.091 M = 19,640 Ω


So,

(tc − to) = (19,640)(10 × 10−6) ln4 − 0.1081 − 0.108

= 0.289 s

The flash lasts for 0.289 s.

P 7.104 [a] At t = 0 we have

τ = (800)(25) × 10−3 = 20 sec; 1/τ = 0.05

vc(∞) = 40 V; vc(0) = 5 V

vc = 40 − 35e−0.05t V, 0 ≤ t ≤ to

40 − 35e−0.05to = 15; ·. . e0.05to = 1.4

to = 20 ln 1.4 s = 6.73 s

At t = to we have

The Thévenin equivalent with respect to the capacitor is

τ =(800

81

)(25) × 10−3 =

2081

s;1τ

=8120

= 4.05

vc(to) = 15 V; vc(∞) =4081

V

vc(t) =4081

+(15 − 40

81

)e−4.05(t−to) V =

4081

+117581

e−4.05(t−to)

·. .4081

+117581

e−4.05(t−to) = 5

Problems 7–91

117581

e−4.05(t−to) =36581

e4.05(t−to) =1175365

= 3.22

t − to =1

4.05ln 3.22 ∼= 0.29 s

One cycle = 7.02 seconds.

N = 60/7.02 = 8.55 flashes per minute

[b] At t = 0 we have

τ = 25R × 10−3; 1/τ = 40/R

vc = 40 − 35e−(40/R)t

40 − 35e−(40/R)to = 15

·. . to =R

40ln 1.4, R in kΩ

At t = to:

vTh =10

R + 10(40) =

400R + 10

; RTh =10R

R + 10kΩ

τ =(25)(10R) × 10−3

R + 10=

0.25RR + 10

;1τ

=4(R + 10)

R

vc =400

R + 10+(15 − 400

R + 10

)e− 4(R+10)

R(t−to)

·. .400

R + 10+[15R − 250

R + 10

]e− 4(R+10)

R(t−to) = 5

or(15R − 250

R + 10

)e− 4(R+10)

R(t−to) =

5R − 350(R + 10)


·. . e4(R+10)

R(t−to) =

3R − 50R − 70

·. . t − to =R

4(R + 10)ln(3R − 50

R − 70

)

At 12 flashes per minute to + (t − to) = 5 s

·. .R

40ln 1.4︸︷︷︸+

R

4(R + 10)ln(3R − 50

R − 70

)= 5

dominantterm

Start the trial-and-error procedure by setting (R/40) ln 1.4 = 5, thenR = 200/(ln 1.4) or 594.40 kΩ. If R = 594.40 kΩ then t − to ∼= 0.29 s.Second trial set (R/40) ln 1.4 = 4.7 s or R = 558.74 kΩ.

With R = 558.74 kΩ, t − to ∼= 0.30 s

The procedure converges to R = 559.3 kΩ

P 7.105 [a] to = RC ln(

Vmin − Vs

Vmax − Vs

)= (3700)(250 × 10−6) ln

(−700−100

)

= 1.80 s

tc − to =RCRL

R + RLln(

Vmax − VTh

Vmin − VTh

)

RL

R + RL=

1.31.3 + 3.7

= 0.26 RC = (3700)(250 × 10−6) = 0.925 s

VTh =1000(1.3)1.3 + 3.7

= 260 V RTh = 3.7 k||1.3 k = 962 Ω

·. . tc − to = (0.925)(0.26) ln(640/40) = 0.67 s

·. . tc = 1.8 + 0.67 = 2.47 s

flashes/min =60

2.47= 24.32

[b] 0 ≤ t ≤ to:

vL = 1000 − 700e−t/τ1

τ1 = RC = 0.925 s

to ≤ t ≤ tc:

vL = 260 + 640e−(t−to)/τ2

τ2 = RThC = 962(250) × 10−6 = 0.2405 s

Problems 7–93

0 ≤ t ≤ to : i =1000 − vL

3700=

737

e−t/0.925 A

to ≤ t ≤ tc : i =1000 − vL

3700=

74370

− 64370

e−(t−to)/0.2405

Graphically, i versus t is

The average value of i will equal the areas (A1 + A2) divided by tc.

·. . iavg =A1 + A2

tc

A1 =737

∫ to

0e−t/0.925 dt

=6.47537

(1 − e− ln 7) = 0.15 A–s

A2 =∫ tc

to

74 − 64e−(t−to)/0.2405

370dt

=74370

(tc − to) +15.392370

(e− ln 16 − 1)

=17.797370

ln 16 − 15.392370

(1 − e− ln 16)

= 0.09436 A–s

iavg =(0.15 + 0.09436)

0.925 ln 7 + 0.2405 ln 16(1000) = 99.06 mA

[c] Pavg = (1000)(99.06 × 10−3) = 99.06 W

No. of kw hrs/yr =(99.06)(24)(365)

1000= 867.77

Cost/year = (867.77)(0.05) = 43.39 dollars/year

P 7.106 [a] Replace the circuit attached to the capacitor with its Thévenin equivalent, wherethe equivalent resistance is the parallel combination of the two resistors, andthe open-circuit voltage is obtained by voltage division across the lampresistance. The resulting circuit is


RTh = R‖RL =RRL

R + RL; VTh =

RL

R + RLVs

From this circuit,

vC(∞) = VTh; vC(0) = Vmax; τ = RThC

Thus,

vC(t) = VTh + (Vmax − VTh)e−(t−to)/τ

where

τ =RRLC

R + RL

[b] Now, set vC(tc) = Vmin and solve for (tc − to):

VTh + (Vmax − VTh)e−(tc−to)/τ = Vmin

e−(tc−to)/τ =Vmin − VTh

Vmax − VTh

−(tc − to)τ

= lnVmin − VTh

Vmax − VTh

(tc − to) = − RRLC

R + RLln

Vmin − VTh

Vmax − VTh

(tc − to) =RRLC

R + RLln

Vmax − VTh

Vmin − VTh

P 7.107 [a] 0 ≤ t ≤ 0.5:

i =2160

+(30

60− 21

60

)e−t/τ where τ = L/R.

i = 0.35 + 0.15e−60t/L

i(0.5) = 0.35 + 0.15e−30/L = 0.40

·. . e30/L = 3; L =30ln 3

= 27.31 H

Problems 7–95

[b] 0 ≤ t ≤ tr, where tr is the time the relay releases:

i = 0 +(30

60− 0

)e−60t/L = 0.5e−60t/L

·. . 0.4 = 0.5e−60tr/L; e60tr/L = 1.25

tr =27.31 ln 1.25

60∼= 0.10 s

8Natural and Step Responses of RLC

Circuits

Assessment Problems

AP 8.1 [a]1

(2RC)2 =1

LC, therefore C = 500 nF

[b] α = 5000 =1

2RC, therefore C = 1 µF

s1,2 = −5000 ±√

25 × 106 − (103)(106)20

= (−5000 ± j5000) rad/s

[c]1√LC

= 20,000, therefore C = 125 nF

s1,2 =[−40 ±

√(40)2 − 202

]103,

s1 = −5.36 krad/s, s2 = −74.64 krad/s

AP 8.2 iL =1

50 × 10−3

∫ t

0[−14e−5000x + 26e−20,000x] dx + 30 × 10−3

= 20−14e−5000x

−5000

∣∣∣∣t0+

26e−20,000t

−20,000

∣∣∣∣t0

+ 30 × 10−3

= 56 × 10−3(e−5000t − 1) − 26 × 10−3(e−20,000t − 1) + 30 × 10−3

= [56e−5000t − 56 − 26e−20,000t + 26 + 30] mA

= 56e−5000t − 26e−20,000t mA, t ≥ 0

AP 8.3 From the given values of R, L, and C, s1 = −10 krad/s and s2 = −40 krad/s.

[a] v(0−) = v(0+) = 0, therefore iR(0+) = 0

8–1

8–2 CHAPTER 8. Natural and Step Responses of RLC Circuits

[b] iC(0+) = −(iL(0+) + iR(0+)) = −(−4 + 0) = 4 A

[c] Cdvc(0+)

dt= ic(0+) = 4, therefore

dvc(0+)dt

=4C

= 4 × 108 V/s

[d] v = [A1e−10,000t + A2e

−40,000t] V, t ≥ 0+

v(0+) = A1 + A2,dv(0+)

dt= −10,000A1 − 40,000A2

Therefore A1 + A2 = 0, −A1 − 4A2 = 40,000; A1 = 40,000/3 V

[e] A2 = −40,000/3 V

[f] v = [40,000/3][e−10,000t − e−40,000t] V, t ≥ 0

AP 8.4 [a]1

2RC= 8000, therefore R = 62.5 Ω

[b] iR(0+) =10 V

62.5 Ω= 160 mA

iC(0+) = −(iL(0+) + iR(0+)) = −80 − 160 = −240 mA = Cdv(0+)

dt

Thereforedv(0+)

dt=

−240 mC

= −240 kV/s

[c] B1 = v(0+) = 10 V,dvc(0+)

dt= ωdB2 − αB1

Therefore 6000B2 − 8000B1 = −240,000, B2 = (−80/3) V

[d] iL = −(iR + iC); iR = v/R; iC = Cdv

dt

v = e−8000t[10 cos 6000t − 803

sin 6000t] V

Therefore iR = e−8000t[160 cos 6000t − 12803

sin 6000t] mA

iC = e−8000t[−240 cos 6000t +4603

sin 6000t] mA

iL = 10e−8000t[8 cos 6000t +823

sin 6000t] mA, t ≥ 0

AP 8.5 [a]( 1

2RC

)2

=1

LC=

106

4, therefore

12RC

= 500, R = 100 Ω

[b] 0.5CV 20 = 12.5 × 10−3, therefore V0 = 50 V

[c] 0.5LI20 = 12.5 × 10−3, I0 = 250 mA

Problems 8–3

[d] D2 = v(0+) = 50,dv(0+)

dt= D1 − αD2

iR(0+) =50100

= 500 mA

Therefore iC(0+) = −(500 + 250) = −750 mA

Thereforedv(0+)

dt= −750 × 10−3

C= −75,000 V/s

Therefore D1 − αD2 = −75,000; α =1

2RC= 500, D1 = −50,000 V/s

[e] v = [50e−500t − 50,000te−500t] V

iR =v

R= [0.5e−500t − 500te−500t] A, t ≥ 0+

AP 8.6 [a] iR(0+) =V0

R=

40500

= 0.08 A

[b] iC(0+) = I − iR(0+) − iL(0+) = −1 − 0.08 − 0.5 = −1.58 A

[c]diL(0+)

dt=

Vo

L=

400.64

= 62.5 A/s

[d] α =1

2RC= 1000;

1LC

= 1,562,500; s1,2 = −1000 ± j750 rad/s

[e] iL = if + B′1e

−αt cos ωdt + B′2e

−αt sin ωdt, if = I = −1 A

iL(0+) = 0.5 = if + B′1, therefore B′

1 = 1.5 A

diL(0+)dt

= 62.5 = −αB′1 + ωdB

′2, therefore B′

2 = (25/12) A

Therefore iL(t) = −1 + e−1000t[1.5 cos 750t + (25/12) sin 750t] A, t ≥ 0

[f] v(t) =ŁdiLdt

= 40e−1000t[cos 750t − (154/3) sin 750t]V t ≥ 0

AP 8.7 [a] i(0+) = 0, since there is no source connected to L for t < 0.

[b] vc(0+) = vC(0−) =(

15 k15 k + 9 k

)(80) = 50 V

[c] 50 + 80i(0+) + Ldi(0+)

dt= 100,

di(0+)dt

= 10,000 A/s

[d] α = 8000;1

LC= 100 × 106; s1,2 = −8000 ± j6000 rad/s

[e] i = if + e−αt[B′1 cos ωdt + B′

2 sin ωdt]; if = 0, i(0+) = 0

Therefore B′1 = 0;

di(0+)dt

= 10,000 = −αB′1 + ωdB

′2

Therefore B′2 = 1.67 A; i = 1.67e−8000t sin 6000t A, t ≥ 0


AP 8.8 vc(t) = vf + e−αt[B′1 cos ωdt + B′

2 sin ωdt], vf = 100 V

vc(0+) = 50 V;dvc(0+)

dt= 0; therefore 50 = 100 + B′

1

B′1 = −50 V; 0 = −αB′

1 + ωdB′2

Therefore B′2 =

α

ωd

B′1 =

(80006000

)(−50) = −66.67 V

Therefore vc(t) = 100 − e−8000t[50 cos 6000t + 66.67 sin 6000t] V, t ≥ 0

Problems

P 8.1 [a] α =1

2RC=

12(1000)(2 × 10−6)

= 250

ω2o =

1LC

=1

(12.5)(2 × 10−6)= 40,000

s1,2 = −250 ±√

2502 − 40,000 = −250 ± 150

s1 = −100 rad/s

s2 = −400 rad/s

[b] overdamped

[c] Note — we want ωd = 120 rad/s:

ωd =√

ω2o − α2

·. . α2 = ω2o − ω2

d = 40,000 − (120)2 = 25,600

α = 160

12RC

= 160; ·. . R =1

2(160)(2 × 10−6)= 1562.5 Ω

[d] s1, s2 = −160 ±√

1602 − 40,000 = −160 ± j120 rad/s

[e] α =√

40,000 =1

2RC; ·. . R =

12(200)(2 × 10−6)

= 1250 Ω

Problems 8–5

P 8.2 [a] −α +√

α2 − ω2o = −250

−α −√

α2 − ω2o = −1000

Adding the above equations, − 2α = −1250

α = 625 rad/s

12RC

=1

2R(0.1 × 10−6)= 625

R = 8 kΩ

2√

α2 − ω2o = 750

4(α2 − ω2o) = 562,500

·. . ωo = 500 rad/s

ω2o = 25 × 104 =

1LC

·. . L =1

(25 × 104)(0.1 × 10−6)= 40 H

[b] iR =v(t)R

= −1e−250t + 4e−1000t mA, t ≥ 0+

iC = Cdv(t)dt

= 0.2e−250t − 3.2e−1000t mA, t ≥ 0+

iL = −(iR + iC) = 0.8e−250t − 0.8e−1000t mA, t ≥ 0

P 8.3 [a] iR(0) =15200

= 75mA

iL(0) = −45mA

iC(0) = −iL(0) − iR(0) = 45 − 75 = −30 mA

[b] α =1

2RC=

12(200)(0.2 × 10−6)

= 12,500

ω2o =

1LC

=1

(50 × 10−3)(0.2 × 10−6)= 108

s1,2 = −12,500 ±√

1.5625 × 108 − 108 = −12,500 ± 7500

s1 = −5000 rad/s; s2 = −20,000 rad/s

v = A1e−5000t + A2e

−20,000t

v(0) = A1 + A2 = 15


dv

dt(0) = −5000A1 − 20,000A2 =

−30 × 10−3

0.2 × 10−6 = −15 × 104V/s

Solving, A1 = 10; A2 = 5

v = 10e−5000t + 5e−20,000t V, t ≥ 0

[c] iC = Cdv

dt

= 0.2 × 10−6[−50,000e−5000t − 100,000e−20,000t]

= −10e−5000t − 20e−20,000t mA

iR = 50e−5000t + 25e−20,000t mA

iL = −iC − iR = −40e−5000t − 5e−20,000t mA, t ≥ 0

P 8.41

2RC=

12(312.5)(0.2 × 10−6)

= 8000

1LC

=1

(50 × 10−3)(0.2 × 10−6)= 108

s1,2 = −8000 ±√

80002 − 108 = −8000 ± j6000 rad/s

·. . response is underdamped

v(t) = B1e−8000t cos 6000t + B2e

−8000t sin 6000t

v(0+) = 15 V = B1; iR(0+) =15

312.5= 48 mA

iC(0+) = [−iL(0+) + iR(0+)] = −[−45 + 48] = −3 mA

dv(0+)dt

=−3 × 10−3

0.2 × 10−6 = −15,000 V/s

dv(0)dt

= −8000B1 + 6000B2 = −15,000

6000B2 = 8000(15) − 15,000; ·. . B2 = 17.5 V

v(t) = 15e−8000t cos 6000t + 17.5e−8000t sin 6000t V, t ≥ 0

Problems 8–7

P 8.5 α =1

2RC=

12(250)(0.2 × 10−6)

= 104

α2 = 108; ·. . α2 = ω2o

Critical damping:

v = D1te−αt + D2e

−αt

iR(0+) =15250

= 60 mA

iC(0+) = −[iL(0+) + iR(0+)] = −[−45 + 60] = −15 mA

v(0) = D2 = 15

dv

dt= D1[t(−αe−αt) + e−αt] − αD2e

−αt

dv

dt(0) = D1 − αD2 =

iC(0)C

=−15 × 10−3

0.2 × 10−6 = −75,000

D1 = αD2 − 75,000 = (104)(15) − 75,000 = 75,000

v = (75,000t + 15)e−10,000t V, t ≥ 0

P 8.6 α = 1000/2 = 500

R =1

2αC=

12(500)(2.5 × 10−6)

= 400 Ω

v(0+) = 3(1 + 1) = 6 V

iR(0+) =6

400= 15 mA

dv

dt= −300e−100t − 2700e−900t

dv(0+)dt

= −300 − 2700 = −3000 V/s

iC(0+) = 2.5 × 10−6(−3000) = −7.5 mA

iL(0+) = −[iR(0+) + iC(0+)] = −[15 − 7.5] = −7.5 mA


P 8.7 [a] α = 20,000; ωd = 15,000

ωd =√

ω2o − α2

·. . ω2o = ω2

d + α2 = 225 × 106 + 400 × 106 = 625 × 106

1LC

= 625 × 106

L =1

(625 × 106)(40 × 10−9)= 40 mH

[b] α =1

2RC

·. . R =1

2αC=

12(20,000)(40 × 10−9)

= 625 Ω

[c] Vo = v(0) = 100 V

[d] Io = iL(0) = −iR(0) − iC(0)

iR(0) =Vo

R=

100625

= 160 mA

iC(0) = Cdv

dt(0)

dv

dt= 100e−20,000t[−15,000 sin 15,000t − 30,000 cos 15,000t]−

20,000e−20,000t[cos 15,000t − 2 sin 15,000t]

dv

dt(0) = 1001(−30,000) − 20,000 = −500 × 104

Cdv

dt(0) = −500 × 104(40 × 10−9) = −200 mA

·. . Io = 200 − 160 = 40 mA

[e]dv

dt= 100e−20,000t[25,000 sin 15,000t − 50,000 cos 15,000t]

= 25 × 105e−20,000t[sin 15,000t − 2 cos 15,000t]

Cdv

dt= 0.1e−20,000t(sin 15,000t − 2 cos 15,000t)

iC(t) = 0.1e−20,000t(sin 15,000t − 2 cos 15,000t) A

iR(t) = 0.16e−20,000t(cos 15,000t − 2 sin 15,000t) A

iL(t) = −iR(t) − iC(t)

= e−20,000t(40 cos 15,000t + 220 sin 15,000t) mA, t ≥ 0

Problems 8–9

P 8.8 [a] 2α = 1000; α = 500 rad/s

2√

α2 − ω2o = 600; ωo = 400 rad/s

C =1

2αR=

12(500)(250)

= 4 µF

L =1

ω2oC

=1

(400)2(4 × 10−6)= 1.5625 H

iC(0+) = A1 + A2 = 45 mA

diCdt

+diLdt

+diRdt

= 0

diC(0)dt

= −diL(0)dt

− diR(0)dt

diL(0)dt

=15

1.5625= 9.6 A/s

diR(0)dt

=1R

dv(0)dt

=1R

iC(0)C

=45 × 10−3

(250)(4 × 10−6)= 45 A/s

·. .diC(0)

dt= −9.6 − 45 = −54.6 A/s

·. . 200A1 + 800A2 = 54.6 A1 + A2 = 0.045

Solving, A1 = −31 mA; A2 = 76 mA

·. . iC = −31e−200t + 76e−800t mA, t ≥ 0+

[b] By hypothesis

v = A3e−200t + A4e

−800t, t ≥ 0

v(0) = A3 + A4 = 15

dv(0)dt

=45 × 10−3

4 × 10−6 = 11,250 V/s

−200A3 − 800A4 = 11,250; ·. . A3 = 38.75 V; A4 = −23.75 V

v = 38.75e−200t − 23.75e−800t V, t ≥ 0

[c] iR(t) =v

250= 155e−200t − 95e−800t mA, t ≥ 0+

[d] iL = −iR − iC

iL = −124e−200t + 19e−800t mA, t ≥ 0


P 8.9 [a]( 1

2RC

)2

=1

LC= (500)2

·. . C =1

(500)2(4)= 1 µF

12RC

= 500

·. . R =1

2(500)(10−6)= 1 kΩ

v(0) = D2 = 8 V

iR(0) =8

1000= 8mA

iC(0) = −8 + 10 = 2 mA

dv

dt(0) = D1 − 500D2 =

2 × 10−3

10−6 = 2000 V/s

·. . D1 = 2000 + 500(8) = 6000 V/s

[b] v = 6000te−500t + 8e−500t V, t ≥ 0

dv

dt= [−3 × 106t + 2000]e−500t

iC = Cdv

dt= (−3000t + 2)e−500t mA, t ≥ 0+

P 8.10 [a] α =1

2RC= 0.5 rad/s

ω2o =

1LC

= 25.25

ωd =√

25.25 − (0.5)2 = 5 rad/s

·. . v = B1e−t/2 cos 5t + B2e

−t/2 sin 5t

v(0) = B1 = 0; v = B2e−t/2 sin 5t

iR(0+) = 0 A; iC(0+) = 4 A;dv

dt(0+) =

40.08

= 50 V/s

50 = −αB1 + ωdB2 = −0.5(0) + 5B2

·. . B2 = 10

·. . v = 10e−t/2 sin 5t V, t ≥ 0

Problems 8–11

[b]dv

dt= −5e−t/2 sin 5t + 10e−t/2(5 cos 5t)

dv

dt= 0 when 10 cos 5t = sin 5t or tan 5t = 10

·. . 5t1 = 1.47, t1 = 294.23 ms

5t2 = 1.47 + π, t2 = 922.54 ms

5t3 = 1.47 + 2π, t3 = 1550.86 ms

[c] t3 − t1 = 1256.6 ms; Td =2πωd

=2π5

= 1256.6 ms

[d] t2 − t1 = 628.3 ms;Td

2=

1256.62

= 628.3 ms

[e] v(t1) = 10e−(0.147115) sin 5(0.29423) = 8.59 V

v(t2) = 10e−(0.46127) sin 5(0.92254) = −6.27 V

v(t3) = 10e−(0.77543) sin 5(1.55086) = 4.58 V

[f]

P 8.11 [a] α = 0; ωd = ωo =√

25.25 = 5.02 rad/s

v = B1 cos ωot + B2 sin ωot; v(0) = B1 = 0; v = B2 sin ωot

Cdv

dt(0) = −iL(0) = 4

50 = −αB1 + ωdB2 = −0 +√

25.25B2

·. . B2 = 50/√

25.25 = 9.95 V

v = 9.95 sin 5.02t V, t ≥ 0

[b] 2πf = 5.02; f =5.022π

∼= 0.80 Hz


[c] 9.95 V

P 8.12 [a] ω2o =

1LC

=1

(12.5)(3.2 × 10−9)= 25 × 106

ωo = 5000 rad/s

12RC

= 5000; R =1

2(5000)(3.2 × 10−9)= 31.25 kΩ

[b] v(t) = D1te−5000t + D2e

−5000t

v(0) = 100 V = D2

dv

dt= (D1t + 100)(−5000e−5000t) + D1e

−5000t

dv

dt(0) = −500 × 103 + D1 =

iC(0)C

iC(0) = −iR(0) − iL(0)

iR(0) =100

31,500= 3.2 mA

·. . iC(0) = −(3.2 + 6.4) = −9.6 mA

·. .dv

dt(0) = −9.6 × 10−3

3.2 × 10−9 = −3 × 106

·. . −500 × 103 + D1 = −3 × 106

D1 = −25 × 105V/s

·. . v(t) = (−25 × 105t + 100)e−5000t V, t ≥ 0

[c] iC(t) = 0 whendv

dt(t) = 0

dv

dt= (−25 × 105t + 100)(−5000)e−5000t + e−5000t(−25 × 105)

= (125 × 108t − 30 × 105)e−5000t

dv

dt= 0 when 125 × 108t1 = 3 × 106; ·. . t1 = 240 µs

v(240µs) = e−1.2[(−25 × 105)(240 × 10−6) + 100] = −150.6 V

Problems 8–13

[d] iL(240µs) = −iR(240µs) =−150.631,250

= −4.82 mA

ωC(240µs) =12(3.2 × 10−9)(−150.6)2 = 36.29 µJ

ωL(240µs) =12(12.5)(−4.82 × 10−3)2 = 145.2 µJ

ω(240µs) = ωC + ωL = 181.49 µJ

ω(0) =12(12.5)(6.4 × 10−3)2 +

12(3.2 × 10−9)(100)2 = 272 µJ

% remaining =181.49272

(100) = 66.72%

P 8.13 [a] α =1

2RC= 1250, ωo = 103, therefore overdamped

s1 = −500, s2 = −2000

therefore v = A1e−500t + A2e

−2000t

v(0+) = 0 = A1 + A2;[dv(0+)

dt

]=

iC(0+)C

= 98,000 V/s

Therefore − 500A1 − 2000A2 = 98,000

A1 =+98015

, A2 =−98015

v(t) =[980

15

][e−500t − e−2000t] V, t ≥ 0

[b]

Example 8.4: vmax∼= 74.1 V at 1.4 ms


Example 8.5: vmax∼= 36.1 V at 1.0 ms

Problem 8.13: vmax∼= 30.9 at 0.92 ms

P 8.14 From the form of the solution we have

v(0) = A1 + A2

dv(0+)dt

= −α(A1 + A2) + jωd(A1 − A2)

We know both v(0) and dv(0+)/dt will be real numbers. To facilitate the algebra welet these numbers be K1 and K2, respectively. Then our two simultaneous equationsare

K1 = A1 + A2

K2 = (−α + jωd)A1 + (−α − jωd)A2

The characteristic determinate is

∆ =

∣∣∣∣∣∣∣1 1

(−α + jωd) (−α − jωd)

∣∣∣∣∣∣∣ = −j2ωd

The numerator determinates are

N1 =

∣∣∣∣∣∣∣K1 1

K2 (−α − jωd)

∣∣∣∣∣∣∣ = −(α + jωd)K1 − K2

and N2 =

∣∣∣∣∣∣∣1 K1

(−α + jωd) K2

∣∣∣∣∣∣∣ = K2 + (α − jωd)K1

It follows that A1 =N1

∆=

ωdK1 − j(αK1 + K2)2ωd

and A2 =N2

∆=

ωdK1 + j(αK1 + K2)2ωd

We see from these expressions that A1 = A∗2

Problems 8–15

P 8.15 By definition, B1 = A1 + A2. From the solution to Problem 8.14 we have

A1 + A2 =2ωdK1

2ωd

= K1

But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. (8.30).By definition, B2 = j(A1 − A2). From Problem 8.14 we have

B2 = j(A1 − A2) =j[−2j(αK1 + K2)]

2ωd

=αK1 + K2

ωd

It follows that

K2 = −αK1 + ωdB2, but K2 =dv(0+)

dtand K1 = B1

Thus we have

dv

dt(0+) = −αB1 + ωdB2,

which is identical to Eq. (8.31).

P 8.16 t < 0 : Vo = 15 V, Io = −60 mA

t > 0:

iR(0) =15100

= 150 mA; iL(0) = −60 mA

iC(0) = −150 − (−60) = −90 mA

α =1

2RC=

12(100)(10−6)

= 5000 rad/s


ω2o =

1LC

=1

(62.5 × 10−3)(10−6)= 16 × 106

s1,2 = −5000 ±√

25 × 106 − 16 × 106 = −5000 ± 3000

s1 = −2000 rad/s; s2 = −8000 rad/s

·. . vo = A1e−2000t + A2e

−8000t

A1 + A2 = vo(0) = 15

dvo

dt(0) = −2000A1 − 8000A2 =

−90 × 10−3

10−6 = −90,000

Solving, A1 = 5 V, A2 = 10 V

·. . vo = 5e−2000t + 10e−8000t V, t ≥ 0

P 8.17 ω2o =

1LC

=1

(62.5 × 10−3)(10−6)= 16 × 106

α =1

2RC=

12(250)(10−6)

= 2500

s1,2 = −2500 ±√

25002 − 16 × 106 = −2500 ± j3122.5rad/s

vo(t) = B1e−2500t cos 3122.5t + B2e

−2500t sin 3122.5t

vo(0) = B1 = 15 V

iR(0) =15200

= 75 mA

iL(0) = −60 mA

iC(0) = −iR(0) − iL(0) = −15 mA ·. .iC(0)

C= −15,000 V/s

dvo

dt(0) = −2500B1 + 3122.5B2 = −15,000 V/s

·. . 3122.5B2 = 2500(15) − 15,000 ·. . B2 = 7.21 V

vo(t) = 15e−2500t cos 3122.5t + 7.21e−2500t sin 3122.5t V, t ≥ 0

Problems 8–17

P 8.18 ω2o =

1LC

=1

(62.5 × 10−3)(10−6)= 16 × 106

α =1

2RC=

12(125)(10−6)

= 4000

·. . α2 = ω2o (critical damping)

vo(t) = D1te−4000t + D2e

−4000t

vo(0) = D2 = 15 V

iR(0) =15125

= 120 mA

iL(0) = −60 mA

iC(0) = −60 mA

dvo

dt(0) = −4000D2 + D1

iC(0)C

=−60 × 10−3

10−6 = −60,000

D1 − 4000D2 = −60,000; D1 = 0

vo(t) = 15e−4000t V, t ≥ 0

P 8.19

vT = −2 × 104iφ + 16 × 103iT ; iφ =20100

(−iT )

= 4000it + 16,000iT = 20,000iT

vT

iT= 20 kΩ


Vo =30005000

(50) = 30 V; Io = 0

iC(0) = −iR(0) − iL(0) = − 3020,000

= −1.5 mA

iC(0)C

=−1.5 × 10−3

0.25 × 10−6 = −6000

ω2o =

1LC

=1

(40)(0.25 × 10−6)= 105

α =1

2RC=

12(20 × 103)(0.25 × 10−6)

= 100 rad/s

ωd =√

105 − 1002 = 300 rad/s

vo = B1e−100t cos 300t + B2e

−100t sin 300t

vo(0) = B1 = 30 V

dvo

dt(0) = 300B2 − 100B1 = −6000

·. . 300B2 = 100(30) − 6000; ·. . B2 = −10 V

vo = 30e−100t cos 300t − 10e−100t sin 300t V, t ≥ 0

P 8.20 [a] v = L

(diLdt

)= 16[e−20,000t − e−80,000t] V, t ≥ 0

[b] iR =v

R= 40[e−20,000t − e−80,000t] mA, t ≥ 0+

[c] iC = I − iL − iR = [−8e−20,000t + 32e−80,000t] mA, t ≥ 0+

P 8.21 [a] v = L

(diLdt

)= 40e−32,000t sin 24,000t V, t ≥ 0

[b] iC(t) = I − iR − iL = 24 × 10−3 − v

625− iL

= [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA, t ≥ 0+

P 8.22 v = L

(diLdt

)= 960,000te−40,000t V, t ≥ 0

Problems 8–19

P 8.23 t < 0 : iL = 9/3000 = 3 mAt > 0:

6 k‖3 k = 2 kΩ

iL(0) = 3 mA, iL(∞) = 9 mA

ω2o =

1LC

=1

(62.5)(2.5 × 10−6)= 6400; ωo = 80 rad/s

α =1

2RC=

12(2000)(2.5 × 10−6)

= 100; α2 = 104

α2 − ω2o = 104 − 6400 = 3600

s1,2 = −100 ± 60 rad/s

s1 = −40 rad/s; s2 = −160 rad/s

iL = If + A′1e

−40t + A′2e

−160t

iL(∞) = If = 9mA

iL(0) = A′1 + A′

2 + If = 3 mA

·. . A′1 + A′

2 + 9 m = 3 m so A′1 + A′

2 = −6 mA

diLdt

(0) = 0 = −40A′1 − 160A′

2

Solving, A′1 = −8 mA, A′

2 = 2 mA

iL = 9 − 8e−40t + 2e−160t mA, t ≥ 0


P 8.24 ω2o =

1LC

=1

(50 × 10−3)(0.2 × 10−6)= 108; ωo = 104 rad/s

α =1

2RC=

12(200)(0.2 × 10−6)

= 12,500 rad/s ·. . overdamped

s1,2 = −12,500 ±√

(12,500)2 − 108 = −12,500 ± 7500 rad/s

s1 = −5000 rad/s; s2 = −20,000 rad/s

If = 60 mA

iL = 60 × 10−3 + A′1e

−5000t + A′2e

−20,000t

·. . −45 × 10−3 = 60 × 10−3 + A′1 + A′

2; A′1 + A′

2 = −105 × 10−3

diLdt

= −5000A′1 − 20,000A′

2 =15

0.05= 300

Solving, A′1 = −120 mA; A′

2 = 15 mA

iL = 60 − 120e−5000t + 15e−20,000t mA, t ≥ 0

P 8.25 α =1

2RC=

12(312.5)(0.2 × 10−6)

= 8000; α2 = 64 × 106

ωo = 104 underdamped

s1,2 = −8000 ± j√

80002 − 108 = −8000 ± j6000 rad/s

iL = 60 × 10−3 + B′1e

−8000t cos 6000t + B′2e

−8000t sin 6000t

−45 × 10−3 = 60 × 10−3 + B′1

·. . B′1 = −105 mA

diLdt

(0) = −8000B′1 + 6000B′

2 = 300

·. . B′2 = −90 mA

iL = 60 − 105e−8000t cos 6000t − 90e−8000t sin 6000t mA, t ≥ 0

Problems 8–21

P 8.26 α =1

2RC=

12(250)(0.2 × 10−6)

= 104

α2 = 108 = ω2o critical damping

iL = If + D′1te

−104t + D′2e

−104t = 60 × 10−3 + D′1te

−104t + D′2e

−104t

iL(0) = −45 × 10−3 = 60 × 10−3 + D′2; ·. . D′

2 = −105 mA

diLdt

(0) = −104D′2 + D′

1 = 300 A/s

·. . D′1 = 300 + 104(−105 × 10−3) = −750 A/s

iL = 60 − 750,000te−104t − 105e−104t mA, t ≥ 0

P 8.27 For t > 0

α =1

2RC= 1000;

1LC

= 64 × 104

s1,2 = −1000 ± 600 rad/s

s1 = −400 rad/s; s2 = −1600 rad/s

vo = Vf + A′1e

−400t + A′2e

−1600t

Vf = 0; vo(0+) = 0; iC(0+) = 30 mA

·. . A′1 + A′

2 = 0

dvo(0+)dt

=iC(0+)

1.25 × 10−6 = 24,000 V/s

dvo(0+)dt

= −400A′1 − 1600A′

2 = 24,000

Solving,

A′1 = 20 V; A′

2 = −20 V

vo = 20e−400t − 20e−1600t V, t ≥ 0


P 8.28 [a] From the solution to Prob. 8.27 s1 = −400 rad/s and s2 = −1600 rad/s,therefore

io = If + A′1e

−400t + A′2e

−1600t

If = 30 mA; io(0+) = 0;dio(0+)

dt= 0

·. . 0 = 30 × 10−3 + A′1 + A′

2; −400A′1 − 1600A′

2 = 0

Solving

A′1 = −40 mA; A′

2 = 10 mA

·. . io = 30 − 40e−400t + 10e−1600t mA, t ≥ 0

[b]diodt

= 16e−400t − 16e−1600t

vo = Ldiodt

= 20e−400t − 20e−1600t V, t ≥ 0

This agrees with the solution to Problem 8.27

P 8.29 α =1

2RC=

12(400)(1.25 × 10−6)

= 1000

ω2o =

1LC

=1

(1.25 × 10−6)(1.25)= 64 × 104

s1,2 = −1000 ±√

10002 − 64 × 104 = −1000 ± 600 rad/s

s1 = −400 rad/s; s2 = −1600 rad/s

vo(∞) = 0 = Vf

·. . vo = A′1e

−400t + A′2e

−1600t

vo(0) = 12 = A′1 + A′

2

Note: iC(0+) = 0

·. .dvo

dt(0) = 0 = −400A′

1 − 1600A′2

Solving, A′1 = 16 V, A′

2 = −4 V

vo(t) = 16e−400t − 4e−1600t V, t > 0

Problems 8–23

P 8.30 [a] io = If + A′1e

−400t + A′2e

−1600t

If =12400

= 30mA; io(0) = 0

0 = 30 × 10−3 + A′1 + A′

2,·. . A′

1 + A′2 = −30 × 10−3

diodt

(0) =12

1.25= −400A′

1 − 1600A′2

Solving, A′1 = −32 mA; A′

2 = 2 mA

io = 30 − 32e−400t + 2e−1600t mA, t ≥ 0

[b]diodt

= [12.8e−400t − 3.2e−1600t]

vo = Ldiodt

= 16e−400t − 4e−1600t V, t ≥ 0

This agrees with the solution to Problem 8.29.

P 8.31 iL(0−) = iL(0+) = 37.5 mA

For t > 0

iL(0−) = iL(0+) = 37.5 mA

α =1

2RC= 100 rad/s; ω2

o =1

LC= 6400

s1 = −40 rad/s s2 = −160 rad/s

vo(∞) = 0 = Vf

vo = A′1e

−40t + A′2e

−160t

iC(0+) = −37.5 + 37.5 + 0 = 0

·. .dvo

dt= 0


dvo

dt(0) = −40A′

1 − 160A′2

·. . A′1 + 4A′

2 = 0; A′1 + A′

2 = 0

·. . A′1 = 0; A′

2 = 0

·. . vo = 0 for t ≥ 0

Note: vo(0) = 0; vo(∞) = 0;dvo(0)

dt= 0

Hence the 37.5 mA current circulates between the current source and the idealinductor in the equivalent circuit. In the original circuit the 7.5 V source sustains acurrent of 37.5 mA in the inductor. This is an example of a circuit going directly intosteady state when the switch is closed. There is no transient period, or interval.

P 8.32 t < 0:

vo(0−) = vo(0+) =625

781.25(25) = 20 V

iL(0−) = iL(0+) = 0

t > 0

−160 × 10−3 +20125

+ iC(0+) + 0 = 0; ·. . iC(0+) = 0

12RC

=1

2(125)(5 × 10−6)= 800 rad/s

ω2o =

1LC

=1

(312.5 × 10−3)(5 × 10−6)= 64 × 104

·. . α2 = ω2o critically damped

Problems 8–25

[a] vo = Vf + D′1te

−800t + D′2e

−800t

Vf = 0

dvo(0)dt

= −800D′2 + D′

1 = 0

vo(0+) = 20 = D′2

D′1 = 800D′

2 = 16,000 V/s

·. . vo = 16,000te−800t + 20e−800t V, t ≥ 0+

[b] iL = If + D′3te

−800t + D′4e

−800t

iL(0+) = 0; If = 160 mA;diL(0+)

dt=

20312.5 × 10−3 = 64 A/s

·. . 0 = 160 + D′4; D′

4 = −160 mA;

−800D′4 + D′

3 = 64; D′3 = −64 A/s

·. . iL = 160 − 64,000te−800t − 160e−800t mA t ≥ 0

P 8.33 [a] wL =∫ ∞

0pdt =

∫ ∞

0voiL dt

vo = 16,000te−800t + 20e−800t V

iL = 0.16 − 64te−800t − 0.16e−800t A

p = 3.2e−800t + 2560te−800t − 3840te−1600t

−1,024,000t2e−1600t − 3.2e−1600t W

wL = 3.2∫ ∞

0e−800t dt + 2560

∫ ∞

0te−800t dt − 3480

∫ ∞

0te−1600t dt

−1,024,000∫ ∞

0t2e−1600t dt − 3.2

∫ ∞

0e−1600t dt

= 3.2e−800t

−800

∣∣∣∣∣∞

0

+2560

(800)2 e−800t(−800t − 1)∣∣∣∣∣∞

0

− 3840(1600)2 e−1600t(−1600t − 1)

∣∣∣∣∣∞

0

− 1,024,000(−1600)3 e−1600t(16002t2 + 3200t + 2)

∣∣∣∣∣∞

0

− 3.2e−1600t

(−1600)

∣∣∣∣∣∞

0


All the upper limits evaluate to zero hence

wL =3.2800

+25608002 − 3840

16002 − (1,024,000)(2)16003 − 3.2

1600= 4 mJ

Note this value corresponds to the final energy stored in the inductor, i.e.

wL(∞) =12(312.5 × 10−3)(0.16)2 = 4 mJ.

[b] v = 16,000te−800t + 20e−800t V

iR =v

125= 128te−800t + 0.16e−800t A

pR = viR = 2,048,000t2e−1600t + 5120te−1600t + 3.2e−1600t

wR =∫ ∞

0pR dt

= 2,048,000∫ ∞

0t2e−1600t dt + 5120

∫ ∞

0te−1600t dt + 3.2

∫ ∞

0e−1600t dt

=2,048,000e−1600t

−16003 [16002t2 + 3200t + 2]∣∣∣∣∞0

+

5120e−1600t

16002 (−1600t − 1)∣∣∣∣∞0

+3.2e−1600t

(−1600)

∣∣∣∣∞0

Since all the upper limits evaluate to zero we have

wR =2,048,000(2)

16003 +512016002 +

3.21600

= 5 mJ

[c] 160 = iR + iC + iL (mA)

iR + iL = 160 + 64,000te−800t mA

·. . iC = 160 − (iR + iL) = −64,000te−800t mA = −64te−800t A

pC = viC = [16,000te−800t + 20e−800t][−64te−800t]

= −1,024,000t2e−1600t − 1280e−1600t

wC = −1,024,000∫ ∞

0t2e−1600t dt − 1280

∫ ∞

0te−1600t dt

wC =−1,024,000e−1600t

−16003 [16002t2 + 3200t + 2]∣∣∣∣∞0

−1280e−1600t

16002 (−1600t − 1)∣∣∣∣∞0

Since all upper limits evaluate to zero we have

wC =−1,024,000(2)

16003 − 1280(1)16002 = −1 mJ

Problems 8–27

Note this 1 mJ corresponds to the initial energy stored in the capacitor, i.e.,

wC(0) =12(5 × 10−6)(20)2 = 1 mJ.

Thus wC(∞) = 0 mJ which agrees with the final value of v = 0.

[d] is = 160 mA

ps(del) = 160v mW

= 0.16[16,000te−800t + 20e−800t]

= 3.2e−800t + 2560te−800t W

ws = 3.2∫ ∞

0e−800t dt +

∫ ∞

02560te−800t dt

=3.2e−800t

−800

∣∣∣∣∞0

+2560e−800t

8002 (−800t − 1)∣∣∣∣∞0

=3.2800

+2560800

= 8 mJ

[e] wL = 4 mJ (absorbed)

wR = 5 mJ (absorbed)

wC = 1 mJ (delivered)

wS = 8 mJ (delivered)∑

wdel = wabs = 9 mJ.

P 8.34 vC(0+) =3.75 × 103

11.25 × 103 (150) = 50 V

iL(0+) = 100 mA; iL(∞) =1507500

= 20 mA

α =1

2RC=

12(2500)(0.25 × 10−6)

= 800

ω2o =

1LC

=1

(4)(0.25 × 10−6)= 106

α2 = 64 × 104; α2 < ω2o ; ·. . underdamped

s1,2 = −800 ± j√

8002 − 106 = −800 ± j600 rad/s


iL = If + B′1e


−αt sin ωdt

= 20 + B′1e

−800t cos 600t + B′2e

−800t sin 600t

iL(0) = 20 × 10−3 + B′1; B′

1 = 100 m − 20 m = 80 mA

diLdt

(0) = 600B′2 − 800B′

1 =504

= 12.5

·. . 600B2 = 800(80 × 10−3) + 12.5; B′2 = 127.5 mA

·. . iL = 20 + 80e−800t cos 600t + 127.5e−800t sin 600t mA, t ≥ 0

P 8.35 [a] 2α = 5000; α = 2500 rad/s√α2 − ω2

o = 1500; ω2o = 4 × 106; ωo = 2000 rad/s

α =R

2L= 2500; R = 5000L

ω2o =

1LC

= 4 × 106; L =109

4 × 106(50)= 5H

R = 25,000 Ω

[b] i(0) = 0

Ldi(0)dt

= vc(0);12(50) × 10−9v2

c (0) = 90 × 10−6

·. . v2c (0) = 3600; vc(0) = 60 V

di(0)dt

=605

= 12 A/s

[c] i(t) = A1e−1000t + A2e

−4000t

i(0) = A1 + A2 = 0

di(0)dt

= −1000A1 − 4000A2 = 12

Solving,

·. . A1 = 4 mA; A2 = −4 mA

i(t) = 4e−1000t − 4e−4000t mA t ≥ 0

Problems 8–29

[d]di(t)dt

= −4e−1000t + 16e−4000t

di

dt= 0 when 16e−4000t = 4e−1000t

or e3000t = 4

·. . t =ln 43000

µs = 462.10 µs

[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA

[f] vL(t) = 5di

dt= [−20e−1000t + 80e−4000t] V, t ≥ 0+

P 8.36 α = 2000 rad/s; ωd = 1500 rad/s

ω2o − α2 = 225 × 104; ω2

o = 625 × 104; wo = 2500 rad/s

α =R

2L= 2000; R = 4000L

1LC

= 625 × 104; L =1

(625 × 104)(80 × 10−9)= 2 H

·. . R = 8 kΩ

i(0+) = B1 = 7.5 mA; at t = 0+

60 + vL(0+) − 30 = 0; ·. . vL(0+) = −30 V

di(0+)dt

=−302

= −15 A/s

·. .di(0+)

dt= 1500B2 − 2000B1 = −15

·. . 1500B2 = 2000(7.5 × 10−3) − 15; ·. . B2 = 0 A

·. . i = 7.5e−2000t sin 1500t mA, t ≥ 0


P 8.37 From Prob. 8.36 we know vc will be of the form

vc = B3e−2000t cos 1500t + B4e

−2000t sin 1500t

From Prob. 8.36 we have

vc(0) = −30 V = B3

and

dvc(0)dt

=iC(0)

C=

7.5 × 10−3

80 × 10−9 = 93.75 × 103

dvc(0)dt

= 1500B4 − 2000B3 = 93,750

·. . 1500B4 = 2000(−30) + 93,750; B4 = 22.5 V

vc(t) = −30e−2000t cos 1500t + 22.5e−2000t sin 1500t V t ≥ 0

P 8.38 [a] ω2o =

1LC

=1

(80 × 10−3)(0.5 × 10−6)= 25 × 106

α =R

2L= ωo = 5000 rad/s

·. . R = (5000)(2)L = 800 Ω

[b] i(0) = iL(0) = 30 mA

vc(0) = 800i(0) + 80 × 10−3di(0)dt

20 − 800(30 × 10−3)80 × 10−3 =

di(0)dt

·. .di(0)dt

= −50 A/s

[c] vC = D1te−5000t + D2e

−5000t

vC(0) = D2 = 20 V

dvC

dt(0) = D1 − 5000D2 =

iC(0)C

=−iL(0)

C

D1 − 100,000 = − 30 × 10−3

0.5 × 10−6 = −60,000 ·. . D1 = 40,000 V/s

vC = 40,000te−5000t + 20e−5000t V, t ≥ 0

Problems 8–31

P 8.39 [a] For t > 0:

Since i(0−) = i(0+) = 0

va(0+) = 72 V

[b] va = 5000i +1

0.1 × 10−6

∫ t

0i dx + 72

dva

dt= 5000

di

dt+ 10 × 106i

dva(0+)dt

= 5000di(0+)

dt+ 10 × 106i(0+) = 5000

di(0+)dt

−Ldi(0+)

dt= 72

di(0+)dt

= − 722.5

= −28.8 A/s

·. .dva(0+)

dt= −144,000 V/s

[c] α =R

2L=

12,5002(2.5)

= 2500 rad/s

ω2o =

1LC

=1

(2.5)(0.1 × 10−6)= 4 × 106

s1,2 = −2500 ±√

25002 − 4 × 106 = −2500 ± 1500 rad/s

Overdamped:

va = A1e−1000t + A2e

−4000t

va(0) = 72 = A1 + A2

dva(0)dt

= −144,000 = −1000A1 − 4000A2

Solving, A1 = 48; A2 = 24

va = 48e−1000t + 24e−4000t V, t ≥ 0+


P 8.40 ω2o =

1LC

=1

(10)(4 × 10−3)= 25

α =R

2L=

802(10)

= 4; α2 = 16

α2 < ω2o

·. . underdamped

s1,2 = −4 ± j√

9 = −4 ± j3 rad/s

i = B1e−4t cos 3t + B2e

−4t sin 3t

i(0) = B1 = −240/100 = −2.4 A

di

dt(0) = 3B2 − 4B1 = 0

·. . B2 = −3.2 A

i = −2.4e−4t cos 3t − 3.2 sin 3t A, t ≥ 0

P 8.41 t < 0:

i(0) =240

8 + 30‖70 + 11=

24040

= 6 A

vo(0) = 240 − 8(6) − 70100

(6)(20) = 108 V

Problems 8–33

t > 0:

α =R

2L=

202(1)

= 10, α2 = 100

ω2o =

1LC

=1

(1)(5 × 10−3)= 200

ω2o > α2 underdamped

s1,2 = −100 ± √100 − 200 = −10 ± j10 rad/s

vo = B1e−10t cos 10t + B2e

−10t sin 10t

vo(0) = B1 = 108 V

Cdvo

dt(0) = −6,

dvo

dt=

−65 × 10−3 = −1200 V/s

dvo

dt(0) = −10B1 + 10B2 = −1200

10B2 = −1200 + 10B1 = −1200 + 1080; B2 = −120/10 = −12 V

·. . vo = 108e−10t cos 10t − 12e−10t sin 10t V, t ≥ 0

P 8.42 [a] t < 0:

io =80800

= 100 mA; vo = 500io = (500)(0.01) = 50 V

t > 0:

α =R

2L=

5002(2.5 × 10−3)

= 105 rad/s

ω2o =

1LC

=1

(2.5 × 10−3)(40 × 10−9)= 100 × 108

α2 = ω2o

·. . critically damped


·. . io(t) = D1te−105t + D2e

−105t

io(0) = D2 = 100 mA

diodt

(0) = −αD2 + D1 = 0

·. . D1 = 105(100 × 10−3) = 10,000

io(t) = 10,000te−105t + 0.1e−105t A, t ≥ 0

[b] vo(t) = D3te−105t + D4e

−105t

vo(0) = D4 = 50

Cdvo

dt(0) = −0.1

dvo

dt(0) =

−0.140 × 10−9 = −25 × 105 V/s = −αD4 + D3

·. . D3 = 105(50) − 25 × 105 = 25 × 105

vo(t) = 25 × 105te−105t + 50e−105t V, t ≥ 0

P 8.43 α =R

2L=

80002(1)

= 4000 rad/s

ω2o =

1LC

=1

(1)(50 × 10−9)= 20 × 106

s1,2 = −4000 ±√

40002 − 20 × 106 = −4000 ± j2000 rad/s

vo = Vf + B′1e

−4000t cos 2000t + B′2e

−4000t sin 2000t

vo(0) = 0 = Vf + B′1

vo(∞) = 80 V; ·. . B′1 = −80 V

dvo(0)dt

= 0 = 2000B′2 − 4000B′

1

·. . 2000B′2 = 4000(−80) ·. . B′

2 = −160 V

vo = 80 − 80e−4000t cos 2000t − 160e−4000t sin 2000t V, t ≥ 0

Problems 8–35

P 8.44 t < 0:

iL(0) =−15030

= −5 A

vC(0) = 18iL(0) = −90 V

t > 0:

α =R

2L=

102(0.1)

= 50 rad/s

ω2o =

1LC

=1

(0.1)(2 × 10−3)= 5000

ωo > α2 ·. . underdamped

s1,2 = −50 ±√

502 − 5000 = −50 ± j50

vc = 60 + B′1e

−50t cos 50t + B′2e

−50t sin 50t

vc(0) = −90 = 60 + B′1

·. . B′1 = −150

Cdvc

dt(0) = −5;

dvc

dt(0) =

−52 × 10−3 = −2500

dvc

dt(0) = −50B′

1 + 50B2 = −2500 ·. . B′2 = −200

vc = 60 − 150e−50t cos 50t − 200e−50t sin 50t V, t ≥ 0


P 8.45 iC(0) = 0; vo(0) = 50 V

α =R

2L=

80002(160 × 10−3)

= 25,000 rad/s

ω2o =

1LC

=1

(160 × 10−3)(10 × 10−9)= 625 × 106

·. . α2 = ω2o ; critical damping

vo(t) = Vf + D′1te

−25,000t + D′2e

−25,000t

Vf = 250 V

vo(0) = 250 + D′2 = 50; D′

2 = −200 V

dvo

dt(0) = −25,000D′

2 + D′1 = 0

D′1 = 25,000D′

2 − 5 × 106 V/s

vo = 250 − 5 × 106te−25,000t − 200e−25,000t V, t ≥ 0

P 8.46 [a] t < 0:

io(0−) =120

30,000= 4 mA

vC(0−) = 80 − (10,000)(0.004) = 40 V

t = 0+:

5 kΩ‖20 kΩ = 4 kΩ

·. . vo(0+) = −(0.004)(4000) + 40 = 40 − 16 = 24 V

Problems 8–37

[b] vo(t) = vc − 4000io

dvo

dt(0+) =

dvc

dt(0+) − 4000

diodt

(0+)

dvc

dt(0+) =

−4 × 10−3

(125/64) × 10−6 = −2048 V/s

−vL(0+) + vo(0+) + 40 = 0 vL = 64 V

diodt

(0+) =645

= 12.8 A/s

dvo

dt(0+) = −2048 − 4000(12.8) = −53,248 V/s

[c] ω2o =

1LC

=1

(5)[(125/64) × 10−6]= 10.24 × 104

α =R

2L=

40002(5)

= 400 rad/s; α2 = 16 × 104

α2 > ω2o overdamped

s1,2 = −400 ± 240 rad/s

vo(t) = Vf + A′1e

−160t + A′2e

−640t

Vf = vo(∞) = −40 V

−40 + A′1 + A′

2 = 24

−160A′1 − 640A′

2 = −53,248

Solving, A′1 = −25.6; A′

2 = 89.6

·. . vo(t) = −40 − 25.6e−160t + 89.6e−640t V, t ≥ 0+

P 8.47 [a] vc = Vf + [B′1 cos ωdt + B′

2 sin ωdt] e−αt

dvc

dt= [(ωdB

′2 − αB′

1) cos ωdt − (αB′2 + ωdB

′1) sin ωdt]e−αt

Since the initial stored energy is zero,

vc(0+) = 0 anddvc(0+)

dt= 0

It follows that B′1 = −Vf and B′

2 =αB′

1

ωd


When these values are substituted into the expression for [dvc/dt], we get

dvc

dt=(

α2

ωd

+ ωd

)Vfe

−αt sin ωdt

But Vf = V andα2

ωd

+ ωd =α2 + ω2

d

ωd

=ω2

o

ωd

Thereforedvc

dt=(

ω2o

ωd

)V e−αt sin ωdt

[b]dvc

dt= 0 when sin ωdt = 0, or ωdt = nπ

where n = 0, 1, 2, 3, . . .

Therefore t =nπ

ωd

[c] When tn =nπ

ωd

, cos ωdtn = cos nπ = (−1)n

and sin ωdt = sin nπ = 0

Therefore vc(tn) = V [1 − (−1)ne−αnπ/ωd ]

[d] It follows from [c] that

v(t1) = V + V e−(απ/ωd) and vc(t3) = V + V e−(3απ/ωd)

Thereforevc(t1) − V

vc(t3) − V=

e−(απ/ωd)

e−(3απ/ωd) = e(2απ/ωd)

But2πωd

= t3 − t1 = Td, thus α =1Td

ln[vc(t1) − V ][vc(t3) − V ]

P 8.48 α =1Td

ln

vc(t1) − V

vc(t3) − V

; Td = t3 − t1 =

3π7

− π

7=

2π7

ms

α =70002π

ln[63.8426.02

]≈ 1000; ωd =

2πTd

= 7000 rad/s

ω2o = ω2

d + α2 = 49 × 106 + 106 = 50 × 106

L =1

(50 × 106)(0.1 × 10−6)= 200 mH; R = 2αL = 400 Ω

Problems 8–39

P 8.49 [a] Let i be the current in the direction of the voltage drop vo(t). Then by hypothesis

i = if + B′1e


−αt sin ωdt

if = i(∞) = 0, i(0) =Vg

R= B′

1

Therefore i = B′1e


−αt sin ωdt

Ldi(0)dt

= 0, thereforedi(0)dt

= 0

di

dt= [(ωdB

′2 − αB′

1) cos ωdt − (αB′2 + ωdB

′1) sin ωdt] e−αt

Therefore ωdB′2 − αB′

1 = 0; B′2 =

α

ωd

B′1 =

α

ωd

Vg

R

Therefore

vo = Ldi

dt= −

L

(α2Vg

ωdR+

ωdVg

R

)sin ωdt

e−αt

= −

LVg

R

(α2

ωd

+ ωd

)sin ωdt

e−αt

= −VgL

R

(α2 + ω2

d

ωd

)e−αt sin ωdt

vo = − Vg

RCωd

e−αt sin ωdt V, t ≥ 0+

[b]dvo

dt= − Vg

ωdRCωd cos ωdt − α sin ωdte−αt

dvo

dt= 0 when tan ωdt =

ωd

α

Therefore ωdt = tan−1(ωd/α) (smallest t)

t =1ωd

tan−1(

ωd

α

)

P 8.50 [a] From Problem 8.49 we have

vo =−Vg

RCωd

e−αt sin ωdt

α =R

2L=

48002(64 × 10−3)

= 37,500 rad/s

ω2o =

1LC

=1

(64 × 10−3)(4 × 10−9)= 3906.25 × 106


ωd =√

ω2o − α2 = 50 krad/s

−Vg

RCωd

=−(−72)

(4800)(4 × 10−9)(50 × 103)= 75

·. . vo = 75e−37,500t sin 50,000t V, t ≥ 0

[b] From Problem 8.49

td =1ωd

tan−1(

ωd

α

)=

150,000

tan−1

(50,00037,500

)

td = 18.55 µs

[c] vmax = 75e−0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V

[d] R = 480 Ω; α = 3750 rad/s

ωd = 62,387.4 rad/s

vo = 601.08e−3750t sin 62,387.4t V, t ≥ 0

td = 24.22 µs

vmax = 547.92 V

P 8.51 [a]d2vo

dt2=

1R1C1R2C2

vg

1R1C1R2C2

=10−6

(100)(400)(0.5)(0.2) × 10−6 × 10−6 = 250

·. .d2vo

dt2= 250vg

0 ≤ t ≤ 0.5−:

vg = 80 mV

d2vo

dt2= 20

Let g(t) =dvo

dt, then

dg

dt= 20 or dg = 20 dt

∫ g(t)

g(0)dx = 20

∫ t

0dy

g(t) − g(0) = 20t, g(0) =dvo

dt(0) = 0

g(t) =dvo

dt= 20t

Problems 8–41

dvo = 20t dt

∫ vo(t)

vo(0)dx = 20

∫ t

0x dx; vo(t) − vo(0) = 10t2, vo(0) = 0

vo(t) = 10t2 V, 0 ≤ t ≤ 0.5−

dvo1

dt= − 1

R1C1vg = −20vg = −1.6

dvo1 = −1.6 dt

∫ vo1(t)

vo1(0)dx = −1.6

∫ t

0dy

vo1(t) − vo1(0) = −1.6t, vo1(0) = 0

vo1(t) = −1.6t V, 0 ≤ t ≤ 0.5−

0.5+ ≤ t ≤ tsat:

d2vo

dt2= −10, let g(t) =

dvo

dt

dg(t)dt

= −10; dg(t) = −10 dt

∫ g(t)

g(0.5+)dx = −10

∫ t

0.5dy

g(t) − g(0.5+) = −10(t − 0.5) = −10t + 5

g(0.5+) =dvo(0.5+)

dt

Cdvo

dt(0.5+) =

0 − vo1(0.5+)400 × 103

vo1(0.5+) = vo(0.5−) = −1.6(0.5) = −0.80 V

·. . Cdvo(0.5+)

dt=

0.800.4 × 106 = 2 µA

dvo

dt(0.5+) =

2 × 10−6

0.2 × 10−6 = 10 V/s

·. . g(t) = −10t + 5 + 10 = −10t + 15 =dvo

dt

·. . dvo = −10t dt + 15 dt

∫ vo(t)

vo(0.5+)dx =

∫ t

0.5+−10y dy +

∫ t

0.5+15 dy


vo(t) − vo(0.5+) = −5y2∣∣∣∣t0.5

+ 15y∣∣∣∣t0.5

vo(t) = vo(0.5+) − 5t2 + 1.25 + 15t − 7.5

vo(0.5+) = vo(0.5−) = 2.5 V

·. . vo(t) = −5t2 + 15t − 3.75 V, 0.5+ ≤ t ≤ tsat

dvo1

dt= −20(−0.04) = 0.8, 0.5+ ≤ t ≤ tsat

dvo1 = 0.8 dt;∫ vo1(t)

vo1(0.5+)dx = 0.8

∫ t

0.5+dy

vo1(t) − vo1(0.5+) = 0.8t − 0.4; vo1(0.5+) = vo1(0.5−) = −0.8 V

·. . vo1(t) = 0.8t − 1.2 V, 0.5+ ≤ t ≤ tsat

Summary:

0 ≤ t ≤ 0.5−s : vo1 = −1.6t V, vo = 10t2 V

0.5+s ≤ t ≤ tsat : vo1 = 0.8t − 1.2 V, vo = −5t2 + 15t − 3.75 V

[b] −12.5 = −5t2sat + 15tsat − 3.75

·. . 5t2sat − 15tsat − 8.75 = 0

Solving, tsat = 3.5 sec

vo1(tsat) = 0.8(3.5) − 1.2 = 1.6 V

P 8.52 τ1 = (106)(0.5 × 10−6) = 0.50 s

1τ1

= 2; τ2 = (5 × 106)(0.2 × 10−6) = 1 s; ·. .1τ2

= 1

·. .d2vo

dt2+ 3

dvo

dt+ 2vo = 20

s2 + 3s + 2 = 0

(s + 1)(s + 2) = 0; s1 = −1, s2 = −2

vo = Vf + A′1e

−t + A′2e

−2t; Vf =202

= 10 V

vo = 10 + A′1e

−t + A′2e

−2t

Problems 8–43

vo(0) = 0 = 10 + A′1 + A′

2;dvo

dt(0) = 0 = −A′

1 − 2A′2

·. . A′1 = −20, A′

2 = 10 V

vo(t) = 10 − 20e−t + 10e−2t V, 0 ≤ t ≤ 0.5 s

dvo1

dt+ 2vo1 = −1.6; ·. . vo1 = −0.8 + 0.8e−2t V, 0 ≤ t ≤ 0.5 s

vo(0.5) = 10 − 20e−0.5 + 10e−1 = 1.55 V

vo1(0.5) = −0.8 + 0.8e−1 = −0.51 V

At t = 0.5 s

iC =0 + 0.51400 × 103 − 1.55 − 0

5 × 106 = 0.954 µA

Cdvo

dt= 0.954 µA;

dvo

dt=

0.9540.2

= 4.773 V/s

t ≥ 0.5 s

d2vo

dt2+ 3

dvo

dt+ 2vo = −10

vo(∞) = −5

·. . vo = −5 + A′1e

−(t−0.5) + A′2e

−2(t−0.5)

1.55 = −5 + A′1 + A′

2


dvo

dt(0.5) = 4.773 = −A′

1 − 2A′2

·. . A′1 + A′

2 = 6.55; −A′1 − 2A′

2 = 4.773

Solving,

A′1 = 17.87 V; A′

2 = −11.32 V

·. . vo = −5 + 17.87e−(t−0.5) − 11.32e−2(t−0.5) V, t ≥ 0.5 s

dvo1

dt+ 2vo1 = 0.8

·. . vo1 = 0.4 + (−0.51 − 0.4)e−2(t−0.5) = 0.4 − 0.91e−2(t−0.5) V, t ≥ 0.5 s

P 8.53 At t = 0 the voltage across each capacitor is zero. It follows that since theoperational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore therecannot be an instantaneous change in the current in the 1 µF capacitor. Since thecapacitor current equals C(dvo/dt), the derivative must be zero.

P 8.54 [a] From Example 8.13d2vo

dt2= 2

thereforedg(t)dt

= 2, g(t) =dvo

dt

g(t) − g(0) = 2t; g(t) = 2t + g(0); g(0) =dvo(0)

dt

iR =5

500× 10−3 = 1 µA = iC = −C

dvo(0)dt

dvo(0)dt

=−1 × 10−6

1 × 10−6 = −1 = g(0)

dvo

dt= 2t − 1

dvo = 2t dt − dt

vo − vo(0) = t2 − t; vo(0) = 8 V

vo = t2 − t + 8, 0 ≤ t ≤ tsat

Problems 8–45

[b] t2 − t + 8 = 9

t2 − t − 1 = 0

t = (1/2) ± (√

5/2) ∼= 1.62 s, tsat∼= 1.62 s

(Negative value has no physical significance.)

P 8.55 Part (1) — Example 8.14, with R1 and R2 removed:

[a] Ra = 100 kΩ; C1 = 0.1 µF; Rb = 25 kΩ; C2 = 1 µF

d2vo

dt2=( 1

RaC1

)( 1RbC2

)vg;

1RaC1

= 1001

RbC2= 40

vg = 250 × 10−3; therefored2vo

dt2= 1000

[b] Since vo(0) = 0 =dvo(0)

dt, our solution is vo = 500t2 V

The second op-amp will saturate when

vo = 6 V, or tsat =√

6/500 ∼= 0.1095 s

[c]dvo1

dt= − 1

RaC1vg = −25

[d] Since vo1(0) = 0, vo1 = −25t V

At t = 0.1095 s, vo1∼= −2.74 V

Therefore the second amplifier saturates before the first amplifier saturates.Our expressions are valid for 0 ≤ t ≤ 0.1095 s. Once the second op-ampsaturates, our linear model is no longer valid.

Part (2) — Example 8.14 with vo1(0) = −2 V and vo(0) = 4 V:

[a] Initial conditions will not change the differential equation; hence the equation isthe same as Example 8.14.

[b] vo = 5 + A′1e

−10t + A′2e

−20t (from Example 8.14)

vo(0) = 4 = 5 + A′1 + A′

2


4100

+ iC(0+) − 225

= 0

iC(0+) =4

100mA = C

dvo(0+)dt

dvo(0+)dt

=0.04 × 10−3

10−6 = 40 V/s

dvo

dt= −10A′

1e−10t − 20A′

2e−20t

dvo

dt(0+) = −10A′

1 − 20A′2 = 40

Therefore −A′1 − 2A′

2 = 4 and A′1 + A′

2 = −1Thus, A′

1 = 2 and A′2 = −3

vo = 5 + 2e−10t − 3e−20t V, t ≥ 0

[c] Same as Example 8.14:

dvo1

dt+ 20vo1 = −25

[d] From Example 8.14:

vo1(∞) = −1.25 V; v1(0) = −2 V (given)

Therefore

vo1 = −1.25 + (−2 + 1.25)e−20t = −1.25 − 0.75e−20t V, t ≥ 0

P 8.56 [a]

2Cdva

dt+

va − vg

R+

va

R= 0

(1) Thereforedva

dt+

va

RC=

vg

2RC;

0 − va

R+ C

d(0 − vb)dt

= 0

(2) Thereforedvb

dt+

va

RC= 0, va = −RC

dvb

dt

Problems 8–47

2vb

R+ C

dvb

dt+ C

d(vb − vo)dt

= 0

(3) Thereforedvb

dt+

vb

RC=

12

dvo

dt

From (2) we havedva

dt= −RC

d2vb

dt2and va = −RC

dvb

dt

When these are substituted into (1) we get

(4) − RCd2vb

dt2− dvb

dt=

vg

2RC

Now differentiate (3) to get

(5)d2vb

dt2+

1RC

dvb

dt=

12

d2vo

dt2

But from (4) we have

(6)d2vb

dt2+

1RC

dvb

dt= − vg

2R2C2

Now substitute (6) into (5)

d2vo

dt2= − vg

R2C2

[b] When R1C1 = R2C2 = RC :d2vo

dt2=

vg

R2C2

The two equations are the same except for a reversal in algebraic sign.

[c] Two integrations of the input signal with one operational amplifier.

P 8.57 [a] f(t) = inertial force + frictional force + spring force

= M [d2x/dt2] + D[dx/dt] + Kx

[b]d2x

dt2=

f

M−(

D

M

)(dx

dt

)−(

K

M

)x

Given vA =d2x

dt2, then

vB = − 1R1C1

∫ t

0

(d2x

dy2

)dy = − 1

R1C1

dx

dt

vC = − 1R2C2

∫ t

0vB dy =

1R1R2C1C2

x

vD = −R3

R4· vB =

R3

R4R1C1

dx

dt


vE =[R5 + R6

R6

]vC =

[R5 + R6

R6

]· 1R1R2C1C2

· x

vF =[−R8

R7

]f(t), vA = −(vD + vE + vF )

Therefored2x

dt2=[R8

R7

]f(t) −

[R3

R4R1C1

]dx

dt−[

R5 + R6

R6R1R2C1C2

]x

Therefore M =R7

R8, D =

R3R7

R8R4R1C1and K =

R7(R5 + R6)R8R6R1R2C1C2

Box Number Function

1 inverting and scaling


3 integrating and scaling

4 integrating and scaling


6 noninverting and scaling

P 8.58 [a] Given that the current response is underdamped we know i will be of the form

i = If + [B′1 cos ωdt + B′

2 sin ωdt]e−αt

where α =R

2L

and ωd =√

ω2o − α2 =

√1

LC− α2

The capacitor will force the final value of i to be zero, therefore If = 0.By hypothesis i(0+) = Vdc/R therefore B′

1 = Vdc/R.At t = 0+ the voltage across the primary winding is zero hence di(0+)/dt = 0.From our equation for i we have

di

dt= [(ωdB

′2 − αB′

1) cos ωdt − (ωdB′1 + αB′

2) sin ωdt]e−αt

Hencedi(0+)

dt= ωdB

′2 − αB′

1 = 0

Thus

B′2 =

α

ωd

B′1 =

αVdc

ωdR

It follows directly that

i =Vdc

R

[cos ωdt +

α

ωd

sin ωdt]e−αt

Problems 8–49

[b] Since ωdB′1 − αB′

1 = 0 it follows that

di

dt= −(ωdB

′1 + αB′

2)e−αt sin ωdt

But αB′2 =

α2Vdc

ωdRand ωdB

′1 =

ωdVdc

R

Therefore

ωdB′1 + αB′

2 =ωdVdc

R+

α2Vdc

ωdR=

Vdc

R

[ω2

d + α2

ωd

]

But ω2d + α2 = ω2

o =1

LC

Hence

ωdB′1 + αB′

2 =Vdc

ωdRLC

Now since v1 = Ldi

dtwe get

v1 = −LVdc

ωdRLCe−αt sin ωdt = − Vdc

ωdRCe−αt sin ωdt

[c] vc = Vdc − iR − Ldi

dt

iR = Vdc

(cos ωdt +

α

ωd

sin ωdt)

e−αt

vc = Vdc − Vdc

(cos ωdt +

α

ωd

sin ωdt)

e−αt +Vdc


= Vdc − Vdce−αt cos ωdt +

(Vdc

ωdRC− αVdc

ωd

)e−αt sin ωdt

= Vdc

[1 − e−αt cos ωdt +

1ωd

( 1RC

− α)

e−αt sin ωdt]

= Vdc [1 − e−αt cos ωdt + Ke−αt sin ωdt]

P 8.59 vsp = Vdc

[1 − a


]

dvsp

dt=

−aVdc

ωdRC

d

dt[e−αt sin ωdt]

=−aVdc

ωdRC[−αe−αt sin ωdt + ωd cos ωdte

−αt]

=aVdce

−αt

ωdRC[α sin ωdt − ωd cos ωdt]


dvsp

dt= 0 when α sin ωdt = ωd cos ωdt

or tan ωdt =ωd

α; ωdt = tan−1

(ωd

α

)

·. . tmax =1ωd

tan−1(

ωd

α

)

Note that because tan θ is periodic, i.e., tan θ = tan(θ ± nπ), where n is an integer,there are an infinite number of solutions for t where dvsp/dt = 0, that is

t =tan−1(ωd/α) ± nπ

ωd

Because of e−αt in the expression for vsp and knowing t ≥ 0 we know vsp will bemaximum when t has its smallest positive value. Hence

tmax =tan−1(ωd/α)

ωd

.

P 8.60 [a] vc = Vdc[1 − e−αt cos ωdt + Ke−αt sin ωdt]

dvc

dt= Vdc

d

dt[1 + e−αt(K sin ωdt − cos ωdt)]

= Vdc(−αe−αt)(K sin ωdt − cos ωdt)+

e−αt[ωdK cos ωdt + ωd sin ωdt]= Vdce

−αt[(ωd − αK) sin ωdt + (α + ωdK) cos ωdt]

dvc

dt= 0 when (ωd − αK) sin ωdt = −(α + ωdK) cos ωdt

or tan ωdt =[α + ωdK

αK − ωd

]

·. . ωdt ± nπ = tan−1[α + ωdK

αK − ωd

]

tc =1ωd

tan−1

(α + ωdK

αK − ωd

)± nπ

α =R

2L=

4 × 103

6= 666.67 rad/s

ωd =

√109

1.2− (666.67)2 = 28,859.81 rad/s

Problems 8–51

K =1ωd

( 1RC

− α)

= 21.63

tc =1ωd

tan−1(−43.29) + nπ

=

1ωd

−1.55 + nπ

The smallest positive value of t occurs when n = 1, therefore

tc max = 55.23 µs

[b] vc(tc max) = 12[1 − e−αtc max cos ωdtc max + Ke−αtc max sin ωdtc max]

= 262.42 V

[c] From the text example the voltage across the spark plug reaches its maximumvalue in 53.63 µs. If the spark plug does not fire the capacitor voltage peaks in55.23 µs. When vsp is maximum the voltage across the capacitor is 262.15 V.If the spark plug does not fire the capacitor voltage reaches 262.42 V.

P 8.61 [a] w =12L[i(0+)]2 =

12(5)(16) × 10−3 = 40 mJ

[b] α =R

2L=

3 × 103

10= 300 rad/s

ωd =

√109

1.25− (300)2 = 28,282.68 rad/s

1RC

=106

0.75=

4 × 106

3

tmax =1ωd

tan−1(

ωd

α

)= 55.16 µs

vsp (tmax) = 12 − 12(50)(4 × 106)3(28,282.68)

e−αtmax sin ωdtmax = −27,808.04 V

[c] vc (tmax) = 12[1 − e−αtmax cos ωdtmax + Ke−αtmax sin ωdtmax]

K =1ωd

[ 1RC

− α]

= 47.13

vc (tmax) = 568.15 V

9Sinusoidal Steady State Analysis

Assessment Problems

AP 9.1 [a] V = 170/−40 V

[b] 10 sin(1000t + 20) = 10 cos(1000t − 70)

·. . I = 10/−70 A

[c] I = 5/36.87 + 10/−53.13

= 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57 A

[d] sin(20,000πt + 30) = cos(20,000πt − 60)Thus,

V = 300/45 − 100/−60 = 212.13 + j212.13 − (50 − j86.60)

= 162.13 + j298.73 = 339.90/61.51 mV

AP 9.2 [a] v = 18.6 cos(ωt − 54) V

[b] I = 20/45 − 50/− 30 = 14.14 + j14.14 − 43.3 + j25

= −29.16 + j39.14 = 48.81/126.68

Therefore i = 48.81 cos(ωt + 126.68) mA

[c] V = 20 + j80 − 30/15 = 20 + j80 − 28.98 − j7.76

= −8.98 + j72.24 = 72.79/97.08

v = 72.79 cos(ωt + 97.08) V

AP 9.3 [a] ωL = (104)(20 × 10−3) = 200 Ω

[b] ZL = jωL = j200 Ω

9–1

9–2 CHAPTER 9. Sinusoidal Steady State Analysis

[c] VL = IZL = (10/30)(200/90) × 10−3 = 2/120 V

[d] vL = 2 cos(10,000t + 120) V

AP 9.4 [a] XC =−1ωC

=−1

4000(5 × 10−6)= −50 Ω

[b] ZC = jXC = −j50 Ω

[c] I =VZC

=30/25

50/−90 = 0.6/115 A

[d] i = 0.6 cos(4000t + 115) A

AP 9.5 I1 = 100/25 = 90.63 + j42.26

I2 = 100/145 = −81.92 + j57.36

I3 = 100/−95 = −8.72 − j99.62

I4 = −(I1 + I2 + I3) = (0 + j0) A, therefore i4 = 0 A

AP 9.6 [a] I =125/−60

|Z|/θz

=125|Z| /(−60 − θZ)

But −60 − θZ = −105 ·. . θZ = 45

Z = 90 + j160 + jXC

·. . XC = −70 Ω; XC = − 1ωC

= −70

·. . C =1

(70)(5000)= 2.86 µF

[b] I =Vs

Z=

125/−60

(90 + j90)= 0.982/−105A; ·. . |I| = 0.982 A

AP 9.7 [a]

ω = 2000 rad/s

ωL = 10 Ω,−1ωC

= −20 Ω

Zxy = 20‖j10 + 5 + j20 =20(j10)

(20 + j10)+ 5 − j20

= 4 + j8 + 5 − j20 = (9 − j12) Ω

Problems 9–3

[b] ωL = 40 Ω,−1ωC

= −5 Ω

Zxy = 5 − j5 + 20‖j40 = 5 − j5 +[(20)(j40)20 + j40

]

= 5 − j5 + 16 + j8 = (21 + j3) Ω

[c] Zxy =[

20(jωL)20 + jωL

]+(

5 − j106

25ω

)

=20ω2L2

400 + ω2L2 +j400ωL

400 + ω2L2 + 5 − j106

25ωThe impedance will be purely resistive when the j terms cancel, i.e.,

400ωL

400 + ω2L2 =106

25ωSolving for ω yields ω = 4000 rad/s.

[d] Zxy =20ω2L2

400 + ω2L2 + 5 = 10 + 5 = 15 Ω

AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 Ω in Assessment Problem 9.7.Thus,

V = 150/0, Is =V

Zxy=

150/0

15= 10/0 A


IL =20

20 + j20(10) = 5 − j5 = 7.07/−45 A

iL = 7.07 cos(4000t − 45) A, Im = 7.07 A

AP 9.9 After replacing the delta made up of the 50 Ω, 40 Ω, and 10 Ω resistors with itsequivalent wye, the circuit becomes


The circuit is further simplified by combining the parallel branches,

(20 + j40)‖(5 − j15) = (12 − j16) Ω

Therefore I =136/0

14 + 12 − j16 + 4= 4/28.07 A

AP 9.10 V1 = 240/53.13 = 144 + j192 V

V2 = 96/−90 = −j96 V

jωL = j(4000)(15 × 10−3) = j60 Ω

1jωC

= −j6 × 106

(4000)(25)= −j60 Ω

Perform source transformations:

V1

j60=

144 + j192j60

= 3.2 − j2.4 A

V2

20= −j

9620

= −j4.8 A

Combine the parallel impedances:

Y =1

j60+

130

+1

−j60+

120

=j5j60

=112

Z =1Y

= 12 Ω

Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87 V

vo = 48 cos(4000t + 36.87) V

Problems 9–5

AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the 20 Ωresistor and VTh = node voltage across the capacitor. Writing the node voltageequations gives us

V1

20− 2/45 +

V1 − 10Ix

j10= 0 and VTh =

−j1010 − j10

(10Ix)

We also have

Ix =V1

20

Solving these equations for VTh gives VTh = 10/45V. To find the Théveninimpedance, we remove the independent current source and apply a test voltagesource at the terminals a, b. Thus

It follows from the circuit that

10Ix = (20 + j10)Ix

Therefore

Ix = 0 and IT =VT

−j10+

VT

10

ZTh =VT

IT

, therefore ZTh = (5 − j5) Ω

AP 9.12 The phasor domain circuit is as shown in the following diagram:



−10 +V5

+V

−j(20/9)+

Vj5

+V − 100/−90

20= 0

Therefore V = 10 − j30 = 31.62/−71.57

Therefore v = 31.62 cos(50,000t − 71.57) V

AP 9.13 Let Ia, Ib, and Ic be the three clockwise mesh currents going from left to right.Summing the voltages around meshes a and b gives

33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib)

and

0 = (3 − j5)(Ib − Ia) + 2(Ib − Ic).

But

Vx = −j5(Ia − Ib),

therefore

Ic = −0.75[−j5(Ia − Ib)].

Solving for I = Ia = 29 + j2 = 29.07/3.95 A.

AP 9.14 [a] M = 0.4√

0.0625 = 0.1 H, ωM = 80 Ω

Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω

Therefore |Z22| = 500 Ω, Z∗22 = (400 − j300) Ω

Zr =( 80

500

)2

(400 − j300) = (10.24 − j7.68) Ω

[b] I1 =245.20

184 + 100 + j400 + Zr

= 0.50/− 53.13 A

i1 = 0.5 cos(800t − 53.13) A

[c] I2 =(

jωM

Z22

)I1 =

j80500/36.87 (0.5/− 53.13) = 0.08/0 A

i2 = 80 cos 800t mA

Problems 9–7

AP 9.15 I1 =Vs

Z1 + Z2/a2 =25 × 103/0

1500 + j6000 + (25)2(4 − j14.4)

= 4 + j3 = 5/36.87 A

V1 = Vs − Z1I1 = 25,000/0 − (4 + j3)(1500 + j6000)

= 37,000 − j28,500

V2 = − 125

V1 = −1480 + j1140 = 1868.15/142.39 V

I2 =V2

Z2=

1868.15/142.39

4 − j14.4= 125/− 143.13 A

Also, I2 = −25I1


Problems

P 9.1 [a] ω = 2πf = 3769.91 rad/s, f =ω

2π= 600 Hz

[b] T = 1/f = 1.67 ms

[c] Vm = 10 V

[d] v(0) = 10 cos(−53.13) = 6 V

[e] φ = −53.13; φ =−53.13(2π)

360 = −0.9273 rad

[f] V = 0 when 3769.91t − 53.13 = 90. Now resolve the units:

(3769.91 rad/s)t =143.13

(180/π)= 2.498 rad, t = 662.64 µs

[g] (dv/dt) = (−10)3769.91 sin(3769.91t − 53.13)

(dv/dt) = 0 when 3769.91t − 53.13 = 0

or 3769.91t =53.13

57.3/rad= 0.9273 rad

Therefore t = 245.97 µs

P 9.2 Vrms =

√1T

∫ T/2

0V 2

m sin2 2πT

t dt

∫ T/2

0V 2

m sin2(2π

T

)t dt =

V 2m

2

∫ T/2

0

(1 − cos

4πT

t)

dt =V 2

mT

4

Therefore Vrms =

√1T

V 2mT

4=

Vm

2

P 9.3 [a] 40 V

[b] 2πf = 100π; f = 50Hz

[c] ω = 100π = 314.159 rad/s

[d] θ(rad) =2π

360 (60) =π

3= 1.05 rad

[e] θ = 60

[f] T =1f

=150

= 20 ms

[g] v = −40 when

100πt +π

3= π; ·. . t = 6.67 ms

Problems 9–9

[h] v = 40 cos[100π

(t − 0.01

3

)+

π

3

]

= 40 cos[100πt − (π/3) + (π/3)]

= 40 cos 100πt V

[i] 100π(t − to) + (π/3) = 100πt − (π/2)

·. . 100πto =5π6

; to = 8.33 ms

[j] 100π(t + to) + (π/3) = 100πt + 2π

·. . 100πto =5π3

; to = 16.67 ms

16.67 ms to the left

P 9.4

[a] Left as φ becomes more positive

[b] Left

P 9.5 [a] By hypothesis

v = 80 cos(ωt + θ)

dv

dt= −80ω sin(ωt + θ)

·. . 80ω = 80,000; ω = 1000 rad/s

[b] f =ω

2π= 159.155 Hz; T =

1f

= 6.28 ms

−2π/36.28

= −0.3333, ·. . θ = −90 − (−0.3333)(360) = 30

·. . v = 80 cos(1000t + 30) V


P 9.6 [a]T

2= 8 + 2 = 10 ms; T = 20 ms

f =1T

=1

20 × 10−3 = 50Hz

[b] v = Vm sin(ωt + θ)

ω = 2πf = 100π rad/s

100π(−2 × 10−3) + θ = 0; ·. . θ =π

5rad = 36

v = Vm sin[100πt + 36]

80.9 = Vm sin 36; Vm = 137.64 V

v = 137.64 sin[100πt + 36] = 137.64 cos[100πt − 54] V

P 9.7 u =∫ to+T

toV 2

m cos2(ωt + φ) dt

= V 2m

∫ to+T

to

12

+12

cos(2ωt + 2φ) dt

=V 2

m

2

∫ to+Tto

dt +∫ to+T

tocos(2ωt + 2φ) dt

=V 2

m

2

T +

12ω

[sin(2ωt + 2φ) |to+T

to

]

=V 2

m

2

T +

12ω

[sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)]

= V 2m

(T

2

)+

12ω

(0) = V 2m

(T

2

)

P 9.8 Vm =√

2Vrms =√

2(120) = 169.71 V

P 9.9 [a] The numerical values of the terms in Eq. 9.8 are

Vm = 20, R/L = 1066.67, ωL = 60√

R2 + ω2L2 = 100

φ = 25, θ = tan−1 60/80, θ = 36.87

Substitute these values into Equation 9.9:

i =[−195.72e−1066.67t + 200 cos(800t − 11.87)

]mA, t ≥ 0

[b] Transient component = −195.72e−1066.67t mASteady-state component = 200 cos(800t − 11.87) mA

[c] By direct substitution into Eq 9.9 in part (a), i(1.875 ms) = 28.39 mA

[d] 200 mA, 800 rad/s, −11.87

Problems 9–11

[e] The current lags the voltage by 36.87.

P 9.10 [a] From Eq. 9.9 we have

Ldi

dt=

VmR cos(φ − θ)√R2 + ω2L2

e−(R/L)t − ωLVm sin(ωt + φ − θ)√R2 + ω2L2

Ri =−VmR cos(φ − θ)e−(R/L)t

√R2 + ω2L2

+VmR cos(ωt + φ − θ)√

R2 + ω2L2

Ldi

dt+ Ri = Vm

[R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√

R2 + ω2L2

]

ButR√

R2 + ω2L2= cos θ and

ωL√R2 + ω2L2

= sin θ

Therefore the right-hand side reduces to

Vm cos(ωt + φ)

At t = 0, Eq. 9.9 reduces to

i(0) =−Vm cos(φ − θ)√

R2 − ω2L2+

Vm cos(φ − θ)√R2 + ω2L2

= 0

[b] iss =Vm√

R2 + ω2L2cos(ωt + φ − θ)

Therefore

Ldissdt

=−ωLVm√R2 + ω2L2

sin(ωt + φ − θ)

and

Riss =VmR√

R2 + ω2L2cos(ωt + φ − θ)

Ldissdt

+ Riss = Vm

[R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√

R2 + ω2L2

]

= Vm cos(ωt + φ)

P 9.11 [a] Y = 50/60 + 100/− 30 = 111.8/− 3.43

y = 111.8 cos(500t − 3.43)

[b] Y = 200/50 − 100/60 = 102.99/40.29

y = 102.99 cos(377t + 40.29)

[c] Y = 80/30 − 100/− 225 + 50/− 90 = 161.59/− 29.96

y = 161.59 cos(100t − 29.96)


[d] Y = 250/0 + 250/120 + 250/− 120 = 0

y = 0

P 9.12 [a] 1000Hz

[b] θv = 0

[c] I =200/0

jωL=

200ωL

/− 90 = 25/− 90; θi = −90

[d]200ωL

= 25; ωL =20025

= 8 Ω

[e] L =8

2π(1000)= 1.27 mH

[f] ZL = jωL = j8 Ω

P 9.13 [a] ω = 2πf = 314,159.27 rad/s

[b] I =VZC

=10 × 10−3/0

1/jωC= jωC(10 × 10−3)/0 = 10 × 10−3ωC/90

·. . θi = 90

[c] 628.32 × 10−6 = 10 × 10−3 ωC

1ωC

=10 × 10−3

628.32 × 10−6 = 15.92 Ω, ·. . XC = −15.92 Ω

[d] C =1

15.92(ω)=

1(15.92)(100π × 103)

C = 0.2 µF

[e] Zc = j(−1

ωC

)= −j15.92 Ω

P 9.14 [a] jωL = j(2 × 104)(300 × 10−6) = j6 Ω

1jωC

= −j1

(2 × 104)(5 × 10−6)= −j10 Ω; Ig = 922/30 A

Problems 9–13

[b] Vo = 922/30Ze

Ze =1Ye

; Ye =110

+ j110

+1

8 + j6

Ye = 0.18 + j0.04 S

Ze =1

0.18 + j0.04= 5.42/− 12.53 Ω

Vo = (922/30 )(5.42/− 12.53 ) = 5000.25/17.47 V

[c] vo = 5000.25 cos(2 × 104t + 17.47) V

P 9.15 [a] ZL = j(8000)(5 × 10−3) = j40 Ω

ZC =−j

(8000)(1.25 × 10−6)= −j100 Ω

[b] I =600/20

40 + j40 − j100= 8.32/76.31 A

[c] i = 8.32 cos(8000t + 76.31) A

P 9.16 Z = 4 + j(50)(0.24) − j1

(50)(0.0025)= 4 + j4 = 5.66/45 Ω

Io =VZ

=0.1/− 90

5.66/45 = 17.68/− 135 mA

io(t) = 17.68 cos(50t − 135) mA

P 9.17 [a] Y =1

3 + j4+

116 − j12

+1

−j4

= 0.12 − j0.16 + 0.04 + j0.03 + j0.25

= 0.16 + j0.12 = 200/36.87 mS

[b] G = 160 mS

[c] B = 120 mS


[d] I = 8/0 A, V =IY

=8

0.2/36.87 = 40/−36.87 V

IC =VZC

=40/−36.87

4/−90 = 10/53.13 A

iC = 10 cos(ωt + 53.13) A, Im = 10 A

P 9.18 ZL = j(2000)(60 × 10−3) = j120 Ω

ZC =−j

(2000)(12.5 × 10−6)= −j40 Ω

Construct the phasor domain equivalent circuit:

Using current division:

I =(120 − j40)

120 − j40 + 40 + j120(0.5) = 0.25 − j0.25 A

Vo = j120I = 30 + j30 = 42.43/45 V

vo = 42.43 cos(2000t + 45) V

P 9.19 [a] Vg = 300/78; Ig = 6/33

·. . Z =Vg

Ig

=300/78

6/33 = 50/45 Ω

[b] ig lags vg by 45:

2πf = 5000π; f = 2500 Hz; T = 1/f = 400 µs

·. . ig lags vg by45

360 (400 µs) = 50 µs

Problems 9–15

P 9.201

jωC=

1(1 × 10−6)(50 × 103)

= −j20 Ω

jωL = j50 × 103(1.2 × 10−3) = j60 Ω

Vg = 40/0 V

Ze = −j20 + 30‖j60 = 24 − j8 Ω

Ig =40/0

24 − j8= 1.5 + j0.5 mA

Vo = (30‖j60)Ig =30(j60)30 + j60

(1.5 + j0.5) = 30 + j30 = 42.43/45 V

vo = 42.43 cos(50,000t + 45) V

P 9.21 [a] Z1 = R1 − j1

ωC1

Z2 =R2/jωC2

R2 + (1/jωC2)=

R2

1 + jωR2C2=

R2 − jωR22C2

1 + ω2R22C

22

Z1 = Z2 when R1 =R2

1 + ω2R22C

22

and

1ωC1

=ωR2

2C2

1 + ω2R22C

22

or C1 =1 + ω2R2

2C22

ω2R22C2

[b] R1 =1000

1 + (40 × 103)2(1000)2(50 × 10−9)2 = 200 Ω

C1 =1 + (40 × 103)2(1000)2(50 × 10−9)2

(40 × 103)2(1000)2(50 × 10−9)= 62.5 nF


P 9.22 [a] Y2 =1R2

+ jωC2

Y1 =1

R1 + (1/jωC1)=

jωC1

1 + jωR1C1=

ω2R1C21 + jωC1

1 + ω2R21C

21

Therefore Y1 = Y2 when

R2 =1 + ω2R2

1C21

ω2R1C21

and C2 =C1

1 + ω2R21C

21

[b] R2 =1 + (50 × 103)2(1000)2(40 × 10−9)2

(50 × 103)2(1000)(40 × 10−9)2 = 1250 Ω

C2 =40 × 10−9

1 + (50 × 103)2(1000)2(40 × 10−9)2 = 8 nF

P 9.23 [a] Z1 = R1 + jωL1

Z2 =R2(jωL2)R2 + jωL2

=ω2L2

2R2 + jωL2R22

R2 + ω2L22

Z1 = Z2 when R1 =ω2L2

2R2

R22 + ω2L2

2and L1 =

R22L2

R22 + ω2L2

2

[b] R1 =(4000)2(1.25)2(5000)50002 + 40002(1.25)2 = 2500 Ω

L1 =(5000)2(1.25)

50002 + 40002(1.25)2 = 625 mH

P 9.24 [a] Y2 =1R2

− j

ωL2

Y1 =1

R1 + jωL1=

R1 − jωL1

R21 + ω2L2

1

Therefore Y2 = Y1 when

R2 =R2

1 + ω2L21

R1and L2 =

R21 + ω2L2

1

ω2L1

[b] R2 =80002 + 10002(4)2

8000= 10 kΩ

L2 =80002 + 10002(4)2

10002(4)= 20 H

Problems 9–17

P 9.25 Vg = 500/30 V; Ig = 0.1/83.13 mA

Z =Vg

Ig

= 5000/− 53.13 Ω = 3000 − j4000 Ω

z = 3000 + j

(ω − 32 × 103

ω

)

ω − 32 × 103

ω= −4000

ω2 + 4000ω − 32 × 103 = 0

ω = 7.984 rad/s

P 9.26 [a] Zeq =50,000

3+

−j20 × 106

ω‖(1200 + j0.2ω)

=50,000

3+

−j20 × 106

ω

(1200 + j0.2ω)1200 + j[0.2ω − 20×106

ω]

=50,000

3+

−j20×106

ω(1200 + j0.2ω)

[1200 − j

(0.2ω − 20×106

ω

)]12002 +

(0.2ω − 20×106

ω

)2

Im(Zeq) = −20 × 106

ω(1200)2 − 20 × 106

ω

[0.2ω

(0.2ω − 20 × 106

ω

)]= 0

−20 × 106(1200)2 − 20 × 106

[0.2ω

(0.2ω − 20 × 106

ω

)]= 0

−(1200)2 = 0.2ω(

0.2ω − 20 × 106

ω

)

0.22ω2 − 0.2(20 × 106) + 12002 = 0

ω2 = 64 × 106 ·. . ω = 8000 rad/s

·. . f = 1273.24 Hz

[b] Zeq =50,000

3+ −j2500‖(1200 + j1600)

=50,000

3+

(−j2500)(1200 + j1600)1200 − j900

= 20,000 Ω

Ig =30/0

20,000= 1.5/0 mA

ig(t) = 1.5 cos 8000t mA


P 9.27 [a] Find the equivalent impedance seen by the source, as a function of L, and set theimaginary part of the equivalent impedance to 0, solving for L:

ZC =−j

(500)(2 × 10−6)= −j1000 Ω

Zeq = −j1000 + j500L‖2000 = −j1000 +2000(j500L)2000 + j500L

= −j1000 +2000(j500L)(2000 − j500L)

20002 + (500L)2

Im(Zeq) = −1000 +20002(500L)

20002 + (500L)2 = 0

·. .20002(500L)

20002 + (500L)2 = 1000

·. . 5002L2 − 1220002L + 20002 = 0

Solving the quadratic equation, L = 4 H

[b] Ig =100/0

−j1000 + j2000‖2000=

100/0

1000= 0.1/0 A

ig(t) = 0.1 cos 500t A

P 9.28 [a] jωL + R‖(−j/ωC) = jωL +−jR/ωC

R − j/ωC

= jωL +−jR

ωCR − j1

= jωL +−jR(ωCR + j1)

ω2C2R2 + 1

Im(Zab) = ωL − ωCR2

ω2C2R2 + 1= 0

·. . L =CR2

ω2C2R2 + 1

·. . ω2C2R2 + 1 =CR2

L

·. . ω2 =(CR2/L) − 1

C2R2 =(25×10−9)(100)2

160×10−6 − 1(25 × 10−9)2(100)2 = 900 × 108

ω = 300 krad/s

Problems 9–19

[b] Zab(300 × 103) = j48 +(100)(−j133.33)100 − j133.33

= 64 Ω

P 9.29 jωL = j100 × 103(0.6 × 10−3) = j60 Ω

1jωC

=−j

(100 × 103)(0.4 × 10−6)= −j25 Ω

VT = −j25IT + 5I∆ − 30I∆

I∆ =−j60

30 + j60IT

VT = −j25IT + 25j60

30 + j60IT

VT

IT

= Zab = 20 − j15 = 25/− 36.87 Ω

P 9.30 [a] Z1 = 400 − j106

500(2.5)= 400 − j800 Ω

Z2 = 2000‖j500L =j106L

2000 + j500L

ZT = Z1 + Z2 = 400 − j800 +j106L

2000 + j500L

= 400 +500 × 106L2

20002 + 5002L2 − j800 + j2 × 109L

20002 + 5002L2

ZT is resistive when

2 × 109L

20002 + 5002L2 = 800 or 5002L2 − 25 × 105L + 20002 = 0

Solving, L1 = 8 H and L2 = 2 H.


[b] When L = 8 H:

ZT = 400 +500 × 106(8)2

20002 + 5002(8)2 = 2000 Ω

Ig =200/0

2000= 100/0 mA

ig = 100 cos 500t mA

When L = 2 H:

ZT = 400 +500 × 106(2)2

20002 + 500(2)2 = 800 Ω

Ig =200/0

800= 250/0 mA

ig = 250 cos 500t mA

P 9.31 [a] Y1 =11

2500 × 103 = 4.4 × 10−6 S

Y2 =1

14,000 + j5ω

=14,000

196 × 106 + 25ω2 − j5ω

196 × 106 + 25ω2

Y3 = jω2 × 10−9

YT = Y1 + Y2 + Y3

For ig and vo to be in phase the j component of YT must be zero; thus,

ω2 × 10−9 =5ω

196 × 106 + 25ω2

or

25ω2 + 196 × 106 =5

2 × 10−9

·. . 25ω2 = 2304 × 106 ·. . ω = 9600 rad/s

[b] YT = 4.4 × 10−6 +14,000

196 × 106 + 25(9600)2 = 10 × 10−6 S

·. . ZT = 100 kΩ

Vo = (0.25 × 10−3/0)(100 × 103) = 25/0 V

vo = 25 cos 9600t V

Problems 9–21

P 9.32 [a] Zg = 500 − j106

ω+

103(j0.5ω)103 + j0.5ω

= 500 − j106

ω+

500jω(1000 − j0.5ω)106 + 0.25ω2

= 500 − j106

ω+

250ω2

106 + 0.25ω2 + j5 × 105ω

106 + 0.25ω2

·. . If Zg is purely real,106

ω=

5 × 105ω

106 + 0.25ω2

2(106 + 0.25ω2) = ω2 ·. . 4 × 106 = ω2

·. . ω = 2000 rad/s

[b] When ω = 2000 rad/s

Zg = 500 − j500 + (j1000‖1000) = 1000 Ω

·. . Ig =20/0

1000= 20/0 mA

Vo = Vg − IgZ1

Z1 = 500 − j500 Ω

Vo = 20/0 − (0.02/0)(500 − j500) = 10 + j10 = 14.14/45 V

vo = 14.14 cos(2000t + 45) V

P 9.33 Zab = 1 − j8 + (2 + j4)‖(10 − j20) + (40‖j20)

= 1 − j8 + 3 + j4 + 8 + j16 = 12 + j12 Ω = 16.971/45 Ω

P 9.34 First find the admittance of the parallel branches

Yp =1

2 − j6+

112 + j4

+12

+1

j0.5= 0.625 − j1.875 S

Zp =1Yp

=1

0.625 − j1.875= 0.16 + j0.48 Ω

Zab = −j4.48 + 0.16 + j0.48 + 2.84 = 3 − j4 Ω

Yab =1

Zab=

13 − j4

= 120 + j160 mS

= 200/53.13 mS


P 9.35 Simplify the top triangle using series and parallel combinations:

(1 + j1)‖(1 − j1) = 1 Ω

Convert the lower left delta to a wye:

Z1 =(j1)(1)

1 + j1 − j1= j1 Ω

Z2 =(−j1)(1)

1 + j1 − j1= −j1 Ω

Z3 =(j1)(−j1)1 + j1 − j1

= 1 Ω

Convert the lower right delta to a wye:

Z4 =(−j1)(1)

1 + j1 − j1= −j1 Ω

Z5 =(−j1)(j1)1 + j1 − j1

= 1 Ω

Z6 =(j1)(1)

1 + j1 − j1= j1 Ω


Simplify the middle portion of the circuit by making series and parallelcombinations:

(1 + j1 − j1)‖(1 + 1) = 1‖2 = 2/3 Ω

Zab = −j1 + 2/3 + j1 = 2/3 Ω

Problems 9–23

P 9.36 Vo = VgZo

ZT

=500 − j1000

300 + j1600 + 500 − j1000(100/0) = 111.8/− 100.3 V

vo = 111.8 cos(8000t − 100.3) V

P 9.371

jωC= −j400 Ω

jωL = j1200 Ω

Let Z1 = 200 − j400 Ω; Z2 = 600 + j1200 Ω

Ig = 400/0 mA

Io =Z2

Z1 + Z2Ig =

600 + j1200800 + j800

(0.4/0)

= 450 + j150 mA = 474.34/18.43 mA

io = 474.34 cos(20,000t + 18.43) mA

P 9.38

V1 = j5(−j2) = 10 V

−25 + 10 + (4 − j3)I1 = 0 ·. . I1 =15

4 − j3= 2.4 + j1.8 A

I2 = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A


VZ = −j5I2 + (4 − j3)I1 = −j5(2.4 − j3.2) + (4 − j3)(2.4 + j1.8) = −1 − j12 V

−25 + (1 + j3)I3 + (−1 − j12) = 0 ·. . I3 = 6.2 − j6.6 A

IZ = I3 − I2 = (6.2 − j6.6) − (2.4 − j3.2) = 3.8 − j3.4 A

Z =VZ

IZ

=−1 − j123.8 − j3.4

= 1.42 − j1.88 Ω

P 9.39 Is = 3/0 mA

1jωC

= −j0.4 Ω

jωL = j0.4 Ω

After source transformation we have

Vo =−j0.4‖j0.4‖5

28 + −j0.4‖j0.4‖5(66 × 10−3) = 10 mV

vo = 10 cos 200t mV

P 9.40 [a]

Va = (50 + j150)(2/0) = 100 + j300 V

Ib =100 + j300120 − j40

= j2.5 A = 2.5/90 A

Ic = 2/0 + j2.5 + 6 + j3.5 = 8 + j6 A = 10/36.87 A

Vg = 5Ic + Va = 5(8 + j6) + 100 + j300 = 140 + j330 V = 358.47/67.01 V

Problems 9–25

[b] ib = 2.5 cos(800t + 90) A

ic = 10 cos(800t + 36.87) A

vg = 358.47 cos(800t + 67.01) V

P 9.41 [a] jωL = j(1000)(100) × 10−3 = j100 Ω

1jωC

= −j106

(1000)(10)= −j100 Ω


Vab =(100 + j100)‖(−j100)

j100 + (100 + j100)‖(−j100)(247.49/45) = 350/0

VTh = Vab = 350/0 V

[b] Remove the voltage source and combine impedances in parallel to findZTh = Zab:

Yab =1

j100+

1100 + j100

+1

−j100= 5 − j5 mS

ZTh = Zab =1

Yab= 100 + j100 Ω

[c]


P 9.42 Using voltage division:

VTh =36

36 + j60 − j48(240) = 216 − j72 = 227.68/− 18.43 V

Remove the source and combine impedances in series and in parallel:

ZTh = 36‖(j60 − j48) = 3.6 + j10.8 Ω

P 9.43 Open circuit voltage:

V2

10+ 88Iφ +

V2 − 15V2

−j50= 0

Iφ =5 − (V2/5)

200

Solving,

V2 = −66 + j88 = 110/126.87 V = VTh

Find the Thévenin equivalent impedance using a test source:

IT =VT

10+ 88Iφ +

0.8Vt

−j50

Iφ =−VT /5

200

IT = VT

(110

− 881/5200

+0.8

−j50

)

Problems 9–27

·. .VT

IT

= 30 − j40 = ZTh

IN =VTh

ZTh=

−66 + j8830 − j40

= −2.2 + j0 A = 2.2/180 A

The Norton equivalent circuit:

P 9.44 Short circuit current

Iβ =−6Iβ

2

2Iβ = −6Iβ; ·. . Iβ = 0

I1 = 0; ·. . Isc = 10/−45 A = IN

The Norton impedance is the same as the Thévenin impedance. Find it using a testsource

VT = 6Iβ + 2Iβ = 8Iβ, Iβ =j1

2 + j1IT


ZTh =VT

IT

=8Iβ

[(2 + j1)/j1]Iβ

=j8

2 + j1= 1.6 + j3.2 Ω

P 9.45 Using current division:

IN = Isc =50

80 + j60(4) = 1.6 − j1.2 = 2/− 36.87 A

ZN = −j100‖(80 + j60) = 100 − j50 Ω

The Norton equivalent circuit:

P 9.46 ω = 2π(200/π) = 400 rad/s

Zc =−j

400(10−6)= −j2500 Ω

VT = (10,000 − j2500)IT + 100(200)IT

ZTh =VT

IT

= 30 − j2.5 kΩ

P 9.47

IN =5 − j15

ZN+ (1 − j3) mA, ZN in kΩ

Problems 9–29

IN =−18 − j13.5

ZN

+ 4.5 − j6 mA, ZN in kΩ

5 − j15ZN

+ 1 − j3 =−18 − j13.5

ZN

+ (4.5 − j6)

23 − j1.5ZN

= 3.5 − j3 ·. . ZN = 4 + j3 kΩ

IN =5 − j154 + j3

+ 1 − j3 = −j6 mA = 6/− 90 mA


V1 − 25020 + j10

− 0.03Vo +V1

50 − j100= 0

·. . Vo =−j100

50 − j100V1


V1

20 + j10+

j3V1

50 − j100+

V1

50 − j100=

25020 + j10

V1 = 500 − j250 V; Vo = 300 − j400 V = VTh = 500/− 53.13 V


Isc =250/0

70 + j10= 3.5 − j0.5 A

ZTh =VTh

Isc=

300 − j4003.5 − j0.5

= 100 − j100 Ω

The Thévenin equivalent circuit:


(9 + j4)Ia − Ib = −60/0

Problems 9–31

−Ia + (9 − j4)Ib = 60/0

Solving,

Ia = −5 + j2.5 A; Ib = 5 + j2.5 A

VTh = 4Ia + (4 − j4)Ib = 10/0 V


(9 + j4)Ia − 1Ib − 4Isc = −60

−1Ia + (9 − j4)Ib − (4 − j4)Isc = 60

−4Ia − (4 − j4)Ib + (8 − j4)Isc = 0

Solving,

Isc = 2.07/0

ZTh =VTh

Isc=

10/0

2.07/0 = 4.83 Ω


Alternate calculation for ZTh:

∑Z = 4 + 1 + 4 − j4 = 9 − j4

Z1 =4

9 − j4

Z2 =4 − j49 − j4

Z3 =16 − j169 − j4

Za = 4 + j4 +4

9 − j4=

56 + j209 − j4

Zb = 4 +4 − j49 − j4

=40 − j209 − j4

Za‖Zb =2640 − j320864 − j384

Z3 + Za‖Zb =16 − j169 − j4

+2640 − j320864 − j384

=4176 − j1856864 − j384

= 4.83 Ω

Problems 9–33

P 9.50 [a]

IT =VT

1000+

VT − αVT

−j1000

IT

VT

=1

1000− (1 − α)

j1000=

j − 1 + α

j1000

·. . ZTh =VT

IT

=j1000

α − 1 + j

ZTh is real when α = 1.

[b] ZTh = 1000 Ω

[c] ZTh = 500 − j500 =j1000

α − 1 + j

=1000

(α − 1)2 + 1+ j

1000(α − 1)(α − 1)2 + 1

Equate the real parts:

1000(α − 1)2 + 1

= 500 ·. . (α − 1)2 + 1 = 2

·. . (α − 1)2 = 1 so α = 0

Check the imaginary parts:

(α − 1)1000(α − 1)2 + 1

∣∣∣∣α=1

= −500

Thus, α = 0.

[d] ZTh =1000

(α − 1)2 + 1+ j

1000(α − 1)(α − 1)2 + 1

For Im(ZTh) > 0, α must be greater than 1. So ZTh is inductive for1 < α ≤ 10.


P 9.51

V1 − 240j10

+V1

50+

V1

30 + j10= 0

Solving for V1 yields

V1 = 198.63/− 24.44 V

Vo =30

30 + j10(V1) = 188.43/− 42.88 V

P 9.52 jωL = j(2000)(1 × 10−3) = j2 Ω

1jωC

= −j106

(2000)(100)= −j5 Ω

Vg1 = 20/− 36.87 = 16 − j12 V

Vg2 = 50/−106.26 = −14 − j48 V

Vo − (16 − j12)j2

+Vo

10+

Vo − (−14 − j48)−j5

= 0

Solving,

Vo = 36/0 V

vo(t) = 36 cos 2000t V

Problems 9–35

P 9.53 From the solution to Problem 9.52 the phasor-domain circuit is

Making two source transformations yields

Ig1 =16 − j12

j2= −6 − j8 A

Ig2 =−14 − j48

−j5= 9.6 − j2.8 A

Y =1j2

+110

+1

−j5= (0.1 − j0.3) S

Z =1Y

= 1 + j3 Ω

Ie = Ig1 + Ig2 = 3.6 − j10.8 A

Hence the circuit reduces to

Vo = ZIe = (1 + j3)(3.6 − j10.8) = 36/0 V

·. . vo(t) = 36 cos 2000t V


P 9.54 The circuit with the mesh currents identified is shown below:


−20/− 36.87 + j2I1 + 10(I1 − I2) = 0

50/− 106.26 + 10(I2 − I1) − j5I2 = 0

In standard form:

I1(10 + j2) + I2(−10) = 20/− 36.87

I1(−10) + I2(10 − j5) = −50/− 106.26 = 50/73.74

Solving on a calculator yields:

I1 = −6 + j10A; I2 = −9.6 + j10A

Thus,

Vo = 10(I1 − I2) = 36V

and

vo(t) = 36 cos 2000tV

P 9.55 From the solution to Problem 9.52 the phasor-domain circuit with the right-handsource removed is

V′o =

10‖ − j5j2 + 10‖ − j5

(16 − j12) = 18 − j26 V

Problems 9–37

With the left hand source removed

V′′o =

10‖j2−j5 + 10‖j2

(−14 − j48) = 18 + j26 V

Vo = V′o + V′′

o = 18 − j26 + 18 + j26 = 36 V

vo(t) = 36 cos 2000t V

P 9.56 Write a KCL equation at the top node:

Vo

−j8+

Vo − 2.4I∆

j4+

Vo

5− (10 + j10) = 0


I∆ =Vo

−j8

Solving,

Vo = j80 = 80/90 V

P 9.57

Write node voltage equations:

Left Node:

V1

40+

V1 − Vo/8j20

= 0.025/0

Right Node:

Vo

50+

Vo

j25+ 16Io = 0


The constraint equation is

Io =V1 − Vo/8

j20

Solution:

Vo = (4 + j4) = 5.66/45 VV1 = (0.8 + j0.6) = 1.0/36.87 VIo = (5 − j15) = 15.81/− 71.57 mA

P 9.58

(10 + j5)Ia − j5Ib = 100/0

−j5Ia − j5Ib = j100

Solving,

Ia = −j10 A; Ib = −20 + j10 A

Io = Ia − Ib = 20 − j20 = 28.28/− 45 A

io(t) = 28.28 cos(50,000t − 45) A

P 9.59

(12 − j12)Ia − 12Ig − 5(−j8) = 0

Problems 9–39

−12Ia + (12 + j4)Ig + j20 − 5(j4) = 0

Solving,

Ig = 4 − j2 = 4.47/− 26.57 A

P 9.60 Set up the frequency domain circuit to use the node voltage method:

At V1: − 5/0 +V1 − V2

−j8+

V1 − 20/90

−j4= 0

At V2:V2 − V1

−j8+

V2

j4+

V2 − 20/90

12= 0

In standard form:

V1

(1

−j8+

1−j4

)+ V2

(− 1

−j8

)= 5/0 +

20/90

−j4

V1

(− 1

−j8

)+ V2

(1

−j8+

1j4

+112

)=

20/90

12

Solving on a calculator:

V1 = −83

+ j43

V V2 = −8 + j4 V

Thus

V0 = V1 − 20/90 = −83

− j563

= 18.86/− 98.13 V


P 9.61 jωL = j5000(60 × 10−3) = j300 Ω

1jωC

=−j

(5000)(2 × 10−6)= −j100 Ω

−400/0 + (50 + j300)Ia − 50Ib − 150(Ia − Ib) = 0

(150 − j100)Ib − 50Ia + 150(Ia − Ib) = 0

Solving,

Ia = −0.8 − j1.6 A; Ib = −1.6 + j0.8 A

Vo = 100Ib = −160 + j80 = 178.89/153.43 V

vo = 178.89 cos(5000t + 153.43) V

P 9.62

10/0 = (1 − j1)I1 − 1I2 + j1I3

−5/0 = −1I1 + (1 + j1)I2 − j1I3

Problems 9–41

1 = j1I1 − j1I2 + I3

Solving,

I1 = 11 + j10 A; I2 = 11 + j5 A; I3 = 6 A

Ia = I3 − 1 = 5 A = 5/0 A

Ib = I1 − I3 = 5 + j10 A = 11.18/63.43 A

Ic = I2 − I3 = 5 + j5 A = 7.07/45 A

Id = I1 − I2 = j5 A = 5/90 A

P 9.63

Va − (100 − j50)20

+Va

j5+

Va − (140 + j30)12 + j16

= 0

Solving,

Va = 40 + j30 V

IZ + (30 + j20) − 140 + j30−j10

+(40 + j30) − (140 + j30)

12 + j16= 0

Solving,

IZ = −30 − j10 A

Z =(100 − j50) − (140 + j30)

−30 − j10= 2 + j2 Ω


P 9.64 [a]1

jωC= −j50 Ω

jωL = j120 Ω

Ze = 100‖ − j50 = 20 − j40 Ω

Ig = 2/0

Vg = IgZe = 2(20 − j40) = 40 − j80 V

Vo =j120

80 + j80(40 − j80) = 90 − j30 = 94.87/− 18.43 V

vo = 94.87 cos(16 × 105t − 18.43) V

[b] ω = 2πf = 16 × 105; f =8 × 105

π

T =1f

=π

8 × 105 = 1.25π µs

·. .18.43360

(1.25π µs) = 201.09 ns

·. . vo lags ig by 201.09 ns

P 9.65 jωL = j106(10 × 10−6) = j10 Ω

1jωC

=−j

(106)(0.1 × 10−6)= −j10 Ω

Va = 50/− 90 = −j50 V

Vb = 25/90 = j25 V

(10 − j10)I1 + j10I2 − 10I3 = −j50

Problems 9–43

j10I1 + 10I2 − 10I3 = 0

−10I1 − 10I2 + 20I3 = j25

Solving,

I1 = 0.5 − j1.5 A; I3 = −1 + j0.5 A I2 = −2.5 A

Ia = −I1 = −0.5 + j1.5 = 1.58/108.43 A

Ib = −I3 = 1 − j0.5 = 1.12/− 26.57 A

ia = 1.58 cos(106t + 108.43) A

ib = 1.12 cos(106t − 26.57) A

P 9.66 [a] jωL1 = j(5000)(2 × 10−3) = j10 Ω

jωL2 = j(5000)(8 × 10−3) = j40 Ω

jωM = j10 Ω

70 = (10 + j10)Ig + j10IL

0 = j10Ig + (30 + j40)IL

Solving,

Ig = 4 − j3 A; IL = −1 A

ig = 5 cos(5000t − 36.87) A

iL = 1 cos(5000t − 180) A

[b] k =M√L1L2

=2√16

= 0.5


[c] When t = 100π µs,

5000t = (5000)(100π) × 10−6 = 0.5π = π/2 rad = 90

ig(100πµs) = 5 cos(53.13) = 3 A

iL(100πµs) = 1 cos(−90) = 0 A

w =12L1i

21 +

12L2i

22 + Mi1i2 =

12(2 × 10−3)(9) + 0 + 0 = 9 mJ

When t = 200π µs,

5000t = π rad = 180

ig(200πµs) = 5 cos(180 − 36.87) = −4 A

iL(200πµs) = 1 cos(180 − 180) = 1 A

w =12(2 × 10−3)(16) +

12(8 × 10−3)(1) + 2 × 10−3(−4)(1) = 12 mJ

P 9.67 Remove the voltage source to find the equivalent impedance:

ZTh = 45 + j125 +(

20|5 + j5|

)2

(5 − j5) = 85 + j85 Ω

Using voltage division:

VTh = Vcd = j20I1 = j20(

4255 + j5

)= 850 + j850 V = 1202.1/45 V

P 9.68 [a] jωL1 = j(200 × 103)(10−3) = j200 Ω

jωL2 = j(200 × 103)(4 × 10−3) = j800 Ω

1jωC

=−j

(200 × 103)(12.5 × 10−9)= −j400 Ω

·. . Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω

·. . Z∗22 = 300 − j400 Ω

M = k√

L1L2 = 2k × 10−3

Problems 9–45

ωM = (200 × 103)(2k × 10−3) = 400k

Zr =[400k500

]2

(300 − j400) = k2(192 − j256) Ω

Zin = 200 + j200 + 192k2 − j256k2

|Zin| = [(200 + 192k2)2 + (200 − 256k2)2]12

d|Zin|dk

=12[(200 + 192k2)2 + (200 − 256k2)2]−

12 ×

[2(200 + 192k2)384k + 2(200 − 256k2)(−512k)]

d|Zin|dk

= 0 when

768k(200 + 192k2) − 1024k(200 − 256k2) = 0

·. . k2 = 0.125; ·. . k =√

0.125 = 0.3536

[b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)]

= 224 + j168 = 280/36.87 Ω

I1 (max) =560/0

224 + j168= 2/− 36.87 A

·. . i1 (peak) = 2 A

Note — You can test that the k value obtained from setting d|Zin|/dk = 0leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1,

Zin = 392 − j56 = 395.98/− 8.13 Ω

Thus,

|Zin|k=1 > |Zin|k=√

0.125

If k = 0,

Zin = 200 + j200 = 282.84/45 Ω

Thus,

|Zin|k=0 > |Zin|k=√

0.125

P 9.69 jωL1 = j50 Ω

jωL2 = j32 Ω


1jωC

= −j20 Ω

jωM = j(4 × 103)k√

(12.5)(8) × 10−3 = j40k Ω

Z22 = 5 + j32 − j20 = 5 + j12 Ω

Z∗22 = 5 − j12 Ω

Zr =[

40k|5 + j12|

]2

(5 − j12) = 47.337k2 − j113.609k2

Zab = 20 + j50 + 47.337k2 − j113.609k2 = (20 + 47.337k2) + j(50 − 113.609k2)

Zab is resistive when

50 − 113.609k2 = 0 or k2 = 0.44 so k = 0.66

·. . Zab = 20 + (47.337)(0.44) = 40.83 Ω

P 9.70 [a] jωLL = j100 Ω

jωL2 = j500 Ω

Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω

Z∗22 = 800 − j600 Ω

ωM = 270 Ω

Zr =( 270

1000

)2

[800 − j600] = 58.32 − j43.74 Ω

[b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω

P 9.71

ZL =V3

I3= 80/60 Ω

Problems 9–47

V2

10=

V3

1; 10I2 = 1I3

V1

8= −V2

1; 8I1 = −1I2

Zab =V1

I1

Substituting,

Zab =V1

I1=

−8V2

−I2/8=

82V2

I2

=82(10V3)

I3/10=

(8)2(10)2V3

I3= (8)2(10)2ZL = 512, 000/60 Ω

P 9.72 In Eq. 9.69 replace ω2M2 with k2ω2L1L2 and then write Xab as

Xab = ωL1 − k2ω2L1L2(ωL2 + ωLL)R2

22 + (ωL2 + ωLL)2

= ωL1

1 − k2ωL2(ωL2 + ωLL)

R222 + (ωL2 + ωLL)2

For Xab to be negative requires

R222 + (ωL2 + ωLL)2 < k2ωL2(ωL2 + ωLL)

or

R222 + (ωL2 + ωLL)2 − k2ωL2(ωL2 + ωLL) < 0

which reduces to

R222 + ω2L2

2(1 − k2) + ωL2ωLL(2 − k2) + ω2L2L < 0

But k ≤ 1 hence it is impossible to satisfy the inequality. Therefore Xab can neverbe negative if XL is an inductive reactance.


P 9.73 [a]

Zab =Vab

I1 + I2=

V2

I1 + I2=

V2

(1 + N1/N2)I1

N1I1 = N2I2, I2 =N1

N2I1

V1

V2=

N1

N2, V1 =

N1

N2V2

V1 + V2 = ZLI1 =(

N1

N2+ 1

)V2

Zab =I1ZL

(N1/N2 + 1)(1 + N1/N2)I1

·. . Zab =ZL

[1 + (N1/N2)]2Q.E.D.

[b] Assume dot on the N2 coil is moved to the lower terminal. Then

V1 = −N1

N2V2 and I2 = −N1

N2I1

As before

Zab =V2

I1 + I2and V1 + V2 = ZLI1

·. . Zab =V2

(1 − N1/N2)I1=

ZLI1

[1 − (N1/N2)]2I1

Zab =ZL

[1 − (N1/N2)]2Q.E.D.

Problems 9–49

P 9.74 [a]

Zab =Vab

I1=

V1 + V2

I1

V1

N1=

V2

N2, V2 =

N2

N1V1

N1I1 = N2I2, I2 =N1

N2I1

V2 = (I1 + I2)ZL = I1

(1 +

N1

N2

)ZL

V1 + V2 =(

N1

N2+ 1

)V2 =

(1 +

N1

N2

)2

ZLI1

·. . Zab =(1 + N1/N2)2ZLI1

I1

Zab =(1 +

N1

N2

)2

ZL Q.E.D.

[b] Assume dot on N2 is moved to the lower terminal, then

V1

N1=

−V2

N2, V1 =

−N1

N2V2

N1I1 = −N2I2, I2 =−N1

N2I1

As in part [a]

V2 = (I2 + I1)ZL and Zab =V1 + V2

I1

Zab =(1 − N1/N2)V2

I1=

(1 − N1/N2)(1 − N1/N2)ZLI1

I1

Zab = [1 − (N1/N2)]2 ZL Q.E.D.


P 9.75 [a] I =24024

+240j32

= (10 − j7.5) A

Vs = 240/0 + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68 V

[b] Use the capacitor to eliminate the j component of I, therefore

Ic = j7.5 A, Zc =240j7.5

= −j32 Ω

Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90 V

[c] Let Ic denote the magnitude of the current in the capacitor branch. Then

I = (10 − j7.5 + jIc) = 10 + j(Ic − 7.5) A

Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)]

= (247 − 0.8Ic) + j(7.25 + 0.1Ic)

It follows that

240 cos α = (247 − 0.8Ic) and 240 sin α = (7.25 + 0.1Ic)

Now square each term and then add to generate the quadratic equation

I2c − 605.77Ic + 5325.48 = 0; Ic = 302.88 ± 293.96

Therefore

Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω.

P 9.76 The phasor domain equivalent circuit is

Vo =Vm/0

2− IRx; I =

Vm/0

Rx − jXC

As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angledecreases from 0 to −180, as shown in the following phasor diagram:

Problems 9–51

P 9.77 [a]

I =1207.5

+120j12

= 16 − j10 A

V = (0.15 + j6)(16 − j10) = 62.4 + j94.5 = 113.24/56.56 V

Vs = 120/0 + V = 205.43/27.39 V

[b]

[c] I =1202.5

+120j4

= 48 − j30 A

V = (0.15 + j6)(48 − j30) = 339.73/56.56 V

Vs = 120 + V = 418.02/42.7 V

The amplitude of Vs must be increased from 205.43 V to 418.02 V (more thandoubled) to maintain the load voltage at 120 V.


[d] I =1202.5

+120j4

+120−j2

= 48 + j30 A

V = (0.15 + j6)(48 + j30) = 339.73/120.57 V

Vs = 120 + V = 297.23/100.23 V

The amplitude of Vs must be increased from 205.43 V to 297.23 V to maintainthe load voltage at 120 V.

P 9.78 Vg = 4/0 V;1

jωC= −j20 kΩ

Let Va = voltage across the capacitor, positive at upper terminalThen:

Va − 4/0

20,000+

Va

−j20,000+

Va

20,000= 0; ·. . Va = (1.6 − j0.8) V

0 − Va

20,000+

0 − Vo

10,000= 0; Vo = −Va

2

·. . Vo = −0.8 + j0.4 = 0.89/153.43 V

vo = 0.89 cos(200t + 153.43) V

P 9.79 [a]

Va − 4/0

20,000+ jωCoVa +

Va

20,000= 0

Va =4

2 + j20,000ωCo

Vo = −Va

2(see solution to Prob. 9.78)

Problems 9–53

Vo =−2

2 + j4 × 106Co

=2/180

2 + j4 × 106Co

·. . denominator angle = 45

so 4 × 106Co = 2 ·. . Co = 0.5 µF

[b] Vo =2/180

2 + j2= 0.707/135 V

vo = 0.707 cos(200t + 135) V

P 9.801

jωC1= −j10 kΩ

1jωC2

= −j100 kΩ

Va − 25000

+Va

−j10,000+

Va

20,000+

Va − Vo

100,000= 0

20Va − 40 + j10Va + 5Va + Va − Vo = 0

·. . (26 + j10)Va − Vo = 40

0 − Va

20,000+

0 − Vo

−j100,000= 0

j5Va − Vo = 0

Solving,

Vo = 1.43 + j7.42 = 7.55/79.11 V

vo(t) = 7.55 cos(106t + 79.11) V


P 9.81 [a] Vg = 25/0 V

Vp =20100

Vg = 5/0; Vn = Vp = 5/0 V

580,000

+5 − Vo

Zp= 0

Zp = −j80,000‖40,000 = 32,000 − j16,000 Ω

Vo =5Zp

80,000+ 5 = 7 − j1 = 7.07/− 8.13 V

vo = 7.07 cos(50,000t − 8.13) V

[b] Vp = 0.2Vm/0; Vn = Vp = 0.2Vm/0

0.2Vm

80,000+

0.2Vm − Vo

32,000 − j16,000= 0

·. . Vo = 0.2Vm +32,000 − j16,000

80,000Vm(0.2) = 0.2Vm(1.4 − j0.2)

·. . |0.2Vm(1.4 − j0.2)| ≤ 10

·. . Vm ≤ 35.36 V

P 9.82 [a]1

jωC= −j20 Ω

Vn

20+

Vn − Vo

−j20= 0

Vo

−j20=

Vn

20+

Vn

−j20

Vo = −j1Vn + Vn = (1 − j1)Vn

Vp =Vg(1/jωCo)5 + (1/jωCo)

=Vg

1 + j(5)(105)Co

Vg = 6/0 V

Vp =6/0

1 + j5 × 105Co

= Vn

·. . Vo =(1 − j1)6/0

1 + j5 × 105Co

|Vo| =√

2(6)√1 + 25 × 1010C2

o

= 6

Solving,

Co = 2 µF

Problems 9–55

[b] Vo =6(1 − j1)1 + j1

= −j6 V

vo = 6 cos(105t − 90) V

P 9.83 [a]

Because the op-amps are ideal Iin = Io, thus

Zab =Vab

Iin=

Vab

Io

; Io =Vab − Vo

Z

Vo1 = Vab; Vo2 = −(

R2

R1

)Vo1 = −KVo1 = −KVab

Vo = Vo2 = −KVab

·. . Io =Vab − (−KVab)

Z=

(1 + K)Vab

Z

·. . Zab =Vab

(1 + K)VabZ =

Z

(1 + K)

[b] Z =1

jωC; Zab =

1jωC(1 + K)

; ·. . Cab = C(1 + K)

P 9.84 [a] I1 =12024

+240

8.4 + j6.3= 23.29 − j13.71 = 27.02/−30.5 A

I2 =12012

− 12024

= 5/0 A

I3 =12012

+240

8.4 + j6= 28.29 − j13.71 = 31.44/−25.87 A

I4 =12024

= 5/0 A; I5 =12012

= 10/0 A

I6 =240

8.4 + j6.3= 18.29 − j13.71 = 22.86/−36.87 A


[b]I1 = 0 I3 = 15 A I5 = 10 A

I2 = 10 + 5 = 15 A I4 = −5 A I6 = 5 A

[c] The clock and television set were fed from the uninterrupted side of the circuit,that is, the 12 Ω load includes the clock and the TV set.

[d] No, the motor current drops to 5 A, well below its normal running value of22.86 A.

[e] After fuse A opens, the current in fuse B is only 15 A.

P 9.85 [a] The circuit is redrawn, with mesh currents identified:


120/0 = 23Ia − 2Ib − 20Ic

120/0 = −2Ia + 43Ib − 40Ic

0 = −20Ia − 40Ib + 70Ic

Solving,

Ia = 24/0 A Ib = 21.96/0 A Ic = 19.40/0 A

The branch currents are:

I1 = Ia = 24/0 A

I2 = Ia − Ib = 2.04/0 A

I3 = Ib = 21.96/0 A

I4 = Ic = 19.40/0 A

I5 = Ia − Ic = 4.6/0 A

I6 = Ib − Ic = 2.55/0 A

Problems 9–57

[b] Let N1 be the number of turns on the primary winding; because the secondarywinding is center-tapped, let 2N2 be the total turns on the secondary. FromFig. 9.58,

13,200N1

=2402N2

orN2

N1=

1110

The ampere turn balance requires

N1Ip = N2I1 + N2I3

Therefore,

Ip =N2

N1(I1 + I3) =

1110

(24 + 21.96) = 0.42/0 A

Check voltages —

V4 = 10I4 = 194/0 VV5 = 20I5 = 92/0 VV6 = 40I6 = 102/0 V

All of these voltages are low for a reasonable distribution circuit.

P 9.86 [a]

The three mesh current equations are

120/0 = 23Ia − 2Ib − 20Ic

120/0 = −2Ia + 23Ib − 20Ic

0 = −20Ia − 20Ib + 50Ic

Solving,

Ia = 24/0 A; Ib = 24/0 A; Ic = 19.2/0 A

·. . I2 = Ia − Ib = 0 A

[b] Ip =N2

N1(I1 + I3) =

N2

N1(Ia + Ib

=1

110(24 + 24) = 0.436/0 A


[c] Check voltages —

V4 = 10I4 = 10Ic = 192/0 VV5 = 20I5 = 20(Ia − Ic) = 96/0 VV6 = 40I6 = 20(Ib − Ic) = 96/0 V

Where the two loads are equal, the current in the neutral conductor (I2) is zero,and the voltages V5 and V6 are equal. The voltages V4, V5, and V6 are too lowfor a reasonable dirtribution circuit.

P 9.87 [a]

125 = (R + 0.05 + j0.05)I1 − (0.03 + j0.03)I2 − RI3

125 = −(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2 − RI3

Subtracting the above two equations gives

0 = (R + 0.08 + j0.08)I1 − (R + 0.08 + j0.08)I2

·. . I1 = I2 so In = I1 − I2 = 0 A

[b] V1 = R(I1 − I3); V2 = R(I2 − I3)

Since I1 = I2 (from part [a]) V1 = V2

[c]

Problems 9–59

250 = (660.04 + j0.04)Ia − 660Ib

0 = −660Ia + 670Ib

Solving,

Ia = 25.28/− 0.23 = 25.28 − j0.10 A

Ib = 24.90/− 0.23 = 24.90 − j0.10 A

I1 = Ia − Ib = 0.377 − j0.00153 A

V1 = 60I1 = 22.63 − j0.0195 = 22.64/− 0.23 V

V2 = 600I1 = 226.3 − j0.915 = 226.4/− 0.23 V

[d]

125 = (60.05 + j0.05)I1 − (0.03 + j0.03)I2 − 60I3

125 = −(0.03 + j0.03)I1 + (600.05 + j0.05)I2 − 600I3

0 = −60I1 − 600I2 + 670I3

Solving,

I1 = 26.97/− 0.24 = 26.97 − j0.113 A

I2 = 25.10/− 0.24 = 25, 10 − j0.104 A

I3 = 24.90/− 0.24 = 24.90 − j0.104 A

V1 = 60(I1 − I3) = 124.4/− 0.27 V

V2 = 600(I2 − I3) = 124.6/− 0.20 V

[e] Because an open neutral can result in severely unbalanced voltages across the125 V loads.


P 9.88 [a] Let N1 = primary winding turns and 2N2 = secondary winding turns. Then

14,000N1

=2502N2

; ·. .N2

N1=

1112

= a

In part c),

Ip = 2aIa

·. . Ip =2N2Ia

N1=

156

Ia

=156

(25.28 − j0.10)

Ip = 451.4 − j1.8 mA = 451.4/− 0.23 mA

In part d),

IpN1 = I1N2 + I2N2

·. . Ip =N2

N1(I1 + I2)

=1

112(26.97 − j0.11 + 25.10 − j0.10)

=1

112(52.07 − j0.22)

Ip = 464.9 − j1.9 mA = 464.9/− 0.24 mA

[b] Yes, because the neutral conductor carries non-zero current whenever the load isnot balanced.

10Sinusoidal Steady State Power

Calculations

Assessment Problems

AP 10.1 [a] V = 100/− 45 V, I = 20/15 A

Therefore

P =12(100)(20) cos[−45 − (15)] = 500 W, A → B

Q = 1000 sin −60 = −866.03 VAR, B → A

[b] V = 100/− 45, I = 20/165

P = 1000 cos(−210) = −866.03 W, B → A

Q = 1000 sin(−210) = 500 VAR, A → B

[c] V = 100/− 45, I = 20/− 105

P = 1000 cos(60) = 500 W, A → B

Q = 1000 sin(60) = 866.03 VAR, A → B

[d] V = 100/0, I = 20/120

P = 1000 cos(−120) = −500 W, B → A

Q = 1000 sin(−120) = −866.03 VAR, B → A

AP 10.2 pf = cos(θv − θi) = cos[15 − (75)] = cos(−60) = 0.5 leading

rf = sin(θv − θi) = sin(−60) = −0.866

10–1

10–2 CHAPTER 10. Sinusoidal Steady State Power Calculations

AP 10.3 From Ex. 9.4 Ieff =Iρ√3

=0.18√

3A

P = I2effR =

(0.03243

)(5000) = 54 W

AP 10.4 [a] Z = (39 + j26)‖(−j52) = 48 − j20 = 52/− 22.62 Ω

Therefore I =250/0

48 − j20 + 1 + j4= 4.85/18.08 A(rms)

VL = ZI = (52/− 22.62)(4.85/18.08) = 252.20/− 4.54 V(rms)

IL =VL

39 + j26= 5.38/− 38.23 A(rms)

[b] SL = VLI∗L = (252.20/− 4.54)(5.38/+ 38.23) = 1357/33.69

= (1129.09 + j752.73) VA

PL = 1129.09 W; QL = 752.73 VAR

[c] P = |I|21 = (4.85)2 · 1 = 23.52 W; Q = |I|24 = 94.09 VAR

[d] Sg(delivering) = 250I∗ = (1152.62 − j376.36) VA

Therefore the source is delivering 1152.62 W and absorbing 376.36magnetizing VAR.

[e] Qcap =|VL|2−52

=(252.20)2

−52= −1223.18 VAR

Therefore the capacitor is delivering 1223.18 magnetizing VAR.

Check: 94.09 + 752.73 + 376.36 = 1223.18 VAR and

1129.09 + 23.52 = 1152.62 W

AP 10.5 Series circuit derivation:

S = 250I∗ = (40,000 − j30,000)

Therefore I∗ = 160 − j120 = 200/− 36.87 A(rms)

I = 200/36.87 A(rms)

Z =VI

=250

200/36.87 = 1.25/− 36.87 = (1 − j0.75) Ω

Therefore R = 1 Ω, XC = −0.75 Ω

Problems 10–3

Parallel circuit derivation:

P =(250)2

R; therefore R =

(250)2

40,000= 1.5625 Ω

Q =(250)2

XC; therefore XC =

(250)2

−30,000= −2.083 Ω

AP 10.6 S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA

S2 = 6000(0.8) + j6000(0.6) = 4800 − j3600 VA

ST = S1 + S2 = 13,800 + j8400 VA

ST = 200I∗; therefore I∗ = 69 + j42 I = 69 − j42 A

Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91 V(rms)

AP 10.7 [a] The phasor domain equivalent circuit and the Thévenin equivalent are shownbelow:Phasor domain equivalent circuit:

Thévenin equivalent:

VTh = 3−j800

20 − j40= 48 − j24 = 53.67/− 26.57 V

ZTh = 4 + j18 +−j800

20 − j40= 20 + j10 = 22.36/26.57 Ω

For maximum power transfer, ZL = (20 − j10) Ω


[b] I =53.67/− 26.57

40= 1.34/− 26.57 A

Therefore P =(

1.34√2

)2

20 = 18 W

[c] RL = |ZTh| = 22.36 Ω

[d] I =53.67/− 26.57

42.36 + j10= 1.23/− 39.85 A

Therefore P =(

1.23√2

)2

(22.36) = 17 W

AP 10.8

Mesh current equations:

660 = (34 + j50)I1 + j100(I1 − I2) + j40I1 + j40(I1 − I2)

0 = j100(I2 − I1) − j40I1 + 100I2

Solving,

I1 = 3.536/− 45 A,

I2 = 3.5/0 A; ·. . P =12(3.5)2(100) = 612.50 W

AP 10.9 [a]

248 = j400I1 − j500I2 + 375(I1 − I2)

0 = 375(I2 − I1) + j1000I2 − j500I1 + 400I2

Solving,

I1 = 0.80 − j0.62 A; I2 = 0.4 − j0.3 = 0.5/− 36.87 A

·. . P =12(0.25)(400) = 50 W

Problems 10–5

[b] I1 − I2 = 0.4 − j0.32 A

P375 =12|I1 − I2|2(375) = 49.20 W

[c] Pg =12(248)(0.8) = 99.20 W

∑Pabs = 50 + 49.2 = 99.20 W (checks)

AP 10.10 [a] VTh = 210/0 V; V2 = 14V1; I1 = 1

4I2

Short circuit equations:

840 = 80I1 − 20I2 + V1

0 = 20(I2 − I1) − V2

·. . I2 = 14 A; RTh =21014

= 15 Ω

[b] Pmax =(210

30

)2

15 = 735 W

AP 10.11 [a] VTh = −4(146/0) = −584/0 V(rms) = 584/180 V(rms)

V2 = 4V1; I1 = −4I2

Short circuit equations:

146/0 = 80I1 − 20I2 + V1

0 = 20(I2 − I1) + V2

·. . I2 = −146/365 = −0.40 A; RTh =−584−0.4

= 1460 Ω

[b] P =(−584

2920

)2

1460 = 58.40 W


Problems

P 10.1 [a] P =12(100)(10) cos(50 − 15) = 500 cos 35 = 409.58 W (abs)

Q = 500 sin 35 = 286.79 VAR (abs)

[b] P =12(40)(20) cos(−15 − 60) = 400 cos(−75) = 103.53 W (abs)

Q = 400 sin(−75) = −386.37 VAR (del)

[c] P =12(400)(10) cos(30 − 150) = 2000 cos(−120) = −1000 W (del)

Q = 2000 sin(−120) = −1732.05 VAR (del)

[d] P =12(200)(5) cos(160 − 40) = 500 cos(120) = −250 W (del)

Q = 500 sin(120) = 433.01 VAR (abs)

P 10.2 p = P + P cos 2ωt − Q sin 2ωt;dp

dt= −2ωP sin 2ωt − 2ωQ cos 2ωt

dp

dt= 0 when − 2ωP sin 2ωt = 2ωQ cos 2ωt or tan 2ωt = −Q

P

cos 2ωt =P√

P 2 + Q2; sin 2ωt = − Q√

P 2 + Q2

Let θ = tan−1(−Q/P ), then p is maximum when 2ωt = θ and p is minimum when2ωt = (θ + π).

Therefore pmax = P + P · P√P 2 + Q2

− Q(−Q)√P 2 + Q2

= P +√

P 2 + Q2

and pmin = P − P · P√P 2 + Q2

− Q · Q√P 2 + Q2

= P −√

P 2 + Q2

Problems 10–7

P 10.3 [a] hair dryer = 600 W vacuum = 630 W

sun lamp = 279 W air conditioner = 860 W

television = 240 W∑

P = 2609 W

Therefore Ieff =2609120

= 21.74 A

Yes, the breaker will trip.

[b]∑

P = 2609 − 909 = 1700 W; Ieff =1700120

= 14.17 A

Yes, the breaker will not trip if the current is reduced to 14.17 A.

P 10.4 [a] Ieff = 40/115 ∼= 0.35 A; [b] Ieff = 130/115 ∼= 1.13 A

P 10.5 Wdc =V 2

dc

RT ; Ws =

∫ to+T

to

v2s

Rdt

·. .V 2

dc

RT =

∫ to+T

to

v2s

Rdt

V 2dc =

1T

∫ to+T

tov2

s dt

Vdc =

√1T

∫ to+T

tov2

s dt = Vrms = Veff

P 10.6 [a] Area under one cycle of v2g :

A = (52)(2)(30 × 10−6) + 22(2)(37.5 × 10−6)

= 1800 × 10−6

Mean value of v2g :

M.V. =A

200 × 10−6 =1800 × 10−6

200 × 10−6 = 9

·. . Vrms =√

9 = 3 V(rms)

[b] P =V 2

rms

R=

32

2.25= 4 W

P 10.7 i(t) = 200t 0 ≤ t ≤ 75 ms

i(t) = 60 − 600t 75 ms ≤ t ≤ 100 ms

Irms =

√1

0.1

∫ 0.075

0(200)2t2 dt +

∫ 0.1

0.075(60 − 600t)2 dt

=√

10(5.625) + 10(1.875) =√

75 = 8.66 A(rms)


P 10.8 P = I2rmsR

·. . R =3 × 103

75= 40 Ω

P 10.9 Ig = 40/0 mA

jωL = j10,000 Ω;1

jωC= −j10,000 Ω

Io =j10,0005000

(40/0) = 80/90 mA

P =12|Io|2(5000) =

12(0.08)2(5000) = 16 W

Q =12|Io|2(−10,000) = −32 VAR

S = P + jQ = 16 − j32 VA

|S| = 35.78 VA

P 10.10 Ig = 4/0 mA;1

jωC= −j1250 Ω; jωL = j500 Ω

Zeq = 500 + [−j1250‖(1000 + j500)] = 1500 − j500 Ω

Pg = −12|I|2ReZeq = −1

2(0.004)2(1500) = −12 mW

The source delivers 12 mW of power to the circuit.

Problems 10–9

P 10.11 jωL = j105(0.5 × 10−3) = j50 Ω;1

jωC=

1j105[(1/3) × 10−6]

= −j30 Ω

−4 +Vo

j50+

Vo − 50I∆

40 − j30= 0

I∆ =Vo

j50

Place the equations in standard form:

Vo

(1

j50+

140 − j30

)+ I∆

( −5040 − j30

)= 4

Vo

(1

j50

)+ I∆(−1) = 0

Solving,

Vo = 200 − j400 V; I∆ = −8 − j4 A

Io = 4 − (−8 − j4) = 12 + j4 A

P40Ω =12|Io|2(40) =

12(160)(40) = 3200 W

P 10.12 [a] line loss = 7500 − 2500 = 5 kW

line loss = |Ig|220 ·. . |Ig|2 = 250

|Ig| =√

250 A


|Ig|2RL = 2500 ·. . RL = 10 Ω

|Ig|2XL = −5000 ·. . XL = −20 Ω

Thus,

|Z| =√

(30)2 + (X − 20)2 |Ig| =500√

900 + (X − 20)2

·. . 900 + (X − 20)2 =25 × 104

250= 1000

Solving, (X − 20) = ±10.

Thus, X = 10 Ω or X = 30 Ω

[b] If X = 30 Ω:

Ig =500

30 + j10= 15 − j5 A

Sg = −500I∗g = −7500 − j2500 VA

Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars.

Qj30 = |Ig|2X = 250(30) = 7500 VAR

Therefore the line reactance is absorbing 7500 magnetizing vars.

Q−j20 = |Ig|2XL = 250(−20) = −5000 VAR

Therefore the load reactance is generating 5000 magnetizing vars.∑Qgen = 7500 VAR =

∑Qabs

If X = 10 Ω:

Ig =500

30 − j10= 15 + j5 A

Sg = −500I∗g = −7500 + j2500 VA

Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizingvars.

Qj10 = |Ig|2(10) = 250(10) = 2500 VAR

Therefore the line reactance is absorbing 2500 magnetizing vars. The loadcontinues to generate 5000 magnetizing vars.∑

Qgen = 5000 VAR =∑

Qabs

Problems 10–11

P 10.13 Zf = −j10,000‖20,000 = 4000 − j8000 Ω

Zi = 2000 − j2000 Ω

·. .Zf

Zi=

4000 − j80002000 − j2000

= 3 − j1

Vo = −Zf

ZiVg; Vg = 1/0 V

Vo = (3 − j1)(1) = 3 − j1 = 3.16/− 18.43 V

P =12

V 2m

R=

12

(10)1000

= 5 × 10−3 = 5 mW

P 10.14 [a] P =12

(240)2

480= 60 W

− 1ωC

=−9 × 106

(5000)(5)= −360 Ω

Q =12

(240)2

(−360)= −80 VAR

pmax = P +√

P 2 + Q2 = 60 +√

(60)2 + (80)2 = 160 W(del)

[b] pmin = 60 −√

602 + 802 = −40 W(abs)

[c] P = 60 W from (a)

[d] Q = −80 VAR from (a)

[e] generate, because Q < 0

[f] pf = cos(θv − θi)

I =240480

+240

−j360= 0.5 + j0.67 = 0.83/53.13 A

·. . pf = cos(0 − 53.13) = 0.6 leading

[g] rf = sin(−53.13) = −0.8


P 10.15 [a]

The mesh equations are:

(10 − j20)I1 + (j20)I2 = 170

(j20)I1 + (12 − j4)I2 = 0

Solving,

I1 = 4 + j1 A; I2 = 3.5 − j5.5 A

S = −VgI∗1 = −(170)(4 − j1) = −680 + j170 VA

[b] Source is delivering 680 W.

[c] Source is absorbing 170 magnetizing VAR.

[d] P10Ω = (√

17)2(10) = 170 W

P12Ω = (√

42.5)2(12) = 510 W (I1 − I2) = 0.5 + j6.5 A

Q−j20Ω = (√

42.5)2(20) = −850 VAR |I1 − I2| =√

42.5

Qj16Ω = (√

42.5)2(16) = 680 VAR

[e]∑

Pdel = 680 W

∑Pdiss = 170 + 510 = 680 W

·. .∑

Pdel =∑

Pdiss = 680 W

[f]∑

Qabs = 170 + 680 = 850 VAR∑

Qdev = 850 VAR

·. .∑

mag VAR dev =∑

mag VAR abs = 850

Problems 10–13

P 10.16 [a]1

jωC= −j40 Ω; jωL = j80 Ω

Zeq = 40‖ − j40 + j80 + 60 = 80 + j60 Ω

Ig =40/0

80 + j60= 0.32 − j0.24 A

Sg = −12

VgI∗g = −1

240(0.32 + j0.24) = −6.4 − j4.8 VA

P = 6.4 W(del); Q = 4.8 VAR(del)

|S| = |Sg| = 8 VA

[b] I1 =−j40

40 − j40Ig = 0.04 − j0.28 A

P40Ω =12|I1|2(40) = 1.6 W

P60Ω =12|Ig|2(60) = 4.8 W

∑Pdiss = 1.6 + 4.8 = 6.4 W =

∑Pdev

[c] I−j40Ω = Ig − I1 = 0.28 + j0.04 A

Q−j40Ω =12|I−j40Ω|2(−40) = −1.6 VAR(del)

Qj80Ω =12|Ig|2(80) = 6.4 VAR(abs)

∑Qabs = 6.4 − 1.6 = 4.8 VAR =

∑Qdev

P 10.17 [a] Z1 = 240 + j70 = 250/16.26 Ω

pf = cos(16.26) = 0.96 lagging

rf = sin(16.26) = 0.28


Z2 = 160 − j120 = 200/− 36.87 Ω

pf = cos(−36.87) = 0.80 leading

rf = sin(−36.87) = −0.60

Z3 = 30 − j40 = 50/− 53.13 Ω

pf = cos(−53.13) = 0.6 leading

rf = sin(−53.13) = −0.8

[b] Y = Y1 + Y2 + Y3

Y1 =1

250/16.26 ; Y2 =1

200/− 36.87 ; Y3 =1

50/− 53.13

Y = 19.84 + j17.88 mS

Z =1Y

= 37.44/− 42.03 Ω

pf = cos(−42.03) = 0.74 leading

rf = sin(−42.03) = −0.67

P 10.18 [a] S1 = 16 + j18 kVA; S2 = 6 − j8 kVA; S3 = 8 + j0 kVA

ST = S1 + S2 + S3 = 30 + j10 kVA

250I∗ = (30 + j10) × 103; ·. . I = 120 − j40 A

Z =250

120 − j40= 1.875 + j0.625 Ω = 1.98/18.43 Ω

[b] pf = cos(18.43) = 0.9487 lagging


IL = 120 − j40 A(rms)

·. . Vs = 250/0 + (120 − j40)(0.01 + j0.08) = 254.4 + j9.2

= 254.57/2.07 V(rms)

[b] |IL| =√

16,000

P = (16,000)(0.01) = 160 W Q = (16,000)(0.08) = 1280 VAR

[c] Ps = 30,000 + 160 = 30.16 kW Qs = 10,000 + 1280 = 11.28 kVAR

[d] η =30

30.16(100) = 99.47%

Problems 10–15

P 10.20 ST = 4500 − j45000.96

(0.28) = 4500 − j1312.5 VA

S1 =27000.8

(0.8 + j0.6) = 2700 + j2025 VA

S2 = ST − S1 = 1800 − j3337.5 = 3791.95/− 61.66 VA

pf = cos(−61.66) = 0.4747 leading

P 10.21

2400I∗1 = 60,000 + j40,000

I∗1 = 25 + j16.67; ·. . I1 = 25 − j16.67 A(rms)

2400I∗2 = 20,000 − j10,000

I∗2 = 8.33 − j4, 167; ·. . I2 = 8.33 + j4.167 A(rms)

I3 =2400/0

144= 16.67 + j0 A; I4 =

2400/0

j96= 0 − j25 A

Ig = I1 + I2 + I3 + I4 = 50 − j37.5 A

Vg = 2400 + (j4)(50 − j37.5) = 2550 + j200 = 2557.83/4.48 V(rms)

P 10.22 [a] S1 = 60,000 − j70,000 VA

S2 =|VL|2Z∗

2=

(2500)2

24 − j7= 240,000 + j70,000 VA

S1 + S2 = 300,000 VA

2500I∗L = 300,000; ·. . IL = 120/0 A(rms)

Vg = VL + IL(0.1 + j1) = 2500 + (120)(0.1 + j1)

= 2512 + j120 = 2514.86/2.735 V(rms)


[b] T =1f

=160

= 16.67 ms

2.735

360 =t

16.67 ms; ·. . t = 126.62 µs

[c] VL lags Vg by 2.735 or 126.62 µs

P 10.23 [a] From the solution to Problem 9.56 we have:

Vo = j80 = 80/90 V

Sg = −12

VoI∗g = −1

2(j80)(10 − j10) = −400 − j400 VA

Therefore, the independent current source is delivering 400 W and 400magnetizing vars.

I1 =Vo

5= j16 A

P5Ω =12(16)2(5) = 640 W

Therefore, the 8 Ω resistor is absorbing 640 W.

I∆ =Vo

−j8= −10 A

Qcap =12(10)2(−8) = −400 VAR

Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars.

2.4I∆ = −24 V

I2 =Vo − 2.4I∆

j4=

j80 + 24j4

= 20 − j6 A = 20.88/− 16.7 A

Problems 10–17

Qj4 =12|I2|2(4) = 872 VAR

Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars.

Sd.s. = 12(2.4I∆)I∗

2 = 12(−24)(20 + j6)

= −240 − j72 VA

Thus the dependent source is delivering 240 W and 72 magnetizing vars.

[b]∑

Pgen = 400 + 240 = 640 W =∑

Pabs

[c]∑

Qgen = 400 + 400 + 72 = 872 VAR =∑

Qabs


Ia = −j10 A; Ib = −20 + j10 A; Io = 20 − j20 A

S100V = −12(100)I∗

a = −50(j10) = −j500 VA

Thus, the 100 V source is developing 500 magnetizing vars.

Sj100V = −12(j100)I∗

b = −j50(−20 − j10)

= −500 + j1000 VA

Thus, the j100 V source is developing 500 W and absorbing 1000 magnetizingvars.

P10Ω =12|Ia|2(10) = 500 W

Thus the 10 Ω resistor is absorbing 500 W.

Q−j10Ω =12|Ib|2(−10) = −2500 VAR

Thus the −j10 Ω capacitor is developing 2500 magnetizing vars.

Qj5Ω =12|Io|2(5) = 2000 VAR

Thus the j5 Ω inductor is absorbing 2000 magnetizing vars.

[b]∑

Pdev = 500 W =∑

Pabs


[c]∑

Qdev = 500 + 2500 = 3000 VAR∑

Qabs = 1000 + 2000 = 3000 VAR =∑

Qdev

P 10.25 [a] I =465/0

124 + j93= 2.4 − j1.8 = 3/− 36.87 A(rms)

P = (3)2(4) = 36 W

[b] YL =1

120 + j90= 5.33 − j4 mS

·. . XC =1

−4 × 10−3 = −250 Ω

[c] ZL =1

5.33 × 10−3 = 187.5 Ω

[d] I =465/0

191.5 + j3= 2.43/− 0.9 A

P = (2.43)2(4) = 23.58 W

[e] % =23.5836

(100) = 65.5%

Thus the power loss after the capacitor is added is 65.6% of the power lossbefore the capacitor is added.

P 10.26 [a]

250I∗1 = 7500 + j2500; ·. . I1 = 30 − j10 A(rms)

250I∗2 = 2800 − j9600; ·. . I2 = 11.2 + j38.4 A(rms)

I3 =50012.5

+500j50

= 40 − j10 A(rms)

Ig1 = I1 + I3 = 70 − j20 A

Sg1 = 250(70 + j20) = 17,500 + j5000 VA

Problems 10–19

Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars.

Ig2 = I2 + I3 = 51.2 + j28.4 A(rms)

Sg2 = 250(51.2 − j28.4) = 12,800 − j7100 VA

Thus the Vg2 source is delivering 12.8 kW and absorbing 7100 magnetizingvars.

[b]∑

Pgen = 17.5 + 12.8 = 30.3 kW

∑Pabs = 7500 + 2800 +

(500)2

12.5= 30.3kW =

∑Pgen

∑Qdel = 9600 + 5000 = 14.6 kVAR

∑Qabs = 2500 + 7100 +

(500)2

50= 14.6 kVAR =

∑Qdel

P 10.27 S1 = 1200 + 1196 = 2396 + j0 VA

·. . I1 =2396120

= 19.97 A

S2 = 860 + 600 + 240 = 1700 + j0 VA

·. . I2 =1700120

= 14.167 A

S3 = 4474 + 12,200 = 16,674 + j0 VA

·. . I3 =16,674240

= 69.48 A

Ig1 = I1 + I3 = 89.44 A

Ig2 = I2 + I3 = 83.64 A

Breakers will not trip since both feeder currents are less than 100 A.

P 10.28 [a]

I1 =4000 − j1000

125= 32 − j8 A (rms)


I2 =5000 − j2000

125= 40 − j16 A (rms)

I3 =10,000 + j0

250= 40 + j0 A (rms)

·. . Ig1 = I1 + I3 = 72 − j8 A (rms)

In = I1 − I2 = −8 + j8 A (rms)

Ig2 = I2 + I3 = 80 − j16 A(rms)

Vg1 = 0.05Ig1 + 125 + 0.15In = 127.4 + j0.8 V(rms)

Vg2 = −0.15In + 125 + 0.05Ig2 = 130.2 − j2 V(rms)

Sg1 = [(127.4 + j0.8)(72 + j8)] = [9166.4 + j1076.8] VA

Sg2 = [(130.2 − j2)(80 + j16)] = [10,448 + j1923.2] VA

Note: Both sources are delivering average power and magnetizing VAR to thecircuit.

[b] P0.05 = |Ig1|2(0.05) = 262.4 W

P0.15 = |In|2(0.15) = 19.2 W

P0.05 = |Ig2|2(0.05) = 332.8 W∑

Pdis = 262.4 + 19.2 + 332.8 + 4000 + 5000 + 10,000 = 19,614.4 W

∑Pdev = 9166.4 + 10,448 = 19,614.4 W =

∑Pdis

∑Qabs = 1000 + 2000 = 3000 VAR

∑Qdel = 1076.8 + 1923.2 = 3000 VAR =

∑Qabs

P 10.29 [a] Let VL = Vm/0:

SL = 600(0.8 + j0.6) = 480 + j360 VA

I∗ =

480Vm

+ j360Vm

; I =480Vm

− j360Vm

Problems 10–21

120/θ = Vm +(480

Vm

− j360Vm

)(1 + j2)

120Vm/θ = V 2m + (480 − j360)(1 + j2) = V 2

m + 1200 + j600

120Vm cos θ = V 2m + 1200; 120Vm sin θ = 600

(120)2V 2m = (V 2

m + 1200)2 + 6002

14,400V 2m = V 4

m + 2400V 2m + 18 × 105

or

V 4m − 12,000V 2

m + 18 × 105 = 0

Solving,

Vm = 108.85 V and Vm = 12.326 V

If Vm = 108.85 V:

sin θ =600

(108.85)(120)= 0.045935; ·. . θ = 2.63

If Vm = 12.326 V:

sin θ =600

(12.326)(120)= 0.405647; ·. . θ = 23.93

[b]

P 10.30 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA

125I∗L = (17,000 + j10,535.65); I∗

L = 136 + j84.29 A(rms)

·. . IL = 136 − j84.29 A(rms)

Vs = 125 + (136 − j84.29)(0.01 + j0.08) = 133.10 + j10.04

= 133.48/4.31 V(rms)

|Vs| = 133.48 V(rms)

[b] P = |I|2(0.01) = (160)2(0.01) = 256 W


[c](125)2

XC= −10,535.65; XC = −1.483 Ω

− 1ωC

= −1.48; C =1

(1.48)(120π)= 1788.59 µF

[d] I = 136 + j0 A(rms)

Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88

= 126.83/4.92 V(rms)

|Vs| = 126.83 V(rms)

[e] P = (136)2(0.01) = 184.96 W

P 10.31

IL =153,600 − j115,200

4800= 32 − j24 A(rms)

IC =4800−jXC

= j4800XC

= jIC

I = 32 − j24 + jIC = 32 + j(IC − 24)

Vs = 4800 + (2 + j10)[32 + j(IC − 24)]

= (5104 − 10IC) + j(272 + 2IC)

|Vs|2 = (5104 − 10IC)2 + (272 + 2IC)2 = (4800)2

·. . 104I2C − 100,992IC + 3,084,800 = 0

Solving, IC = 31.57 A(rms); IC = 939.51 A(rms)

*Select the smaller value of IC to minimize the magnitude of I.

·. . XC = − 480031.57

= −152.04

·. . C =1

(152.04)(120π)= 17.45 µF

Problems 10–23

P 10.32 ZL = |ZL|/θ = |ZL| cos θ + j|ZL| sin θ

Thus |I| =|VTh|√

(RTh + |ZL| cos θ)2 + (XTh + |ZL| sin θ)2

Therefore P =0.5|VTh|2|ZL| cos θ

(RTh + |ZL| cos θ)2 + (XTh + |ZL| sin θ)2

Let D = demoninator in the expression for P, then

dP

d|ZL| =(0.5|VTh|2 cos θ)(D · 1 − |ZL|dD/d|ZL|)

D2

dD

d|ZL| = 2(RTh + |ZL| cos θ) cos θ + 2(XTh + |ZL| sin θ) sin θ

dP

d|ZL| = 0 when D = |ZL|(

dD

d|ZL|)

Substituting the expressions for D and (dD/d|ZL|) into this equation gives us therelationship R2

Th + X2Th = |ZL|2 or |ZTh| = |ZL|.

P 10.33 [a] ZTh = j40‖40 − j40 = 20 − j20

·. . ZL = Z∗Th = 20 + j20 Ω

[b] VTh =40

40 + j40(120) = 60 − j60 V

I =60 − j60

40= 1.5 − j1.5 A

Pload =12|I|2(20) = 45 W

P 10.34 [a]115.2 + j33.6 − 240

ZTh+

115.2 + j33.680 − j60

= 0

·. . ZTh = 40 − j100 Ω

·. . ZL = 40 + j100 Ω


[b] I =24080

= 3 A(rms)

P = (3)2(40) = 360 W

P 10.35 [a] ZTh = [(3 + j4)‖ − j8] + 7.32 − j17.24 = 15 − j15 Ω

·. . R = |ZTh| = 21.21 Ω

[b] VTh =−j8

3 − j4(112.5) = 144 − j108 V(rms)

I =144 − j10835.21 − j15

= 4.45 − j1.14

P = |I|2(21.21) = 447.35 W

P 10.36 [a] Open circuit voltage:

V1 = 5Iφ = 5100 − 5Iφ

25 + j10

(25 + j10)Iφ = 100 − 5Iφ

Iφ =100

30 + j10= 3 − j1 A

VTh =j3

1 + j3(5Iφ) = 15/0 V

Problems 10–25


V2 = 5Iφ =100 − 5Iφ

25 + j10

Iφ = 3 − j1 A

Isc =5Iφ

1= 15 − j5 A

ZTh =15

15 − j5= 0.9 + j0.3 Ω

ZL = Z∗Th = 0.9 − j0.3 Ω

IL =151.8

= 8.33 A(rms)

P = |IL|2(0.9) = 62.5 W

[b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms)

I1 =VL

j3= −0.833 − j2.5 A(rms)


I2 = I1 + IL = 7.5 − j2.5 A(rms)

5Iφ = I2 + VL·. . Iφ = 3 − j1 A

Id.s. = Iφ − I2 = −4.5 + j1.5 A

Sg = −100(3 + j1) = −300 − j100 VA

Sd.s. = 5(3 − j1)(−4.5 − j1.5) = −75 + j0 VA

Pdev = 300 + 75 = 375 W

% developed =62.5375

(100) = 16.67%

Checks:

P25Ω = (10)(25) = 250 W

P1Ω = (62.5)(1) = 62.5 W

P0.9Ω = 62.5 W∑Pabs = 250 + 62.5 + 62.5 = 375 W =

∑Pdev

Qj10 = (10)(10) = 100 VAR

Qj3 = (6.94)(3) = 20.83 VAR

Q−j0.3 = (69.4)(−0.3) = −20.83 VAR

Qsource = −100 VAR∑Q = 100 + 20.83 − 20.83 − 100 = 0


Vφ − 1005

+Vφ

j5− 0.1Vφ = 0

·. . Vφ = 40 + j80 V(rms)

Problems 10–27

VTh = Vφ + 0.1Vφ(−j5) = Vφ(1 − j0.5) = 80 + j60 V(rms)


Isc = 0.1Vφ +Vφ

−j5= (0.1 + j0.2)Vφ

Vφ − 1005

+Vφ

j5+

Vφ

−j5= 0

·. . Vφ = 100 V(rms)

Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms)

ZTh =VTh

Isc=

80 + j6010 + j20

= 4 − j2 Ω

·. . Ro = |ZTh| = 4.47 Ω

[b]

I =80 + j60

4 +√

20 − j2= 7.36 + j8.82 A(rms)

P = (11.49)2(√

20) = 590.17 W


[c]

I =80 + j60

8= 10 + j7.5 A(rms)

P = (102 + 7.52)(4) = 625 W

[d]

Vφ − 1005

+Vφ

j5+

Vφ − (25 + j50)−j5

= 0

Vφ = 50 + j25 V(rms)

0.1Vφ = 5 + j2.5

5 + j2.5 + IC = 10 + j7.5

IC = 5 + j5 A(rms)

IL =Vφ

j5= 5 − j10 A(rms)

IR = IC + IL = 10 − j5 A(rms)

Ig = IR + 0.1Vφ = 15 − j2.5 A(rms)

Sg = −100I∗g = −1500 − j250 VA

100 = 5(5 + j2.5) + Vcs + 25 + j50 ·. . Vcs = 50 − j62.5 V(rms)

Scs = (50 − j62.5)(5 − j2.5) = 93.75 − j437.5 VA

Thus,∑Pdev = 1500

% delivered to Ro =6251500

(100) = 41.67%

Problems 10–29

P 10.38 [a] First find the Thévenin equivalent:

jωL = j3000 Ω

ZTh = 6000‖12,000 + j3000 = 4000 + j3000 Ω

VTh =12,000

6000 + 12,000(180) = 120/0 V

−j

ωC= −j1000 Ω

I =120

6000 + j2000= 18 − j6 mA

P =12|I|2(2000) = 360 mW

[b] Set Co = 0.1 µF so −j/ωC = −j2000 Ω j3000 − j2000 = j1000 ΩSet Ro as close as possible to

Ro =√

40002 + 10002 = 4123.1 Ω

·. . Ro = 4000 Ω

[c] I =120

8000 + j1000= 14.77 − j1.85 mA

P =12|I|2(4000) = 443.1 mW

Yes; 443.1 mW > 360 mW

[d] I =1208000

= 15 mA

P =12(0.015)2(4000) = 450 mW

[e] Ro = 4000 Ω; Co = 66.67 nF

[f] Yes; 450 mW > 443.1 mW


P 10.39 [a] Set Co = 0.1 µF, so −j/ωC = −j2000 Ω; also set Ro = 4123.1 Ω

I =120

8123.1 + j1000= 14.55 − j1.79 mA

P =12|I|2(4123.1) = 443.18 mW

[b] Yes; 443.18 mW > 360 mW

[c] Yes; 443.18 mW < 450 mW

P 10.40 [a]1

ωC= 100 Ω; C =

1(100)(120π)

= 26.53 µF

[b] Iwo =13,800300

+13,800j100

= 46 − j138 A(rms)

Vswo = 13,800 + (46 − j138)(1.5 + j12) = 15,525 + j345

= 15,528.83/1.27 V(rms)

Iw =13,800300

= 46 A(rms)

Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28 V(rms)

% increase =(

15,528.8213,879.98

− 1)

(100) = 11.88%

[c] Pwo = |46 − j138|21.5 = 31.74 kW

Pw = 462(1.5) = 3174 W

% increase =(31,740

3174− 1

)(100) = 900%

P 10.41 [a] So = original load = 1600 + j16000.8

(0.6) = 1600 + j1200 kVA

Sf = final load = 1920 + j19200.96

(0.28) = 1920 + j560 kVA

·. . Qadded = 560 − 1200 = −640 kVAR

[b] deliver

[c] Sa = added load = 320 − j640 = 715.54/− 63.43 kVA

pf = cos(−63.43) = 0.4472 leading

Problems 10–31

[d] I∗L =

(1600 + j1200) × 103

2400= 666.67 + j500 A

IL = 666.67 − j500 = 833.33/− 36.87 A(rms)

|IL| = 833.33 A(rms)

[e] I∗L =

(1920 + j560) × 103

2400= 800 + j233.33

IL = 800 − j233.33 = 833.33/− 16.26 A(rms)

|IL| = 833.33 A(rms)

P 10.42 [a] Pbefore = Pafter = (833.33)2(0.05) = 34,722.22 W

[b] Vs(before) = 2400 + (666.67 − j500)(0.05 + j0.4)

= 2633.33 + j241.67 = 2644.4/5.24 V(rms)

|Vs(before)| = 2644.4 V(rms)

Vs(after) = 2400 + (800 + j233.33)(0.05 + j0.4)

= 2346.67 + j331.67 = 2369.99/8.04 V(rms)

|Vs(after)| = 2369.99 V(rms)

P 10.43 [a]

180 = 3I1 + j4I1 + j3(I2 − I1) + j9(I1 − I2) − j3I1

0 = 9I2 + j9(I2 − I1) + j3I1

Solving,

I1 = 18 − j18 A(rms); I2 = 12/0 A(rms)

·. . Vo = (12)(9) = 108/0 V(rms)

[b] P = (12)2(9) = 1296 W

[c] Sg = −(180)(18 + j18) = −3240 − j3240 VA ·. . Pg = −3240 W

% delivered =12963240

(100) = 40%



180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1

·. . I1 =180

3 + j7= 9.31 − j21.72 A(rms)

VTh = j9I1 − j3I1 = j6I1 = 130.34 + j55.86 V = 141.81/23.20 V(rms)


180 = 3I1 + j4I1 + j3(Isc − I1) + j9(I1 − Isc) − j3I1

0 = j9(Isc − I1) + j3I1

Solving,

Isc = 20 − j20 A I1 = 30 − j20 A

ZTh =VTh

Isc=

130.34 + j55.8620 − j20

= 1.86 + j4.66 Ω

IL =130.34 + j55.86

3.72= 35 + j15 = 38.08/23.20 A

Problems 10–33

PL = (38.12)2(1.86) = 2700 W

[b] I1 =Zo + j9

j6I2 =

1.86 − j4.66 + j9j6

(35 + j15) = 30/0 A(rms)

Pdev = (180)(30) = 5400 W

P 10.45 [a]

54 = I1 + j2(I1 − I2) + j3I2

0 = 7I2 + j2(I2 − I1) − j3I2 + j8I2 + j3(I1 − I2)

Solving,

I1 = 12 − j21 A(rms); I2 = −3 A(rms)

Vo = 7I2 = −21/180 V(rms)

[b] P = |I2|2(7) = 63 W

[c] Pg = (54)(12) = 648 W

% delivered =63648

(100) = 9.72%

P 10.46 [a]

Open circuit:

VTh = −j3I1 + j2I1 = −jI1

I1 =54

1 + j2= 10.8 − j21.6 A

VTh = −21.6 − j10.8 V


Short circuit:

54 = I1 + j2(I1 − Isc) + j3Isc

0 = j2(Isc − I1) − j3Isc + j8Isc + j3(I1 − Isc)

Solving,

Isc = −3.32 + j5.82

ZTh =VTh

Isc=

−21.6 − j10.8−3.32 + j5.82

= 0.2 + 3.6j = 3.6/86.82 Ω

·. . RL = |ZTh| = 3.606 Ω

[b]

I =−21.6 − j10.83.806 + j3.6

= 4.610/163.2 A

P = |I|2(3.6) = 76.6 W, which is greater than when RL = 7 Ω

P 10.47 [a]

54 = I1 + j2(I1 − I2) + j4kI2

0 = 7I2 + j2(I2 − I1) − j4kI2 + j8I2 + j4k(I1 − I2)

Place the equations in standard form:

54 = (1 + j2)I1 + j(4k − 2)I2

0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2

I1 =54 − I2j(4k − 2)

(1 + j2)

Substituting,

I2 = − j54(4k − 2)[7 + j(10 − 8k)](1 + j2) + (4k − 2)2

For Vo = 0, I2 = 0, so if 4k − 2 = 0, then k = 0.5.

Problems 10–35

[b] When I2 = 0

I1 =54

1 + j2= 10.8 − j21.6 A(rms)

Pg = (54)(10.8) = 583.2 W

Check:

Ploss = |I1|2(1) = 583.2 W

P 10.48 [a] From Problem 9.67, ZTh = 85 + j85 Ω and VTh = 850 + j850 V. Thus, formaximum power transfer, ZL = Z∗

Th = 85 − j85 Ω:

I2 =850 + j850

170= 5 + j5 A

425/0 = (5 + j5)I1 − j20(5 + j5)

·. . I1 =325 + j100

5 + j5= 42.5 − j22.5 A

Sg(del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA

Pg = 18,062.5 W

[b] Ploss = |I1|2(5) + |I2|2(45) = 11,562.5 + 2250 = 13,812.5 W

% loss in transformer =18,062.5 − 13,812.5

18,062.5(100) = 23.53%


P 10.49 [a] From Problem 9.70,

Zab = 100 + j136.26 so

I1 =50

100 + j13.74 + 100 + 136.26=

50200 + j150

= 160 − j120 mA

I2 =jωM

Z22I1 =

j270800 + j600

(0.16 − j0.12) = 51.84 + j15.12 mA

VL = (300 + j100)(51.84 + j15.12)103 = 14.04 + j9.72 V

|VL| = 17.08 V

[b] Pg(ideal) = 50(0.16) = 8 W

Pg(practical) = 8 − |I1|2(100) = 4 W

PL = |I2|2(300) = 874.8 mW

% delivered =0.8748

4(100) = 21.87%

P 10.50 [a]

Open circuit:

VTh =120

16 + j12(j10) = 36 + j48 V

Short circuit:

(16 + j12)I1 − j10Isc = 120

−j10I1 + (11 + j23)Isc = 0

Solving,

Isc = 2.4/0 A

ZTh =36 + j48

2.4= 15 + j20 Ω

·. . ZL = Z∗Th = 15 − j20 Ω

IL =VTh

ZTh + ZL

=36 + j48

30= 1.2 + j1.6 A(rms) = 2.0/53.13 A(rms)

PL = |IL|2(15) = 60 W

Problems 10–37

[b] I1 =Z22I2

jωM=

26 + j3j10

(1.2 + j1.6) = 5.23/− 30.29 A)rms)

Ptransformer = (120)(5.23) cos(−30.29) − (5.23)2(4) = 432.8 W

% delivered =60

432.8(100) = 13.86%

P 10.51 [a] jωL1 = j(10,000)(1 × 10−3) = j10 Ω

jωL2 = j(10,000)(1 × 10−3) = j10 Ω

jωM = j10 Ω

200 = (5 + j10)Ig + j5IL

0 = j5Ig + (15 + j10)IL

Solving,

Ig = 10 − j15 A; IL = −5 A

Thus,

ig = 18.03 cos(10,000t − 56.31) A

iL = 5 cos(10,000t − 180) A

[b] k =M√L1L2

=0.5√

1= 0.5

[c] When t = 50π µs:

10,000t = (10,000)(50π) × 10−6 = 0.5π rad = 90

ig(50π µs) = 18.03 cos(90 − 56.31) = 15 A

iL(50π µs) = 5 cos(90 − 180) = 0 A

w =12L1i

21 +

12L2i

22 + Mi1i2 =

12(10−3)(15)2 + 0 + 0 = 112.5 mJ

When t = 100π µs:

10,000t = (104)(100π) × 10−6 = π = 180

ig(100π µs) = 18.03 cos(180 − 56.31) = −10 A

iL(100π µs) = 5 cos(180 − 180) = 5 A

w =12(10−3)(10)2 +

12(10−3)(5)2 + 0.5 × 10−3(−10)(5) = 37.5 mJ


[d] From (a), Im = 5 A,

·. . P =12(5)2(15) = 187.5 W

[e] Open circuit:

VTh =200

5 + j10(−j5) = −80 − j40 V

Short circuit:

200 = (5 + j10)I1 + j5Isc

0 = j10Isc + j5I1

Solving,

Isc = −11.094/123.69 A; I1 = 22.188/− 56.31 A

ZTh =VTh

Isc=

−80 − j4011.094/123.69 = 1 + j8 Ω

·. . RL = 8.962 Ω

[f]

I =−80 − j409.062 + j8

= 7.399/165.13 A

P =12(7.399)2(8.062) = 220.70 W

[g] ZL = Z∗Th = 1 − j8 Ω

[h] I =−80 − j40

2= 44.72/− 153.43 A

P =12(44.72)2(1) = 1000 W

Problems 10–39

P 10.52 [a]

10 = j1(I1 − I2) + j1(I3 − I2) − j1(I1 − I3)

0 = 1I2 + j2(I2 − I3) + j1(I2 − I1) + j1(I2 − I1) + j1(I2 − I3)

0 = 1I3 − j1(I3 − I1) + j2(I3 − I2) + j1(I1 − I2)

Solving,

I1 = 6.25 + j7.5 A(rms); I2 = 5 + j2.5 A(rms); I3 = 5 − j2.5 A(rms)

Ia = I1 = 6.25 + j7.5 A Ib = I1 − I2 = 1.25 + j5 A

Ic = I2 = 5 + j2.5 A Id = I3 − I2 = −j5 A

Ie = I1 − I3 = 1.25 + j10 A If = I3 = 5 − j2.5 A

[b]

Va = 10 V Vb = j1Ib + j1Id = j1.25 V

Vc = 1Ic = 5 + j2.5 V Vd = j2Id + j1Ib = 5 + j1.25 V

Ve = −j1Ie = 10 − j1.25 V Vf = 1If = 5 − j2.5 V

Sa = −10I∗a = −62.5 + j75 VA

Sb = VbI∗b = 6.25 + j1.5625 VA


Sc = VcI∗c = 31.25 + j0 VA

Sd = VdI∗d = −6.25 + j25 VA

Se = VeI∗e = 0 − j101.5625 VA

Sf = VfI∗f = 31.25 VA

[c]∑

Pdev = 62.5 W∑

Pabs = 6.25 + 31.25 − 6.25 + 31.25 = 62.5 W

Note that the total power absorbed by the coupled coils is zero:6.25 − 6.25 = 0 = Pb + Pd

[d]∑

Qdev = 101.5625 VARThe capacitor is developing magnetizing vars.∑

Qabs = 75 + 1.5625 + 25 = 101.5625 VAR∑

Q absorbed by the coupled coils is Qb + Qd = 26.5625 VAR


I1 =10/0

1 + j2= 2 − j4 A

VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57 V


10/0 = (1 + j2)I1 − j3.2Isc

Problems 10–41

0 = −j3.2I1 + j5.4Isc

Solving,

Isc = 5.89/− 5.92 A

ZTh =14.31/26.57

5.89/− 5.92 = 2.43/32.49 = 2.048 + j1.304 Ω

·. . I2 =14.31/26.57

4.096= 3.49/26.57 A

10/0 = (1 + j2)I1 − j3.2I2

·. . I1 =10 + j3.2I2

1 + j2=

10 + j3.2(3.49/26.57)1 + j2

= 5/0 A

Zg =10/0

5/0 = 2 + j0 = 2/0 Ω

P 10.54 [a]

272/0 = 2Ig + j10Ig + j14(Ig − I2) − j6I2

+j14Ig − j8I2 + j20(Ig − I2)

0 = j20(I2 − Ig) − j14Ig + j8I2 + j4I2

+j8(I2 − Ig) − j6Ig + 8I2


Solving,

Ig = 20 − j4 A(rms); I2 = 24/0 A(rms)

P8Ω = (24)2(8) = 4608 W

[b] Pg(developed) = (272)(20) = 5440 W

[c] Zab =Vg

Ig

− 2 =272

20 − j4− 2 = 11.08 + j2.62 = 11.38/13.28 Ω

[d] P2Ω = |Ig|2(2) = 832 W∑

Pdiss = 832 + 4608 = 5440 W =∑

Pdev

P 10.55 [a]

300 = 60I1 + V1 + 20(I1 − I2)

0 = 20(I2 − I1) + V2 + 40I2

V2 =14

V1; I2 = −4I1

Solving,

V1 = 260 V(rms); V2 = 65 V(rms)

I1 = 0.25 A(rms); I2 = −1.0 A(rms)

V5A = V1 + 20(I1 − I2) = 285 V(rms)

·. . P = −(285)(5) = −1425 W

Thus 1425 W is delivered by the current source to the circuit.

[b] I20Ω = I1 − I2 = 1.25 A(rms)

·. . P20Ω = (1.25)2(20) = 31.25 W

Problems 10–43

P 10.56

30Vo = Va;Io

30= Ia; Vo = 10Io therefore

Va

Ia= 9 kΩ

Vb

1=

−Va

20; Ib = −20Ia; therefore

Vb

Ib=

9000400

= 22.5 Ω

Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers arelossless, P10Ω = P22.5Ω, and the power delivered to the 22.5 Ω resistor is 22(22.5) or90 W.

P 10.57 [a]Vb

Ib=

a210400

= 2.5 Ω; therefore a2 = 100, a = 10

[b] Ib =505

= 10 A; P = (100)(2.5) = 250 W

P 10.58 [a] ZTh = 720 + j1500 +(200

50

)2

(40 − j30) = 1360 + j1020 = 1700/36.87 Ω

·. . Zab = 1700 Ω

Zab =ZL

(1 + N1/N2)2

(1 + N1/N2)2 = 6800/1700 = 4

·. . N1/N2 = 1 or N2 = N1 = 1000 turns

[b] VTh =255/0

40 + j30(j200) = 1020/53.13 V

IL =1020/53.13

3060 + j1020= 0.316/34.7 A(rms)

Since the transformer is ideal, P6800 = P1700.

P = |IL|2(1700) = 170 W


[c]

255/0 = (40 + j30)I1 − j200(0.26 + j0.18)

·. . I1 = 4.13 − j1.80 A(rms)

Pgen = (255)(4.13) = 1053 W

Ptrans = 1053 − 170 = 883 W

% transmitted =8831053

(100) = 83.85%

P 10.59 [a]

For maximum power transfer, Zab = 90 kΩ

Zab =(1 +

N1

N2

)2

ZL

·. .(1 +

N1

N2

)2

=90,000400

= 225

1 +N1

N2= ±15;

N1

N2= 15 − 1 = 14

[b] P = |Ii|2(90,000) =(

180180,000

)2

(90,000) = 90 mW

[c] V1 = RiIi = (90,000)(

180180,000

)= 90 V

[d]

Vg = (2.25 × 10−3)(100,000‖80,000) = 100 V

Problems 10–45

Pg(del) = (2.25 × 10−3)(100) = 225 mW

% delivered =90225

(100) = 40%

P 10.60 [a] Zab =(1 +

N1

N2

)2

(1 − j2) = 25 − j50 Ω

·. . I1 =100/0

15 + j50 + 25 − j50= 2.5/0 A

I2 =N1

N2I1 = 10/0 A

·. . IL = I1 + I2 = 12.5/0 A(rms)

P1Ω = (12.5)2(1) = 156.25 W

P15Ω = (2.5)2(15) = 93.75 W

[b] Pg = −100(2.5/0) = −250 W∑

Pabs = 156.25 + 93.75 = 250 W =∑

Pdev

P 10.61 [a] 25a21 + 4a2

2 = 500

I25 = a1I; P25 = a21I2(25)

I4 = a2I; P4 = a22I2(4)

P4 = 4P25; a22I24 = 100a2

1I2

·. . 100a21 = 4a2

2

25a21 + 100a2

1 = 500; a1 = 2

25(4) + 4a22 = 500; a2 = 10

[b] I =2000/0

500 + 500= 2/0 A(rms)

I25 = a1I = 4 A

P25Ω = (16)(25) = 400 W

[c] I4 = a2I = 10(2) = 20 A(rms)

V4 = (20)(4) = 80/0 V(rms)



500 = 100I1 + V1

V2 = 400I2

V1

1=

V2

2·. . V2 = 2V1

I1 = 2I2

Substitute and solve:

2V1 = 400I1/2 = 200I1·. . V1 = 100I1

500 = 100I1 + 100I1·. . I1 = 500/200 = 2.5 A

·. . I2 =12

I1 = 1.25 A

V1 = 100(2.5) = 250 V; V2 = 2V1 = 500 V

VTh = 20I1 + V1 − V2 + 40I2 = −150 V(rms)


500 = 80(Isc + I1) + 360(Isc + 0.5I1)

2V1 = 40I1

2+ 360(Isc + 0.5I1)

500 = 80(I1 + Isc) + 20I1 + V1

Problems 10–47

Solving,

Isc = −1.47 A; I1 = 4.41 A; V1 = 176.47 V

RTh =VTh

Isc=

−150−1.47

= 102 Ω

P =752

102= 55.15 W

[b]

500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)]

575 +6000102

+27,000102

= 80I1 + 180I1

·. . I1 = 3.456 A

Psource = (500)[3.456 − (75/102)] = 1360.29 W

% delivered =55.15

1360.29(100) = 4.05%

[c] P80Ω = 80(

I1 − 75102

)2

= 592.13 W

P20Ω = 20I21 = 238.86 W

P40Ω = 40I22 = 119.43 W

P102Ω =752

102= 55.15 W

P360Ω = 360(

I2 − 75102

)2

= 354.73 W

∑Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W =

∑Pdev



40/0 = 4(I1 + I3) + 12I3 + VTh

I1

4= −I3; I1 = −4I3

Solving,

VTh = 40/0 V


40/0 = 4I1 + 4I3 + I1 + V1

4V1 = 16(I1/4) = 4I1; ·. . V1 = I1

·. . 40/0 = 6I1 + 4I3

Also,

40/0 = 4(I1 + I3) + 12I3

Solving,

I1 = 6 A; I3 = 1 A; Isc = I1/4 + I3 = 2.5 A

RTh =VTh

Isc=

402.5

= 16 Ω

Problems 10–49

I =40/0

32= 1.25/0 A(rms)

P = (1.25)2(16) = 25 W

[b]

40 = 4(I1 + I3) + 12I3 + 20

4V1 = 4I1 + 16(I1/4 + I3); ·. . V1 = 2I1 + 4I3

40 = 4I1 + 4I3 + I1 + V1

·. . I1 = 6 A; I3 = −0.25 A; I1 + I3 = 5.75/0 A; V1 = 11/0 V

P40V (developed) = 40(5.75) = 230 W

·. . % delivered =25230

(100) = 10.87%

[c] PRL= 25 W; P16Ω = (1.5)2(16) = 36 W

P4Ω = (5.75)2(4) = 132.25 W; P1Ω = (6)2(1) = 36 W

P12Ω = (−0.25)2(12) = 0.75 W∑

Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W =∑

Pdev

P 10.64 [a] Replace the circuit to the left of the primary winding with a Thévenin equivalent:

VTh = (15)(20‖j10) = 60 + j120 V

ZTh = 2 + 20‖j10 = 6 + j8 Ω


Transfer the secondary impedance to the primary side:

Zp =125

(100 + jXC) = 4 + jXC

25Ω

Now maximize I by setting (XC/25) = −8 Ω:

·. . C =1

200(20 × 103)= 0.25 µF

[b] I =60 + j120

10= 6 + j12 A

P = |I|2(4) = 720 W

[c]Ro

25= 6 Ω; ·. . Ro = 150 Ω

[d] I =60 + j120

12= 5 + j10 A

P = |I|2(6) = 750 W

P 10.65 [a] Zab = 50 − j400 =(1 − N1

N2

)2

ZL =(1 − 2800

700

)2

ZL = 9ZL

·. . ZL =19(50 − j400) = 5.556 − j44.444 Ω

[b]

I1 =24100

= 240/0 mA

Problems 10–51

N1I1 = −N2I2

I2 = −4I1 = 960/180 mA

IL = I1 + I2 = 720/180 mA(rms)

VL = (5.556 − j44.444)IL = −4 + j32 = 32.25/97.13 V(rms)

P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves only theresistor R2. Then,

Pmed = 500 W =V 2

R2=

1202

R2

Thus,

R2 =1202

500= 28.8 Ω

[b] Now move to the LOW setting, as shown in Fig. 10.30, which involves theresistors R1 and R2 connected in series:

Plow =V 2

R1 + R2=

V 2

R1 + 28.8= 250 W

Thus,

R1 =1202

250− 28.8 = 28.8 Ω

[c] Note that the HIGH setting has R1 and R2 in parallel:

Phigh =V 2

R1‖R2=

1202

28.8‖28.8= 1000 W

If the HIGH setting has required power other than 1000 W, this problem souldnot have been solved. In other words, the HIGH power setting was chosen insuch a way that it would be satisfied once the two resistor values werecalculated to satisfy the LOW and MEDIUM power settings.


P 10.67 [a] PL =V 2

R1 + R2; R1 + R2 =

V 2

PL

PM =V 2

R2; R2 =

V 2

PM

PH =V 2(R1 + R2)

R1R2

R1 + R2 =V 2

PL; R1 =

V 2

PL− V 2

PM

PH =V 2V 2/PL(

V 2

PL− V 2

PM

) (V 2

PM

) =PMPLPM

PL(PM − PL)

PH =P 2

M

PM − PL

[b] PH =(750)2

(750 − 250)= 1125 W

P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH. Thus

PM − PL =P 2

M

PHor

P 2M

PH− PM + PL = 0

P 2M − PMPH + PLPH = 0

·. . PM =PH

2±√(

PH

2

)2

− PLPH

=PH

2± PH

√14

−(

PL

PH

)

For the specified values of PL and PH

PM = 500 ± 1000√

0.25 − 0.24 = 500 ± 100

·. . PM1 = 600 W; PM2 = 400 W

Note in this case we design for two medium power ratingsIf PM1 = 600 W

R2 =(120)2

600= 24 Ω

Problems 10–53

R1 + R2 =(120)2

240= 60 Ω

R1 = 60 − 24 = 36 Ω

CHECK: PH =(120)2(60)(36)(24)

= 1000 W

If PM2 = 400 W

R2 =(120)2

400= 36 Ω

R1 + R2 = 60 Ω (as before)

R1 = 24 Ω

CHECK: PH = 1000 W

P 10.69 R1 + R2 + R3 =(120)2

600= 24 Ω

R2 + R3 =(120)2

900= 16 Ω

·. . R1 = 24 − 16 = 8 Ω

R3 + R1‖R2 =(120)2

1200= 12 Ω

·. . 16 − R2 +8R2

8 + R2= 12

R2 − 8R2

8 + R2= 4

8R2 + R22 − 8R2 = 32 + 4R2

R22 − 4R2 − 32 = 0

R2 = 2 ± √4 + 32 = 2 ± 6

·. . R2 = 8 Ω; ·. . R3 = 8 Ω


P 10.70 R2 =(220)2

500= 96.8 Ω

R1 + R2 =(220)2

250= 193.6 Ω

·. . R1 = 96.8 Ω

CHECK: R1‖R2 = 48.4 Ω

PH =(220)2

48.4= 1000 W

11Balanced Three-Phase Circuits

Assessment Problems

AP 11.1 Make a sketch:

We know VAN and wish to find VBC. To do this, write a KVL equation to find VAB,and use the known phase angle relationship between VAB and VBC to find VBC.

VAB = VAN + VNB = VAN − VBN

Since VAN, VBN, and VCN form a balanced set, and VAN = 240/− 30V, and thephase sequence is positive,

VBN = |VAN|//VAN − 120 = 240/− 30 − 120 = 240/− 150 V

Then,

VAB = VAN − VBN = (240/− 30) − (240/− 150) = 415.46/0 V

Since VAB, VBC, and VCA form a balanced set with a positive phase sequence, wecan find VBC from VAB:

VBC = |VAB|/(/VAB − 120) = 415.69/0 − 120 = 415.69/− 120 V

Thus,

VBC = 415.69/− 120 V

11–1

11–2 CHAPTER 11. Balanced Three-Phase Circuits

AP 11.2 Make a sketch:

We know VCN and wish to find VAB. To do this, write a KVL equation to find VBC,and use the known phase angle relationship between VAB and VBC to find VAB.

VBC = VBN + VNC = VBN − VCN

Since VAN, VBN, and VCN form a balanced set, and VCN = 450/− 25 V, and thephase sequence is negative,

VBN = |VCN|//VCN − 120 = 450/− 23 − 120 = 450/− 145 V

Then,

VBC = VBN − VCN = (450/− 145) − (450/− 25) = 779.42/− 175 V

Since VAB, VBC, and VCA form a balanced set with a negative phase sequence, wecan find VAB from VBC:

VAB = |VBC|//VBC − 120 = 779.42/− 295 V

But we normally want phase angle values between +180 and −180. We add 360

to the phase angle computed above. Thus,

VAB = 779.42/65 V

AP 11.3 Sketch the a-phase circuit:

Problems 11–3

[a] We can find the line current using Ohm’s law, since the a-phase line current isthe current in the a-phase load. Then we can use the fact that IaA, IbB, and IcC

form a balanced set to find the remaining line currents. Note that since wewere not given any phase angles in the problem statement, we can assume thatthe phase voltage given, VAN, has a phase angle of 0.

2400/0 = IaA(16 + j12)

so

IaA =2400/0

16 + j12= 96 − j72 = 120/− 36.87 A

With an acb phase sequence,

/IbB = /IaA + 120 and /IcC = /IaA − 120

so

IaA = 120/− 36.87 A

IbB = 120/83.13 A

IcC = 120/− 156.87 A

[b] The line voltages at the source are Vab Vbc, and Vca. They form a balanced set.To find Vab, use the a-phase circuit to find VAN, and use the relationshipbetween phase voltages and line voltages for a y-connection (see Fig. 11.9[b]).From the a-phase circuit, use KVL:

Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0

= (0.1 + j0.8)(96 − j72) + 2400/0 = 2467.2 + j69.6

2468.18/1.62 V

From Fig. 11.9(b),

Vab = Van(√

3/− 30) = 4275.02/− 28.38 V


/Vbc = /Vab + 120 and /Vca = /Vab − 120

so

Vab = 4275.02/− 28.38 V

Vbc = 4275.02/91.62 V

Vca = 4275.02/− 148.38 V


[c] Using KVL on the a-phase circuit

Va′n = Va′a + Van = (0.2 + j0.16)IaA + Van

= (0.02 + j0.16)(96 − j72) + (2467.2 + j69.9)

= 2480.64 + j83.52 = 2482.05/1.93 V


/Vb′n = /Va′n + 120 and /Vc′n = /Va′n − 120

so

Va′n = 2482.05/1.93 V

Vb′n = 2482.05/121.93 V

Vc′n = 2482.05/− 118.07 V

AP 11.4 IcC = (√

3/− 30)ICA = (√

3/− 30) · 8/− 15 = 13.86/− 45 A

AP 11.5 IaA = 12/(65 − 120) = 12/− 55

IAB =[(

1√3

)/− 30

]IaA =

(/− 30

√3

)· 12/− 55

= 6.93/− 85 A

AP 11.6 [a] IAB =[(

1√3

)/30

][69.28/− 10] = 40/20 A

Therefore Zφ =4160/0

40/20 = 104/− 20 Ω

[b] IAB =[(

1√3

)/− 30

][69.28/− 10] = 40/− 40 A

Therefore Zφ = 104/40 Ω

AP 11.7 Iφ =110

3.667+

110j2.75

= 30 − j40 = 50/− 53.13 A

Therefore |IaA| =√

3Iφ =√

3(50) = 86.60 A

Problems 11–5

AP 11.8 [a] |S| =√

3(208)(73.8) = 26,587.67 VA

Q =√

(26,587.67)2 − (22,659)2 = 13,909.50 VAR

[b] pf =22,659

26,587.67= 0.8522 lagging

AP 11.9 [a] VAN =(

2450√3

)/0 V; VANI∗

aA = Sφ = 144 + j192 kVA

Therefore

I∗aA =

(144 + j192)10002450/

√3

= (101.8 + j135.7) A

IaA = 101.8 − j135.7 = 169.67/− 53.13 A

|IaA| = 169.67 A

[b] P =(2450)2

R; therefore R =

(2450)2

144,000= 41.68 Ω

Q =(2450)2

X; therefore X =

(2450)2

192,000= 31.26 Ω

[c] Zφ =VAN

IaA=

2450/√

3169.67/− 53.13 = 8.34/53.13 = (5 + j6.67) Ω

·. . R = 5 Ω, X = 6.67 Ω


Problems

P 11.1 [a] First, convert the cosine waveforms to phasors:

Va = 208/27; Vb = 208/147; Vc = 208/− 93

Subtract the phase angle of the a-phase from all phase angles:

/V′a = 27 − 27 = 0

/V′b = 147 − 27 = 120

/V′c = −93 − 27 = −120

Compare the result to Eqs. 11.1 and 11.2:

Therefore acb

[b] First, convert the cosine waveforms to phasors:

Va = 4160/− 18; Vb = 4160/− 138; Vc = 4160/+ 102

Subtract the phase angle of the a-phase from all phase angles:

/V′a = −18 + 18 = 0

/V′b = −138 + 18 = −120

/V′c = 102 + 18 = 120

Compare the result to Eqs. 11.1 and 11.2:

Therefore abc

P 11.2 [a] Va = 180/0 V

Vb = 180/− 120 V

Vc = 180/− 240 = 180/120 V

Balanced, positive phase sequence

[b] Va = 180/− 90 V

Vb = 180/30 V

Vc = 180/− 210 V = 180/150 V

Balanced, negative phase sequence

[c] Va = 400/− 270 V = 400/90 V

Vb = 400/120 V

Vc = 400/− 30 V

Unbalanced, phase angle in b-phase

Problems 11–7

[d] Va = 200/30 V

Vb = 201/150 V

Vc = 200/270 V = 200/− 90 V

Unbalanced, unequal amplitude in the b-phase

[e] Va = 208/42 V

Vb = 208/− 78 V

Vc = 208/− 201 V = 208/159 V

Unbalanced, phase angle in the c-phase

[f] Unbalanced; the frequencies of the waveforms are not the same for the positivesequence of Eq. 11.1

P 11.3 Va = Vm/0 = Vm + j0

Vb = Vm/− 120 = −Vm(0.5 + j0.866)

Vc = Vm/120 = Vm(−0.5 + j0.866)

Va + Vb + Vc = (Vm)(1 + j0 − 0.5 − j0.866 − 0.5 + j0.866)

= Vm(0) = 0

For the negative sequences of Eq. 11.2, Vb and Vc are interchanged, but the sum isstill zero.

P 11.4 I =Va + Vb + Vc

3(RW + jXW)= 0

P 11.5 [a] IaA =20025

= 8/0 A

IbB =200/− 120

30 − j40= 4/− 66.87 A

IcC =200/120

80 + j60= 2/83.13 A

The magnitudes are unequal and the phase angles are not 120 apart.

b] Io = IaA + IbB + IcC = 9.96/− 9.79 A


P 11.6 [a] IaA =277/0

80 + j60= 2.77/− 36.87 A

IbB =277/− 120

80 + j60= 2.77/− 156.87 A

IcC =277/120

80 + j60= 2.77/83.13 A

Io = IaA + IbB + IcC = 0

[b] VAN = (78 + j54)IaA = 262.79/− 2.17 V

[c] VAB = VAN − VBN

VBN = (77 + j56)IbB = 263.73/− 120.84 V

VAB = 262.79/− 2.17 − 263.73/− 120.84 = 452.89/28.55 V

[d] Unbalanced — see conditions for a balanced circuit on p. 504 of the text!

P 11.7 Zga + Zla + ZLa = 60 + j80 Ω

Zgb + Zlb + ZLb = 40 + j30Ω

Zgc + Zlc + ZLc = 20 + j15Ω

VN − 24060 + j80

+VN − 240/120

40 + j30+

VN − 240/− 120

20 + j15+

VN

10= 0

Solving for VN yields

VN = 42.94/− 156.32 V

Io =VN

10= 4.29/− 156.32 A

P 11.8 Make a sketch of the load in the frequency domain. Note that we convert the timedomain line-to-neutral voltages to phasors:

Problems 11–9

Note that these three voltages form a balanced set with an abc phase sequence. First,use KVL to find VAB:

VAB = VAN + VNB = VAN − VBN

= (169.71/26) − (169.71/− 94) = 293.95/56 V

With an abc phase sequence,

/VBC = /VAB − 120 and /VCA = /VAB + 120

so

VAB = 293.95/56 V

VBC = 293.95/− 64 V

VCA = 293.95/176 V

To get back to the time domain, perform an inverse phasor transform of the threeline voltages, using a frequency of ω:

vAB(t) = 293.95 cos(ωt + 56) V

vBC(t) = 293.95 cos(ωt − 64) V

vCA(t) = 293.95 cos(ωt + 176) V

P 11.9 Make a sketch of the three-phase line and load:

Z = 0.25 + j2 Ω/φ

ZL = 30.48 + j22.86 Ω/φ


[a] The line currents are IaA, IbB, and IcC. To find IaA, first find VAN and use Ohm’slaw for the a-phase load impedance. Since we are only concerned with findingvoltage and current magnitudes, the phase sequence doesn’t matter and wearbitrarily assume a positive phase sequence. Since we are not given any phaseangles in the problem statement, we can assume the angle of VAB is 0. UseFig. 11.9(a) to find VAN from VAB.

VAN =660√

3/(0 − 30) = 381.05/− 30 V

Now find IaA using Ohm’s law:

IaA =VAN

ZL=

381.05/− 30

30.48 + j22.86= 3.993 − j9.20 = 10/− 66.87 V

Thus, the magnitude of the line current is

|IaA| = 10 A

[b] The line voltage at the source is Vab. From KVL on the top loop of thethree-phase circuit,

Vab = VaA + VAB + VBb

= ZIaA + VAB + ZIBb

= ZIaA + VAB − ZIbB

= (0.25 + j2)(10/− 66.87) + 660/0 − (0.25 + j2)(10/− 173.13)

= 684.71/2.10 V

Thus, the magnitude of the line voltage at the source is

|Vab| = 684.71 V

P 11.10 Make a sketch of the a-phase:

[a] Find the a-phase line current from the a-phase circuit:

IaA =125/0

0.1 + j0.8 + 19.9 + j14.2=

125/0

20 + j15

= 4 − j3 = 4/− 36.87 A

Problems 11–11

Find the other line currents using the acb phase sequence:

IbB = 5/− 36.87 + 120 = 5/83.13 A

IcC = 5/− 36.87 − 120 = 5/− 156.87 A

[b] The phase voltage at the source is Van = 125/0 V. Use Fig. 11.9(b) to find theline voltage, Van, from the phase voltage:

Vab = Van(√

3/− 30) = 216.51/− 30 V

Find the other line voltages using the acb phase sequence:

Vbc = 216.51/− 30 + 120 = 216.51/90 V

Vca = 216.51/− 30 − 120 = 216.51/− 150 V

[c] The phase voltage at the load in the a-phase is VAN. Calculate its value usingIaA and the load impedance:

VAN = IaAZL = (4 − j3)(19.9 + j14.2) = 122.2 − j2.9 = 122.23/− 1.36 V

Find the phase voltage at the load for the b- and c-phases using the acbsequence:

VBN = 122.23/− 1.36 + 120 = 122.23/118.64 V

VCN = 122.23/− 1.36 − 120 = 122.23/− 121.36 V

[d] The line voltage at the load in the a-phase is VAB. Find this line voltage fromthe phase voltage at the load in the a-phase, VAN, using Fig, 11.9(b):

VAB = VAN(√

3/− 30) = 211.71/− 31.36 V

Find the line voltage at the load for the b- and c-phases using the acb sequence:

VBC = 211.71/− 31.36 + 120 = 211.71/88.69 V

VCA = 211.71/− 31.36 − 120 = 211.71/− 151.36 V

P 11.11 [a] IAB =480

60 + j45= 6.4/− 36.87 A

IBC = 6.4/− 156.87 A

ICA = 6.4/83.13 A

[b] IaA =√

3/− 30IAB = 11.09/− 66.87 A

IbB = 11.09/173.13 A

IcC = 11.09/53.13 A


[c] Transform the ∆-connected load to a Y-connected load:

ZY =Z∆

3=

60 + j453

= 20 + j15 Ω

The single-phase equivalent circuit is:

Van = 277.13/− 30 + (0.8 + j0.6)(11.09/− 66.87)

= 288.21/− 30 V

Vab =√

3/30Van = 499.20/0 V

Vbc = 499.20/− 120 V

Vca = 499.20/120 V

P 11.12 [a]

IaA =7650

72 + j21+

765050

= 252.54/− 6.49 A

|IaA| = 252.54 A

[b] IAB =7650

√3/30

150= 88.33/30 A

|IAB| = 88.33 A

Problems 11–13

[c] IAN =7650/0

72 + j21= 102/− 16.26 A

|IAN| = 102 A

[d] Van = (252.54/− 6.49)(j1) + 7650/0 = 7682.66/1.87 V

|Vab| =√

3(7682.66) = 13,306.76 V

P 11.13 [a] Since the phase sequence is acb (negative) we have:

Van = 2399.47/30 V

Vbn = 2399.47/150 V

Vcn = 2399.47/− 90 V

ZY =13Z∆ = 0.9 + j4.5 Ω/φ

[b] Vab = 2399.47/30 − 2399.47/150 = 2399.47√

3/0 = 4156/0 V

Since the phase sequence is negative, it follows that

Vbc = 4156/120 V

Vca = 4156/− 120 V


[c]

Iba =4156

2.7 + j13.5= 301.87/− 78.69 A

Iac = 301.87/− 198.69 A

IaA = Iba − Iac = 522.86/− 48.69 A

Since we have a balanced three-phase circuit and a negative phase sequence wehave:

IbB = 522.86/71.31 A

IcC = 522.86/− 168.69 A

Problems 11–15

[d]

IaA =2399.47/30

0.9 + j4.5= 522.86/− 48.69 A

Since we have a balanced three-phase circuit and a negative phase sequence wehave:

IbB = 522.86/71.31 A

IcC = 522.86/− 168.69 A

P 11.14 [a]

[b] IaA =2399.47/30

1920 − j560= 1.2/46.26 A

VAN = (1910 − j636)(1.2/46.26) = 2415.19/27.84 V

|VAB| =√

3(2415.19) = 4183.23 V


[c] |Iab| =1.2√

3= 0.69 A

[d] Van = (1919.1 − j564.5)(1.2/46.26) = 2400/29.87 V

|Vab| =√

3(2400) = 4156.92 V

P 11.15 [a]

[b] IaA =13,800√

3(2.375 + j1.349)= 2917/− 29.6 A

|IaA| = 2917 A

[c] VAN = (2.352 + j1.139)(2917/− 29.6) = 7622.94/− 3.76 V

|VAB| =√

3|VAN| = 13,203.31 V

[d] Van = (2.372 + j1.319)(2917/− 29.6) = 7616.93/− 0.52 V

|Vab| =√

3|Van| = 13,712.52 V

[e] |IAB| =|IaA|√

3= 1684.13 A

[f] |Iab| = |IAB| = 1684.13 A

P 11.16 [a] IAB =4160/0

160 + j120= 20.8/− 36.87 A

IBC = 20.8/83.13 A

ICA = 20.8/− 156.87 A

[b] IaA =√

3/30IAB = 36.03/− 6.87 A

IbB = 36.03/113.13 A

IcC = 36.03/− 126.87 A

[c] Iba = IAB = 20.8/− 36.87 A;

Icb = IBC = 20.8/83.13 A;

Iac = ICA = 20.8/− 156.87 A;

Problems 11–17

P 11.17 [a] IAB =480/0

2.4 − j0.7= 192/16.26 A

IBC =480/120

8 + j6= 48/83.13 A

ICA =480/− 120

20= 24/− 120 A

[b] IaA = IAB − ICA

= 210/20.79 A

IbB = IBC − IAB

= 178.68/− 178.04 A

IcC = ICA − IBC

= 70.7/− 104.53 A

P 11.18 From the solution to Problem 11.17 we have:

SAB = (480/0)(192/− 16.26) = 88,473.6 − j25,804.8 VA

SBC = (480/120)(48/− 83.13) = 18,432.0 + j13,824.0 VA

SCA = (480/− 120)(24/120) = 11,520 + j0 VA

P 11.19 [a]

I1 =24,000

√3/0

400 + j300= 66.5 − j49.9 A

I2 =24,000

√3/0

800 − j600= 33.3 + j24.9 A

I∗3 =

57,600 + j734,40024,000

√3

= 1.4 + j17.7

I3 = 1.4 − j17.7 A


IaA = I1 + I2 + I3 = 101.2 − j42.7 A = 109.8/− 22.8 A

Van = (2 + j16)(101.2 − j42.7) + 24,000√

3 = 42,456.2 + j1533.2 V

Sφ = VanI∗aA = (42,456.2 + j1533.8)(101.2 + j42.7)

= 4,229.2 + j1964.0 kVA

ST = 3Sφ = 12,687.7 + j9892.1 kVA

[b] S1/φ = 24,000√

3(66.5 + j49.9) = 2765.0 + j2073.8 kVA

S2/φ = 24,000√

3(33.3 − j24.9) = 1382.5 − j1036.9 kVA

S3/φ = 57.6 + j734.4 kVA

Sφ(load) = 4205.1 + j1771.3 kVA

% delivered =(4205.1

4229.2

)(100) = 99.4%

P 11.20 [a]

IaA =1365/0

30 + j40= 27.3/− 53.13 A

ICA =IaA√

3/150 = 15.76/96.87 A

[b] Sg/φ = −1365I∗aA = −22,358.7 − j29,811.6 VA

·. . Pdeveloped/phase = 22.359 kW

Pabsorbed/phase = |IaA|228.5 = 21.241 kW

% delivered =21.24122.359

(100) = 95%

Problems 11–19

P 11.21 Let pa, pb, and pc represent the instantaneous power of phases a, b, and c,respectively. Then assuming a positive phase sequence, we have

pa = vaniaA = [Vm cos ωt][Im cos(ωt − θφ)]

pb = vbnibB = [Vm cos(ωt − 120)][Im cos(ωt − θφ − 120)]

pc = vcnicC = [Vm cos(ωt + 120)][Im cos(ωt − θφ + 120)]

The total instantaneous power is pT = pa + pb + pc, so

pT = VmIm[cos ωt cos(ωt − θφ) + cos(ωt − 120) cos(ωt − θφ − 120)

+ cos(ωt + 120) cos(ωt − θφ + 120)]

Now simplify using trigonometric identities. In simplifying, collect the coefficientsof cos(ωt − θφ) and sin(ωt − θφ). We get

pT = VmIm[cos ωt(1 + 2 cos2 120) cos(ωt − θφ)

+2 sin ωt sin2 120 sin(ωt − θφ)]

= 1.5VmIm[cos ωt cos(ωt − θφ) + sin ωt sin(ωt − θφ)]

= 1.5VmIm cos θφ

P 11.22 [a] S1/φ = 40,000(0.96) − j40,000(0.28) = 38,400 − j11,200 VA

S2/φ = 60,000(0.8) + j60,000(0.6) = 48,000 + j36,000 VA

S3/φ = 33,600 + j5200 VA

ST/φ = S1 + S2 + S3 = 120,000 + j30,000 VA


·. . I∗aA =

120,000 + j30,0002400

= 50 + j12.5

·. . IaA = 50 − j12.5 A

Van = 2400 + (50 − j12.5)(1 + j8) = 2550 + j387.5 = 2579.27/8.64 V

|Vab| =√

3(2579.27) = 4467.43 V

[b] Sg/φ = (2550 + j387.5)(50 + j12.5) = 122,656.25 + j51,250 VA

% efficiency =120,000

122,656.25(100) = 97.83%

P 11.23 [a] S1 = (4.864 + j3.775) kVA

S2 = 17.636(0.96) + j17.636(0.28) = (16.931 + j4.938) kVA√

3VLIL sin θ3 = 13,853; sin θ3 =13,853√

3(208)(73.8)= 0.521

Therefore cos θ3 = 0.854

Therefore

P3 =13,8530.521

× 0.854 = 22,693.58 W

S3 = 22.694 + j13.853 kVA

ST = S1 + S2 + S3 = 44.49 + j22.57 kVA

ST/φ =13ST = 14.83 + j7.52 kVA

208√3

I∗aA = (14.83 + j7.52)103; I∗

aA = 123.49 + j62.64 A

IaA = 123.49 − j62.64 = 138.46/− 26.90 A (rms)

[b] pf = cos(−26.90) = 0.892 lagging

P 11.24

4000I∗1 = (210 + j280)103

Problems 11–21

I∗1 =

2104

+ j2804

= 52.5 + j70 A

I1 = 52.5 − j70 A

I2 =4000/0

15.36 − j4.48= 240 + j70 A

·. . IaA = I1 + I2 = 292.5 + j0 A

Van = 4000 + j0 + 292.5(0.1 + j0.8) = 4036.04/3.32 V

|Vab| =√

3|Van| = 6990.62 V

P 11.25 [a] POUT = 746 × 100 = 74,600 W

PIN = 74,600/(0.97) = 76,907.22 W√

3VLIL cos θ = 76,907.22

IL =76,907.22√3(208)(0.88)

= 242.58 A

[b] Q =√

3VLIL sin φ =√

3(208)(242.58)(0.475) = 41,510.12 VAR

P 11.26 [a] I∗aA =

(160 + j46.67)103

1200= 133.3 + j38.9

IaA = 133.3 − j38.9 A

Van = 1200 + (133.3 − j38.9)(0.18 + j1.44) = 1280 + j185 V

IC =1280 + j185

−j60= −3.1 + j21.3 A

Ina = (IaA + IC) = −130.3 − j17.6 = 131.4/7.7 A


[b] Sg/φ = (1280 + j185)(−130.3 − j17.6) = −163,472 − j46,567.4 VA

SgT = 3Sg/φ = −490.4 − j139.7 kVA

Therefore, the source is delivering 490.4 kW and 139.7 kvars.

[c] Pdel = 490.4 kW

Pabs = 3(160,000) + 3|IaA|2(0.18)

= 490.4 kW = Pdel

[d] Qdel = 3|IC|2(60) + 139.7 × 103 = 223.3 kVAR

Qabs = 3(46,666) + 3|IaA|2(1.44)

= 223.3 kVAR = Qdel

P 11.27 [a]

Ss/φ =13(60)(0.96 − j0.28) × 103 = 19.2 − j5.6 kVA

S1/φ = 15 kVA

S2/φ = Ss/φ − S1/φ = 4.2 − j5.6 kVA

·. . I∗2 =

4200 − j5600630/

√3

= 11.547 − j15.396 A

I2 = 11.547 + j15.396 A

Zy =630/0 /

√3

I2= 11.34 − j15.12 Ω

Z∆ = 3Zy = 34.02 − j45.36 Ω

Problems 11–23

[b] R =(630/

√3)2

4200= 31.5 Ω; R∆ = 3R = 94.5 Ω

XL =(630/

√3)2

−5600= −23.625 Ω; X∆ = 3XL = −70.875 Ω

P 11.28 Assume a ∆-connect load (series):

Sφ =13(96 × 103)(0.8 + j0.6) = 25,600 + j19,200 VA

Z∗∆φ =

|480|225,600 + j19,200

= 5.76 − j4.32 Ω

Z∆φ = 5.76 + 4.32 Ω

Now assume a Y-connected load (series):

ZY φ =13Z∆φ = 1.92 + j1.44 Ω


Now assume a ∆-connected load (parallel):

Pφ =|480|2R∆

R∆φ =|480|225,600

= 9 Ω

Qφ =|480|2X∆

X∆φ =|480|219,200

= 12 Ω

Now assume a Y-connected load (parallel):

RY φ =13R∆φ = 3 Ω

XY φ =13X∆φ = 4 Ω

Problems 11–25

P 11.29 Sg/φ =13(41.6)(0.707 + j0.707) × 103 = 9803.73 + j9806.69 VA

I∗aA =

9803.73 + j9803.73240/

√3

= 70.75 + j70.77 A

IaA = 70.75 − j70.77 A

VAN =240√

3− (0.04 + j0.03)(70.75 − j70.77)

= 133.61 + j0.71 = 133.61/0.30 V

|VAB| =√

3(133.61) = 231.42 V

[b] SL/φ = (133.61 + j0.71)(70.76 + j70.76) = 9403.1 + j9506.3 VA

SL = 3SL/φ = 28,209 + j28,519 VA

Check:

Sg = 41,600(0.707 + j0.707) = 29,411 + j29,420 VA

P = 3|IaA|2(0.04) = 1202 W

Pg = PL + P = 28,209 + 1202 = 29,411 W (checks)

Q = 3|IaA|2(0.03) = 901 VAR

Qg = QL + Q = 28,519 + 901 = 29,420 VAR (checks)


P 11.30 [a]

SL/φ =13

[720 + j

7200.8

(0.6)]103 = 240,000 + j180,000 VA

I∗aA =

240,000 + j180,0002400

= 100 + j75 A

IaA = 100 − j75 A

Van = 2400 + (0.8 + j6.4)(100 − j75)

= 2960 + j580 = 3016.29/11.09 V

|Vab| =√

3(3016.29) = 5224.37 V

[b]

I1 = 100 − j75 A (from part [a])

S2 = 0 − j13(576) × 103 = −j192,000 VAR

I∗2 =

−j192,0002400

= −j80 A

·. . I2 = j80 A

IaA = 100 − j75 + j80 = 100 + j5 A

Van = 2400 + (100 + j5)(0.8 + j6.4)

= 2448 + j644 = 2531.29/14.74 V

|Vab| =√

3(2531.29) = 4384.33 V

Problems 11–27

[c] |IaA| = 125 A

Ploss/φ = (125)2(0.8) = 12,500 W

Pg/φ = 240,000 + 12,500 = 252.5 kW

% η =240

252.5(100) = 95.05%

[d] |IaA| = 100.125 A

P/φ = (100.125)2(0.8) = 8020 W

% η =240,000248,200

(100) = 96.77%

[e] Zcap/Y = −j24002

−192,000= −j30 Ω

Zcap/∆ = 3Zcap/Y = −j90 Ω

·. .1

ωC= 90; C =

1(90)(120π)

= 29.47 µF

P 11.31 [a] From Assessment Problem 11.9, IaA = (101.8 − j135.7) A

Therefore Icap = j135.7 A

Therefore ZCY =2450/

√3

j135.7= −j10.42 Ω

Therefore CY =1

(10.42)(2π)(60)= 254.5 µF

ZC∆ = (−j10.42)(3) = −j31.26 Ω

Therefore C∆ =254.5

3= 84.84 µF

[b] CY = 254.5 µF

[c] |IaA| = 101.8 A


P 11.32 Zφ = |Z|/θ =VAN

IaA

θ = /VAN − /IaA

θ1 = /VAB − /IaA

For a positive phase sequence,

/VAB = /VAN + 30

Thus,

θ1 = /VAN + 30 − /IaA = θ + 30

Similarly,

Zφ = |Z|/θ =VCN

IcC

θ = /VCN − /IcC

θ2 = /VCB − /IcC

For a positive phase sequence,

/VCB = /VBA − 120 = /VAB + 60

/IcC = /IaA + 120

Thus,

θ2 = /VAB + 60 − /IaA − 120 = θ1 − 60

= θ + 30 − 60 = θ − 30

P 11.33 Use values from the negative sequence part of Example 11.1 — part (g):

VAB = 199.58/− 31.19 V

IaA = 2.5/− 36.87 A

Wm1 = |VAB||IaA| cos(/VAB − /IaA) = (199.58)(2.4) cos(5.68) = 476.63 W

Wm2 = |VCB||IcC| cos(/VCB − /IcC) = (199.58)(2.4) cos(65.68) = 197.29 W

CHECK: W1 + W2 = 673.9 = (2.4)2(39)(3) = 673.9 W

Problems 11–29

P 11.34 [a] W2 − W1 = VLIL[cos(θ − 30) − cos(θ + 30)]

= VLIL[cos θ cos 30 + sin θ sin 30

− cos θ cos 30 + sin θ sin 30]

= 2VLIL sin θ sin 30 = VLIL sin θ,

therefore√

3(W2 − W1) =√

3VLIL sin θ = QT

[b] Zφ = (8 + j6) Ω

QT =√

3[2476.25 − 979.75] = 2592 VAR,

QT = 3(12)2(6) = 2592 VAR;

Zφ = (8 − j6) Ω

QT =√

3[979.75 − 2476.25] = −2592 VAR,

QT = 3(12)2(−6) = −2592 VAR;

Zφ = 5(1 + j√

3) Ω

QT =√

3[2160 − 0] = 3741.23 VAR,

QT = 3(12)2(5√

3) = 3741.23 VAR;

Zφ = 10/− 75 Ω

QT =√

3[−645.53 − 1763.63] = −4172.80 VAR,

QT = 3(12)2[−10 sin 75] = −4172.80 VAR

P 11.35 IaA = (VAN/Zφ) = |IL|/−θφ A,

Zφ = |Z|/θφ, VBC = |VL|/− 90 V,

Wm = |VL| |IL| cos[−90 − (−θφ)]

= |VL| |IL| cos(θφ − 90)

= |VL| |IL| sin θφ,

therefore√

3Wm =√

3|VL| |IL| sin θφ = Qtotal


P 11.36 [a] Z = 16 − j12 = 20/− 36.87 Ω

VAN = 680/0 V; ·. . IaA = 34/36.87 A

VBC = VBN − VCN = 680√

3/− 90 V

Wm = (680√

3)(34) cos(−90 − 36.87) = −24,027.0 W√

3Wm = −41,616.0 VAR

[b] Qφ = (342)(−12) = −13,872 VAR

QT = 3Qφ = −41,616 VAR =√

3Wm

P 11.37 [a] Zφ = 160 + j120 = 200/36.87 Ω

Sφ =41602

160 − j120= 69,222.4 + j51,916.8 VA

ST = 3Sφ = 207,667.2 + j155,750.4 VA

[b] Wm1 = (4160)(36.03) cos(0 + 6.87) = 148,808.64 W

Wm2 = (4160)(36.03) cos(−60 + 126.87) = 58,877.55 W

Check: PT = 207.7 kW = Wm1 + Wm2.

P 11.38 [a] I∗aA =

144(0.96 − j0.28)103

7200= 20/− 16.26 A

VBN = 7200/− 120 V; VCN = 7200/120 V

VBC = VBN − VCN = 7200√

3/− 90 V

IbB = 20/− 103.74 A

Wm1 = (7200√

3)(20) cos(−90 + 103.74) = 242,278.14 W

[b] Current coil in line aA, measure IaA.Voltage coil across AC, measure VAC.

[c] IaA = 20/16.76 A

VAC = VAN − VCN = 7200√

3/− 30 V

Wm2 = (7200√

3)(20) cos(−30 − 16.26) = 172,441.86 W

[d] Wm1 + Wm2 = 414.72kW

PT = 432,000(0.96) = 414.72 kW = Wm1 + Wm2

Problems 11–31

P 11.39 [a] W1 = |VBA||IbB| cos θ1

Negative phase sequence:

VBA = 240√

3/150 V

IaA =240/0

13.33/− 30 = 18/30 A

IbB = 18/150 A

W1 = (18)(240)√

3 cos 0 = 7482.46 W

W2 = |VCA||IcC| cos θ2

VCA = 240√

3/− 150 V

IcC = 18/− 90 A

W2 = (18)(240)√

3 cos(−60) = 3741.23 W

[b] Pφ = (18)2(40/3) cos(−30) = 3741.23 W

PT = 3Pφ = 11,223.69 W

W1 + W2 = 7482.46 + 3741.23 = 11,223.69 W

·. . W1 + W2 = PT (checks)

P 11.40 [a] Negative phase sequence:

VAB = 240√

3/− 30 V

VBC = 240√

3/90 V

VCA = 240√

3/− 150 V

IAB =240

√3/− 30

20/30 = 20.78/− 60 A

IBC =240

√3/90

60/0 = 6.93/90 A

ICA =240

√3/− 150

40/− 30 = 10.39/− 120 A

IaA = IAB + IAC = 18/− 30 A

IcC = ICB + ICA = ICA − IBC = 16.75/− 108.06 A

Wm1 = 240√

3(18) cos(−30 + 30) = 7482.46 W

Wm2 = 240√

3(16.75) cos(−90 + 108.07) = 6621.23 W


[b] Wm1 + Wm2 = 14,103.69 W

PA = (12√

3)2(20 cos 30) = 7482.46 W

PB = (4√

3)2(60) = 2880 W

PC = (6√

3)2[40 cos(−30)] = 3741.23 W

PA + PB + PC = 14,103.69 = Wm1 + Wm2

P 11.41 tan φ =√

3(W2 − W1)W1 + W2

= 0.7498

·. . φ = 36.86

·. . 2400|IL| cos 66.87 = 40,823.09

|IL| = 43.3 A

|Zφ| =2400/

√3

43.3= 32 Ω ·. . Zφ = 32/36.87 Ω

P 11.42 [a] Z =13Z∆ = 4.48 + j15.36 = 16/73.74 Ω

IaA =600/0

16/73.74 = 37.5/− 73.74 A

IbB = 37.5/− 193.74 A

VAC = 600√

3/− 30 V

VBC = 600√

3/− 90 V

W1 = (600√

3)(37.5) cos(−30 + 73.74) = 28,156.15 W

W2 = (600√

3)(37.5) cos(−90 + 193.74) = −9256.15 W

[b] W1 + W2 = 18,900 W

PT = 3(37.5)2(13.44/3) = 18,900 W

[c]√

3(W1 − W2) = 64,800 VAR

QT = 3(37.5)2(46.08/3) = 64,800 VAR

Problems 11–33

P 11.43 From the solution to Prob. 11.17 we have

IaA = 210/20.79 A and IbB = 178.68/− 178.04 A

[a] W1 = |Vac| |IaA| cos(θac − θaA)

= 480(210) cos(60 − 20.79) = 78,103.2 W

[b] W2 = |Vbc| |IbB| cos(θbc − θbB)

= 480(178.68) cos(120 + 178.04) = 40,317.7 W

[c] W1 + W2 = 118,421 W

PAB = (192)2(2.4) = 88,473.6 W

PBC = (48)2(8) = 18,432 W

PCA = (24)2(20) = 11,520 W

PAB + PBC + PCA = 118,425.7

therefore W1 + W2 ≈ Ptotal (round-off differences)

P 11.44 [a] For one phase,

[b]

[c]


[d]

P 11.45 [a] Q =|V|2XC

·. . |XC| =(13,800)2

1.2 × 106 = 158.70 Ω

·. .1

ωC= 158.70; C =

12π(60)(158.70)

= 16.71 µF

[b] |XC| =(13,800/

√3)2

1.2 × 106 =13(158.70)

·. . C = 3(16.71) = 50.14 µF

P 11.46 If the capacitors remain connected when the substation drops its load, the expressionfor the line current becomes

13,800√3

I∗aA = −j1.2 × 106

or I∗aA = −j150.61 A

Hence IaA = j150.61 A

Now,

Van =13,800√

3/0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71 V

The magnitude of the line-to-line voltage at the generating plant is

|Vab| =√

3(7245.05) = 12,548.80 V.

This is a problem because the voltage is below the acceptable minimum of 13 kV.Thus when the load at the substation drops off, the capacitors must be switched off.

Problems 11–35

P 11.47 Before the capacitors are added the total line loss is

PL = 3|150.61 + j150.61|2(0.6) = 81.66 kW

After the capacitors are added the total line loss is

PL = 3|150.61|2(0.6) = 40.83 kW

Note that adding the capacitors to control the voltage level also reduces the amountof power loss in the lines, which in this example is cut in half.

P 11.48 [a]13,800√

3I∗aA = 80 × 103 + j200 × 103 − j1200 × 103

I∗aA =

80√

3 − j1000√

313.8

= 10.04 − j125.51 A

·. . IaA = 10.04 + j125.51 A

Van =13,800√

3/0 + (0.6 + j4.8)(10.04 + j125.51)

= 7371.01 + j123.50 = 7372.04/0.96 V

·. . |Vab| =√

3(7372.04) = 12,768.75 V

[b] Yes, the magnitude of the line-to-line voltage at the power plant is less than theallowable minimum of 13 kV.

P 11.49 [a]13,800√

3I∗aA = (80 + j200) × 103

I∗aA =

80√

3 + j200√

313.8

= 10.04 + j25.1 A

·. . IaA = 10.04 − j25.1 A

Van =13,800√

3/0 + (0.6 + j4.8)(10.04 − j25.1)

= 8093.95 + j33.13 = 8094.02/0.23 V

·. . |Vab| =√

3(8094.02) = 14,019.25 V

[b] Yes: 13 kV < 14,019.25 < 14.6 kV

[c] Ploss = 3|10.04 + j125.51|2(0.6) = 28.54 kW

[d] Ploss = 3|10.04 + j25.1|2(0.6) = 1.32 kW

[e] Yes, the voltage at the generating plant is at an acceptable level and the line lossis greatly reduced.

12Introduction to the Laplace Transform

Assessment Problems

AP 12.1 [a] cosh βt =eβt + e−βt

2Therefore,

Lcosh βt =12

∫ ∞

0−[e−(s−β)t + e−(s+β)t]dt

=12

[e−(s−β)t

−(s − β)

∣∣∣∣∞0−

+e−(s+β)t

−(s + β)

∣∣∣∣∞0−

]

=12

(1

s − β+

1s + β

)=

s

s2 − β2

[b] sinh βt =eβt − e−βt

2Therefore,

Lsinh βt =12

∫ ∞

0−

[e−(s−β)t − e−(s+β)t

]dt

=12

[e−(s−β)t

−(s − β)

]∞

0−− 1

2

[e−(s+β)t

−(s + β)

]∞

0−

=12

(1

s − β− 1

s + β

)=

β

(s2 − β2)

AP 12.2 [a] Let f(t) = te−at:

F (s) = Lte−at =1

(s + a)2

Now, Ltf(t) = −dF (s)ds

12–1

12–2 CHAPTER 12. Introduction to the Laplace Transform

So, Lt · te−at = − d

ds

[1

(s + a)2

]=

2(s + a)2

[b] Let f(t) = e−at sinh βt, then

Lf(t) = F (s) =β

(s + a)2 − β2

L

df(t)dt

= sF (s) − f(0−) =

s(β)(s + a)2 − β2 − 0 =

βs

(s + a)2 − β2

[c] Let f(t) = cos ωt. Then

F (s) =s

(s2 + ω2)and

dF (s)ds

=−(s2 − ω2)(s2 + ω2)2

Therefore Lt cos ωt = −dF (s)ds

=s2 − ω2

(s2 + ω2)2

AP 12.3 F (s) =6s2 + 26s + 26

(s + 1)(s + 2)(s + 3)=

K1

s + 1+

K2

s + 2+

K3

s + 3

K1 =6 − 26 + 26

(1)(2)= 3; K2 =

24 − 52 + 26(−1)(1)

= 2

K3 =54 − 78 + 26

(−2)(−1)= 1

Therefore f(t) = [3e−t + 2e−2t + e−3t] u(t)

AP 12.4 F (s) =7s2 + 63s + 134

(s + 3)(s + 4)(s + 5)=

K1

s + 3+

K2

s + 4+

K3

s + 5

K1 =63 − 189 + 134

1(2)= 4; K2 =

112 − 252 + 134(−1)(1)

= 6

K3 =175 − 315 + 134

(−2)(−1)= −3

f(t) = [4e−3t + 6e−4t − 3e−5t]u(t)

AP 12.5 F (s) =10(s2 + 119)

(s + 5)(s2 + 10s + 169)

s1,2 = −5 +√

25 − 169 = −5 + j12

Problems 12–3

F (s) =K1

s + 5+

K2

s + 5 − j12+

K∗2

s + 5 + j12

K1 =10(25 + 119)25 − 50 + 169

= 10

K2 =10[(−5 + j12)2 + 119]

(j12)(j24)= j4.167 = 4.167/90

Therefore

f(t) = [10e−5t + 8.33e−5t cos(12t + 90)] u(t)

= [10e−5t − 8.33e−5t sin 12t] u(t)

AP 12.6 F (s) =4s2 + 7s + 1

s(s + 1)2 =K0

s+

K1

(s + 1)2 +K2

s + 1

K0 =1

(1)2 = 1; K1 =4 − 7 + 1

−1= 2

K2 =d

ds

[4s2 + 7s + 1

s

]s=−1

=s(8s + 7) − (4s2 + 7s + 1)

s2

∣∣∣∣∣s=−1

=1 + 2

1= 3

Therefore f(t) = [1 + 2te−t + 3e−t] u(t)

AP 12.7 F (s) =40

(s2 + 4s + 5)2 =40

(s + 2 − j1)2(s + 2 + j1)2

=K1

(s + 2 − j1)2 +K2

(s + 2 − j1)+

K∗1

(s + 2 + j1)2

+K∗

2

(s + 2 + j1)

K1 =40

(j2)2 = −10 = 10/180 and K∗1 = −10

K2 =d

ds

[40

(s + 2 + j1)2

]s=−2+j1

=−80(j2)3 = −j10 = 10/− 90

K∗2 = j10


Therefore

f(t) = [20te−2t cos(t + 180) + 20e−2t cos(t − 90)] u(t)

= 20e−2t[sin t − t cos t] u(t)

AP 12.8 F (s) =5s2 + 29s + 32(s + 2)(s + 4)

=5s2 + 29s + 32

s2 + 6s + 8= 5 − s + 8

(s + 2)(s + 4)

s + 8(s + 2)(s + 4)

=K1

s + 2+

K2

s + 4

K1 =−2 + 8

2= 3; K2 =

−4 + 8−2

= −2

Therefore,

F (s) = 5 − 3s + 2

+2

s + 4

f(t) = 5δ(t) + [−3e−2t + 2e−4t]u(t)

AP 12.9 F (s) =2s3 + 8s2 + 2s − 4

s2 + 5s + 4= 2s − 2 +

4(s + 1)(s + 1)(s + 4)

= 2s − 2 +4

s + 4

f(t) = 2dδ(t)dt

− 2δ(t) + 4e−4t u(t)

AP 12.10

lims→∞ sF (s) = lim

s→∞

[7s3[1 + (9/s) + (134/7s2)]

s3[1 + (3/s)][1 + (4/s)][1 + (5/s)]

]= 7

·. . f(0+) = 7

lims→0

sF (s) = lims→0

[7s3 + 63s2 + 134s

(s + 3)(s + 4)(s + 5)

]= 0

·. . f(∞) = 0


s→∞

[s3[4 + (7/s) + (1/s2)]

s3[1 + (1/s)]2

]= 4

·. . f(0+) = 4

Problems 12–5

lims→0

sF (s) = lims→0

[4s2 + 7s + 1

(s + 1)2

]= 1

·. . f(∞) = 1


s→∞

[40s

s4[1 + (4/s) + (5/s2)]2

]= 0

·. . f(0+) = 0

lims→0

sF (s) = lims→0

[40s

(s2 + 4s + 5)2

]= 0

·. . f(∞) = 0


Problems

P 12.1 [a] f(t) = 5t[u(t) − u(t − 2)] + 10[u(t − 2) − u(t − 6)]+

(−5t + 40)[u(t − 6) − u(t − 8)]

[b] f(t) = (10 sin πt)[u(t) − u(t − 2)]

[c] f(t) = 4t[u(t) − u(t − 5)]

P 12.2 [a] (10 + t)[u(t + 10) − u(t)] + (10 − t)[u(t) − u(t − 10)]

= (t + 10)u(t + 10) − 2tu(t) + (t − 10)u(t − 10)

[b] (−24 − 8t)[u(t + 3) − u(t + 2)] − 8[u(t + 2) − u(t + 1)] + 8t[u(t + 1) − u(t − 1)]

+8[u(t − 1) − u(t − 2)] + (24 − 8t)[u(t − 2) − u(t − 3)]

= −8(t + 3)u(t + 3) + 8(t + 2)u(t + 2) + 8(t + 1)u(t + 1) − 8(t − 1)u(t − 1)

−8(t − 2)u(t − 2) + 8(t − 3)u(t − 3)

P 12.3

P 12.4 [a]

Problems 12–7

[b] f(t) = −20t[u(t) − u(t − 1)] − 20[u(t − 1) − u(t − 2)]

+20 cos(π2 t)[u(t − 2) − u(t − 4)]

+(100 − 20t)[u(t − 4) − u(t − 5)]

P 12.5 [a] A =(1

2

)bh =

(12

)(2ε)

(1ε

)= 1

[b] 0; [c] ∞

P 12.6 [a] I =∫ 3

−1(t3 + 2)δ(t) dt +

∫ 3

−18(t3 + 2)δ(t − 1) dt

= (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26

[b] I =∫ 2

−2t2δ(t) dt +

∫ 2

−2t2δ(t + 1.5) dt +

∫ 2

−2t2δ(t − 3) dt

= 02 + (−1.5)2 + 0 = 2.25

P 12.7 f(t) =12π

∫ ∞

−∞(4 + jω)(9 + jω)

· πδ(ω) · ejtω dω =( 1

2π

)(4 + j09 + j0

πejt0

)=

29

P 12.8 As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent of ε,i.e.,

A =∫ ∞

−∞ε

π

1ε2 + t2

dt = 1

P 12.9 F (s) =∫ ε

−ε

12ε

e−st dt =esε − e−sε

2εs

F (s) =12s

limε→0

[sesε + se−sε

1

]=

12s

· 2s1

= 1

P 12.10 [a] Let dv = δ′(t − a) dt, v = δ(t − a)

u = f(t), du = f ′(t) dt

Therefore∫ ∞

−∞f(t)δ′(t − a) dt = f(t)δ(t − a)

∣∣∣∣∞−∞−∫ ∞

−∞δ(t − a)f ′(t) dt

= 0 − f ′(a)

[b] Lδ′(t) =∫ ∞

0−δ′(t)e−st dt = −

[d(e−st)

dt

]t=0

= −[−se−st

]t=0

= s


P 12.11

F (s) =∫ −ε/2

−ε

4ε3 e−st dt +

∫ ε/2

−ε/2

(−4ε3

)e−st dt +

∫ ε

ε/2

4ε3 e−st dt

Therefore F (s) =4

sε3 [esε − 2esε/2 + 2e−sε/2 − e−sε]

Lδ′′(t) = limε→0

F (s)

After applying L’Hopital’s rule three times, we have

limε→0

2s3

[sesε − s

4esε/2 − s

4e−sε/2 + se−sε

]=

2s3

(3s2

)

Therefore Lδ′′(t) = s2

P 12.12 L

dnf(t)dtn

= snF (s) − sn−1f(0−) − sn−2f ′(0−) − · · · ,

Therefore

Lδ(n)(t) = sn(1) − sn−1δ(0−) − sn−2δ′(0−) − · · · = sn

P 12.13 [a] Lt =1s2 ; therefore Lte−at =

1(s + a)2

[b] sin ωt =ejωt − e−jωt

j2Therefore

Lsin ωt =(

1j2

)(1

s − jω− 1

s + jω

)=(

1j2

)( 2jωs2 + ω2

)

=ω

s2 + ω2

Problems 12–9

[c] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ)Therefore

Lsin(ωt + θ) = cos θLsin ωt + sin θLcos ωt=

ω cos θ + s sin θ

s2 + ω2

[d] Lt =∫ ∞

0te−st dt =

e−st

s2 (−st − 1)∣∣∣∣∞0

= 0 − 1s2 (0 − 1) =

1s2

[e] f(t) = cosh t cosh θ + sinh t sinh θ

From Assessment Problem 12.1(a)

Lcosh t =s

s2 − 1From Assessment Problem 12.1(b)

Lsinh t =1

s2 − 1

·. . Lcosh(t + θ) = cosh θ[

s

s2 − 1

]+ sinh θ

[ 1s2 − 1

]

=sinh θ + s[cosh θ]

s2 − 1

P 12.14 [a] Lte−at =∫ ∞

0−te−(s+a)t dt

=e−(s+a)t

(s + a)2

[− (s + a)t − 1

]∞0−

= 0 +1

(s + a)2

·. . Lte−at =1

(s + a)2

[b]

L

d

dt(te−at)

=

s

(s + a)2 − 0

=s

(s + a)2

[c]d

dt(te−at) = −ate−at + e−at

L−ate−at + e−at =−a

(s + a)2 +1

(s + a)=

−a

(s + a)2 +s + a

(s + a)2

·. . L

d

dt(te−at)

=

s

(s + a)2 CHECKS


P 12.15 [a] Lf ′(t) =∫ ε

−ε

e−st

2εdt +

∫ ∞

ε− ae−a(t−ε)e−st dt

=1

2sε(esε − e−sε) −

(a

s + a

)e−sε = F (s)

limε→0

F (s) = 1 − a

s + a=

s

s + a

[b] Le−at =1

s + a

Therefore Lf ′(t) = sF (s) − f(0−) =s

s + a− 0 =

s

s + a

P 12.16 Le−atf(t) =∫ ∞

0−[e−atf(t)]e−st dt =

∫ ∞

0−f(t)e−(s+a)t dt = F (s + a)

P 12.17 [a] L∫ t

0−e−ax dx

=

F (s)s

=1

s(s + a)

[b] L∫ t

0−y dy

=

1s

( 1s2

)=

1s3

[c]∫ t

0−e−ax dx =

1a

− e−at

a

L

1a

− e−at

a

=

1a

[1s

− 1s + a

]=

1s(s + a)

∫ t

0−y dy =

t2

2; L

t2

2

=

12

· 2s3 =

1s3

P 12.18 [a] L

d sin ωt

dt

=

sω

s2 + ω2 − 0

[b] L

d cos ωt

dt

=

s2

s2 + ω2 − 0

[c] L

d3(t2)dt3

= s3

( 2s3

)− s2(0) − s(0) − 2(0) = 2

[d]d sin ωt

dt= (cos ωt) · ω, Lω cos ωt =

ωs

s2 + ω2

d cos ωt

dt= −ω sin ωt + δ(t)

L−ω sin ωt + δ(t) = − ω2

s2 + ω2 + 1 =s2

s2 + ω2

d2(t2)dt2

= 2u(t);d3(t2)dt3

= 2δ(t); L2δ(t) = 2

Problems 12–11

P 12.19 [a] f(t) = 5t[u(t) − u(t − 2)]

+(20 − 5t)[u(t − 2) − u(t − 6)]

+(5t − 40)[u(t − 6) − u(t − 8)]

= 5tu(t) − 10(t − 2)u(t − 2)

+10(t − 6)u(t − 6) − 5(t − 8)u(t − 8)

·. . F (s) =5[1 − 2e−2s + 2e−6s − e−8s]

s2

[b]

f ′(t) = 5[u(t) − u(t − 2)] − 5[u(t − 2) − u(t − 6)]

+5[u(t − 6) − u(t − 8)]

= 5u(t) − 10u(t − 2) + 10u(t − 6) − 5u(t − 8)

Lf ′(t) =5[1 − 2e−2s + 2e−6s − e−8s]

s

[c]

f ′′(t) = 5δ(t) − 10δ(t − 2) + 10δ(t − 6) − 5δ(t − 8)

Lf ′′(t) = 5[1 − 2e−2s + 2e−6s − e−8s]


P 12.20 [a]∫ t

0−x dx =

t2

2

L

t2

2

=

12

∫ ∞

0−t2e−st dt

=12

[e−st

−s3 (s2t2 + 2st + 2)∣∣∣∣∣∞

0−

]

=1

2s3 (2) =1s3

·. . L∫ t

0−x dx

=

1s3

[b] L∫ t

0−x dx

=

Lts

=1/s2

s=

1s3

·. . L∫ t

0−x dx

=

1s3 CHECKS

P 12.21 [a] L40e−8(t−3)u(t − 3) =40e−3s

(s + 8)[b] First rewrite f(t) as

f(t) = (5t − 10)u(t − 2) + (40 − 10t)u(t − 4)

+(10t − 80)u(t − 8) + (50 − 5t)u(t − 10)

= 5(t − 2)u(t − 2) − 10(t − 4)u(t − 4)

+10(t − 8)u(t − 8) − 5(t − 10)u(t − 10)

·. . F (s) =5[e−2s − 2e−4s + 2e−8s − e−10s]

s2

P 12.22 Lf(at) =∫ ∞

0−f(at)e−st dt

Let u = at, du = a dt, u = 0− when t = 0−

and u = ∞ when t = ∞

Therefore Lf(at) =∫ ∞

0−f(u)e−(u/a)s du

a=

1aF (s/a)

P 12.23 [a] f1(t) = e−at sin ωt; F1(s) =ω

(s + a)2 + ω2

F (s) = sF1(s) − f1(0−) =sω

(s + a)2 + ω2 − 0

Problems 12–13

[b] f1(t) = e−at cos ωt; F1(s) =s + a

(s + a)2 + ω2

F (s) =F1(s)

s=

s + a

s[(s + a)2 + ω2]

[c]d

dt[e−at sin ωt] = ωe−at cos ωt − ae−at sin ωt

Therefore F (s) =ω(s + a) − ωa

(s + a)2 + ω2 =ωs

(s + a)2 + ω2

∫ t

0−e−ax cos ωx dx =

−ae−at cos ωt + ωe−at sin ωt + a

a2 + ω2

Therefore

F (s) =1

a2 + ω2

[ −a(s + a)(s + a)2 + ω2 +

ω2

(s + a)2 + ω2 +a

s

]

=s + a

s[(s + a)2 + ω2]

P 12.24 [a]dF (s)

ds=

d

ds

[∫ ∞

0−f(t)e−st dt

]= −

∫ ∞

0−tf(t)e−st dt

Therefore Ltf(t) = −dF (s)ds

[b]d2F (s)

ds2 =∫ ∞

0−t2f(t)e−st dt;

d3F (s)ds3 =

∫ ∞

0−−t3f(t)e−st dt

ThereforednF (s)

dsn= (−1)n

∫ ∞

0−tnf(t)e−st dt = (−1)nLtnf(t)

[c] Lt5 = Lt4t = (−1)4 d4

ds4

( 1s2

)=

120s6

Lt sin βt = (−1)1 d

ds

(β

s2 + β2

)=

2βs

(s2 + β2)2

Lte−t cosh t:

From Assessment Problem 12.1(a),

F (s) = Lcosh t =s

s2 − 1

dF

ds=

(s2 − 1)1 − s(2s)(s2 − 1)2 = − s2 + 1

(s2 − 1)2

Therefore − dF

ds=

s2 + 1(s2 − 1)2


Thus

Lt cosh t =s2 + 1

(s2 − 1)2

Le−tt cosh t =(s + 1)2 + 1

[(s + 1)2 − 1]2=

s2 + 2s + 2s2(s + 2)2

P 12.25 [a]∫ ∞

sF (u)du =

∫ ∞

s

[∫ ∞

0−f(t)e−ut dt

]du =

∫ ∞

0−

[∫ ∞

sf(t)e−ut du

]dt

=∫ ∞

0−f(t)

∫ ∞

se−ut du dt =

∫ ∞

0−f(t)

[e−tu

−t

∣∣∣∣∞s

]dt

=∫ ∞

0−f(t)

[−e−st

−t

]dt = L

f(t)

t

[b] Lt sin βt =2βs

(s2 + β2)2

therefore L

t sin βt

t

=∫ ∞

s

[2βu

(u2 + β2)2

]du

Let ω = u2 + β2, then ω = s2 + β2 when u = s, and ω = ∞ when u = ∞;also dω = 2u du, thus

L

t sin βt

t

= β

∫ ∞

s2+β2

[dω

ω2

]= β

(−1ω

) ∣∣∣∣∞s2+β2

=β

s2 + β2

P 12.26 Ig(s) =1.2s

s2 + 1;

1RC

= 1.6;1

LC= 1;

1C

= 1.6

V (s)R

+1L

V (s)s

+ C[sV (s) − v(0−)] = Ig(s)

V (s)[ 1R

+1Ls

+ sC]

= Ig(s)

V (s) =Ig(s)

1R

+ 1Ls

+ sC=

LsIg(s)LRs + 1 + s2LC

=1CsIg(s)

s2 + 1RC

s + 1LC

=(1.6)(1.2)s2

(s2 + 1.6s + 1)(s2 + 1)=

1.92s2

(s2 + 1.6s + 1)(s2 + 1)

P 12.27 [a]vo − Vdc

R+

1L

∫ t

0vo dx + C

dvo

dt= 0

·. . vo +R

L

∫ t

0vo dx + RC

dvo

dt= Vdc

Problems 12–15

[b] Vo +R

L

Vo

s+ RCsVo =

Vdc

s

·. . sLVo + RVo + RCLs2Vo = LVdc

·. . Vo(s) =(1/RC)Vdc

s2 + (1/RC)s + (1/LC)

[c] io =1L

∫ t

0vo dx

Io(s) =Vo

sL=

(1/RCL)Vdc

s[s2 + (1/RC)s + (1/LC)]

P 12.28 [a]1

LC=

1(200 × 10−3)(100 × 10−9)

= 50 × 106

1RC

=1

(5000)(100 × 10−9)= 2000

Vo(s) =70,000

s2 + 2000s + 50 × 106

s1,2 = −1000 ± j7000 rad/s

Vo(s) =70,000

(s + 1000 − j7000)(s + 1000 + j7000)

=K1

s + 1000 − j7000+

K∗1

s + 1000 + j7000

K1 =70,000j14,000

= 5/− 90

vo(t) = 10e−1000t cos(7000t − 90)u(t) V

= 10e−1000t sin(7000t)u(t) V

[b] Io(s) =35(10,000)

s(s + 1000 − j7000)(s + 1000 + j7000)

=K1

s+

K2

s + 1000 − j7000+

K∗2

s + 1000 + j7000

K1 =35(10,000)50 × 106 = 7 mA

K2 =35(10,000)

(−1000 + j7000)(j14,000)= 3.54/171.87 mA

io(t) = [7 + 7.07e−1000t cos(7000t + 171.87)]u(t) mA


P 12.29 [a] Idc =1L

∫ t

0vo dx +

vo

R+ C

dvo

dt

[b]Idc

s=

Vo(s)sL

+Vo(s)

R+ sCVo(s)

·. . Vo(s) =Idc/C

s2 + (1/RC)s + (1/LC)

[c] io = Cdvo

dt

·. . Io(s) = sCVo(s) =sIdc

s2 + (1/RC)s + (1/LC)

P 12.30 [a]1

RC=

1(1 × 103)(2 × 10−6)

= 500

1LC

=1

(12.5)(2 × 10−6)= 40,000

Vo(s) =500,000Idc

s + 500s + 40,000

=500,000Idc

(s + 100)(s + 400)

=15,000

(s + 100)(s + 400)

=K1

s + 100+

K2

s + 400

K1 =15,000300

= 50; K2 =15,000−300

= −50

Vo(s) =50

s + 100− 50

s + 400

vo(t) = [50e−100t − 50e−400t]u(t) V

[b] Io(s) =0.03s

(s + 100)(s + 400)

=K1

s + 100+

K2

s + 400

K1 =0.03(−100)

300= −0.01

K2 =0.03(−400)

−300= 0.04

Problems 12–17

Io(s) =−0.01s + 100

+0.04

s + 400

io(t) = (40e−400t − 10e−100t)u(t) mA

[c] io(0) = 40 − 10 = 30 mA

Yes. The initial inductor current is zero by hypothesis, the initial resistorcurrent is zero because the initial capacitor voltage is zero by hypothesis. Thusat t = 0 the source current appears in the capacitor.

P 12.31 [a] Cdv1

dt+

v1 − v2

R= ig

1L

∫ t

0v2 dτ +

v2 − v1

R= 0

or

Cdv1

dt+

v1

R− v2

R= ig

−v1

R+

v2

R+

1L

∫ t

0v2 dτ = 0

[b] CsV1(s) +V1(s)

R− V2(s)

R= Ig(s)

−V1(s)R

+V2(s)

R+

V2(s)sL

= 0

or

(RCs + 1)V1(s) − V2(s) = RIg(s)

−sLV1(s) + (R + sL)V2(s) = 0

Solving,

V2(s) =sIg(s)

C[s2 + (R/L)s + (1/LC)]

P 12.321C

= 5 × 106;1

LC= 25 × 106;

R

L= 8000

V2(s) =(6 × 10−3)(5 × 106)

s2 + 8000s + 25 × 106

s1,2 = −4000 ± j3000

V2(s) =30,000

(s + 4000 − j3000)(s + 4000 + j3000)

=K1

s + 4000 − j3000+

K∗1

s + 4000 + j3000


K1 =30,000j6000

= −j5 = 5/− 90

v2(t) = 10e−4000t cos(3000t − 90)

= [10e−4000t sin 3000t]u(t) V

P 12.33 [a] For t ≥ 0+:

vo

R+ C

dvo

dt+ io = 0

vo = Ldiodt

;dvo

dt= L

d2iodt2

·. .L

R

diodt

+ LCd2iodt2

ord2iodt2

+1

RC

diodt

+1

LCio = 0

[b] s2Io(s) − sIdc − 0 +1

RC[sIo(s) − Idc] +

1LC

Io(s) = 0

Io(s)[s2 +

1RC

s +1

LC

]= Idc(s + 1/RC)

Io(s) =Idc[s + (1/RC)]

[s2 + (1/RC)s + (1/LC)]

P 12.341

RC= 8000;

1LC

= 16 × 106

Io(s) =0.005(s + 8000)

s2 + 8000s + 16 × 106

s1,2 = −4000

Io(s) =0.005(s + 8000)

(s + 4000)2 =K1

(s + 4000)2 +K2

s + 4000

K1 = 0.005(s + 8000)∣∣∣∣s=−4000

= 20

K2 =d

ds[0.005(s + 8000)]s=−4000 = 0.005

Io(s) =20

(s + 4000)2 +0.005

s + 4000

io(t) = [20te−4000t + 0.005e−4000t]u(t) V

Problems 12–19

P 12.35 [a] 300 = 60i1 + 25di1dt

+ 10d

dt(i2 − i1) + 5

d

dt(i1 − i2) − 10

di1dt

0 = 5d

dt(i2 − i1) + 10

di1dt

+ 40i2

Simplifying the above equations gives:

300 = 60i1 + 10di1dt

+ 5di2dt

0 = 40i2 + 5di1dt

+ 5di2dt

[b]300s

= (10s + 60)I1(s) + 5sI2(s)

0 = 5sI1(s) + (5s + 40)I2(s)

[c] Solving the equations in (b),

I1(s) =60(s + 8)

s(s + 4)(s + 24)

I2(s) =−60

(s + 4)(s + 24)

[d] I1(s) =K1

s+

K2

s + 4+

K3

s + 24

K1 =(60)(8)(4)(24)

= 5; K2 =(60)(4)

(−4)(20)= −3

K3 =(60)(−16)

(−24)(−20)= −2

I1(s) =(5

s− 3

s + 4− 2

s + 24

)

i1(t) = (5 − 3e−4t − 2e−24t)u(t) A

I2(s) =K1

s + 4+

K2

s + 24

K1 =−6020

= −3; K2 =−60−20

= 3

I2(s) =( −3

s + 4+

3s + 24

)

i2(t) = (3e−24t − 3e−4t)u(t) A

[e] i1(∞) = 5 A; i2(∞) = 0 A


[f] Yes, at t = ∞

i1 =30060

= 5 A

Since i1 is a dc current at t = ∞ there is no voltage induced in the 10 Hinductor; hence, i2 = 0. Also note that i1(0) = 0 and i2(0) = 0. Thus oursolutions satisfy the condition of no initial energy stored in the circuit.

P 12.36 From Problem 12.26:

V (s) =1.92s2

(s2 + 1.6s + 1)(s2 + 1)

s2 + 1.6s + 1 = (s + 0.8 + j0.6)(s + 0.8 − j0.6); s2 + 1 = (s − j1)(s + j1)

Therefore

V (s) =1.92s2

(s + 0.8 + j0.6)(s + 0.8 − j0.6)(s − j1)(s + j1)

=K1

s + 0.8 − j0.6+

K∗1

s + 0.8 + j0.6+

K2

s − j1+

K∗2

s + j1

K1 =1.92s2

(s + 0.8 + j0.6)(s2 + 1)

∣∣∣∣s=−0.8+j0.6

= 1/− 126.87

K2 =1.92s2

(s + j1)(s2 + 1.6s + 1)

∣∣∣∣s=j1

= 0.6/0

Therefore

v(t) = [2e−0.8t cos(0.6t − 126.87) + 1.2 cos(t)]u(t) V

P 12.37 [a] F (s) =K1

s + 1+

K2

s + 2+

K3

s + 4

K1 =8s2 + 37s + 32(s + 2)(s + 4)

∣∣∣∣s=−1

= 1

K2 =8s2 + 37s + 32(s + 1)(s + 4)

∣∣∣∣s=−2

= 5

K3 =8s2 + 37s + 32(s + 1)(s + 2)

∣∣∣∣s=−4

= 2

f(t) = [e−t + 5e−2t + 2e−4t]u(t)

Problems 12–21

[b] F (s) =K1

s+

K2

s + 2+

K3

s + 3+

K4

s + 5

K1 =8s3 + 89s2 + 311s + 300

(s + 2)(s + 3)(s + 5)

∣∣∣∣s=0

= 10

K2 =8s3 + 89s2 + 311s + 300

s(s + 3)(s + 5)

∣∣∣∣s=−2

= 5

K3 =8s3 + 89s2 + 311s + 300

s(s + 2)(s + 5)

∣∣∣∣s=−3

= −8

K4 =8s3 + 89s2 + 311s + 300

s(s + 2)(s + 3)

∣∣∣∣s=−5

= 1

f(t) = [10 + 5e−2t − 8e−3t + e−5t]u(t)

[c] F (s) =K1

s + 1+

K2

s + 2 − j+

K∗2

s + 2 + j

K1 =22s2 + 60s + 58

s2 + 4s + 5

∣∣∣∣s=−1

= 10

K2 =22s2 + 60s + 58

(s + 1)(s + 2 + j)

∣∣∣∣s=−2+j

= 6 + j8 = 10/53.13

f(t) = [10e−t + 20e−2t cos(t + 53.13)]u(t)

[d] F (s) =K1

s+

K2

s + 7 − j+

K∗2

s + 7 + j

K1 =250(s + 7)(s + 14)

s2 + 14s + 50

∣∣∣∣s=0

= 490

K2 =250(s + 7)(s + 14)

s(s + 7 + j)

∣∣∣∣s=−7+j

= 125/− 163.74

f(t) = [490 + 250e−7t cos(t − 163.74)]u(t)

P 12.38 [a] F (s) =K1

s2 +K2

s+

K3

s + 5

K1 =100

s + 5

∣∣∣∣s=0

= 20

K2 =d

ds

[ 100s + 5

]=

−100(s + 5)2

∣∣∣∣s=0

= −4

K3 =100s2

∣∣∣∣s=−5

= 4

f(t) = [20t − 4 + 4e−5t]u(t)


[b] F (s) =K1

s+

K2

(s + 1)2 +K3

s + 1

K1 =50(s + 5)(s + 1)2

∣∣∣∣s=0

= 250

K2 =50(s + 5)

s

∣∣∣∣s=−1

= −200

K3 =d

ds

[50(s + 5)

s

]=[50s

− 50(s + 5)s2

]s=−1

= −250

f(t) = [250 − 200te−t − 250e−t]u(t)

[c] F (s) =K1

s2 +K2

s+

K3

s + 3 − j+

K∗3

s + 3 + j

K1 =100(s + 3)

s2 + 6s + 10

∣∣∣∣s=0

= 30

K2 =d

ds

[100(s + 3)

s2 + 6s + 10

]

=[

100s2 + 6s + 10

− 100(s + 3)(2s + 6)(s2 + 6s + 10)2

]s=0

= 10 − 18 = −8

K3 =100(s + 3)

s2(s + 3 + j)

∣∣∣∣s=−3+j

= 4 + j3 = 5/36.87

f(t) = [30t − 8 + 10e−3t cos(t + 36.87)]u(t)

[d] F (s) =K1

s+

K2

(s + 1)3 +K3

(s + 1)2 +K4

s + 1

K1 =5(s + 2)2

(s + 1)3

∣∣∣∣s=0

= 20

K2 =5(s + 2)2

s

∣∣∣∣s=−1

= −5

K3 =d

ds

[5(s + 2)2

s

]=[10(s + 2)

s− 5(s + 2)2

s2

]s=−1

= −10 − 5 = −15

K4 =12

d

ds

[10(s + 2)

s− 5(s + 2)2

s2

]

=12

[10s

− 10(s + 2)s2 − 10(s + 2)

s2 +10(s + 2)2

s3

]s=−1

Problems 12–23

=12(−10 − 10 − 10 − 10) = −20

f(t) = [20 − 2.5t2e−t − 15te−t − 20e−t]u(t)

[e] F (s) =K1

s+

K2

(s + 2 − j)2 +K∗

2

(s + 2 + j)2 +K3

s + 2 − j+

K∗3

s + 2 − j

K1 =400

(s2 + 4s + 5)2

∣∣∣∣s=0

= 16

K2 =400

s(s + 2 + j)2

∣∣∣∣s=−2+j

= 44.72/26.57

K3 =d

ds

[400

s(s + 2 + j)2

]=[ −400s2(s + 2 + j)2 +

−800s(s + 2 + j)3

]s=−2+j

= 12 + j16 − 20 + j40 = −8 + j56 = 56.57/98.13

f(t) = [16 + 89.44te−2t cos(t + 26.57) + 113.14e−2t cos(t + 98.13)]u(t)

P 12.39 [a] 5

F (s) = s2 + 6s + 8 5s2 + 38s + 80

5s2 + 30s + 40

8s + 40

F (s) = 5 +8s + 40

s2 + 6s + 8= 5 +

K1

s + 2+

K2

s + 4

K1 =8s + 40s + 4

∣∣∣∣s=−2

= 12

K2 =8s + 40s + 2

∣∣∣∣s=−4

= −4

f(t) = 5δ(t) + [12e−2t − 4e−4t]u(t)

[b] 10

F (s) = s2 + 48s + 625 10s2 + 512s + 7186

10s2 + 480s + 6250

32s + 936

F (s) = 10 +32s + 936

s2 + 48s + 625= 10 +

K1

s + 24 − j7+

K∗2

s + 24 + j7

K1 =32s + 936

s + 24 + j7

∣∣∣∣s=−24+j7

= 16 − j12 = 20/− 36.87

f(t) = 10δ(t) + [40e−24t cos(7t − 36.87)]u(t)


[c] s − 10

F (s) = s2 + 15s + 50 s3 + 5s2 − 50s − 100

s3 + 15s2 + 50s

−10s2 − 100s − 100

−10s2 − 150s − 500

50s + 400

F (s) = s − 10 +K1

s + 5+

K2

s + 10

K1 =50s + 400

s + 10

∣∣∣∣s=−5

= 30

K2 =50s + 400

s + 5

∣∣∣∣s=−10

= 20

f(t) = δ′(t) − 10δ(t) + [30e−5t + 20e−10t]u(t)

P 12.40 [a] F (s) =K1

s2 +K2

s+

K3

s + 1 − j2+

K∗3

s + 1 + j2

K1 =100(s + 1)s2 + 2s + 5

∣∣∣∣s=0

= 20

K2 =d

ds

[100(s + 1)s2 + 2s + 5

]=[

100s2 + 2s + 5

− 100(s + 1)(2s + 2)(s2 + 2s + 5)2

]s=0

= 20 − 8 = 12

K3 =100(s + 1)

s2(s + 1 + j2)

∣∣∣∣s=−1+j2

= −6 + j8 = 10/126.87

f(t) = [20t + 12 + 20e−t cos(2t + 126.87)]u(t)

[b] F (s) =K1

s+

K2

(s + 5)3 +K3

(s + 5)2 +K4

s + 5

K1 =500

(s + 5)3

∣∣∣∣s=0

= 4

K2 =500s

∣∣∣∣s=−5

= −100

K3 =d

ds

[500s

]=

−500s2

∣∣∣∣s=−5

= −20

K4 =12

d

ds

[−500s2

]=

12

1000(s3)

∣∣∣∣s=−5

= −4

f(t) = [4 − 50t2e−5t − 20te−5t − 4e−5t]u(t)

Problems 12–25

[c] F (s) =K1

s+

K2

(s + 1)3 +K3

(s + 1)2 +K4

s + 1

K1 =40(s + 2)(s + 1)3

∣∣∣∣s=0

= 80

K2 =40(s + 2)

s

∣∣∣∣s=−1

= −40

K3 =d

ds

[40(s + 2)

s

]=[40s

− 40(s + 2)s2

]s=−1

= −40 − 40 = −80

K4 =12

d

ds

[40s

− 40(s + 2)s2

]

=12

[−40s2 − 40

s2 +80(s + 2)

s3

]s=−1

=12(−40 − 40 − 80) = −80

f(t) = [80 − 20t2e−t − 80te−t − 80e−t]u(t)

[d] F (s) =K1

s+

K2

(s + 1)4 +K3

(s + 1)3 +K4

(s + 1)2 +K5

s + 1

K1 =(s + 5)2

(s + 1)4

∣∣∣∣s=0

= 25

K2 =(s + 5)2

s

∣∣∣∣s=−1

= −16

K3 =d

ds

[(s + 5)2

s

]=[2(s + 5)

s− (s + 5)2

s2

]s=−1

= −8 − 16 = −24

K4 =12

d

ds

[2(s + 5)

s− (s + 5)2

s2

]

=12

[2s

− 2(s + 5)s2 − 2(s + 5)

s2 +2(s + 5)2

s3

]s=−1

=12(−2 − 8 − 8 − 32) = −25

K5 =16

d

ds

[2s

− 2(s + 5)s2 − 2(s + 5)

s2 +2(s + 5)2

s3

]

=16

[−2s2 − 2

s2 +4(s + 5)

s3 − 2s2 +

4(s + 5)s3 +

4(s + 5)s3 − 6(s + 5)2

s4

]s=−1

=16(−2 − 2 − 16 − 2 − 16 − 16 − 96) = −25

f(t) = [25 − (8/3)t3e−t − 12t2e−t − 25te−t − 25e−t]u(t)


P 12.41 f(t) = L−1

K

s + α − jβ+

K∗

s + α + jβ

= Ke−αtejβt + K∗e−αte−jβt

= |K|e−αt[ejθejβt + e−jθe−jβt]

= |K|e−αt[ej(βt+θ) + e−j(βt+θ)]

= 2|K|e−αt cos(βt + θ)

P 12.42 [a] Ltnf(t) = (−1)n

[dnF (s)

dsn

]

Let f(t) = 1, then F (s) =1s, thus

dnF (s)dsn

=(−1)nn!s(n+1)

Therefore Ltn = (−1)n

[(−1)nn!s(n+1)

]=

n!s(n+1)

It follows that Lt(r−1) =(r − 1)!

sr

and Lt(r−1)e−at =(r − 1)!(s + a)r

ThereforeK

(r − 1)!Ltr−1e−at =

K

(s + a)r= L

Ktr−1e−at

(r − 1)!

[b] f(t) = L−1

K

(s + α − jβ)r+

K∗

(s + α + jβ)r

Therefore

f(t) =Ktr−1

(r − 1)!e−(α−jβ)t +

K∗tr−1

(r − 1)!e−(α+jβ)t

=|K|tr−1e−αt

(r − 1)!

[ejθejβt + e−jθe−jβt

]

=[2|K|tr−1e−αt

(r − 1)!

]cos(βt + θ)

P 12.43 [a] lims→∞ sV (s) = lim

s→∞

[1.92s3

s4[1 + (1.6/s) + (1/s2)][1 + (1/s2)]

]= 0

Therefore v(0+) = 0

[b] No, V has a pair of poles on the imaginary axis.

Problems 12–27

P 12.44 [a] sF (s) =8s3 + 37s2 + 32s

(s + 1)(s + 2)(s + 4)

lims→0

sF (s) = 0, ·. . f(∞) = 0

lims→∞ sF (s) = 8, ·. . f(0+) = 8

[b] sF (s) =8s3 + 89s2 + 311s + 300

(s + 2)(s2 + 8s + 15)

lims→0

sF (s) = 10; ·. . f(∞) = 10

lims→∞ sF (s) = 8, ·. . f(0+) = 8

[c] sF (s) =22s3 + 60s2 + 58s

(s + 1)(s2 + 4s + 5)

lims→0

sF (s) = 0, ·. . f(∞) = 0

lims→∞ sF (s) = 22, ·. . f(0+) = 22

[d] sF (s) =250(s + 7)(s + 14)

(s2 + 14s + 50)

lims→0

sF (s) =250(7)(14)

50= 490, ·. . f(∞) = 490

lims→∞ sF (s) = 250, ·. . f(0+) = 250

P 12.45 [a] sF (s) =100

s(s + 5)F (s) has a second-order pole at the origin so we cannot use the final valuetheorem.

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[b] sF (s) =50(s + 5)(s + 1)2

lims→0

sF (s) = 250, ·. . f(∞) = 250

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[c] sF (s) =100(s + 3)

s(s2 + 6s + 10)F (s) has a second-order pole at the origin so we cannot use the final valuetheorem.

lims→∞ sF (s) = 0, ·. . f(0+) = 0


[d] sF (s) =5(s + 2)2

(s + 1)3

lims→0

sF (s) = 20, ·. . f(∞) = 20

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[e] sF (s) =400

(s2 + 4s + 5)2

lims→0

sF (s) = 16, ·. . f(∞) = 16

lims→∞ sF (s) = 0, ·. . f(0+) = 0

P 12.46 All of the F (s) functions referenced in this problem are improper rational functions,and thus the corresponding f(t) functions contain impulses (δ(t)). Thus, neither theinitial value theorem nor the final value theorem may be applied to these F (s)functions!

P 12.47 sVo(s) =sVdc/RC

s2 + (1/RC)s + (1/LC)

lims→0

sVo(s) = 0, ·. . vo(∞) = 0

lims→∞ sVo(s) = 0, ·. . vo(0+) = 0

sIo(s) =Vdc/RCL)

s2 + (1/RC)s + (1/LC)

lims→0

sIo(s) =Vdc/RLC

1/LC=

Vdc

R, ·. . io(∞) =

Vdc

R

lims→∞ sIo(s) = 0, ·. . io(0+) = 0

P 12.48 sVo(s) =(Idc/C)s

s2 + (1/RC)s + (1/LC)

lims→0

sVo(s) = 0, ·. . vo(∞) = 0

lims→∞ sVo(s) = 0, ·. . vo(0+) = 0

sIo(s) =s2Idc

s2 + (1/RC)s + (1/LC)

lims→0

sIo(s) = 0, ·. . io(∞) = 0

lims→∞ sIo(s) = Idc, ·. . vo(0+) = Idc

Problems 12–29

P 12.49 [a] sF (s) =100(s + 1)

s(s2 + 2s + 5)F (s) has a second-order pole at the origin, so we cannot use the final valuetheorem here.

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[b] sF (s) =500

(s + 5)3

lims→0

sF (s) = 4, ·. . f(∞) = 4

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[c] sF (s) =40(s + 2)(s + 1)3

lims→0

sF (s) = 80, ·. . f(∞) = 80

lims→∞ sF (s) = 0, ·. . f(0+) = 0

[d] sF (s) =(s + 5)2

(s + 1)4

lims→0

sF (s) = 25, ·. . f(∞) = 25

lims→∞ sF (s) = 0, ·. . f(0+) = 0

P 12.50 sIo(s) =Idcs[s + (1/RC)]

s2 + (1/RC)s + (1/LC)

lims→0

sIo(s) = 0, ·. . io(∞) = 0

lims→∞ sIo(s) = Idc, ·. . io(0+) = Idc

13The Laplace Transform in Circuit

Analysis

Assessment Problems

AP 13.1 [a] Y =1R

+1sL

+ sC =C[s2 + (1/RC)s + (1/LC)

s

1RC

=106

(500)(0.025)= 80,000;

1LC

= 25 × 108

Therefore Y =25 × 10−9(s2 + 80,000s + 25 × 108)

s

[b] −z1,2 = −40,000 ±√

16 × 108 − 25 × 108 = −40,000 ± j30,000 rad/s

−z1 = −40,000 − j30,000 rad/s

−z2 = −40,000 + j30,000 rad/s

−p1 = 0 rad/s

AP 13.2 [a] Z = 2000 +1Y

= 2000 +4 × 107s

s2 + 80,000s + 25 × 108

=2000(s2 + 105s + 25 × 108)

s2 + 80,000s + 25 × 108 =2000(s + 50,000)2

s2 + 80,000s + 25 × 108

[b] −z1 = −z2 = −50,000 rad/s

−p1 = −40,000 − j30,000 rad/s

−p2 = −40,000 + j30,000 rad/s

13–1

13–2 CHAPTER 13. The Laplace Transform in Circuit Analysis

AP 13.3 [a] At t = 0−, 0.2v1 = 0.8v2; v1 = 4v2; v1 + v2 = 100 V

Therefore v1(0−) = 80 V = v1(0+); v2(0−) = 20 V = v2(0+)

I =(80/s) + (20/s)

5000 + [(5 × 106)/s] + (1.25 × 106/s)=

20 × 10−3

s + 1250

V1 =80s

− 5 × 106

s

(20 × 10−3

s + 1250

)=

80s + 1250

V2 =20s

− 1.25 × 106

s

(20 × 10−3

s + 1250

)=

20s + 1250

[b] i = 20e−1250tu(t) mA; v1 = 80e−1250tu(t) V

v2 = 20e−1250tu(t) V

AP 13.4 [a]

I =Vdc/s

R + sL + (1/sC)=

Vdc/L

s2 + (R/L)s + (1/LC)

Vdc

L= 40;

R

L= 1.2;

1LC

= 1.0

I =40

s2 + 1.2s + 1

Problems 13–3

[b] I =40

(s + 0.6 − j0.8)(s + 0.6 + j0.8)=

K1

s + 0.6 − j0.8+

K∗1

s + 0.6 + j0.8

K1 =40

j1.6= −j25 = 25/− 90; K∗

1 = 25/90

i = 50e−0.6t cos(0.8t − 90) = [50e−0.6t sin 0.8t]u(t) A

[c] V = sLI =160s

s2 + 1.2s + 1=

160s(s + 0.6 − j0.8)(s + 0.6 + j0.8)

=K1

s + 0.6 − j0.8+

K∗1

s + 0.6 + j0.8

K1 =160(−0.6 + j0.8)

j1.6= 100/36.87

[d] v(t) = [200e−0.6t cos(0.8t + 36.87)]u(t) V

AP 13.5 [a]

The two node voltage equations are

V1 − V2

s+ V1s =

5s

andV2

3+

V2 − V1

s+

V2 − (15/s)15

= 0

Solving for V1 and V2 yields

V1 =5(s + 3)

s(s2 + 2.5s + 1), V2 =

2.5(s2 + 6)s(s2 + 2.5s + 1)

[b] The partial fraction expansions of V1 and V2 are

V1 =15s

− 50/3s + 0.5

+5/3

s + 2and V2 =

15s

− 125/6s + 0.5

+25/3s + 2

It follows that

v1(t) =[15 − 50

3e−0.5t +

53e−2t

]u(t) V and

v2(t) =[15 − 125

6e−0.5t +

253

e−2t]u(t) V

[c] v1(0+) = 15 − 503

+53

= 0

v2(0+) = 15 − 1256

+253

= 2.5 V


[d] v1(∞) = 15 V; v2(∞) = 15 V

AP 13.6 [a]

With no load across terminals a–b, Vx = 20/s:

12

[20s

− VTh

]s +

[1.2

(20s

)− VTh

]= 0

therefore VTh =20(s + 2.4)s(s + 2)

Vx = 5IT and ZTh =VT

IT

Solving for IT gives

IT =(VT − 5IT )s

2+ VT − 6IT

Therefore

14IT = VT s − 5sIT + 2VT ; therefore ZTh =5(s + 2.8)

s + 2[b]

I =VTh

ZTh + 2 + s=

20(s + 2.4)s(s + 3)(s + 6)

Problems 13–5

AP 13.7 [a] i2 = 1.25e−t − 1.25e−3t; thereforedi2dt

= −1.25e−t + 3.75e−3t

Thereforedi2dt

= 0 when

1.25e−t = 3.75e−3t or e2t = 3, t = 0.5(ln 3) = 549.31 ms

i2(max) = 1.25[e−0.549 − e−3(0.549)] = 481.13 mA

[b] From Eqs. 13.68 and 13.69, we have

∆ = 12(s2 + 4s + 3) = 12(s + 1)(s + 3) and N1 = 60(s + 2)

Therefore I1 =N1

∆=

5(s + 2)(s + 1)(s + 3)

A partial fraction expansion leads to the expression

I1 =2.5

s + 1+

2.5s + 3

Therefore we get

i1 = 2.5[e−t + e−3t]u(t) A

[c]di1dt

= −2.5[e−t + 3e−3t];di1(0.54931)

dt= −2.89 A/s

[d] When i2 is at its peak value,

di2dt

= 0

Therefore L2

(di2dt

)= 0 and i2 = −

(M

12

)(di1dt

)

[e] i2(max) =−2(−2.89)

12= 481.13 mA (Checks)

AP 13.8 [a] The s-domain circuit with the voltage source acting alone is

V ′ − (20/s)2

+V ′

1.25s+

V ′s20

= 0


V ′ =200

(s + 2)(s + 8)=

100/3s + 2

− 100/3s + 8

v′ =1003

[e−2t − e−8t]u(t) V

[b] With the current source acting alone,

V ′′

2+

V ′′

1.25s+

V ′′s20

=5s

V ′′ =100

(s + 2)(s + 8)=

50/3s + 2

− 50/3s + 8

v′′ =503

[e−2t − e−8t]u(t) V

[c] v = v′ + v′′ = [50e−2t − 50e−8t]u(t) V

AP 13.9 [a]Vo

s + 2+

Vos

10= Ig; therefore

Vo

Ig

= H(s) =10(s + 2)

s2 + 2s + 10[b] −z1 = −2 rad/s; −p1 = −1 + j3 rad/s; −p2 = −1 − j3 rad/s

AP 13.10 [a] Vo =10(s + 2)

s2 + 2s + 10· 1s

=K0

s+

K1

s + 1 − j3+

K∗1

s + 1 + j3

K0 = 2; K1 = (5/3)/− 126.87; K∗1 = (5/3)/126.87

vo = [2 + (10/3)e−t cos(3t − 126.87)]u(t) V

[b] Vo =10(s + 2)

s2 + 2s + 10· 1 =

K2

s + 1 − j3+

K∗2

s + 1 + j3

K2 = 5.27/− 18.43; K∗2 = 5.27/18.43

vo = [10.54e−t cos(3t − 18.43)]u(t) V

AP 13.11 [a] H(s) = Lh(t) = Lvo(t)

vo(t) = 10,000 cos θe−70t cos 240t − 10,000 sin θe−70t sin 240t

= 9600e−70t cos 240t − 2800e−70t sin 240t

Problems 13–7

Therefore H(s) =9600(s + 70)

(s + 70)2 + (240)2 − 2800(240)(s + 70)2 + (240)2

=9600s

s2 + 140s + 62,500

[b] Vo(s) = H(s) · 1s

=9600

s2 + 140s + 62,500

=K1

s + 70 − j240+

K∗1

s + 70 + j240

K1 =9600j480

= −j20 = 20/− 90

Therefore

vo(t) = [40e−70t cos(240t − 90)]u(t) V = [40e−70t sin 240t]u(t) V

AP 13.12 From Assessment Problem 13.9:

H(s) =10(s + 2)

s2 + 2s + 10

Therefore H(j4) =10(2 + j4)

10 − 16 + j8= 4.47/− 63.43

Thus,

vo = (10)(4.47) cos(4t − 63.43) = 44.7 cos(4t − 63.43) V

AP 13.13 [a] Let R1 = 10 kΩ, R2 = 50 kΩ, C = 400 pF, R2C = 2 × 10−5

then V1 = V2 =VgR2

R2 + (1/sC)

AlsoV1 − Vg

R1+

V1 − Vo

R1= 0

therefore Vo = 2V1 − Vg

Now solving for Vo/Vg, we get H(s) =R2Cs − 1R2Cs + 1

It follows that H(j50,000) =j − 1j + 1

= j1 = 1/90

Therefore vo = 10 cos(50,000t + 90) V


[b] Replacing R2 by Rx gives us H(s) =RxCs − 1RxCs + 1

Therefore

H(j50,000) =j20 × 10−6Rx − 1j20 × 10−6Rx + 1

=Rx + j50,000Rx − j50,000

Thus,

50,000Rx

= tan 60 = 1.7321, Rx = 28,867.51 Ω

Problems 13–9

Problems

P 13.1 Iscab = IN =−LI0

sL=

−I0

s; ZN = sL

Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.

P 13.2 i =1L

∫ t

0−vdτ + I0; therefore I =

( 1L

)(V

s

)+

I0

s=

V

sL+

I0

s

P 13.3 VTh = Vab = CVo

( 1sC

)=

Vo

s; ZTh =

1sC

P 13.4 [a] Z = R + sL +1

sC=

L[s2 + (R/L)s + (1/LC)]s

=0.0025[s2 + 16 × 107s + 1010]

s

[b] Zeros at −62.5 rad/s and −1.6 × 108 rad/sPole at 0.

P 13.5 [a] Y =1R

+1sL

+ sC =C[s2 + (1/RC)s + (1/LC)]

s

Z =1Y

=s/C

s2 + (1/RC)s + (1/LC)=

4 × 106s

s2 + 2000s + 64 × 104

[b] zero at −z1 = 0poles at −p1 = −400 rad/s and −p2 = −1600 rad/s

P 13.6 [a]

Z =(R + sL)(1/sC)R + sL + (1/sC)

=(1/C)(s + R/L)

s2 + (R/L)s + (1/LC)

R

L=

2500.08

= 3125;1

LC=

1(0.08)(0.5 × 10−6)

= 25 × 106

Z =2 × 106(s + 3125)

s2 + 3125s + 25 × 106


[b] Z =2 × 106(s + 3125)

(s + 1562.5 − j4749.6)(s + 1562.5 + j4749.6)

−z1 = −3125 rad/s; −p1 = −1562.5 + j4749.6 rad/s

−p2 = −1562.5 − j4749.6 rad/s

P 13.7 Transform the Y-connection of the two resistors and the capacitor into the equivalentdelta-connection:

where

Za =(1/s)(1) + (1)(1/s) + (1)(1)

1/s= s + 2

Zb = Zc =(1/s)(1) + (1)(1/s) + (1)(1)

1=

s + 2s

Then

Zab = Za‖[(s‖Zc) + (s‖Zb)] = Za‖2(s‖Zb)

s‖Zb =s + 2

s + (s + 2)/s=

s(s + 2)s2 + s + 2

Zab = (s + 2)‖ 2s(s + 2)s2 + s + 2

=2s(s + 2)2

(s + 2)(s2 + s + 2) + 2s(s + 2)

=2s(s + 2)

s2 + 3s + 2=

2ss + 1

One zero at the origin (0 rad/s); one pole at −1 rad/s.

P 13.8 Z1 =16s

+ s‖4 =16s

+4s

s + 4=

4(s2 + 4s + 16)s(s + 4)

Zab = 4‖4(s2 + 4s + 16)s(s + 4)

=16(s2 + 4s + 16)8s2 + 32s + 64

Problems 13–11

=2(s2 + 4s + 16)

s2 + 4s + 8=

2(s + 2 + j3.46)(s + 2 − j3.46)(s + 2 + j2)(s + 2 − j2)

Zeros at −2 + j3.46 rad/s and −2 − j3.46 rad/s; poles at −2 + j2 rad/s and −2 − j2rad/s.

P 13.9 [a] For t > 0:

[b] Vo =2.5s

(16 × 105)/s + 5000 + 2.5s

(−150s

)

=−150s

s2 + 2000s + 64 × 104

=−150s

(s + 400)(s + 1600)

[c] Vo =K1

s + 400+

K2

s + 1600

K1 =−150s

s + 1600

∣∣∣∣s=−400

= 50

K2 =−150ss + 400

∣∣∣∣s=−1600

= −200

Vo =50

s + 400− 200

s + 1600

vo(t) = (50e−400t − 200e−1600t)u(t) V


P 13.10 [a] For t < 0:

1Re

=18

+180

+120

= 0.1875; Re = 5.33 Ω

v1 = (9)(5.33) = 48 V

iL(0−) =4820

= 2.4 A

vC(0−) = −v1 = −48 V

For t = 0+:

s-domain circuit:

where

R = 20 Ω; C = 6.25 µF; γ = −48 V;

L = 6.4 mH; and ρ = −2.4 A

[b]Vo

R+ VosC − γC +

Vo

sL− ρ

s= 0

·. . Vo =γ[s + (ρ/γC)]

s2 + (1/RC)s + (1/LC)

ρ

γC=

−2.4(−48)(6.25 × 10−6)

= 8000

Problems 13–13

1RC

=1

(20)(6.25 × 10−6)= 8000

1LC

=1

(6.4 × 10−3)(6.25 × 10−6)= 25 × 106

Vo =−48(s + 8000)

s2 + 8000s + 25 × 106

[c] IL =Vo

sL− ρ

s=

Vo

0.0064s+

2.4s

=−7500(s + 8000)

s(s2 + 8000s + 25 × 106)+

2.4s

=2.4(s + 4875)

(s2 + 8000s + 25 × 106)

[d] Vo =−48(s + 8000)

s2 + 8000s + 25 × 106

=K1

s + 4000 − j3000+

K∗1

s + 4000 + j3000

K1 =−48(s + 8000)

s + 4000 + j3000

∣∣∣∣s=−4000+j3000

= 40/126.87

vo(t) = [80e−4000t cos(3000t + 126.87)]u(t) V

[e] IL =2.4(s + 4875)

s2 + 8000s + 25 × 106

=K1

s + 4000 − j3000+

K∗1

s + 4000 + j3000

K1 =2.4(s + 4875)

s + 4000 + j3000

∣∣∣∣s=−4000+j3000

= 1.25/− 16.26

iL(t) = [2.5e−4000t cos(3000t − 16.26)]u(t) A

P 13.11 For t < 0:

vo(0−) − 5005

+vo(0−)

25+

vo(0−)100

= 0


25vo(0−) = 10,000 ·. . vo(0−) = 400 V

iL(0−) =vo(0−)

25=

40025

= 16 A

For t > 0 :

Vo + 40025 + 25s

+Vo

100+

Vo − (400/s)100/s

= 0

Vo

( 125 + 25s

+1

100+

s

100

)= 4 − 400

25 + 25s

·. . Vo =400(s − 3)s2 + 2s + 5

Io =Vo − (400/s)

100/s=

−20s − 20s2 + 2s + 5

=K1

s + 1 − j2+

K∗1

s + 1 + j2

K1 =−20(s + 1)s + 1 + j2

∣∣∣∣s=−1+j2

= −10

io(t) = [−20e−t cos 2t]u(t) A

Problems 13–15

P 13.12 [a] For t < 0:

V2 =10

10 + 40(450) = 90 V

For t > 0:

[b] V1 =25(450/s)

(125,000/s) + 25 + 1.25 × 10−3s

=9 × 106

s2 + 20, 000s + 108 =9 × 106

(s + 10,000)2

v1(t) = (9 × 106te−10,000t)u(t) V

[c] V2 =90s

− (25,000/s)(450/s)(125,000/s) + 1.25 × 10−3s + 25

=90(s + 20,000)

s2 + 20,000s + 108

=900,000

(s + 10,000)2 +90

s + 10,000

v2(t) = [9 × 105te−10,000t + 90e−10,000t]u(t) V


P 13.13 [a] For t < 0:

iL(0−) =−100

4 + 10‖40 + 8=

−10020

= −5 A

i1 =1050

(5) = 1 A

vC(0−) = 10(1) + 4(5) − 100 = −70 V

For t > 0:

[b] (20 + 2s + 100/s)I = 10 +70s

·. . I =5(s + 7)

s2 + 10s + 50

Vo =100s

I − 70s

=−70s2 − 200s

s(s2 + 10s + 50)=

−70(s + 20/7)s2 + 10s + 50

=K1

s + 5 − j5+

K∗1

s + 5 + j5

K1 =−70(s + 20/7)

s + 5 + j5

∣∣∣∣s=−5+j5

= 38.1/− 156.8

[c] vo(t) = 76.2e−5t cos(5t − 156.8)u(t) V

Problems 13–17

P 13.14 [a] iL(0−) = iL(0+) =243

= 8 A directed upward

VT = 25Iφ +[

20(10/s)20 + (10/s)

]IT =

25IT (10/s)20 + (10/s)

+( 200

10 + 20s

)IT

VT

IT

= Z =250 + 20020s + 10

=45

2s + 1

Vo

5+

Vo(2s + 1)45

+Vo

5.625s=

8s

[9s + (2s + 1)s + 8]Vo

45s=

8s

Vo[2s2 + 10s + 8] = 360

Vo =360

2s2 + 10s + 8=

180s2 + 5s + 4

[b] Vo =180

(s + 1)(s + 4)=

K1

s + 1+

K2

s + 4

K1 =1803

= 60; K2 =180−3

= −60

Vo =60

s + 1− 60

s + 4

vo(t) = [60e−t − 60e−4t]u(t) V


P 13.15 [a]

Vo − 35/s2

+ 0.4V∆ +Vo − 8Iφ

s + (250/s)= 0

V∆ =[

Vo − 8Iφ

s + (250/s)

]s; Iφ =

(35/s) − Vo

2

Solving for Vo yields:

Vo =29.4s2 + 56s + 1750

s(s2 + 2s + 50)=

29.4s2 + 56s + 1750s(s + 1 − j7)(s + 1 + j7)

Vo =K1

s+

K2

s + 1 − j7+

K∗2

s + 1 + j7

K1 =29.4s2 + 56s + 1750

s2 + 2s + 50

∣∣∣∣s=0

= 35

K2 =29.4s2 + 56s + 1750

s(s + 1 + j7)

∣∣∣∣s=−1+j7

= −2.8 + j0.6 = 2.86/167.91

·. . vo(t) = [35 + 5.73e−t cos(7t + 167.91)]u(t) V

[b] At t = 0+ vo = 35 + 5.73 cos(167.91) = 29.4 V

vo − 352

+ 0.4v∆ = 0; vo − 35 + 0.8v∆ = 0

vo = v∆ + 8iφ = v∆ + 8(0.4v∆) = 4.2v∆

vo + (0.8)vo

4.2= 35; ·. . vo(0+) = 29.4 V(Checks)

Problems 13–19

At t = ∞, the circuit is

v∆ = 0, iφ = 0 ·. . vo = 35 V(Checks)

P 13.16 [a] For t < 0:

Vc − 50400

+Vc

1200+

Vc − 137.5500

= 0

Vc

( 1400

+1

1200+

1500

)=

50400

+137.5500

Vc = 75 V

iL(0−) =75 − 137.5

500= −0.125 A

For t > 0:

[b] Vo =5 × 105

sI +

75s

0 = −137.5s

+ 100I +5 × 105

sI +

75s

− 1.25 × 10−3 + 0.01sI


I

(100 +

5 × 105

s+ 0.01s

)=

62.5s

+ 1.25 × 10−3

·. . I =6250 + 0.125s

s2 + 104s + 5 × 107

Vo =5 × 105

s

( 6250 + 0.125ss2 + 104s + 5 × 107

)+

75s

=75s2 + 812,500s + 6875 × 106

s(s2 + 104s + 5 × 107)

[c] Vo =K1

s+

K2

s + 5000 − j5000+

K∗2

s + 5000 + j5000

K1 =75s2 + 812,500s + 6875 × 106

s2 + 104s + 5 × 107

∣∣∣∣s=0

= 137.5

K2 =75s2 + 812,500s + 6875 × 106

s(s + 5000 + j5000)

∣∣∣∣s=−5000+j5000

= 40.02/141.34

vo(t) = [137.5 + 80.04e−5000t cos(5000t + 141.34)]u(t) V

P 13.17

5 × 10−3

s=

Vo

200 + 4 × 106/s+ 3.75 × 10−3Vφ +

Vo

0.04s

Vφ =4 × 106/s

200 + 4 × 106/sVo =

4 × 106Vo

200s + 4 × 106

·. .5 × 10−3

s=

Vos

200s + 4 × 106 +15,000Vo

200s + 4 × 106 +25Vo

s

·. . Vo =s + 20,000

s2 + 20,000s + 108 =K1

(s + 10,000)2 +K2

s + 10,000

K1 = 10,000; K2 = 1

Vo =10,000

(s + 10,000)2 +1

s + 10,000

vo(t) = [10,000te−10,000t + e−10,000t]u(t) V

Problems 13–21

P 13.18 vo(0−) = vo(0+) = 0

−0.05s

+Vo

1000+

Vo

25− 21

Vo

1000+

Vo

107/s= 0

Vo

( 201000

+s

107

)=

0.05s

·. . Vo =500,000

s(s + 200,000)=

2.5s

− 2.5s + 200,000

vo(t) = [2.5 − 2.5e−200,000t]u(t) V

P 13.19 [a] io(0−) =20

4000= 5 mA

Io =20/s + Lρ

R + sL + 1/sC

=20/L + sρ

s2 + sR/L + 1/LC=

40 + s(0.005)s2 + 8000s + 16 × 106

Vo = −Lρ + sLIo = −0.0025 +0.0025s(s + 8000)

s2 + 8000s + 16 × 106

=−40,000

(s + 4000)2

vo(t) = −40,000te−4000tu(t) V


[b] Io =0.005(s + 8000)

s2 + 8000s + 16 × 106

=K1

(s + 4000)2 +K2

s + 4000

K1 = 20 K2 = 0.005

io(t) = [20te−4000t + 0.005e−4000t]u(t) A

P 13.20

Vo =5 × 106/s

1120 + 0.8s + 5 × 106/s

(240s

)

=12 × 108

s(0.8s2 + 1120s + 5 × 106)

=15 × 108

s(s2 + 1400s + 625 × 104)

=K1

s+

K2

s + 700 − j2400+

K∗2

s + 700 + j2400

K1 = 240; K2 = 125/163.74

vo(t) = [240 + 250e−700t cos(2400t + 163.74)]u(t) V

P 13.21

Vo − 240/s1120 + 0.8s

+Vos

5 × 106 + 14.4 × 10−6 = 0

Problems 13–23

Vo

( 11120 + 0.8s

+s

5 × 106

)=

240/s0.8s + 1120

− 14.4 × 10−6

Vo =−72s2 − 100,800s + 15 × 108

s(s2 + 1400s + 625 × 104)

=240s

+162.5/163.74

s + 700 − j2400+

162.5/− 163.74

s + 700 + j2400

·. . vo(t) = [240 + 325e−700t cos(2400t + 163.74)]u(t) V

P 13.22 [a]

Vo =(1/sC)(LIg)

R + sL + (1/sC)=

Ig/C

s2 + (R/L)s + (1/LC)

Ig

C=

150.1

= 150

R

L= 7;

1LC

= 10

Vo =150

s2 + 7s + 10

[b] sVo =150s

s2 + 7s + 10

lims→0

sVo = 0; ·. . vo(∞) = 0

lims→∞ sVo = 0; ·. . vo(0+) = 0

[c] Vo =150

(s + 2)(s + 5)=

50s + 2

+−50s + 5

vo = [50e−2t − 50e−5t]u(t) V


P 13.23 IL =Ig

s− Vo

1/sC=

Ig

s− sCVo

IL =15s

− 15s(s + 2)(s + 5)

=15s

−[ −10s + 2

+25

s + 5

]

iL(t) = [15 + 10e−2t − 25e−5t]u(t) A

Check:

iL(0+) = 0 (ok); iL(∞) = 15 (ok)

P 13.24 [a]

[b] I1 =25/s

2500 + (125,000/s)=

0.01s + 50

V1 =(100,000/s)(25/s)2500 + (125,000/s)

=1000

s(s + 50)

V2 =(25,000/s)(25/s)

2500 + (125,000/s)=

250s(s + 50)

[c] i1(t) = 10e−50tu(t) mA

V1 =20s

− 20s + 50

·. . v1(t) = (20 − 20e−50t)u(t) V

V2 =5s

− 5s + 50

·. . v2(t) = (5 − 5e−50t)u(t) V

[d] i1(0+) = 10 mA

i1(0+) =25

2.5 × 10−3 = 10 mA(Checks)

v1(0+) = 0; v2(0+) = 0(Checks)

v1(∞) = 20 V; v2(∞) = 5 V(Checks)

Problems 13–25

v1(∞) + v2(∞) = 25 V(Checks)

(10 × 10−6)v1(∞) = 200 µC

(40 × 10−6)v2(∞) = 200 µC(Checks)

P 13.25 [a]

100‖5s =500s

5s + 100=

100ss + 20

Vo =100s

s + 20

[50

(s + 25)2

]=

5000s(s + 20)(s + 25)2

Io =Vo

100=

50s(s + 20)(s + 25)2

IL =Vo

5s=

1000(s + 20)(s + 25)2

[b] Vo =K1

s + 20+

K2

(s + 25)2 +K3

s + 25

K1 =5000s

(s + 25)2

∣∣∣∣s=−20

= −4000

K2 =5000s

(s + 20)

∣∣∣∣s=−25

= 25,000

K3 =d

ds

[ 5000ss + 20

]s=−25

=[

5000s + 20

− 5000s(s + 20)2

]s=−25

= 4000

vo(t) = [−4000e−20t + 25,000te−25t + 4000e−25t]u(t) V

Io =K1

s + 20+

K2

(s + 25)2 +K3

s + 25

K1 =50s

(s + 25)2

∣∣∣∣s=−20

= −40

K2 =50s

(s + 20)

∣∣∣∣s=−25

= 250

K3 =d

ds

[ 50ss + 20

]s=−25

=[

50s + 20

− 50s(s + 20)2

]s=−25

= 40


io(t) = [−40e−20t + 250te−25t + 40e−25t]u(t) V

IL =K1

s + 20+

K2

(s + 25)2 +K3

s + 25

K1 =1000

(s + 25)2

∣∣∣∣s=−20

= 40

K2 =1000

(s + 20)

∣∣∣∣s=−25

= −200

K3 =d

ds

[ 1000s + 20

]s=−25

=[− 1000

(s + 20)2

]s=−25

= −40

iL(t) = [40e−20t − 200te−25t − 40e−25t]u(t) V

P 13.26

10s

= (s + 1)I1 − sI2

0 = −sI1 +(s + 1 +

1s

)I2

In standard form,

s(s + 1)I1 − s2I2 = 10

−s2I1 + (s2 + s + 1)I2 = 0

∆ =

∣∣∣∣∣∣∣s(s + 1) −s2

−s2 (s2 + s + 1)

∣∣∣∣∣∣∣ = 2s(s2 + s + 0.5)

N1 =

∣∣∣∣∣∣∣10 −s2

0 (s2 + s + 1)

∣∣∣∣∣∣∣ = 10(s2 + s + 1)

N2 =

∣∣∣∣∣∣∣s(s + 1) 10

−s2 0

∣∣∣∣∣∣∣ = 10s2

Problems 13–27

I1 =N1

∆; I2 =

N2

∆; I0 = I1 − I2

·. . Io =N1 − N2

∆=

5(s + 1)s(s2 + s + 0.5)

=K1

s+

K2

s + 0.5 − j0.5+

K∗2

s + 0.5 + j0.5

K1 =5

0.5= 10

K2 =5(−0.5 + j0.5 + 1)(−0.5 + j0.5)(j1)

= 5/− 180

io(t) = [10 − 10e−t/2 cos 0.5t]u(t) A

P 13.27 [a]

0 = 2.5s(I1 − 6/s) +5s(I1 − I2) + 10I1

−75s

=5s(I2 − I1) + 5(I2 − 6/s)

or

(s2 + 4s + 2)I1 − 2I2 = 6s

−I1 + (s + 1)I2 = −9

∆ =

∣∣∣∣∣∣∣(s2 + 4s + 2) −2

−1 (s + 1)

∣∣∣∣∣∣∣ = s(s + 2)(s + 3)


N1 =

∣∣∣∣∣∣∣6s −2

−9 (s + 1)

∣∣∣∣∣∣∣ = 6(s2 + s − 3)

I1 =N1

∆=

6(s2 + s − 3)s(s + 2)(s + 3)

N2 =

∣∣∣∣∣∣∣(s2 + 4s + 2) 6s

−1 −9

∣∣∣∣∣∣∣ = −9s2 − 30s − 18

I2 =N2

∆=

−9s2 − 30s − 18s(s + 2)(s + 3)

[b] sI1 =6(s2 + s − 3)(s + 2)(s + 3)

lims→∞ sI1 = i1(0+) = 6 A; lim

s→0sI1 = i1(∞) = −3 A

sI2 =−9s2 − 30s − 18(s + 2)(s + 3)

lims→∞ sI2 = i2(0+) = −9 A; lim

s→0sI2 = i2(∞) = −3 A

[c] I1 =6(s2 + s − 3)

s(s + 2)(s + 3)=

K1

s+

K2

s + 2+

K3

s + 3

K1 =6(−3)

6= −3; K2 =

6(4 − 2 − 3)(−2)(1)

= 3

K3 =6(9 − 3 − 3)(−3)(−1)

= 6

i1(t) = [−3 + 3e−2t + 6e−3t]u(t) A

I2 =−9s2 − 30s − 18s(s + 2)(s + 3)

=K1

s+

K2

s + 2+

K3

s + 3

K1 =−186

= −3; K2 =−36 + 60 − 18

(−2)(1)= −3

K3 =−81 + 90 − 18

(−3)(−1)= −3

i2(t) = [−3 − 3e−2t − 3e−3t]u(t) A

Problems 13–29

P 13.28 [a]

54s

= 2I1 − I2 − I3

0 = −I1 +(2 +

45s

)I2 − I3

0 = −I1 − I2 + (2 + 0.36s)I3

∆ =

∣∣∣∣∣∣∣∣∣∣∣

2 −1 −1

−1 (2s + 45)/s −1

−1 −1 (0.36s + 2)

∣∣∣∣∣∣∣∣∣∣∣=

1.08(s + 5)(s + 25)s

N2 =

∣∣∣∣∣∣∣∣∣∣∣

2 (54/s) −1

−1 0 −1

−1 0 (0.36s + 2)

∣∣∣∣∣∣∣∣∣∣∣=

162s

(0.12s + 1)

N3 =

∣∣∣∣∣∣∣∣∣∣∣

2 −1 (54/s)

−1 (2s + 45)/s 0

−1 −1 0

∣∣∣∣∣∣∣∣∣∣∣=

162s2 (s + 15)

I2 =N2

∆=

150(0.12s + 1)(s + 5)(s + 25)

Vo =45s

I2 =6750(0.12s + 1)s(s + 5)(s + 25)

I3 =N3

∆=

150(s + 15)s(s + 5)(s + 25)

= Io

[b] Vo =K1

s+

K2

s + 5+

K3

s + 25

K1 =6750125

= 54; K2 =6750(−0.6 + 1)

(−5)(20)= −27


K3 =6750(−3 + 1)(−25)(−20)

= −27

·. . vo(t) = [54 − 27e−5t − 27e−25t]u(t) V

Io =K1

s+

K2

s + 5+

K3

s + 25

K1 =150(15)(5)(25)

= 18; K2 =150(10)(−5)(20)

= −15

K3 =150(−10)

(−25)(−20)= −3

·. . io(t) = [18 − 15e−5t − 3e−25t]u(t) A

[c] At t = 0+ the circuit is

Both vo and io are zero, which agrees with our solutions in part (a).At t = ∞ the circuit is

Our solutions predict vo(∞) = 54 V and io(∞) = 18 A.Also observe from the circuit at t = 0+ that the voltage across the inductor is54 V. Our solution predicts

vL(0+) = 0.36dio(0+)

dt= 0.36(75 + 75) = 54 V

At t = 0+ the current in the capacitive branch is (1/2)(54/1.5) = 18 A. Fromour solution we have

sI2 =150(0.12 + 1/s)

(1 + 5/s)(1 + 25/s)and lim

s→∞ sI2 = i2(0+) = 150(0.12) = 18 A

Problems 13–31

P 13.29 [a]

120s

= 50(I1 − 0.05Vφ) +250s

(I1 − I2)

120s

= 50I1 − 2.5(250

s

)(I2 − I1) +

250s

I1 − 250s

I2;

0 =250s

(I2 − I1) + 20s(I2 − 0.05Vφ) + 700I2

0 =250s

(I2 − I1) + 20s[I2 − 0.05

(250s

)(I2 − I1)

]Vφ) + 700I2

Simplifying,

(50s + 875)I1 − 875I2 = 120

250(s − 1)I1 + (20s2 + 450s + 250)I2 = 0

∆ =

∣∣∣∣∣∣∣(50s + 875) −875

250(s − 1) (20s2 + 450s + 250)

∣∣∣∣∣∣∣ = 1000s(s2 + 40s + 625)

N1 =

∣∣∣∣∣∣∣120 −875

0 (20s2 + 450s + 250)

∣∣∣∣∣∣∣ = 1200(2s2 + 45s + 25)

N2 =

∣∣∣∣∣∣∣(50s + 875) 120

250(s − 1) 0

∣∣∣∣∣∣∣ = −30,000(s − 1)

I1 =N1

∆=

1.2(2s2 + 45s + 25)s(s2 + 40s + 625)

I2 =N2

∆=

−30(s − 1)s(s2 + 40s + 625)

Io = I2 − 0.05Vφ = I2 − 0.05[250

s(I2 − I1)

]

I2 − I1 =−2.4s(s + 35)

s(s2 + 40s + 625)


250s

(I2 − I1) =−600(s + 35)

s(s2 + 40s + 625)

·. . Io =−30(s − 1)

s(s2 + 40s + 625)+

30(s + 35)s(s2 + 40s + 625)

=1080

s(s2 + 40s + 625)

[b] sIo =1080

(s2 + 40s + 625)

io(0+) = lims→∞ sIo = 0

io(∞) = lims→0

sIo =1080625

= 1728 mA


i0(0+) = 0 (Checks)

At t = ∞ the circuit is

120 = 50(ia − i1) + 700ia

= 50(ia − 0.05vφ) + 700ia = 750ia − 2.5vφ

vφ = −700ia ·. . 120 = 750ia + 1750ia = 2500ia

ia =1202500

= 48 mA

vφ = −700ia = −33.60 V

io(∞) = 48 × 10−3 − 0.05(−33.60) = 48 × 10−3 + 1.68 = 1728 mA (Checks)

Problems 13–33

[d] Io =1080

s(s2 + 40s + 625)=

K1

s+

K2

s + 20 − j15+

K∗2

s + 20 + j15

K1 =1080625

= 1.728

K2 =1080

(−20 + j15)(j30)= 1.44/126.87

io(t) = [1728 + 2880e−20t cos(15t + 126.87)]u(t) mA

Check: io(0+) = 0 mA; io(∞) = 1728 mA

P 13.30 [a]

V1

10+

V1 − 50/s25/s

+V1 − Vo

4s= 0

−5s

+Vo − V1

4s+

Vo − 50/s30

= 0

Simplifying,

(4s2 + 10s + 25)V1 − 25Vo = 200s

−15V1 + (2s + 15)Vo = 400

∆ =

∣∣∣∣∣∣∣(4s2 + 10s + 25) −25

−15 (2s + 15)

∣∣∣∣∣∣∣ = 8s(s + 5)2

No =

∣∣∣∣∣∣∣(4s2 + 10s + 25) 200s

−15 400

∣∣∣∣∣∣∣ = 200(8s2 + 35s + 50)

Vo =No

∆=

200(8s2 + 35s + 50)8s(s + 5)2 =

K1

s+

K2

(s + 5)2 +K3

s + 5

K1 =(25)(50)

25= 50; K2 =

25(200 − 175 + 50)−5

= −375

K3 = 25d

ds

[8s2 + 35s + 50

s

]s=−5

= 25[s(16s + 35) − (8s2 + 35s + 50)

s2

]s=−5


= −5(−45) − 75 = 150

·. . Vo =50s

− 375(s + 5)2 +

150s + 5

[b] vo(t) = [50 − 375te−5t + 150e−5t]u(t) V

[c] At t = 0+:

vo(0+) = 50 + 150 = 200 V(Checks)

At t = ∞:

vo(∞)10

− 5 +vo(∞) − 50

30= 0

·. . 3vo(∞) − 150 + vo(∞) − 50 = 0; ·. . 4vo(∞) = 200

·. . vo(∞) = 50 V(Checks)

Problems 13–35

P 13.31 [a]

10s

I1 +10s

(I1 − I2) + 10(I1 − 9/s) = 0

10s

(I2 − 9/s) +10s

(I2 − I1) + 10I2 = 0

Simplifying,

(s + 2)I1 − I2 = 9

−I1 + (s + 2)I2 =9s

∆ =

∣∣∣∣∣∣∣(s + 2) −1

−1 (s + 2)

∣∣∣∣∣∣∣ = s2 + 4s + 3 = (s + 1)(s + 3)

N1 =

∣∣∣∣∣∣∣9 −1

9/s (s + 2)

∣∣∣∣∣∣∣ =9s2 + 18s + 9

s=

9s(s + 1)2

I1 =N1

∆=

9s

[(s + 1)2

(s + 1)(s + 3)

]=

9(s + 1)s(s + 3)

N2 =

∣∣∣∣∣∣∣(s + 2) 9

−1 9/s

∣∣∣∣∣∣∣ =18s

(s + 1)

I2 =N2

∆=

18(s + 1)s(s + 1)(s + 3)

=18

s(s + 3)

Ia =9s

− I1 =9s

− 9(s + 1)s(s + 3)

=6s

− 6s + 3

Ib = I1 =9(s + 1)s(s + 3)

=3s

+6

s + 3


[b] ia(t) = 6(1 − e−3t)u(t) A

ib(t) = 3(1 + 2e−3t)u(t) A

[c] Va =10s

Ib =10s

(3s

+6

s + 3

)

=30s2 +

60s(s + 3)

=30s2 +

20s

− 20s + 3

Vb =10s

(I2 − I1) =10s

[(6s

− 6s + 3

)−(3

s+

6s + 3

)]

=10s

[3s

− 12s + 3

]=

30s2 − 40

s+

40s + 3

Vc =10s

(9/s − I2) =10s

(9s

− 6s

+6

s + 3

)

=30s2 +

20s

− 20s + 3

[d] va(t) = [30t + 20 − 20e−3t]u(t) V

vb(t) = [30t − 40 + 40e−3t]u(t) V

vc(t) = [30t + 20 − 20e−3t]u(t) V

[e] Calculating the time when the capacitor voltage drop first reaches 1000 V:

30t + 20 − 20e−3t = 1000 or 30t − 40 + 40e−3t = 1000

Note that in either of these expressions the exponential term over timebecomes is negligible when compared to the other terms. Thus,

30t + 20 = 1000 or 30t − 40 = 1000

Thus,

t =98030

= 32.67 s or t =104030

= 34.67 s

Therefore, the breakdown will occur at t = 32.67 s.

Problems 13–37

P 13.32 [a]

20Iφ + 25s(Io − Iφ) + 25(Io − I1) = 0

50s

Iφ + 5I1 + 25(I1 − Io) + 25s(Iφ − Io) = 0

Iφ − I1 =100s

·. . I1 = Iφ − 100s

Simplifying,

(−25s − 5)Iφ + (25s + 25)Io = −2500/s

(50/s + 25s + 30)Iφ + (−25s − 25)Io = 3000/s

∆ =

∣∣∣∣∣∣∣−5(5s + 1) 25(s + 1)

5s(5s2 + 6s + 10) −25(s + 1)

∣∣∣∣∣∣∣ = −625(s + 1)(1 + 2/s)

N2 =

∣∣∣∣∣∣∣−5(5s + 1) −2500/s

5s(5s2 + 6s + 10) 3000/s

∣∣∣∣∣∣∣ = −12,500s2 − 4.8s − 10

s2

Io =N2

∆=

20(s2 − 4.8s − 10)s(s + 1)(s + 2)

[b] io(0+) = lims→∞ sIo = 20 A

io(∞) = lims→0

sIo =−200

2= −100 A



20Iφ + 5I1 = 0; Iφ − I1 = 100

·. . 20Iφ + 5(Iφ − 100) = 0; 25Iφ = 500

·. . Iφ = Io(0+) = 20 A(Checks)

At t = ∞ the circuit is

Io(∞) = −100 A(Checks)

[d] Io =20(s2 − 4.8s − 10)

s(s + 1)(s + 2)=

K1

s+

K2

s + 1+

K3

s + 2

K1 =−200(1)(2)

= −100; K2 =20(1 + 4.8 − 10)

(−1)(1)= 84

K3 =20(4 + 9.6 − 10)

(−2)(−1)= 36

Io =−100

s+

84s + 1

+36

s + 2

Problems 13–39

io(t) = (−100 + 84e−t + 36e−2t)u(t) A

io(∞) = −100 A(Checks)

io(0+) = −100 + 84 + 36 = 20 A(Checks)

P 13.33 vC = 12 × 105te−5000t V, C = 5 µF; therefore

iC = C

(dvC

dt

)= 6e−5000t(1 − 5000t) A

iC > 0 when 1 > 5000t or iC < 0 when 0 < t < 200 µs

and iC < 0 when t > 200 µs

iC = 0 when 1 − 5000t = 0, or t = 200 µs

dvC

dt= 12 × 105e−5000t[1 − 5000t]

·. . iC = 0 whendvC

dt= 0

P 13.34 [a] The s-domain equivalent circuit is

I =Vg

R + sL=

Vg/L

s + (R/L), Vg =

Vm(ω cos φ + s sin φ)s2 + ω2

I =K0

s + R/L+

K1

s − jω+

K∗1

s + jω

K0 =Vm(ωL cos φ − R sin φ)

R2 + ω2L2 , K1 =Vm/φ − 90 − θ(ω)

2√

R2 + ω2L2

where tan θ(ω) = ωL/R. Therefore, we have

i(t) =Vm(ωL cos φ − R sin φ)

R2 + ω2L2 e−(R/L)t +Vm sin[ωt + φ − θ(ω)]√

R2 + ω2L2

[b] iss(t) =Vm√

R2 + ω2L2sin[ωt + φ − θ(ω)]


[c] itr =Vm(ωL cos φ − R sin φ)

R2 + ω2L2 e−(R/L)t

[d] I =Vg

R + jωL, Vg = Vm/φ − 90

Therefore I =Vm/φ − 90

√R2 + ω2L2/θ(ω)

=Vm√

R2 + ω2L2/φ − 90 − θ(ω)

Therefore iss =Vm√

R2 + ω2L2sin[ωt + φ − θ(ω)]

[e] The transient component vanishes when

ωL cos φ = R sin φ or tan φ =ωL

Ror φ = θ(ω)

P 13.35

VTh =10s

10s + 1000· 40

s=

40010s + 1000

=40

s + 100

ZTh = 1000 + 1000‖10s = 1000 +10,000s

10s + 1000=

2000(s + 50)s + 100

I =40/(s + 100)

(5 × 105)/s + 2000(s + 50)/(s + 100)=

40s2000s2 + 600,000s + 5 × 107

=0.02s

s2 + 300s + 25,000=

K1

s + 150 − j50+

K∗1

s + 150 + j50

K1 =0.02s

s + 150 + j50

∣∣∣∣s=−150+j50

= 31.62 × 10−3/71.57

i(t) = 63.25e−150t cos(50t + 71.57)u(t) mA

Problems 13–41

P 13.36 [a]

180s

= (100 + 15s)I1 + 10sI2

0 = 10sI1 + (20s + 200)I2

∆ =

∣∣∣∣∣∣∣15s + 100 10s

10s 20s + 200

∣∣∣∣∣∣∣ = 200(s + 5)(s + 20)

N2 =

∣∣∣∣∣∣∣15s + 100 180/s

10s 0

∣∣∣∣∣∣∣ = −1800

I2 =N2

∆=

−9(s + 5)(s + 20)

Vo = 160I2 =−1440

(s + 5)(s + 20)

[b] sVo =−1440s

(s + 5)(s + 20)

lims→0

sVo = vo(∞) = 0 V

lims→∞ sVo = vo(0+) = 0 V

[c] Vo =−96s + 5

+96

s + 20

vo(t) = [−96e−5t + 96e−20t]u(t) V

P 13.37

180s

= (100 + 15s)I1 − 10sI2

0 = −10sI1 + (20s + 200)I2


∆ =

∣∣∣∣∣∣∣15s + 100 −10s

−10s 20s + 200

∣∣∣∣∣∣∣ = 200(s + 5)(s + 20)

N2 =

∣∣∣∣∣∣∣15s + 100 180/s

−10s 0

∣∣∣∣∣∣∣ = 1800

I2 =N2

∆=

9(s + 5)(s + 20)

Vo = 160I2 =1440

(s + 5)(s + 20)=

96s + 5

− 96s + 20

vo(t) = [96e−5t − 96e−20t]u(t) V

P 13.38 [a] W =12L1i

21 +

12L2i

22 + Mi1i2

W = 4(15)2 + 9(100) + 150(6) = 2700 J

[b] 120i1 + 8di1dt

− 6di2dt

= 0

270i2 + 18di2dt

− 6di1dt

= 0

Laplace transform the equations to get

120I1 + 8(sI1 − 15) − 6(sI2 + 10) = 0

270I2 + 18(sI2 + 10) − 6(sI1 − 15) = 0

In standard form,

(8s + 120)I1 − 6sI2 = 180

−6sI1 + (18s + 270)I2 = −270

∆ =

∣∣∣∣∣∣∣8s + 120 −6s

−6s 18s + 270

∣∣∣∣∣∣∣ = 108(s + 10)(s + 30)

N1 =

∣∣∣∣∣∣∣180 −6s

−270 18s + 270

∣∣∣∣∣∣∣ = 1620(s + 30)

N2 =

∣∣∣∣∣∣∣8s + 120 180

−6s −270

∣∣∣∣∣∣∣ = −1080(s + 30)

Problems 13–43

I1 =N1

∆=

1620(s + 30)108(s + 10)(s + 30)

=15

s + 10

I2 =N2

∆=

−1080(s + 30)108(s + 10)(s + 30)

=−10

s + 10

[c] i1(t) = 15e−10tu(t) A; i2(t) = −10e−10tu(t) A

[d] W120Ω =∫ ∞

0(225e−20t)(120) dt = 27,000

e−20t

−20

∣∣∣∣∞0

= 1350 J

W270Ω =∫ ∞

0(100e−20t)(270) dt = 27,000

e−20t

−20

∣∣∣∣∞0

= 1350 J

W120Ω + W270Ω = 2700 J (Checks)

[e] W =12L1i

21 +

12L2i

22 + Mi1i2 = 900 + 900 − 900 = 900 J

With the dot reversed the s-domain equations are

(8s + 120)I1 + 6sI2 = 60

6sI1 + (18s + 270)I2 = −90

As before, ∆ = 108(s + 10)(s + 30). Now,

N1 =

∣∣∣∣∣∣∣60 6s

−90 18s + 270

∣∣∣∣∣∣∣ = 1620(s + 10)

N2 =

∣∣∣∣∣∣∣8s + 120 60

6s −90

∣∣∣∣∣∣∣ = −1080(s + 10)

I1 =N1

∆=

15s + 30

; I2 =N2

∆=

−10s + 30

i1(t) = 15e−30tu(t) A; i2(t) = −10e−30tu(t) A

W270Ω =∫ ∞

0(100e−60t)(270) dt = 450 J

W120Ω =∫ ∞

0(225e−60t)(120) dt = 450 J

W120Ω + W270Ω = 900 J (Checks)


P 13.39 [a] s-domain equivalent circuit is

Note: i2(0+) = −2010

= −2 A

[b]24s

= (120 + 3s)I1 + 3sI2 + 6

0 = −6 + 3sI1 + (360 + 15s)I2 + 36

In standard form,

(s + 40)I1 + sI2 = (8/s) − 2

sI1 + (5s + 120)I2 = −10

∆ =

∣∣∣∣∣∣∣s + 40 s

s 5s + 120

∣∣∣∣∣∣∣ = 4(s + 20)(s + 60)

N1 =

∣∣∣∣∣∣∣(8/s) − 2 s

−10 5s + 120

∣∣∣∣∣∣∣ =−200(s − 4.8)

s

I1 =N1

∆=

−50(s − 4.8)s(s + 20)(s + 60)

[c] sI1 =−50(s − 4.8)

(s + 20)(s + 60)

lims→∞ sI1 = i1(0+) = 0 A

lims→0

sI1 = i1(∞) =(−50)(−4.8)

(20)(60)= 0.2 A

[d] I1 =K1

s+

K2

s + 20+

K3

s + 60

K1 =2401200

= 0.2; K2 =−50(−20) + 240

(−20)(40)= −1.55

Problems 13–45

K3 =−50(−60) + 240

(−60)(−40)= 1.35

i1(t) = [0.2 − 1.55e−20t + 1.35e−60t]u(t) A

P 13.40 For t < 0:

For t > 0+:

18 × 4 = 72; 18 × 3 = 54

20I1 − 72 + 4sI1 + s(I2 − I1) + 10(I1 − I2) = 0

−54 + 3sI2 + 10(I2 − I1) + s(I1 − I2) = 0

In standard form,

(3s + 30)I1 + (s − 10)I2 = 72


(s − 10)I1 + (2s + 10)I2 = 54

·. . ∆ =

∣∣∣∣∣∣∣(3s + 30) (s − 10)

(s − 10) (2s + 10)

∣∣∣∣∣∣∣ = 5(s + 2)(s + 20)

N1 =

∣∣∣∣∣∣∣72 (s − 10)

54 (2s + 10)

∣∣∣∣∣∣∣ = 90s + 1260

N2 =

∣∣∣∣∣∣∣(3s + 30) 72

(s − 10) 54

∣∣∣∣∣∣∣ = 90s + 2340

Io = I1 − I2 =N1

∆− N2

∆=

−10805(s + 2)(s + 20)

=−216

(s + 2)(s + 20)− 12

s + 2− 12

s + 20

io(t) = [12e−2t + 12e−20t]u(t) A

P 13.41 The s-domain equivalent circuit is

V1 − 48/s4 + (100/s)

+V1 + 9.6

0.8s+

V1

0.8s + 20= 0

V1 =−1200

s2 + 10s + 125

Vo =20

0.8s + 20V1 =

−30,000(s + 25)(s + 5 − j10)(s + 5 + j10)

=K1

s + 25+

K2

s + 5 − j10+

K∗2

s + 5 + j10

Problems 13–47

K1 =−30,000

s2 + 10s + 125

∣∣∣∣s=−25

= −60

K2 =−30,000

(s + 25)(s + 5 + j10)

∣∣∣∣s=−5+j10

= 67.08/63.43

vo(t) = [−60e−25t + 134.16e−5t cos(10t + 63.43)]u(t) V

P 13.42 [a] Voltage source acting alone:

Vo1 − 60/s10

+Vo1s

80+

Vo1

20 + 10s= 0

·. . Vo1 =480(s + 2)

s(s + 4)(s + 6)


Vo2

10+

Vo2s

80+

Vo2 − 30/s10(s + 2)

= 0

·. . Vo2 =240

s(s + 4)(s + 6)

Vo = Vo1 + Vo2 =480(s + 2) + 240s(s + 4)(s + 6)

=480(s + 2.5)

s(s + 4)(s + 6)


[b] Vo =K1

s+

K2

s + 4+

K3

s + 6

K1 =(480)(2.5)

(4)(6)= 50; K2 =

480(−1.5)(−4)(2)

= 90; K3 =480(−3.5)(−6)(−2)

= −140

vo(t) = [50 + 90e−4t − 140e−6t]u(t) V

P 13.43 ∆ =

∣∣∣∣∣∣∣Y11 Y12

Y12 Y22

∣∣∣∣∣∣∣ = Y11Y22 − Y 212

N2 =

∣∣∣∣∣∣∣Y11 [(Vg/R1) + γC − (ρ/s)]

Y12 (Ig − γC)

∣∣∣∣∣∣∣

V2 =N2

∆

Substitution and simplification lead directly to Eq. 13.90.

P 13.44

Va − 4.8/s0.8

+Va

1/s+

Va − Vo

1/s= 0

0 − Va

1/s+

0 − Vo

1.25= 0

Va =−Vo

1.25s

Va(2s + 1.25) − sVo = 6/s

Problems 13–49

−Vo

[(2s + 1.25)

1.25s+ s

]= 6/s

−Vo

[125s2 + 2s + 1.25

1.25s

]= 6/s

Vo =−7.5

1.25s2 + 2s + 1.25=

−6s2 + 1.6s + 1

=K1

s + 0.8 − j0.6+

K∗1

s + 0.8 + j0.6

K1 =−6

s + 0.8 + j0.6

∣∣∣∣s=−0.8+j0.6

= 5/90

vo(t) = 10e−0.8t cos(0.6t + 90)u(t) V = −10e−0.8t sin(0.6t)u(t) V

P 13.45 [a] Vo = −Zf

Zi

Vg

Zf =107

s‖1000 =

1010/s

107/s + 1000=

1010

1000s + 107 =107

s + 104

Zi =2 × 106

s+ 400 =

400s + 2 × 106

s=

400s

(s + 5000)

Vg =20,000

s2

·. . Vo =−107/(s + 104)

(400/s)(s + 5000)· 20,000

s2 =−5 × 108

s(s + 5000)(s + 10,000)

[b] Vo =K1

s+

K2

s + 5000+

K3

s + 10,000

K1 =−5 × 108

(s + 5000)(s + 10,000)

∣∣∣∣s=0

= −10

K2 =−5 × 108

s(s + 10,000)

∣∣∣∣s=−5000

= 20

K3 =−5 × 108

s(s + 5000)

∣∣∣∣s=−10,000

= −10

·. . vo(t) = [−10 + 20e−5000t − 10e−10,000t]u(t) V


[c] −10 + 20e−5000ts − 10e−10,000ts = −5Let x = e−5000ts . Then

10x2 − 20x + 5 = 0

Solving,

x = 0.292893

e−5000ts = 0.292893 ·. . ts = 245.6 µs

[d] vg = m tu(t); Vg =m

s2

Vo =−107s

400(s + 5000)(s + 10,000)· m

s2

=−25,000m

s(s + 5000)(s + 10,000)

K1 =−25,000m

(5000)(10,000)= −5 × 10−4m

·. . −5 = −5 × 10−4m ·. . m = 10,000 V/s

P 13.46 [a]

Vp =50/s

5 + 50/sVg2 =

505s + 50

Vg2

Vp − 40/s20

+Vp − Vo

5+

Vp − Vo

100/s= 0

Vp

( 120

+15

+s

100

)− Vo

(15

+s

100

)=

2s

s + 25100

( 505s + 50

) 16s

− 2s

= Vo

(15

+s

100

)= Vo

(s + 20100

)

Problems 13–51

Vo =100

s + 20

[16(s + 25)

10(s + 10)(s)− 2

s

]=

−40s + 2000s(s + 10)(s + 20)

=K1

s+

K2

s + 10+

K3

s + 20

K1 = 10; K2 = −24; K3 = 14

·. . vo(t) = [10 − 24e−10t + 14e−20t]u(t) V

[b] 10 − 24e−10t + 14e−20t = 5Let x = e−10ts . Then

10 − 24x + 14x2 = 5

14x2 − 24x + 5 = 0

x = 0.242691

e−10ts = 0.242691 ·. . ts = 141.60 ms

P 13.47 Let vo1 equal the output voltage of the first op amp. Then

Vo1 =−Zf1

ZA1Vg where Zf1 = 25 × 103 Ω

ZA1 = 25,000 +25,000(20 × 104/s)

25,000 + (20 × 104/s)

=25,000(s + 16)

(s + 8)Ω

·. . Vo1 =−(s + 8)(s + 16)

Vg

Also,

Vo =−Zf2

ZA2Vo1 where Zf2 =

2 × 108

sΩ and ZA2 = 25,000 Ω

·. . Vo =−8000

sVo1 =

−8000s

[−(s + 8)(s + 16)

]Vg

=8000(s + 8)s(s + 16)

Vg

vg(t) = 16u(t) mV; ·. . Vg =16 × 10−3

s


Vo =128(s + 8)s2(s + 16)

=K1

s2 +K2

s+

K3

s + 16

K1 =128(8)

16= 64

K2 = 128d

ds

[s + 8s + 16

]s=0

= 4

K3 =128(−8)

256= −4

vo(t) = [64t + 4 − 4e−16t]u(t) V

The op amp will saturate when vo = ±6 V. Hence, saturation will occur when

64t + 4 − 4e−16t = 6 or 16t − 0.5 = e−16t

This equation can be solved by trial and error. First note that t > 0.5/16 ort > 31.25 ms.Try 40 ms:

0.64 − 0.5 = 0.14; e−0.64 = 0.53

Try 50 ms:

0.80 − 0.5 = 0.30; e−0.80 = 0.45

Try 60 ms:

0.96 − 0.5 = 0.46; e−0.96 = 0.38

Further trial and error gives

tsat∼= 56.5 ms

P 13.48 [a] Let va be the voltage across the 0.5 µF capacitor, positive at the upper terminal.Let vb be the voltage across the 100 kΩ resistor, positive at the upper terminal.Also note

106

0.5s=

2 × 106

sand

106

0.25s=

4 × 106

s; Vg =

0.5s

sVa

2 × 106 +Va − (0.5/s)

200,000+

Va

200,000= 0

Problems 13–53

sVa + 10Va − 5s

+ 10Va = 0

Va =5

s(s + 20)

0 − Va

200,000+

(0 − Vb)s4 × 106 = 0

·. . Vb = −20s

Va =−100

s2(s + 20)

Vb

100,000+

(Vb − 0)s4 × 106 +

(Vb − Vo)s4 × 106 = 0

40Vb + sVb + sVb = sVo

·. . Vo =2(s + 20)Vb

s; Vo = 2

(−100s3

)=

−200s3

[b] vo(t) = −100t2u(t) V

[c] −100t2 = −4; t = 0.2 s = 200 ms

P 13.49 [a]Vo

Vi

=1/sC

R + 1/sC

H(s) =(1/RC)

s + (1/RC)=

200s + 200

; −p1 = −200 rad/s

[b]Vo

Vi

=R

R + 1/sC=

RCs

RCs + 1=

s

s + (1/RC)

=s

s + 200; z1 = 0, −p1 = −200 rad/s

[c]Vo

Vi

=sL

R + sL=

s

s + R/L=

s

s + 8000

z1 = 0; −p1 = −8000 rad/s

[d]Vo

Vi

=R

R + sL=

R/L

s + (R/L)=

8000s + 8000

−p1 = −8000 rad/s

[e]

Vos

4 × 106 +Vo

10,000+

Vo − Vi

40,000= 0


sVo + 400Vo + 100Vo = 100Vi

H(s) =Vo

Vi

=100

s + 500

−p1 = −500 rad/s

P 13.50 [a] Let R1 = 250 kΩ; R2 = 125 kΩ; C2 = 1.6 nF; and Cf = 0.4 nF. Then

Zf =(R2 + 1/sC2)1/sCf(

R2 + 1sC2

+ 1sCf

) =(s + 1/R2C2)

Cfs(s + C2+Cf

C2Cf R2

)1

Cf

= 2.5 × 109

1R2C2

=62.5 × 107

125 × 103 = 5000 rad/s

C2 + Cf

C2CfR2=

2 × 10−9

(0.64 × 10−18)(125 × 103)= 25,000 rad/s

·. . Zf =2.5 × 109(s + 5000)

s(s + 25,000)Ω

Zi = R1 = 250 × 103 Ω

H(s) =Vo

Vg

=−Zf

Zi

=−104(s + 5000)s(s + 25,000)

[b] −z1 = −5000 rad/s

−p1 = 0; −p2 = −25,000 rad/s

P 13.51 [a]

Va − Vg

1000+

sVa

5 × 106 +(Va − Vo)s5 × 106 = 0

5000Va − 5000Vg + 2sVa − sVo = 0

Problems 13–55

(5000 + 2s)Va − sVo = 5000Vg

(0 − Va)s5 × 106 +

0 − Vo

5000= 0

−sVa − 1000Vo = 0; ·. . Va =−1000

sVo

(2s + 5000)(−1000

s

)Vo − sVo = 5000Vg

1000Vo(2s + 5000) + s2Vo = −5000sVg

Vo(s2 + 2000s + 5 × 106) = −5000sVg

Vo

Vg

=−5000s

s2 + 2000s + 5 × 106

s1,2 = −1000 ±√

106 − 5 × 106 = −1000 ± j2000

Vo

Vg

=−5000s

(s + 1000 − j2000)(s + 1000 + j2000)

[b] z1 = 0; −p1 = −1000 + j2000; −p2 = −1000 − j2000

P 13.52 [a] Zi = 1000 +5 × 106

s=

1000(s + 5000)s

Zf =40 × 106

s‖40,000 =

40 × 106

s + 1000

H(s) = −Zf

Zi

=−40 × 106/(s + 1000)

1000(s + 5000)/s=

−40,000s(s + 1000)(s + 5000)

[b] Zero at s = 0; Poles at −p1 = −1000 rad/s and −p2 = −5000 rad/s

P 13.53 [a]

Va =Vi

500,000 + [(20 × 106)/s](500,000) =

s

s + 40Vi

0.2Vi = Vo + Va


·. . Vo = 0.2Vi − s

s + 40Vi

Vo

Vi

=0.2(s + 40) − s

s + 40=

−0.8s + 8s + 40

=−0.8(s − 10)

s + 40

[b] −z1 = 10 rad/s

−p1 = −40 rad/s

P 13.54

Vg = 25sI1 − 35sI2

0 = −35sI1 +(

50s + 10,000 +16 × 106

s

)I2

∆ =

∣∣∣∣∣∣∣25s −35s

−35s 50s + 10,000 + 16 × 106/s

∣∣∣∣∣∣∣ = 25(s + 2000)(s + 8000)

N2 =

∣∣∣∣∣∣∣25s Vg

−35s 0

∣∣∣∣∣∣∣ = 35sVg

I2 =N2

∆=

35sVg

25(s + 2000)(s + 8000)

Vo =16 × 106

sI2 =

22.4 × 106Vg

(s + 2000)(s + 8000)

H(s) =Vo

Vg

=22.4 × 106

(s + 2000)(s + 8000)

·. . −p1 = −2000 rad/s; −p2 = −8000 rad/s

Problems 13–57

P 13.55 [a]

Vo

5000+

Vo

0.2s+ Vo(10−7)s = Ig

·. . Vo =10 × 106s

s2 + 2000s + 50 × 106 · Ig

Ig =0.1s

s2 + 108 ; Io =Vos

10 × 106

·. . H(s) =s2

s2 + 2000s + 50 × 106

[b] Io =(s2)(0.1s)

(s + 1000 − j7000)(s + 1000 + j7000)(s2 + 108)

Io =0.1s3

(s + 1000 − j7000)(s + 1000 + j7000)(s + j104)(s − j104)

[c] Damped sinusoid of the form

Me−1000t cos(7000t + θ1)

[d] Steady-state sinusoid of the form

N cos(104t + θ2)

[e] Io =K1

s + 1000 − j7000+

K∗1

s + 1000 + j7000+

K2

s − j104 +K∗

2

s + j104

K1 =0.1(−1000 + j7000)3

(j14,000)(−1000 − j3000)(−1000 + j17,000)= 46.90 × 10−3/− 140.54

K2 =0.1(j104)3

(j20,000)(1000 + j3000)(1000 + j17,000)= 92.85 × 10−3/21.80

io(t) = [93.8e−1000t cos(7000t − 140.54) + 185.7 cos(104t + 21.80)] mA

Test:

io(0) = 93.8 cos(−140.54) + 185.7 cos(21.80) mA = 100 mA

Z =1Y

; Y =1

5000+

1j2000

+1

−j1000=

2 + j510,000

·. . Z =10,0002 + j5

= 1856.95/− 68.2 Ω


Vo = IgZ = (0.1/0)(1856.95/− 68.2) = 185.695/− 68.2 V

Io =Vo

−j1000= 185.7/21.80 mA

ioss = 185.7 cos(104t + 21.80) mA(Checks)

P 13.56 [a]

2000(Io − Ig) + 8000Io + µ(Ig − Io)(2000) + 2sIo = 0

·. . Io =1000(1 − µ)

s + 1000(5 − µ)Ig

·. . H(s) =1000(1 − µ)

s + 1000(5 − µ)

[b] µ < 5

[c]µ H(s) Io

−3 4000/(s + 8000) 20,000/s(s + 8000)

0 1000/(s + 5000) 5000/s(s + 5000)

4 −3000/(s + 1000) −15,000/s(s + 1000)

5 −4000/s −20,000/s2

6 −5000/(s − 1000) −25,000/s(s − 1000)µ = −3:

Io =2.5s

− 2.5(s + 8000)

; io = [2.5 − 2.5e−8000t]u(t) A

µ = 0:

Io =1s

− 1s + 5000

; io = [1 − e−5000t]u(t) A

µ = 4:

Io =−15s

+15

s + 1000; io = [−15 + 15e−1000t]u(t) A

µ = 5:

Io =−20,000

s2 ; io = −20,000t u(t) A

Problems 13–59

µ = 6:

Io =25s

− 25s − 1000

; io = 25[1 − e1000t]u(t) A

P 13.57 H(s) =Vo

Vi

=1

s + 1; h(t) = e−t

For 0 ≤ t ≤ 1:

vo =∫ t

0e−λ dλ = (1 − e−t) V

For 1 ≤ t ≤ ∞:

vo =∫ t

t−1e−λ dλ = (e − 1)e−t V

P 13.58 H(s) =Vo

Vi

=s

s + 1= 1 − 1

s + 1; h(t) = δ(t) − e−t

h(λ) = δ(λ) − e−λ

For 0 ≤ t ≤ 1:

vo =∫ t

0[δ(λ) − e−λ] dλ = 1 + [e−λ] |t0= e−t V

For 1 ≤ t ≤ ∞:

vo =∫ t

t−1(−e−λ) dλ = e−λ

∣∣∣∣tt−1

= (1 − e)e−t V


P 13.59 [a]

t < 0 : y(t) = 0

0 ≤ t ≤ 10 : y(t) =∫ t

0625 dλ = 625t

10 ≤ t ≤ 20 : y(t) =∫ 10

t−10625 dλ = 625(10 − t + 10) = 625(20 − t)

20 ≤ t < ∞ : y(t) = 0

[b]

t < 0 : y(t) = 0

0 ≤ t ≤ 10 : y(t) =∫ t

0312.5 dλ = 312.5t

Problems 13–61

10 ≤ t ≤ 20 : y(t) =∫ t

t−10312.5 dλ = 3125

20 ≤ t ≤ 30 : y(t) =∫ 20

t−10312.5 dλ = 312.5(30 − t)

30 ≤ t < ∞ : y(t) = 0

[c]

t < 0 : y(t) = 0

0 ≤ t ≤ 1 : y(t) =∫ t

0625 dλ = 625t

1 ≤ t ≤ 10 : y(t) =∫ t

t−1625 dλ = 625

10 ≤ t ≤ 11 : y(t) =∫ 10

t−1625 dλ = 625(11 − t)

11 ≤ t < ∞ : y(t) = 0


P 13.60 [a] 0 ≤ t ≤ 40:

y(t) =∫ t

0(10)(1)(dλ) = 10λ

∣∣∣∣t0= 10t

40 ≤ t ≤ 80:

y(t) =∫ 40

t−40(10)(1)(dλ) = 10λ

∣∣∣∣40

t−40= 10(80 − t)

t ≥ 80 : y(t) = 0

Problems 13–63

[b] 0 ≤ t ≤ 10:

y(t) =∫ t

040 dλ = 40λ

∣∣∣∣t0= 40t

10 ≤ t ≤ 40:

y(t) =∫ t

t−1040 dλ = 40λ

∣∣∣∣tt−10

= 400


40 ≤ t ≤ 50:

y(t) =∫ 40

t−1040 dλ = 40λ

∣∣∣∣40

t−10= 40(50 − t)

t ≥ 50 : y(t) = 0

[c] The expressions are

0 ≤ t ≤ 1 : y(t) =∫ t

0400 dλ = 400λ

∣∣∣∣t0= 400t

1 ≤ t ≤ 40 : y(t) =∫ t

t−1400 dλ = 400λ

∣∣∣∣tt−1

= 400

40 ≤ t ≤ 41 : y(t) =∫ 40

t−1400 dλ = 400λ

∣∣∣∣40

t−1= 400(41 − t)

41 ≤ t < ∞ : y(t) = 0

[d]

[e] Yes, note that h(t) is approaching 40δ(t), therefore y(t) must approach 40x(t),i.e.

y(t) =∫ t

0h(t − λ)x(λ) dλ →

∫ t

040δ(t − λ)x(λ) dλ

→ 40x(t)

This can be seen in the plot, e.g., in part (c), y(t) ∼= 40x(t).

Problems 13–65

P 13.61 [a] −1 ≤ t ≤ 4:

vo =∫ t+1

010λ dλ = 5λ2

∣∣∣∣t+1

0= 5t2 + 10t + 5 V

4 ≤ t ≤ 9:

vo =∫ t+1

t−410λ dλ = 5λ2

∣∣∣∣t+1

t−4= 50t − 75 V

9 ≤ t ≤ 14:

vo = 10∫ 10

t−4λ dλ + 10

∫ t+1

1010 dλ

= 5λ2∣∣∣∣10

t−4+100λ

∣∣∣∣t+1

10= −5t2 + 140t − 480 V

14 ≤ t ≤ 19:

vo = 100∫ t+1

t−4dλ = 500 V

19 ≤ t ≤ 24:

vo =∫ 20

t−4100 dλ +

∫ t+1

2010(30 − λ) dλ

= 100λ∣∣∣∣20

t−4+ 300λ

∣∣∣∣t+1

20− 5λ2

∣∣∣∣t+1

20

= −5t2 + 190t − 1305 V

24 ≤ t ≤ 29:

vo = 10∫ t+1

t−4(30 − λ) dλ = 300λ

∣∣∣∣t+1

t−4− 5λ2

∣∣∣∣t+1

t−4

= 1575 − 50t V

29 ≤ t ≤ 34:

vo = 10∫ 30

t−4(30 − λ) dλ = 300λ

∣∣∣∣30

t−4− 5λ2

∣∣∣∣30

t−4

= 5t2 − 340t + 5780 V

Summary:

vo = 0 − ∞ ≤ t ≤ −1

vo = 5t2 + 10t + 5 V − 1 ≤ t ≤ 4

vo = 50t − 75 V 4 ≤ t ≤ 9

vo = −5t2 + 140t − 480 V 9 ≤ t ≤ 14


vo = 500 V 14 ≤ t ≤ 19

vo = −5t2 + 190t − 1305 V 19 ≤ t ≤ 24

vo = 1575 − 50t V 24 ≤ t ≤ 29

vo = 5t2 − 340t + 5780 V 29 ≤ t ≤ 34

vo = 0 V 34 ≤ t ≤ ∞[b]

P 13.62 [a] h(λ) =25λ 0 ≤ λ ≤ 5

h(λ) =(4 − 2

5λ)

5 ≤ λ ≤ 10

0 ≤ t ≤ 5:

vo = 10∫ t

0

25λ dλ = 2t2

Problems 13–67

5 ≤ t ≤ 10:

vo = 10∫ 5

0

25λ dλ + 10

∫ t

5

(4 − 2

5λ)

dλ

=4λ2

2

∣∣∣∣50

+ 40λ∣∣∣∣t5

− 4λ2

2

∣∣∣∣t5

= −100 + 40t − 2t2

10 ≤ t ≤ ∞:

vo = 10∫ 5

0

25λ dλ + 10

∫ 10

5

(4 − 2

5λ)

dλ

=4λ2

2

∣∣∣∣50

+ 40λ∣∣∣∣10

5− 4λ2

2

∣∣∣∣10

5

= 50 + 200 − 150 = 100

Summary:

vo = 2t2 V 0 ≤ t ≤ 5

vo = 40t − 100 − 2t2 V 5 ≤ t ≤ 10

vo = 100 V 10 ≤ t ≤ ∞[b]

[c] Area =12(10)(2) = 10 ·. .

12(4)h = 10 so h = 5

h(λ) =52λ 0 ≤ λ ≤ 2

h(λ) =(10 − 5

2λ)

2 ≤ λ ≤ 4


0 ≤ t ≤ 2:

vo = 10∫ t

0

52λ dλ = 12.5t2

2 ≤ t ≤ 4:

vo = 10∫ 2

0

52λ dλ + 10

∫ t

2

(10 − 5

2λ)

dλ

=25λ2

2

∣∣∣∣20

+ 100λ∣∣∣∣t2

− 25λ2

2

∣∣∣∣t2

= −100 + 100t − 12.5t2

4 ≤ t ≤ ∞:

vo = 10∫ 2

0

52λ dλ + 10

∫ 4

2

(10 − 5

2λ)

dλ

=25λ2

2

∣∣∣∣20

+ 100λ∣∣∣∣42

− 25λ2

2

∣∣∣∣42

= 50 + 200 − 150 = 100

vo = 12.5t2 V 0 ≤ t ≤ 2

vo = 100t − 100 − 12.5t2 V 2 ≤ t ≤ 4

vo = 100 V 4 ≤ t ≤ ∞[d] The waveform in part (c) is closer to replicating the input waveform because in

part (c) h(λ) is closer to being an ideal impulse response. That is, the area waspreserved as the base was shortened.

Problems 13–69

P 13.63 [a]

vo =∫ t

010(10e−4λ) dλ

= 100e−4λ

−4

∣∣∣∣t0= −25[e−4t − 1]

= 25(1 − e−4t) V, 0 ≤ t ≤ ∞

[b]

0 ≤ t ≤ 0.5:

vo =∫ t

0100(1 − 2λ) dλ = 100(λ − λ2)

∣∣∣∣t0= 100t(1 − t)

0.5 ≤ t ≤ ∞:

vo =∫ 0.5

0100(1 − 2λ) dλ = 100(λ − λ2)

∣∣∣∣0.5

0= 25


[c]

P 13.64 [a] From Problem 13.49(a)

H(s) =200

s + 200

h(λ) = 200e−200λ

0 ≤ t ≤ 5 ms:

vo =∫ t

020(200)e−200λ dλ = 20(1 − e−200t) V

5 ms ≤ t ≤ ∞:

vo =∫ t

t−5×10−320(200)e−200λ dλ = 20(e1 − 1)e−200t V

[b]

P 13.65 [a] H(s) =2000

s + 2000·. . h(λ) = 2000e−2000λ

Problems 13–71

0 ≤ t ≤ 5 ms:

vo =∫ t

020(2000)e−2000λ dλ = 20(1 − e−2000t) V

5 ms ≤ t ≤ ∞:

vo =∫ t

t−5×10−320(2000)e−2000λ dλ = 20(e10 − 1)e−2000t V

[b] decrease

[c] The circuit with R = 5 kΩ.

P 13.66 [a] Ig =Vo

105 +Vos

5 × 106 =Vo(s + 50)5 × 106

Vo

Ig

= H(s) =5 × 106

s + 50

h(λ) = 5 × 106e−50λu(λ)


0 ≤ t ≤ 0.1 s:

vo =∫ t

0(50 × 10−6)(5 × 106)e−50λ dλ = 250

e−50λ

−50

∣∣∣∣t0

= 5(1 − e−50t) V

0.1 s ≤ t ≤ 0.2 s:

vo =∫ t−0.1

0(−50 × 10−6)(5 × 106e−50λ dλ)

+∫ t

t−0.1(50 × 10−6)(5 × 106e−50λ dλ)

= −250e−50λ

−50

∣∣∣∣t−0.1

0+ 250

e−50λ

−50

∣∣∣∣tt−0.1

= 5[e−50(t−0.1) − 1

]− 5

[e−50t − e−50(t−0.1)

]vo = [10e−50(t−0.1) − 5e−50t − 5] V

Problems 13–73

0.2 s ≤ t ≤ ∞:

vo =∫ t−0.1

t−0.2−250e−50λ dλ +

∫ t

t−0.1250e−50λ dλ

= 5e−50λ

∣∣∣∣t−0.1

t−0.2− 5e−50λ

∣∣∣∣tt−0.1

vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t] V

Summary:

vo = 5(1 − e−50t) V 0 ≤ t ≤ 0.1 s

vo = [10e−50(t−0.1) − 5e−50t − 5] V 0.1 s ≤ t ≤ 0.2 s

vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t] V 0.2 s ≤ t ≤ ∞

[b] Io =Vos

5 × 106 =s

5 × 106 · 5 × 106Ig

s + 50

Io

Ig

= H(s) =s

s + 50= 1 − 50

s + 50

h(λ) = δ(λ) − 50e−50λ

0 < t < 0.1 s:

io =∫ t

0(50 × 10−6)[δ(λ) − 50e−50λ] dλ

= 50 × 10−6 −[50 × 50 × 10−6 e−50λ

−50

] ∣∣∣∣t0

= 50 × 10−6 + 50 × 10−6[e−50t − 1] = 50e−50t µA


0.1 s < t < 0.2 s:

io =∫ t−0.1

0(−50 × 10−6)[δ(λ) − 50e−50λ] dλ

+∫ t

t−0.1(50 × 10−6)(−50e−50λ) dλ

= −50 × 10−6 + 2500 × 10−6 e−50λ

−50

∣∣∣∣t−0.1

0− 2500 × 10−6 e−50λ

−50

∣∣∣∣tt−0.1

= −50 × 10−6 − 50 × 10−6[e−50(t−0.1) − 1] + 50 × 10−6[e−50t − e−50(t−0.1)]

= 50e−50t − 100e−50(t−0.1) µA

Problems 13–75

0.2 s < t < ∞:

io =∫ t−0.1

t−0.2(−50 × 10−6)(−50e−50λ) dλ

+∫ t

t−0.1(50 × 10−6)(−50e−50λ) dλ

= 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA

Summary:

i0 = 50e−50t µA 0 ≤ t ≤ 0.1 s

i0 = 50e−50t − 100e−50(t−0.1) µA 0.1 s ≤ t ≤ 0.2 s

i0 = 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA 0.2 s ≤ t ≤ ∞[c] At t = 0.1−:

vo = 5(1 − e−5) = 4.97 V; i100kΩ =4.970.1

= 49.66 µA; ig = 50 µA

·. . io = 50 − 49.66 = 0.34 µA

From the solution for io we have io(0.1−) = 50e−5 = 0.34 µA (Checks)At t = 0.1+:

vo(0.1+) = vo(0.1−) = 4.97 µV; i100kΩ = 49.66 µA; ig = −50 µA

·. . io(0.1+) = −(50 + 49.66) = −99.66 µA

From the solution for io we have

io(0.1+) = 50e−5 − 100 = −99.66 µA (Checks)


At t = 0.2−:

vo = 10e−5 − 5e−10 − 5 = −4.93 µV

i100kΩ = −49.33 µA ig = −50 µA

io = ig − i100kΩ = −50 + 49.33 = −0.67 µA

From the solution for io, io(0.2−) = 50e−10 − 100e−5 = −0.67 µA (Checks)At t = 0.2+:

vo(0.2+) = io(0.2−) = −4.93 V; i100kΩ = −49.33 µA; ig = 0

io = ig − i100kΩ = 49.33 µA

From the solution for io,io(0.2+) = 50e−10 − 100e−5 + 50 = 49.33 µA(Checks)

P 13.67 H(s) =Vo

Vi

=5

5 + 2.5s=

2s + 2

h(λ) = 2e−2λ; h(t − λ) = 2e−2(t−λ) = 2e−2te2λ

T

2=

π

2; T = π s; f =

1π

Hz

vi(λ) = (20 sin 2λ)[u(λ) − u(λ − π/2)]

(π/2) s ≤ t ≤ ∞:

vo =∫ π/2

0(2e−2te2λ)(20 sin 2λ) dλ = 40e−2t

∫ π/2

0e2λ sin 2λ dλ

= 40e−2t

[e2λ

8(2 sin 2λ − 2 cos 2λ)

]π/2

0

= 10e−2t[eπ(sin π − cos π) − 1(0 − 1)]

= 10e−2t(eπ + 1) = 10(eπ + 1)e−2t V

vo(2.2) = 241.41e−4.4 = 2.96 V

Problems 13–77

P 13.68 [a] Vo =1620

Vg

·. . H(s) =Vo

Vg

=45

h(λ) = 0.8δ(λ)

[b]

0 < t < 0.5 s : vo =∫ t

075[0.8δ(λ)] dλ = 60 A


0.5 s ≤ t ≤ 1.0 s:

vo =∫ t−0.5

0−75[0.8δ(λ)] dλ = −60 A

1 s < t < ∞ : vo = 0

[c]

Yes, because the circuit has no memory.

P 13.69 [a]

Vo − Vg

5+

Vos

4+

Vo

20= 0

(5s + 5)Vo = 4Vg

H(s) =Vo

Vg

=0.8

s + 1; h(λ) = 0.8e−λu(λ)

Problems 13–79

[b]

0 ≤ t ≤ 0.5 s;

vo =∫ t

075(0.8e−λ) dλ = 60

e−λ

−1

∣∣∣∣t0

vo = 60 − 60e−t V, 0 ≤ t ≤ 0.5 s

0.5 s ≤ t ≤ 1 s:

vo =∫ t−0.5

0(−75)(0.8e−λ) dλ +

∫ t

t−0.575(0.8e−λ) dλ

= −60e−λ

−1

∣∣∣∣t−0.5

0+ 60

e−λ

−1

∣∣∣∣tt−0.5

= 120e−(t−0.5) − 60e−t − 60 V, 0.5 s ≤ t ≤ 1 s


1 s ≤ t ≤ ∞;

vo =∫ t−0.5

t−1(−75)(0.8e−λ) dλ +

∫ t

t−0.575(0.8e−λ) dλ

= −60e−λ

−1

∣∣∣∣t−0.5

t−1+ 60

e−λ

−1

∣∣∣∣tt−0.5

= 120e−(t−0.5) − 60e−(t−1) − 60e−t V, 1 s ≤ t ≤ ∞[c]

[d] No, the circuit has memory because of the capacitive storage element.

P 13.70

Vo =20 × 103

5000 + 25 × 105/s + 20 × 103 (5000Ig)

Vo

Ig

= H(s) =4000s

s + 100

H(s) = 4000[1 − 100

s + 100

]= 4000 − 4 × 105

s + 100

h(λ) = 4000δ(λ) − 400,000e−100λu(λ)

Problems 13–81

vo =∫ 10−3

0(−20 × 10−3)[4000δ(λ) − 400,000e−100λ] dλ

+∫ 5×10−3

10−3(10 × 10−3)[−400,000e−100λ] dλ

= −80 + 8000∫ 10−3

0e−100λ dλ −

∫ 5×10−3

10−34000e−100λ dλ

= −80 − 80(e−0.1 − 1) + 40(e−0.5 − e−0.1)

vo(5 × 10−3) = 40e−0.5 − 120e−0.1 = 24.26 − 108.58 = −84.32 V

Alternate solution (not using the convolution integral):

Ig =∫ 4×10−3

0(10 × 10−3)e−st dt +

∫ 6×10−3

4×10−3(−20 × 10−3)e−st dt

= 10−3 e−st

−s

∣∣∣∣4×10−3

0− 20 × 10−3 e−st

−s

∣∣∣∣6×10−3

4×10−3

= 10 × 10−3

[1s

− e−4×10−3s

s

]+ 20 × 10−3

[e−6×10−3s − e−4×10−3s

s

]

=10 × 10−3

s− 30 × 10−3

se−4×10−3s +

20 × 10−3

se−6×10−3s

Vo = IgH(s) =40

s + 100− 120e−4×10−3s

s + 100+

80e−6×10−3s

s + 100

Now use the operational transform L−1e−asF (s) = f(t − a)u(t − a):

vo = 40e−100t − 120e−100(t−4×10−3)u(t − 4 × 10−3)

+ 80e−100(t−6×10−3)u(t − 6 × 10−3) V

vo(5 × 10−3) = 40e−0.5 − 120e−0.1 + 80(0) = −84.32 V (Checks)

P 13.71 [a] H(s) =Vo

Vi

=1/LC

s2 + (R/L)s + (1/LC)

=100

s2 + 20s + 100=

100(s + 10)2

h(λ) = 100λe−10λu(λ)


0 ≤ t ≤ 0.5:

vo = 500∫ t

0λe−10λ dλ

= 500

e−10λ

100(−10λ − 1)

∣∣∣∣t0

= 5[1 − e−10t(10t + 1)]

0.5 ≤ t ≤ ∞:

vo = 500∫ t

t−0.5λe−10λ dλ

= 500

e−10λ

100(−10λ − 1)

∣∣∣∣tt−0.5

= 5e−10t[e5(10t − 4) − 10t − 1]

[b]

Problems 13–83

P 13.72 H(s) =16s

40 + 4s + 16s=

0.8ss + 2

= 0.8(1 − 2

s + 2

)= 0.8 − 1.6

s + 2

h(λ) = 0.8δ(λ) − 1.6e−2λu(λ)

vo =∫ t

075[0.8δ(λ) − 1.6e−2λ] dλ =

∫ t

060δ(λ) dλ − 120

∫ t

0e−2λ dλ

= 60 − 120e−2λ

−2

∣∣∣∣t0= 60 + 60(e−2t − 1)

= 60e−2tu(t) V

P 13.73 [a] Y (s) =∫ ∞

0y(t)e−st dt

Y (s) =∫ ∞

0e−st

[∫ ∞

0h(λ)x(t − λ) dλ

]dt

=∫ ∞

0

∫ ∞

0e−sth(λ)x(t − λ) dλ dt

=∫ ∞

0h(λ)

∫ ∞

0e−stx(t − λ) dt dλ

But x(t − λ) = 0 when t < λ

Therefore Y (s) =∫ ∞

0h(λ)

∫ ∞

λe−stx(t − λ) dt dλ

Let u = t − λ; du = dt; u = 0, t = λ; u = ∞, t = ∞

Y (s) =∫ ∞

0h(λ)

∫ ∞

0e−s(u+λ)x(u) du dλ

=∫ ∞

0h(λ)e−sλ

∫ ∞

0e−sux(u) du dλ

=∫ ∞

0h(λ)e−sλX(s) dλ = H(s) X(s)

We are using one-sided Laplace transforms; therefore h(t) and X(t) areassumed zero for t < 0.


[b] F (s) =a

s(s + a)2 =1s

· a

(s + a)2 = H(s)X(s)

·. . h(t) = u(t), x(t) = at e−atu(t)

·. . f(t) =∫ t

0(1)aλe−aλ dλ = a

[e−aλ

a2 (−aλ − 1)]∣∣∣∣∣

t

0

=1a[e−at(−at − 1) − 1(−1)] =

1a[1 − e−at − ate−at]

=[1a

− 1ae−at − te−at

]u(t)

Check:

F (s) =a

s(s + a)2 =K0

s+

K1

(s + a)2 +K2

s + a

K0 =1a; K1 = −1; K2 =

d

ds

(a

s

)s=−a

= −1a

f(t) =[1a

− te−at − 1ae−at

]u(t)

P 13.74 [a] The s-domain circuit is

The node-voltage equation isV

sL1+

V

R+

V

sL2=

ρ

s

Therefore V =ρR

s + (R/Le)where Le =

L1L2

L1 + L2

Therefore v = ρRe−(R/Le)tu(t) V

Problems 13–85

[b] I1 =V

R+

V

sL2=

ρ[s + (R/L2)]s[s + (R/Le)]

=K0

s+

K1

s + (R/Le)

K0 =ρL1

L1 + L2; K1 =

ρL2

L1 + L2

Thus we have i1 =ρ

L1 + L2[L1 + L2e

−(R/Le)t]u(t) A

[c] I2 =V

sL2=

(ρR/L2)s[s + (R/Le)]

=K2

s+

K3

s + (R/Le)

K2 =ρL1

L1 + L2; K3 =

−ρL1

L1 + L2

Therefore i2 =ρL1

L1 + L2[1 − e−(R/Le)t]u(t)

[d] λ(t) = L1i1 + L2i2 = ρL1

P 13.75 [a] As R → ∞, v(t) → ρLeδ(t) since the area under the impulse generatingfunction is ρLe.

i1(t) → ρL1

L1 + L2as R → ∞

i2(t) → ρL1

L1 + L2as R → ∞


[b] The s-domain circuit is

V

sL1+

V

sL2=

ρ

s; therefore V =

ρL1L2

L1 + L2= ρLe

Therefore v(t) = ρLeδ(t)

I1 = I2 =V

sL2=(

ρL1

L1 + L2

)(1s

)

Therefore i1 = i2 =ρL1

L1 + L2u(t) A

P 13.76 H(j3) =4(3 + j3)

−9 + j24 + 41= 0.42/8.13

·. . vo(t) = 16.97 cos(3t + 8.13) V

P 13.77 [a] H(s) =−Zf

Zi

Zf =(1/Cf )

s + (1/RfCf )=

108

s + 1000

Zi =Ri[s + (1/RiCi)]

s=

10,000(s + 400)s

H(s) =−104s

(s + 400)(s + 1000)

[b] H(j400) =−104(j400)

(400 + j400)(1000 + j400)= 6.565/− 156.8

vo(t) = 13.13 cos(400t − 156.8) V

Problems 13–87

P 13.78 [a]

Vp =0.01s

80 + 0.01sVg =

s

s + 8000Vg

Vn

5000+

Vn − Vo

25,000+ (Vn − Vo)8 × 10−9s = 0

5Vn + Vn − Vo + (Vn − Vo)2 × 10−4s = 0

6Vn + 2 × 10−4sVn = Vo + 2 × 10−4sVo

2 × 10−4Vn(s + 30,000) = 2 × 10−4Vo(s + 5000)

Vn = Vp

Vo =s + 30,000s + 5000

Vf =(

s + 30,000s + 5000

)(sVg

s + 8000

)

H(s) =Vo

Vg

=s(s + 30,000)

(s + 5000)(s + 8000)

[b] vg = 0.6u(t); Vg =0.6s

Vo =0.6(s + 30,000)

(s + 5000)(s + 8000)=

K1

s + 5000+

K2

s + 8000

K1 =0.6(25,000)

3000= 5; K2 =

0.6(22,000)−3000

= −4.4

·. . vo(t) = (5e−5000t − 4.4e−8000t)u(t) V

[c] Vg = 2 cos 10,000t V

H(jω) =j10,000(30,000 + j10,000)

(5000 + j10,000)(8000 + j10,000)= 2.21/− 6.34

·. . vo = 4.42 cos(10,000t − 6.34) V


P 13.79 Vo =50

s + 8000− 20

s + 5000=

30(s + 3000)(s + 5000)(s + 8000)

Vo = H(s)Vg = H(s)(30

s

)

·. . H(s) =s(s + 3000)

(s + 5000)(s + 8000)

H(j6000) =(j6000)(3000 + j6000)

(5000 + j6000)(8000 + j6000)= 0.52/66.37

·. . vo(t) = 61.84 cos(6000t + 66.37) V

P 13.80 Original charge on C1; q1 = V0C1

The charge transferred to C2; q2 = V0Ce =V0C1C2

C1 + C2

The charge remaining on C1; q′1 = q1 − q2 =

V0C21

C1 + C2

Therefore V2 =q2

C2=

V0C1

C1 + C2and V1 =

q′1

C1=

V0C1

C1 + C2

P 13.81 [a] Z1 =1/C1

s + 1/R1C1=

25 × 1010

s + 20 × 104 Ω

Z2 =1/C2

s + 1/R2C2=

6.25 × 1010

s + 12,500Ω

Vo

Z2+

Vo − 10/sZ1

= 0

Vo(s + 12,500)6.25 × 1010 +

Vo(s + 20 × 104)25 × 1010 =

10s

(s + 20 × 104)25 × 1010

Vo =2(s + 200,000)s(s + 50,000)

=K1

s+

K2

s + 50,000

K1 =2(200,000)

50,000= 8

K2 =2(150,000)−50,000

= −6

·. . vo = [8 − 6e−50,000t]u(t) V

Problems 13–89

[b] I0 =V0

Z2=

2(s + 200,000)(s + 12,500)s(s + 50,000)6.25 × 1010

= 32 × 10−12

[1 +

162,500s + 25 × 108

s(s + 50,000)

]

= 32 × 10−12

[1 +

K1

s+

K2

s + 50,000

]

K1 = 50,000; K2 = 112,500

io = 32δ(t) + [1.6 × 106 + 3.6 × 106e−50,000t]u(t) pA

[c] When C1 = 64 pF

Z1 =156.25 × 108

s + 12,500Ω

V0(s + 12,500)625 × 108 +

V0(s + 12,500)156.25 × 108 =

10s

(s + 12,500)156.25 × 108

·. . V0 + 4V0 =40s

V0 =8s

vo = 8u(t) V

I0 =V0

Z2=

8s

(s + 12,500)6.25 × 1010 = 128 × 10−12

[1 +

12,500s

]

io(t) = 128δ(t) + 1.6 × 10−6u(t) pA

P 13.82 Let a =1

R1C1=

1R2C2

Then Z1 =1

C1(s + a)and Z2 =

1C2(s + a)

Vo

Z2+

Vo

Z1=

10/sZ1

VoC2(s + a) + VoC1(s + a) = (10/s)C1(s + a)

Vo =10s

(C1

C1 + C2

)

Thus, vo is the input scaled by the factorC1

C1 + C2.


P 13.83 [a] For t < 0, 0.5v1 = 2v2; therefore v1 = 4v2

v1 + v2 = 100; therefore v1(0−) = 80 V

[b] v2(0−) = 20 V

[c] v3(0−) = 0 V

[d] For t > 0:

I =100/s

3.125/s× 10−6 = 32 × 10−6

i(t) = 32δ(t) µA

[e] v1(0+) = −106

0.5

∫ 0+

0−32 × 10−6δ(t) dt + 80 = −64 + 80 = 16 V

[f] v2(0+) = −106

2

∫ 0+

0−32 × 10−6δ(t) dt + 20 = −16 + 20 = 4 V

[g] V3 =0.625 × 106

s· 32 × 10−6 =

20s

v3(t) = 20u(t) V; v3(0+) = 20 V

Check: v1(0+) + v2(0+) = v3(0+)

P 13.84 [a] For t < 0:

Req = 0.8 kΩ‖4 kΩ‖16 kΩ = 0.64 kΩ; v = 5(640) = 3200 V

i1(0−) =32004000

= 0.8 A; i2(0−) =3200

16,000= 0.2 A

Problems 13–91

[b] For t > 0:

i1 + i2 = 0

8(∆i1) = 2(∆i2)

i1(0−) + ∆i1 + i2(0−) + ∆i2 = 0; therefore ∆i1 = −0.2 A

∆i2 = −0.8 A; i1(0+) = 0.8 − 0.2 = 0.6 A

[c] i2(0−) = 0.2 A

[d] i2(0+) = 0.2 − 0.8 = −0.6 A

[e] The s-domain equivalent circuit for t > 0 is

I1 =0.006

0.01s + 20,000=

0.6s + 2 × 106

i1(t) = 0.6e−2×106tu(t) A

[f] i2(t) = −i1(t) = −0.6e−2×106tu(t) A

[g] V = −0.0064 + (0.008s + 4000)I1 =−0.0016(s + 6.5 × 106)

s + 2 × 106

= −1.6 × 10−3 − 7200s + 2 × 106

v(t) = [−1.6 × 10−3δ(t)] − [7200e−2×106tu(t)] V

P 13.85 [a]

Vo =0.5

50,000 + 5 × 106/s· 106

s


500,00050,000s + 5 × 106 =

10s + 100

vo = 10e−100tu(t) V

[b] At t = 0 the current in the 1 µF capacitor is 10δ(t) µA

·. . vo(0+) = 106∫ 0+

0−10 × 10−6δ(t) dt = 10 V

After the impulsive current has charged the 1 µF capacitor to 10 V it dischargesthrough the 50 kΩ resistor.

Ce =C1C2

C1 + C2=

0.251.25

= 0.2 µF

τ = (50,000)(0.2 × 10−6) = 10−2

1τ

= 100 (Checks)

Note – after the impulsive current passes the circuit becomes

The solution for vo in this circuit is also

vo = 10e−100tu(t) V

P 13.86 [a] After making a source transformation, the circuit is as shown. The impulsecurrent will pass through the capacitive branch since it appears as a shortcircuit to the impulsive current,

Therefore vo(0+) = 106∫ 0+

0−

[δ(t)1000

]dt = 1000 V

Problems 13–93

Therefore wC = (0.5)Cv2 = 0.5 J

[b] iL(0+) = 0; therefore wL = 0 J

[c] Vo(10−6)s +Vo

250 + 0.05s+

Vo

1000= 10−3

Therefore

Vo =1000(s + 5000)

s2 + 6000s + 25 × 106

=K1

s + 3000 − j4000+

K∗1

s + 3000 + j4000

K1 = 559.02/− 26.57; K∗1 = 559.02/26.57

vo = [1118.03e−3000t cos(4000t − 26.57)]u(t) V

[d] The s-domain circuit is

Vos

106 +Vo

250 + 0.05s+

Vo

1000= 10−3

Note that this equation is identical to that derived in part [c], therefore thesolution for Vo will be the same.

P 13.87 [a]

20 = sI1 − 0.5sI2

0 = −0.5sI1 +(s +

3s

)I2


∆ =

∣∣∣∣∣∣∣s −0.5s

−0.5s (s + 3/s)

∣∣∣∣∣∣∣ = s2 + 3 − 0.25s2 = 0.75(s2 + 4)

N1 =

∣∣∣∣∣∣∣20 −0.5s

0 (s + 3/s)

∣∣∣∣∣∣∣ = 20s +60s

=20s2 + 60

s=

20(s2 + 3)s

I1 =N1

∆=

20(s2 + 3)s(0.75)(s2 + 4)

=803

· s2 + 3s(s2 + 4)

=K0

s+

K1

s − j2+

K∗1

s + j2

K0 =803

(34

)= 20; K1 =

803

[ −4 + 3(j2)(j4)

]=

103

/0

·. . i1 =[20 +

203

cos 2t]u(t) A

[b] N2 =

∣∣∣∣∣∣∣s 20

−0.5s 0

∣∣∣∣∣∣∣ = 10s

I2 =N2

∆=

10s0.75(s2 + 4)

=403

(s

s2 + 4

)=

K1

s − j2+

K∗1

s + j2

K1 =403

(j2j4

)=

203

/0

i2 =403

(cos 2t)u(t) A

[c] V0 =3sI2 =

(3s

) 403

(s

s2 + 4

)=

40s2 + 4

=K1

s − j2=

K∗1

s + j2

K1 =40j4

= −j10 = 10/90

vo = 20 cos(2t − 90) = 20 sin 2t

vo = [20 sin 2t]u(t) V

[d] Let us begin by noting i1 jumps from 0 to (80/3) A between 0− and 0+ and inthis same interval i2 jumps from 0 to (40/3) A. Therefore in the derivatives ofi1 and i2 there will be impulses of (80/3)δ(t) and (40/3)δ(t), respectively.Thus

di1dt

=803

δ(t) − 403

sin 2t A/s

Problems 13–95

di2dt

=403

δ(t) − 803

sin 2t A/s

From the circuit diagram we have

20δ(t) = 1di1dt

− 0.5di2dt

=803

δ(t) − 403

sin 2t − 20δ(t)3

+403

sin 2t

= 20δ(t)

Thus our solutions for i1 and i2 are in agreement with known circuit behavior.Let us also note the impulsive voltage will impart energy into the circuit. Sincethere is no resistance in the circuit, the energy will not dissipate. Thus the factthat i1, i2, and vo exist for all time is consistent with known circuit behavior.Also note that although i1 has a dc component, i2 does not. This follows fromknown transformer behavior.Finally we note the flux linkage prior to the appearance of the impulsivevoltage is zero. Now since v = dλ/dt, the impulsive voltage source must bematched to an instantaneous change in flux linkage at t = 0+ of 20.For the given polarity dots and reference directions of i1 and i2 we have

λ(0+) = L1i1(0+) + Mi1(0+) − L2i2(0+) − Mi2(0+)

λ(0+) = 1(80

3

)+ 0.5

(803

)− 1

(403

)− 0.5

(403

)

=1203

− 603

= 20 (Checks)

P 13.88 [a]

V1

104 +V1

[(2 × 105)/s)] + [(5 × 104)/s]= 10−5


V1

104 +sV1

25 × 104 = 10−5

25V1 + sV1 = 2.5

V1 =2.5

s + 25

Vo =(

sV1

25 × 104

)(5 × 104

s

)=

15V1

·. . Vo =0.5

s + 25; vo = 0.5e−25tu(t) V

[b] vo(0+) = 0.5 V

vo(0+) =106

20

∫ 0+

0−10 × 10−6δ(x) dx = 0.5 V (Checks)

Ce =(5)(20)

25= 4 µF

τ = RCe = (10 × 103)(4 × 10−6) = 4 × 10−2 s;1τ

=1004

= 25 (Checks)

Yes, the impulsive current establishes an instantaneous charge on eachcapacitor. After the impulsive current vanishes the capacitors dischargeexponentially to zero volts.

P 13.89 [a] The circuit parameters are

Ra =1202

1200= 12 Ω Rb =

1202

1800= 8 Ω Xa =

1202

350=

144035

Ω

The branch currents are

I1 =120/0

12= 10/0 A(rms) I2 =

120/0

j1440/35= −j

3512

=3512

/− 90 A(rms)

I3 =120/0

8= 15/0 A(rms)

·. . IL = I1 + I2 + I3 = 25 − j3512

= 25.17/− 6.65 A(rms)

Therefore,

i2 =(35

12

)√2 cos(ωt − 90) A and iL = 25.17

√2 cos(ωt − 6.65) A

Thus,

i2(0−) = i2(0+) = 0 A and iL(0−) = iL(0+) = 25√

2 A

Problems 13–97

[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for V0 symbolically.Write a single node voltage equation:

V0 − (Vg + LI0)sL

+V0

Ra

+V0

sLa

= 0

·. . V0 =(Ra/L)Vg + I0Ra

s + [Ra(La + L)]/LaL

where L = 1/120π H, La = 12/35π H, Ra = 12 Ω, and I0Ra = 300√

2 V.Also,

Vg = V0 + IL(j) = 120 +(25 − j

3512

)j = 122.92 + 25j V(rms)

vg(t) = 122.92√

2 cos ωt − 25√

2 sin ωt V, with ω = 120π rad/s.

Thus,

V0 =1440π(122.92

√2s − 3000π

√2)

(s + 1475π)(s2 + 14,400π2)+

300√

2s + 1475π

=K1

s + 1475π+

K2

s − j120π+

K∗2

s + j120π+

300√

2s + 1475π

The coefficients are

K1 = −121.18√

2 V K2 = 61.03√

2/6.85 V K∗2 = 61.03

√2/− 6.85

Note that K1 + 300√

2 = 178.82√

2 V. Thus, the inverse transform of V0 is

v0 = 178.82√

2e−1475πt + 122.06√

2 cos(120πt + 6.85) V

Initially,

v0(0+) = 178.82√

2 + 122.06√

2 cos 6.85 = 300√

2 V

Note that at t = 0+ the initial value of iL, which is 25√

2 A, exists in the 12 Ωresistor Ra. Thus, the initial value of V0 is (25

√2)(12) = 300

√2 V.

[c] The phasor domain equivalent circuit has a j1 Ω inductive impedance in serieswith the parallel combination of a 12 Ω resistive impedance and a j1440/35 Ωinductive impedance (remember that ω = 120π rad/s). Note thatVg = 120/0 + (25.17/− 6.65)(j1) = 125.43/11.50 V(rms). The nodevoltage equation in the phasor domain circuit is

V0 − 125.43/11.50

j1+

V0

12+

35V0

j1440= 0

·. . V0 = 122.06/6.85 V(rms)

Therefore, v0 = 122.06√

2 cos(120πt + 6.85) V, agreeing with thesteady-state component of the result in part (b).


[d] A plot of v0, generated in Excel, is shown below.

P 13.90 [a] At t = 0− the phasor domain equivalent circuit is

I1 =−j120

12= −j10 = 10/− 90A (rms)

I2 =−j120(35)

j1440= −35

12=

3512

/180A (rms)

I3 =−j120

8= −j15 = 15/− 90A (rms)

IL = I1 + I2 + I3 = −3512

− j25 = 25.17/− 96.65A (rms)

iL = 25.17√

2 cos(120πt − 96.65)A

iL(0−) = iL(0+) = −2.92√

2A

Problems 13–99

i2 =3512

√2 cos(120πt + 180)A

i2(0−) = i2(0+) = −3512

√2 = −2.92

√2A

Vg = Vo + j1IL

Vg = −j120 + 25 − j3512

= 25 − j122.92

vg = 25√

2 cos 120πt + 122.92√

2 sin 120πt

·. . Vg =25

√2s + 122.92

√2(120π)

s2 + (120π)2

Use a variation of the s-domain circuit in Fig.13.60, where

Ll =1

120πH; La =

1235π

H; Ra = 12 Ω

iL(0) = −2.92√

2A; i2(0) = −2.92√

2A


0 =Vo − (Vg + iL(0)Ll)

sLl

+Vo

Ra

+Vo + i2(0)La

sLa

Solving for Vo yields

Vo =VgRa/Ll

[s + Ra(Ll + La)/LaLl]+

Ra[iL(0) − i2(0)][s + Ra(Ll + La)/LlLa]

Ra

Ll

= 1440π

Ra(Ll + La)LlLa

=12( 1

120π+ 12

35π)

( 1235π

)( 1120π

)= 1475π

iL(0) − i2(0) = −2.92√

2 + 2.92√

2 = 0

·. . Vo =1440π[25

√2s + 122.92

√2(120π)]

(s + 1475π)[s2 + (120π)2]

=K1

s + 1475π+

K2

s − j120π+

K∗2

s + j120π

K1 = −14.55√

2 K2 = 61.03√

2/− 83.15

·. . vo(t) = −14.55√

2e−1475πt + 122.06√

2 cos(120πt − 83.15)V

Check:

vo(0) = (−14.55 + 14.55)√

2 = 0


[b]

Problems 13–101


Problems 13–103

[c] In Prob. 13.89 the line-to-neutral voltage spikes at 300√

2 V. In part (a) theline-to-neutral voltage has no spike. Thus the amount of voltage disturbancedepends on what part of the cycle the sinusoidal steady-state voltage isswitched.

P 13.91 [a] First find Vg before Rb is disconnected. The phasor domain circuit is

IL =120/− θ

Ra

+120/− θ

Rb

+120/− θ

jXa

=120/− θ

RaRbXa

[(Ra + Rb)Xa − jRaRb]

Since Xl = 1 Ω we have

Vg = 120/− θ +120/− θ

RaRbXa

[RaRb + j(Ra + Rb)Xa]

Ra = 12 Ω; Rb = 8 Ω; Xa =144035

Ω

Vg =120/− θ

1440(1475 + j300)

=2512

/− θ(59 + j12) = 125.43/(−θ + 11.50)

vg = 125.43√

2 cos(120πt − θ + 11.50)V

Let β = −θ + 11.50. Then

vg = 125.43√

2(cos 120πt cos β − sin 120πt sin β)V

Therefore

Vg =125.43

√2(s cos β − 120π sin β)s2 + (120π)2


The s-domain circuit becomes

where ρ1 = iL(0+) and ρ2 = i2(0+).The s-domain node voltage equation is

Vo − (Vg + ρ1Ll)sLl

+Vo

Ra

+Vo + ρ2La

sLa

= 0

Solving for Vo yields

Vo =VgRa/Ll + (ρ1 − ρ2)Ra

[s + (La+Ll)Ra

LaLl]

Substituting the numerical values

Ll =1

120πH; La =

1235π

H; Ra = 12 Ω; Rb = 8 Ω;

gives

Vo =1440πVg + 12(ρ1 − ρ2)

(s + 1475π)

Now determine the values of ρ1 and ρ2.

ρ1 = iL(0+) and ρ2 = i2(0+)

IL =120/− θ

RaRbXa

[(Ra + Rb)Xa − jRaRb]

=120/− θ

96(1440/35)

[(20)(1440)

35− j96

]

= 25.17/(−θ − 6.65)A(rms)

·. . iL = 25.17√

2 cos(120πt − θ − 6.65)A

iL(0+) = ρ1 = 25.17√

2 cos(−θ − 6.65)A

·. . ρ1 = 25√

2 cos θ − 2.92√

2 sin θA

I2 =120/− θ

j(1440/35)=

3512

/(−θ − 90)

Problems 13–105

i2 =3512

√2 cos(120πt − θ − 90)A

ρ2 = i2(0+) = −3512

√2 sin θ = −2.92

√2 sin θA

·. . ρ1 − ρ2 = 25√

2 cos θ

(ρ1 − ρ2)Ra = 300√

2 cos θ

·. . Vo =1440π

s + 1475π· Vg +

300√

2 cos θ

s + 1475π

=1440π

s + 1475π

[125.43

√2(s cos β − 120π sin β)s2 + 14,400π2

]+

300√

2 cos θ

s + 1475π

=K1 + 300

√2 cos θ

s + 1475π+

K2

s − j120π+

K∗2

s + j120π

Now

K1 =(1440π)(125.43

√2)[−1475π cos β − 120π sin β]

14752π2 + 14,400π2

=−1440(125.43

√2)[1475 cosβ + 120 sin β]

14752 + 14,400

Since β = −θ + 11.50, K1 reduces to

K1 = −121.18√

2 cos θ − 14.55√

2 sin θ

From the partial fraction expansion for Vo we see vo(t) will go directly intosteady state when K1 = −300

√2 cos θ. It follow that

−14.55√

2 sin θ = −178.82√

2 cos θ

or tan θ = 12.29

Therefore, θ = 85.35

[b] When θ = 85.35, β = −73.85

·. . K2 =1440π(125.43

√2)[−120π sin(−73.85) + j120π cos(−73.85)

(1475π + j120π)(j240π)

=720

√2(120.48 + j34.88)−120 + j1475

= 61.03√

2/− 78.50

·. . vo = 122.06√

2 cos(120πt − 78.50)V t > 0

= 172.61 cos(120πt − 78.50)V t > 0


[c] vo1 = 169.71 cos(120πt − 85.35)V t < 0

vo2 = 172.61 cos(120πt − 78.50)V t > 0

Problems 13–107


14Introduction to Frequency-Selective

Circuits

Assessment Problems

AP 14.1 fc = 8 kHz, ωc = 2πfc = 16π krad/s

ωc =1

RC; R = 10 kΩ;

·. . C =1

ωcR=

1(16π × 103)(104)

= 1.99 nF

AP 14.2 [a] ωc = 2πfc = 2π(2000) = 4π krad/s

L =R

ωc

=50004000π

= 0.40 H

[b] H(jω) =ωc

ωc + jω=

4000π4000π + jω

When ω = 2πf = 2π(50,000) = 100,000π rad/s

H(j100,000π) =4000π

4000π + j100,000π=

11 + j25

= 0.04/− 87.71

·. . |H(j100,000π)| = 0.04

[c] ·. . θ(100,000π) = −87.71

AP 14.3 ωc =R

L=

50003.5 × 10−3 = 1.43 Mrad/s

14–1

14–2 CHAPTER 14. Introduction to Frequency-Selective Circuits

AP 14.4 [a] ωc =1

RC=

106

R=

106

100= 10 krad/s

[b] ωc =106

5000= 200 rad/s

[c] ωc =106

3 × 104 = 33.33 rad/s

AP 14.5 Let Z represent the parallel combination of (1/sC) and RL. Then

Z =RL

(RLCs + 1)

Thus H(s) =Z

R + Z=

RL

R(RLCs + 1) + RL

=(1/RC)

s + R+RL

RL

(1

RC

) =(1/RC)

s + 1K

(1

RC

)

where K =RL

R + RL

AP 14.6 ω2o =

1LC

so L =1

ω2oC

=1

(24π × 103)2(0.1 × 10−6)= 1.76 mH

Q =ωo

β=

ωo

R/Lso R =

ωoL

Q=

(24π × 103)(1.76 × 10−3)6

= 22.10 Ω

AP 14.7 ωo = 2π(2000) = 4000π rad/s;

β = 2π(500) = 1000π rad/s; R = 250 Ω

β =1

RCso C =

1βR

=1

(1000π)(250)= 1.27 µF

ω2o =

1LC

so L =1

ω2oC

=106

(4000π)2(1.27)= 4.97 mH

AP 14.8 ω2o =

1LC

so L =1

ω2oC

=1

(104π)2(0.2 × 10−6)= 5.07 mH

β =1

RCso R =

1βC

=1

400π(0.2 × 10−6)= 3.98 kΩ

Problems 14–3

AP 14.9 ω2o =

1LC

so L =1

ω2oC

=1

(4000π)2(0.2 × 10−6)= 31.66 mH

Q =fo

β=

5 × 103

200= 25 = ωoRC

·. . R =Q

ωoC=

25(4000π)(0.2 × 10−6)

= 9.95 kΩ

AP 14.10ωo = 8000π rad/s

C = 500 nF

ω2o =

1LC

so L =1

ω2oC

= 3.17 mH

Q =ωo

β=

ωoL

R=

1ωoCR

·. . R =1

ωoCQ=

1(8000π)(500 × 10−9)(5)

= 15.92 Ω

AP 14.11ωo = 2πfo = 2π(20,000) = 40π krad/s; R = 100 Ω; Q = 5

Q =ωo

β=

ωoL

RCso L =

RQ

ωo

=100

40π × 103 = 3.98 mH

ω2o =

1LC

so C =1

ω2oL

=1

(40π × 103)2(3.98 × 10−3)= 15.92 nF


Problems

P 14.1 [a] ωc =R

L=

12710 × 10−3 = 12.7 krad/s

·. . fc =ωc

2π=

12,7002π

= 2021.27 Hz

[b] H(s) =ωc

s + ωc

=12,700

s + 12,700

H(jω) =12,700

12,700 + jω

H(jωc) =12,700

12,700 + j12,700= 0.7071/− 45

H(j0.2ωc) =12,700

12,700 + j2540= 0.981/− 11.31

H(j5ωc) =12,700

12,700 + j63,500= 0.196/− 78.69

[c] vo(t)|ωc = 7.07 cos(12,700t − 45) V

vo(t)|0.2ωc = 9.81 cos(2540t − 11.31) V

vo(t)|5ωc = 1.96 cos(63,500t − 78.69) V

P 14.2 [a] ωo =R

L= 2000π rad/s

R = Lωo = (0.005)(2000π) = 31.42 Ω

[b] Re = 31.42‖270 = 28.14 Ω

ωloaded =Re

L= 5628 rad/s

·. . floaded =ωloaded

2π= 895.77 Hz

P 14.3 Note: add the resistor to the cirucit in Fig. 14.4(a).

[a] H(s) =Vo

Vi

=R

sL + R + Rl

=(R/L)

s + (R + Rl)/L

Problems 14–5

[b] H(jω) =(R/L)(

R+Rl

L

)+ jω

|H(jω)| =(R/L)√(

R+Rl

L

)2+ ω2

|H(jω)|max occurs when ω = 0

[c] |H(jω)|max =R

R + Rl

[d] |H(jωc)| =R√

2(R + Rl)=

R/L√(R+Rl

L

)2+ ω2

c

·. . ω2c =

(R + Rl

L

)2

; ·. . ωc = (R + Rl)/L

[e] Note – add 75 Ω resistor in series with the 10 mH inductor.

ωc =127 + 75

0.01= 20,200 rad/s

H(jω) =12,700

20,200 + jω

H(j0) = 0.6287

H(j20,200) =0.6287√

2/− 45 = 0.4446/− 45

H(j6060) =12,700

20,200 + j6060= 0.6022/− 16.70

H(j60,600) =12,700

20,200 + j60,600= 0.1988/− 71.57

P 14.4 [a] ωc =1

RC=

1(103)(100 × 10−9)

= 10 krad/s

fc =ωc

2π= 1591.55 Hz

[b] H(jω) =ωc

s + ωc

=10,000

s + 10,000

H(jω) =10,000

10,000 + jω

H(jωc) =10,000

10,000 + j10,000= 0.7071/− 45


H(j0.1ωc) =10,000

10,000 + j1000= 0.9950/− 5.71

H(j10ωc) =10,000

10,000 + j100,000= 0.0995/− 84.29

[c] vo(t)|ωc = 200(0.7071) cos(10,000t − 45)

= 141.42 cos(10,000t − 45) mV

vo(t)|0.1ωc = 200(0.9950) cos(1000t − 5.71)

= 199.01 cos(1000t − 5.71) mV

vo(t)|10ωc = 200(0.0995) cos(100,000t − 84.29)

= 19.90 cos(100,000t − 84.29) mV

P 14.5 [a] Let Z =RL(1/sC)RL + 1/sC

=RL

RLCs + 1

Then H(s) =Z

Z + R

=RL

RRLCs + R + RL

=(1/RC)

s +(

R + RL

RRLC

)

[b] |H(jω)| =(1/RC)√

ω2 + [(R + RL)/RRLC]2

|H(jω)| is maximum at ω = 0

[c] |H(jω)|max =RL

R + RL

[d] |H(jωc)| =RL√

2(R + RL)=

(1/RC)√ω2

c + [(R + RL)/RRLC]2

·. . ωc =R + RL

RRLC=

1RC

(1 + (R/RL))

[e] ωc =1

(103)(10−7)[1 + (103/104)] = 10,000(1 + 0.1) = 11,000 rad/s

H(j0) =10,00011,000

= 0.9091/0

Problems 14–7

H(jωc) =10,000

11,000 + j11,000= 0.6428/− 45

H(j0.1ωc) =10,000

11,000 + j1100= 0.9046/− 5.71

H(j10ωc) =10,000

11,000 + j110,000= 0.0905/− 84.29

P 14.6 [a] fc =ωc

2π=

50,0002π

=502π

× 103 = 7957.75 Hz

[b]1

RC= 50 × 103

R =1

(50 × 103)(0.5 × 10−6)= 40 Ω

[c] ωc =1

RC

(1 +

R

RL

)

·. .R

RL

= 0.05 ·. . RL = 20R = 800 Ω

[d] H(j0) =RL

R + RL

=800840

= 0.9524

P 14.7 [a]1

RC=

1(50 × 103)(5 × 10−9)

= 4000 rad/s

fc =40002π

= 636.62 Hz

[b] H(s) =s

s + ωc

·. . H(jω) =jω

4000 + jω

H(jωc) = H(j4000) =j4000

4000 + j4000= 0.7071/45

H(j0.2ωc) = H(j800) =j800

4000 + j800= 0.1961/78.69

H(j5ωc) = H(j20, 000) =j20,000

4000 + j20,000= 0.9806/11.31

[c] vo(t)|ωc = (0.7071)(500) cos(4000t + 45)

= 353.55 cos(4000t + 45) mV

vo(t)|0.2ωc = (0.1961)(500) cos(800t + 78.69)

= 98.06 cos(800t + 78.69) mV

vo(t)|5ωc = (0.9806)(500) cos(20,000t + 11.31)

= 490.29 cos(20,000t + 11.31) mV


P 14.8 [a] H(s) =Vo

Vi

=R

R + Rc + (1/sC)

=R

R + Rc

· s

[s + (1/(R + Rc)C)]

[b] H(jω) =R

R + Rc

· jω

jω + (1/(R + Rc)C)

|H(jω)| =R

R + Rc

· ω√ω2 + 1

(R+Rc)2C2

The magnitude will be maximum when ω = ∞.

[c] |H(jω)|max =R

R + Rc

[d] |H(jωc)| =Rωc

(R + Rc)√

ω2c + [1/(R + Rc)C]2

·. . |H(jω)| =R√

2(R + Rc)when

·. . ω2c =

1(R + Rc)2C2

or ωc =1

(R + Rc)C

[e] ωc =1

(62.5 × 103)(5 × 10−9)= 3200 rad/s

R

R + Rc

=50

62.5= 0.8

·. . H(jω) =0.8jω

3200 + jω

H(jωc) =(0.8)j3200

3200 + j3200= 0.5657/45

H(j0.2ωc) =(0.8)j640

3200 + j640= 0.1569/78.69

H(j5ωc) =(0.8)j16,000

3200 + j16,000= 0.7845/11.31

P 14.9 [a] ωc =1

RC= 2π(300) = 600π rad/s

·. . R =1

ωcC=

1(600π)(100 × 10−9)

= 5305.16 Ω

Problems 14–9

[b] Re = 5305.16‖47,000 = 4767.08 Ω

ωc =1

ReC=

1(4767.08)(100 × 10−9)

= 2097.7 rad/s

fc =ωc

2π=

2097.72π

= 333.86 Hz

P 14.10 [a] ωc =R

Lso R = ωcL = (25 × 103)(5 × 10−3) = 125 Ω

[b] ωc(loaded) =R

L· RL

R + RL

= 24,000 rad/s

·. .RL

R + RL

=ωc(loaded)

ωc(unloaded)=

24,00025,000

= 0.96

RL = 0.96(R + RL) ·. . 0.04RL = 0.96R = (0.96)(125)

·. . RL =(0.96)(125)

0.04= 3 kΩ

P 14.11 By definition Q = ωo/β therefore β = ωo/Q. Substituting this expression into Eqs.14.34 and 14.35 yields

ωc1 = − ωo

2Q+

√√√√( ωo

2Q

)2

+ ω2o

ωc2 =ωo

2Q+

√√√√( ωo

2Q

)2

+ ω2o

Now factor ωo out to get

ωc1 = ωo

− 1

2Q+

√√√√1 +(

12Q

)2

ωc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2

P 14.12 ωo =√

ωc1ωc2 =√

(121)(100) = 110 krad/s

fo =ωo

2π= 17.51 kHz

β = 121 − 100 = 21 krad/s or 3.34 kHz

Q =ωo

β=

11021

= 5.24


P 14.13 β =ωo

Q=

50,0004

= 12.5 krad/s;12,500

2π= 1.99 kHz

ωc2 = 50,000

1

8+

√1 +

(18

)2 = 56.64 krad/s

fc2 =56.64 k

2π= 9.01 kHz

ωc1 = 50,000

−1

8+

√1 +

(18

)2 = 44.14 krad/s

fc1 =44.14 k

2π= 7.02 kHz

P 14.14 [a] ω2o =

1LC

so L =1

[8000(2π)]2(5 × 10−9)= 79.16 mH

R =ωoL

Q=

8000(2π)(79.16 × 10−3)2

= 1.99 kΩ

[b] fc1 = 8000

−1

4+

√1 +

116

= 6.25 kHz

[c] fc2 = 8000

1

4+

√1 +

116

= 10.25 kHz

[d] β = fc2 − fc1 = 4 kHz

or

β =fo

Q=

80002

= 4 kHz

P 14.15 [a] ω2o =

1LC

=1

(10 × 10−3)(10 × 10−9)= 1010

ωo = 105 rad/s = 100 krad/s

[b] fo =ωo

2π=

105

2π= 15.92 kHz

[c] Q = ωoRC = (100 × 103)(8000)(10 × 10−9) = 8

[d] ωc1 = ωo

− 1

2Q+

√√√√1 +(

12Q

)2 = 105

− 1

16+

√1 +

1256

= 93.95 krad/s

[e] ·. . fc1 =ωc1

2π= 14.96 kHz

Problems 14–11

[f] ωc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2 = 105

1

16+

√1 +

1256

= 106.45 krad/s

[g] ·. . fc2 =ωc2

2π= 16.94 kHz

[h] β =ωo

Q=

105

8= 12.5 krad/s or 1.99 kHz

P 14.16 [a] L =1

ω2oC

=1

(50 × 10−9)(20 × 103)2 = 50 mH

R =Q

ωoC=

5(20 × 103)(50 × 10−9)

= 5 kΩ

[b] ωc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2 = 20,000

1

10+

√1 +

1100

= 22.10 krad/s ·. . fc2 =ωc2

2π= 3.52 kHz

ωc1 = ωo

− 1

2Q+

√√√√1 +(

12Q

)2 = 20,000

− 1

10+

√1 +

1100

= 18.10 krad/s ·. . fc1 =ωc1

2π= 2.88 kHz

[c] β =ωo

Q=

20,0005

= 4000 rad/s or 636.62 Hz

P 14.17 [a] ω2o =

1LC

=1

(40 × 10−3)(40 × 10−9)= 625 × 106

ωo = 25 × 103 rad/s = 25 krad/s

fo =25,000

2π= 3978.87 Hz

[b] Q =ωoL

R + Ri

=(25 × 103)(40 × 10−3)

200= 5

[c] ωc1 = ωo

− 1

2Q+

√√√√1 +(

12Q

)2 = 25,000

− 1

10+

√1 +

1100

= 22.62 krad/s or 3.60 kHz

[d] wc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2 = 25,000

1

10+

√1 +

1100

= 27.62 krad/s or 4.40 kHz


[e] β = ωc2 − ωc1 = 27.62 − 22.62 = 5 krad/sor

β =ωo

Q=

25,0005

= 5 krad/s or 795.77 Hz

P 14.18 [a] H(s) =(R/L)s

s2 + (R+Ri)L

s + 1LC

For the numerical values in Problem 14.17 we have

H(s) =4500s

s2 + 5000s + 625 × 106

·. . H(jω) =4500jω

(625 × 106 − ω2) + j5000ω

H(jωo) =j4500(25 × 103)j5000(25 × 103)

= 0.9/0

·. . vo(t) = 500(0.9) cos 25,000t = 450 cos 25,000t mV

[b] From the solution to Problem 14.17,

ωc1 = 22.62 krad/s

H(j22.62 k) =j4500(22.62 × 103)

(113.12 + j113.12) × 106 = 0.6364/45

·. . vo(t) = 500(0.6364) cos(22,620t + 45) = 318.2 cos(22,620t + 45) mV

[c] From the solution to Problem 14.17,

ωc2 = 27.62 krad/s

H(j27.62 k) =j4500(27.62 × 103)

(−138.12 + j138.12) × 106 = 0.6364/− 45

·. . vo(t) = 500(0.6364) cos(27,620t − 45) = 318.2 cos(27,620t − 45) mV

P 14.19 [a]

[b] L =1

ω2oC

=1

(50 × 103)2(20 × 10−9)= 20 mH

R =ωoL

Q=

(50 × 103)(20 × 10−3)6.25

= 160 Ω

Problems 14–13

[c] Re = 160‖480 = 120 Ω

Re + Ri = 120 + 80 = 200 Ω

Qsystem =ωoL

Re + Ri

=(50 × 103)(20 × 10−3)

200= 5

[d] βsystem =ωo

Qsystem=

50 × 103

5= 10 krad/s

βsystem(Hz) =10,000

2π= 1591.55 Hz

P 14.20 [a]Vo

Vi

=Z

Z + Rwhere Z =

1Y

and Y = sC +1sL

+1

RL

=LCRLs2 + sL + RL

RLLs

H(s) =RLLs

RLRLCs2 + (R + RL)Ls + RRL

=(1/RC)s

s2 +[(

R+RL

RL

) (1

RC

)]s + 1

LC

=

(RL

R+RL

) (R+RL

RL

) (1

RC

)s

s2 +[(

R+RL

RL

) (1

RC

)]s + 1

LC

=Kβs

s2 + βs + ω2o

, K =RL

R + RL

[b] βL =(

R + RL

RL

) 1RC

[c] βU =1

RC

·. . βL =(

R + RL

RL

)βU =

(1 +

R

RL

)βU

[d] QL =ωo

β=

ωoRC(R+RL

RL

)[e] QU = ωoRC

·. . QL =(

RL

R + RL

)QU =

1[1 + (R/RL)]

QU

[f] H(jω) =Kjωβ

ω2o − ω2 + jωβ

H(jωo) = K


Let ωc represent a cutoff frequency. Then

|H(jωc)| =K√2

=Kωcβ√

(ω2o − ω2

c )2 + ω2cβ

2

·. .1√2

=ωcβ√

(ω2o − ω2

c )2 + ω2cβ

2

Squaring both sides leads to

(ω2o − ω2

c )2 = ω2

cβ2 or (ω2

o − ω2c ) = ±ωcβ

·. . ω2c ± ωcβ − ω2

o = 0

or

ωc = ∓β

2±√

β2

4+ ω2

o

The two positive roots are

ωc1 = −β

2+

√β2

4+ ω2

o and ωc2 =β

2+

√β2

4+ ω2

o

where

β =(1 +

R

RL

) 1RC

and ω2o =

1LC

P 14.21 [a] ω2o =

1LC

=1

(5 × 10−3)(200 × 10−12)= 1012

ωo = 1 Mrad/s

[b] β =R + RL

RL

· 1RC

=(

500 × 103

400 × 103

)(1

(100 × 103)(200 × 10−12)

)= 62.5 krad/s

[c] Q =ωo

β=

106

62.5 × 103 = 16

[d] H(jωo) =RL

R + RL

= 0.8/0

·. . vo(t) = 250(0.8) cos(106t) = 200 cos 106t mV

[e] β =(1 +

R

RL

) 1RC

=(1 +

100RL

)(50 × 103) rad/s

ωo = 106 rad/s

Q =ωo

β=

201 + (100/RL)

where RL is in kilohms

Problems 14–15

[f]

P 14.22 ω2o =

1LC

=1

(2 × 10−6)(50 × 10−12)= 1016

ωo = 100 Mrad/s

Qu = ωoRC = (100 × 106)(2.4 × 103)(50 × 10−12) = 12

·. .(

RL

R + RL

)12 = 7.5; ·. . RL =

7.54.5

R = 4 kΩ

P 14.23 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage andwe seek the steady-state nature of the output voltage vo.At zero frequency the inductor provides a direct connection between the inputand the output, hence vo = vi when ω = 0.At infinite frequency the capacitor provides the direct connection, hencevo = vi when ω = ∞.At the resonant frequency of the parallel combination of L and C theimpedance of the combination is infinite and hence the output voltage will bezero when ω = ωo.At frequencies on either side of ωo the amplitude of the output voltage will benonzero but less than the amplitude of the input voltage.Thus the circuit behaves like a band-reject filter.

[b] Let Z represent the impedance of the parallel branches L and C, thus

Z =sL(1/sC)sL + 1/sC

=sL

s2LC + 1

Then

H(s) =Vo

Vi

=R

Z + R=

R(s2LC + 1)sL + R(s2LC + 1)

=[s2 + (1/LC)]

s2 +(

1RC

)s +

(1

LC

)


H(s) =s2 + ω2

o

s2 + βs + ω2o

[c] From part (b) we have

H(jω) =ω2

o − ω2

ω2o − ω2 + jωβ

It follows that H(jω) = 0 when ω = ωo

·. . ωo =1√LC

[d] |H(jω)| =ω2

o − ω2√(ω2

o − ω2)2 + ω2β2

|H(jω)| =1√2

when ω2β2 = (ω2o − ω2)2

or ± ωβ = ω2o − ω2, thus

ω2 ± βω − ω2o = 0

The two positive roots of this quadratic are

ωc1 =−β

2+

√√√√(β

2

)2

+ ω2o

ωc2 =β

2+

√√√√(β

2

)2

+ ω2o

Also note that since β = ωo/Q

ωc1 = ωo

−1

2Q+

√√√√1 +(

12Q

)2

ωc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2

[e] It follows from the equations derived in part (b) that

β = 1/RC

[f] By definition Q = ωo/β = ωoRC

Problems 14–17

P 14.24 [a] ω2o =

1LC

=1

(50 × 10−6)(20 × 10−9)= 1012

·. . ωo = 1 Mrad/s

[b] fo =ωo

2π= 159.15 kHz

[c] Q = ωoRC = (106)(750)(20 × 10−9) = 15

[d] ωc1 = ωo

− 1

2Q+

√√√√1 +(

12Q

)2 = 106

− 1

30+

√1 +

1900

= 967.22 krad/s

[e] fc1 =ωc1

2π= 153.94 kHz

[f] ωc2 = ωo

1

2Q+

√√√√1 +(

12Q

)2 = 106

1

30+

√1 +

1900

= 1.03 Mrad/s

[g] fc2 =ωc2

2π= 164.55 kHz

[h] β = fc2 − fc1 = 10.61 kHz

P 14.25 [a] ωo = 2πfo = 8π krad/s

L =1

ω2oC

=1

(8000π)2(0.5 × 10−6)= 3.17 mH

R =Q

ωoC=

5(8000π)(0.5 × 10−6)

= 397.89 Ω

[b] fc2 = fo

1

2Q+

√√√√1 +(

12Q

)2 = 4000

1

10+

√1 +

1100

= 4.42 kHz

fc1 = fo

− 1

2Q+

√√√√1 +(

12Q

)2 = 4000

− 1

10+

√1 +

1100

= 3.62 kHz

[c] β = fc2 − fc1 = 800 Hz

or

β =fo

Q=

40005

= 800 Hz


P 14.26 [a] Re = 397.89‖1000 = 284.63 Ω

Q = ωoReC = (8000π)(284.63)(0.5 × 10−6) = 3.58

[b] β =fo

Q=

40003.58

= 1.12 kHz

[c] fc2 = 4000

1

7.15+

√1 +

17.152

= 4.60 kHz

[d] fc1 = 4000

− 1

7.15+

√1 +

17.152

= 3.48 kHz

P 14.27 [a] Let Z =RL(sL + (1/sC))RL + sL + (1/sC)

Z =RL(s2LC + 1)

s2LC + RLCs + 1

Then H(s) =Vo

Vi

=s2RLCL + RL

(R + RL)LCs2 + RRLCs + R + RL

Therefore

H(s) =(

RL

R + RL

)· [s2 + (1/LC)][

s2 +(

RRL

R+RL

)sL

+ 1LC

]

=K(s2 + ω2

o)s2 + βs + ω2

o

where K =RL

R + RL

; ω2o =

1LC

; β =(

RRL

R + RL

) 1L

[b] ωo =1√LC

[c] β =(

RRL

R + RL

) 1L

[d] Q =ωo

β=

ωoL

[RRL/(R + RL)]

[e] H(jω) =K(ω2

o − ω2)(ω2

o − ω2) + jβω

H(jωo) = 0

[f] H(j0) =Kω2

o

ω2o

= K

Problems 14–19

[g] H(jω) =K[(ωo/ω)2 − 1

][

(ωo/ω)2 − 1]+ jβ/ω

H(j∞) =−K

−1= K

[h] H(jω) =K(ω2

o − ω2)(ω2

o − ω2) + jβω

H(j0) = H(j∞) = K

Let ωc represent a corner frequency. Then

|H(jωc)| =K√2

·. .K√2

=K(ω2

o − ω2c )√

(ω2o − ω2

c )2 + ω2cβ

2

Squaring both sides leads to

(ω2o − ω2

c )2 = ω2

cβ2 or (ω2

o − ω2c ) = ±ωcβ

·. . ω2c ± ωcβ − ω2

o = 0

or

ωc = ∓β

2±√

β2

4+ ω2

o

The two positive roots are

ωc1 = −β

2+

√β2

4+ ω2

o and ωc2 =β

2+

√β2

4+ ω2

o

where

β =RRL

R + RL

· 1L

and ω2o =

1LC

P 14.28 [a] ω2o =

1LC

=1

(10−6)(4 × 10−12)= 0.25 × 1018 = 25 × 1016

ωo = 5 × 108 = 500 Mrad/s

β =RRL

R + RL

· 1L

=(30)(150)

180· 110−6 = 25 Mrad/s = 3.98 MHz

Q =ωo

β=

500 M25 M

= 20


[b] H(j0) =RL

R + RL

=150180

= 0.8333

H(j∞) =RL

R + RL

= 0.8333

[c] fc2 =250π

1

40+

√1 +

11600

= 81.59 MHz

fc2 =250π

− 1

40+

√1 +

11600

= 77.61 MHz

Check: β = fc2 − fc1 = 3.98 MHz.

[d] Q =ωo

β=

500 × 106

RRL

R+RL· 1

L

=500(R + RL)

RRL

=503

(1 +

30RL

)where RL is in ohms.

[e]

P 14.29 [a] ω2o =

1LC

= 1012

·. . L =1

(1012)(400 × 10−12)= 2.5 mH

RL

R + RL

= 0.96; ·. . 0.04RL = 0.96R

·. . RL = 24R ·. . R =36,000

24= 1.5 kΩ

[b] β =(

RL

R + RL

)R · 1

L= 576 × 103

Q =ωo

β=

106

576 × 103 = 1.74

Problems 14–21

P 14.30 Refer to Sections E.5 and E.7.

[a] ωn = 105

2ζωn = 50,000, ζ = 0.25

ωo =√

2ωp =√

2ωn

√1 − 2ζ2 = 132,287.57 rad/s

·. . ω = 0

ω = 132,287.57 rad/s

[b] ωp = ωn

√1 − 2ζ2 = 93,541.43 rad/s

P 14.31 [a] Use the cutoff frequencies to calculate the bandwidth:

ωc1 = 2π(697) = 4379.38 rad/s ωc2 = 2π(941) = 5912.48 rad/s

Thus β = ωc2 − ωc1 = 1533.10 rad/s

Calculate inductance using Eq. (14.32) and capacitance using Eq. (14.31):

L =R

β=

6001533.10

= 0.39 H

C =1

Lωc1ωc2=

1(0.39)(4379.38)(5912.48)

= 0.10 µF

[b] At the outermost two frequencies in the low-frequency group (687 Hz and 941Hz) the amplitudes are

|V697 Hz| = |V941 Hz| =|Vpeak|√

2= 0.707|Vpeak|

because these are cutoff frequencies. We calculate the amplitudes at the othertwo low frequencies using Eq. (14.32):

|V | = (|Vpeak|)(|H(jω)|) = |Vpeak| ωβ√(ω2

o − ω2)2 + (ωβ)2

Therefore

|V770 Hz| = |Vpeak| (4838.05)(1533.10)√(5088.522 − 4838.052)2 + [(4838.05)(1533.10)]2

= 0.948|Vpeak|and

|V852 Hz| = |Vpeak| (5353.27)(1533.10)√(5088.522 − 5353.272)2 + [(5353.27)(1533.10)]2

= 0.948|Vpeak|


It is not a coincidence that these two magnitudes are the same. The frequenciesin both bands of the DTMF system were carefully chosen to produce this typeof predictable behavior with linear filters. In other words, the frequencies werechosen to be equally far apart with respect to the response produced by a linearfilter. Most musical scales consist of tones designed with this same property –note intervals are selected to place the notes equally far apart. That is why theDTMF tones remind us of musical notes! Unlike musical scales, DTMFfrequencies were selected to be harmonically unrelated, to lower the risk ofmisidentifying a tone’s frequency if the circuit elements are not perfectly linear.

[c] The high-band frequency closest to the low-frequency band is 1209 Hz. Theamplitude of a tone with this frequency is

|V1209 Hz| = |Vpeak| =(7596.37)(1533.10)√

(5088.522 − 7596.372)2 + [(7596.37)(1533.10)]2

= 0.344|Vpeak|This is less than one half the amplitude of the signals with the low-band cutofffrequencies, ensuring adequate separation of the bands.

P 14.32 The cutoff frequencies and bandwidth are

ωc1 = 2π(1209) = 7596 rad/s

ωc2 = 2π(1633) = 10.26 krad/s

β = ωc2 − ωc1 = 2664 rad/s

Telephone circuits always have R = 600 Ω. Therefore, the filter’s inductance andcapacitance values are

L =R

β=

6002664

= 0.225 H

C =1

ωc1ωc2L= 0.057 µF

At the highest of the low-band frequencies, 941 Hz, the amplitude is

|Vω| = |Vpeak| ωβ√(ω2

o − ω2)2 + ω2β2

where ωo = √ωc1ωc2 . Thus,

|Vω| =|Vpeak|(5912)(2664)√

[(8828)2 − (5912)2]2 + [(5912)(2664)]2

= 0.344 |Vpeak|

Problems 14–23

Again it is not coincidental that this result is the same as the response of thelow-band filter to the lowest of the high-band frequencies.

P 14.33 From Problem 14.31 the response to the largest of the DTMF low-band tones is0.948|Vpeak|. The response to the 20 Hz tone is

|V20 Hz| =|Vpeak|(125.6)(1533)

[(50892 − 125.62)2 + [(125.6)(1533)]2]1/2

= 0.00744|Vpeak|

·. .0.00744|Vring-peak|0.948|VDTMF-peak| = 0.5

·. . |Vring-peak| = 63.7|VDTMF-peak|

Thus, the 20 Hz signal can be 63.7 times as large as the DTMF tones.

15Active Filter Circuits

Assessment Problems

AP 15.1 H(s) =−(R2/R1)ss + (1/R1C)

1R1C

= 1 rad/s; R1 = 1 Ω, ·. . C = 1 F

R2

R1= 1, ·. . R2 = R1 = 1 Ω

·. . Hprototype(s) =−s

s + 1

AP 15.2 H(s) =−(1/R1C)

s + (1/R2C)=

−20,000s + 5000

1R1C

= 20,000; C = 5 µF

·. . R1 =1

(20,000)(5 × 10−6)= 10 Ω

1R2C

= 5000

·. . R2 =1

(5000)(5 × 10−6)= 40 Ω

15–1

15–2 CHAPTER 15. Active Filter Circuits

AP 15.3 ωc = 2πfc = 2π × 104 = 20,000π rad/s

·. . kf = 20,000π = 62,831.85

C ′ =C

kfkm

·. . 0.5 × 10−6 =1

kfkm

·. . km =1

(0.5 × 10−6)(62,831.85)= 31.83

AP 15.4 For a 2nd order prototype Butterworth high pass filter

H(s) =s2

s2 +√

2s + 1

For the circuit in Fig. 15.25

H(s) =s2

s2 +(

2R2C

)s +

(1

R1R2C2

)

Equate the transfer functions. For C = 1F,

2R2C

=√

2, ·. . R2 =√

2 = 1.414 Ω

1R1R2C2 = 1, ·. . R1 =

1√2

= 0.707 Ω

AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF

For the circuit in Fig 15.26

H(s) =−( 1

R1C

)s

s2 +( 2

R3C

)s +

(R1 + R2

R1R2R3C2

)

=Kβs

s2 + βs + ω2o

β =2

R3C, ·. . R3 =

2βC

β =ωo

Q=

10008

= 125 rad/s

Problems 15–3

·. . R3 =2 × 106

(125)(1)= 16 kΩ

Kβ =1

R1C

·. . R1 =1

KβC=

15(125)(1 × 10−6)

= 1.6 kΩ

ω2o =

R1 + R2

R1R2R3C2

106 =(1600 + R2)

(1600)(R2)(16,000)(10−6)2

Solving for R2,

R2 =(1600 + R2)106

256 × 105 , 246R2 = 16,000, R2 = 65.04 Ω

AP 15.6 ωo = 1000 rad/s; Q = 4;

C = 2 µF

H(s) =s2 + (1/R2C2)

s2 +[4(1 − σ)

RC

]s +

( 1R2C2

)

=s2 + ω2

o

s2 + βs + ω2o

; ωo =1

RC; β =

4(1 − σ)RC

R =1

ωoC=

1(1000)(2 × 10−6)

= 500 Ω

β =ωo

Q=

10004

= 250

·. .4(1 − σ)

RC= 250

4(1 − σ) = 250RC = 250(500)(2 × 10−6) = 0.25

1 − σ =0.254

= 0.0625; ·. . σ = 0.9375


Problems

P 15.1 Summing the currents at the inverting input node yields

0 − Vi

Zi

+0 − Vo

Zf

= 0

·. .Vo

Zf

= −Vi

Zi

·. . H(s) =Vo

Vi

= −Zf

Zi

P 15.2 [a] Zf =R2(1/sC2)

[R2 + (1/sC2)]=

R2

R2C2s + 1

=(1/C2)

s + (1/R2C2)Likewise

Zi =(1/C1)

s + (1/R1C1)

·. . H(s) =−(1/C2)[s + (1/R1C1)][s + (1/R2C2)](1/C1)

= −C1

C2

[s + (1/R1C1)][s + (1/R2C2)]

[b] H(jω) =−C1

C2

[jω + (1/R1C1)jω + (1/R2C2)

]

H(j0) =−C1

C2

(R2C2

R1C1

)=

−R2

R1

[c] H(j∞) = −C1

C2

(j

j

)=

−C1

C2

[d] As ω → 0 the two capacitor branches become open and the circuit reduces to aresistive inverting amplifier having a gain of −R2/R1.As ω → ∞ the two capacitor branches approach a short circuit and in this casewe encounter an indeterminate situation; namely vn → vi but vn = 0 becauseof the ideal op amp. At the same time the gain of the ideal op amp is infinite sowe have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate wecan reason that for finite large values of ω H(jω) will approach −C1/C2 invalue. In other words, the circuit approaches a purely capacitive invertingamplifier with a gain of (−1/jωC2)/(1/jωC1) or −C1/C2.

Problems 15–5

P 15.3 [a] Zf =(1/C2)

s + (1/R2C2)

Zi = R1 +1

sC1=

R1

s[s + (1/R1C1)]

H(s) = − (1/C2)[s + (1/R2C2)]

· s

R1[s + (1/R1C1)]

= − 1R1C2

s

[s + (1/R1C1)][s + (1/R2C2)]

[b] H(jω) = − 1R1C2

jω(jω + 1

R1C1

) (jω + 1

R2C2

)H(j0) = 0

[c] H(j∞) = 0[d] As ω → 0 the capacitor C1 disconnects vi from the circuit. Therefore

vo = vn = 0.As ω → ∞ the capacitor short circuits the feedback network, thus Zf = 0 andtherefore vo = 0.

P 15.4 [a] K = 10(10/20) = 3.16 =R2

R1

R2 =1

ωcC=

1(2π)(103)(750 × 10−9)

= 212.21 Ω

R1 =R2

K=

212.213.16

= 67.11 Ω

[b]

P 15.5 [a] R1 =1

ωcC=

1(2π)(8 × 103)(3.9 × 10−9)

= 5.10 kΩ

K = 10(14/20) = 5.01 =R2

R1

·. . R2 = 5.01R1 = 25.57 kΩ


[b]

P 15.6 For the RC circuit

H(s) =Vo

Vi

=(1/RC)

s + (1/RC)

R′ = kmR; C ′ =C

kmkf

·. . R′C ′ = kmRC

kmkf

=1kf

RC =1kf

1R′C ′ = kf

H ′(s) =(1/R′C ′)

s + (1/R′C ′)=

kf

s + kf

H ′(s) =1

(s/kf ) + 1

For the RL circuit

H(s) =V0

Vi

=R/L

s + (R/L)

R′ = kmR; L′ =km

kf

L

R′

L′ =kmRkm

kfL

= kf

(R

L

)= kf

H ′(s) =(R′/L′)

s + (R′/L′)=

kf

s + kf

H ′(s) =1

(s/kf ) + 1

Problems 15–7

P 15.7 For the RC circuit

H(s) =Vo

Vi

=s

s + (1/RC)

R′ = kmR; C ′ =C

kmkf

·. . R′C ′ =RC

kf

=1kf

;1

R′C ′ = kf

H ′(s) =s

s + (1/R′C ′)=

s

s + kf

=(s/kf )

(s/kf ) + 1

For the RL circuit

H(s) =s

s + (R/L)

R′ = kmR; L′ =kmL

kf

R′

L′ = kf

(R

L

)= kf

H ′(s) =s

s + (R′/L′)=

s

s + kf

=(s/kf )

(s/kf ) + 1

P 15.8 H(s) =(R/L)s

s2 + (R/L)s + (1/LC)=

βs

s2βs + ω2o

For the prototype circuit ωo = 1 and β = ωo/Q = 1/Q.For the scaled circuit

H ′(s) =(R′/L′)s

s2 + (R′/L′)s + (1/L′C ′)

where R′ = kmR; L′ =km

kf

L; and C ′ =C

kfkm

·. .R′

L′ =kmRkm

kfL

= kf

(R

L

)= kfβ

1L′C ′ =

kfkm

km

kfLC

=k2

f

LC= k2

f


Q′ =ω′

o

β′ =kfωo

kfβ= Q

therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit.Also note β′ = kfβ.

·. . H ′(s) =

(kf

Q

)s

s2 +(

kf

Q

)s + k2

f

H ′(s) =

(1Q

) (skf

)[(

skf

)2+ 1

Q

(skf

)+ 1

]

P 15.9 [a] L = 1 H; C = 1 F

R =1Q

=120

= 0.05 Ω

[b] kf =ω′

o

ωo

= 40,000; km =R′

R=

50000.05

= 100,000

Thus,

R′ = kmR = (0.05)(100,000) = 5 kΩ

L′ =km

kf

L =100,00040,000

(1) = 2.5 H

C ′ =C

kmkf

=1

(40,000)(100,000)= 250 pF

[c]

P 15.10 [a] Since ω2o = 1/LC and ωo = 1 rad/s,

C =1L

=1Q

[b] H(s) =(R/L)s

s2 + (R/L)s + (1/LC)

H(s) =(1/Q)s

s2 + (1/Q)s + 1

Problems 15–9

[c] In the prototype circuit

R = 1 Ω; L = 16 H; C =1L

= 0.0625 F

·. . km =R′

R= 10,000; kf =

ω′o

ωo

= 25,000

Thus

R′ = kmR = 10 kΩ

L′ =km

kf

L =10,00025,000

(16) = 6.4 H

C ′ =C

kmkf

=0.0625

(10,000)(25,000)= 250 pF

[d]

[e] H ′(s) =116

(s

25,000

)(

s25,000

)2+ 1

16

(s

25,000

)+ 1

H ′(s) =1562.5s

s2 + 1562.5s + 625 × 106

P 15.11 [a] Using the first prototype

ωo = 1 rad/s; C = 1 F; L = 1 H; R = 25 Ω

km =R′

R=

40,00025

= 1600; kf =ω′

o

ωo

= 50,000

Thus,

R′ = kmR = 40 kΩ; L′ =km

kf

L =1600

50,000(1) = 32 mH;

C ′ =C

kmkf

=1

(1600)(50,000)= 12.5 nF

Using the second prototype

ωo = 1 rad/s; C = 25 F

L =125

= 40 mH; R = 1 Ω


km =R′

R= 40,000; kf =

ω′o

ωo

= 50,000

Thus,

R′ = kmR = 40 kΩ; L′ =km

kf

L =40,00050,000

(0.04) = 32 mH;

C ′ =C

kmkf

=25

(40,000)(50,000)= 12.5 nF

[b]

P 15.12 For the scaled circuit

H ′(s) =s2 +

(1

L′C′

)s2 +

(R′L′

)s +

(1

L′C′

)

L′ =km

kf

L; C ′ =C

kmkf

·. .1

L′C ′ =k2

f

LC; R′ = kmR

·. .R′

L′ = kf

(R

L

)

It follows then that

H ′(s) =s2 +

(k2

f

LC

)

s2 +(

RL

)kfs +

k2f

LC

=

(skf

)2+(

1LC

)[(

skf

)2+(

RL

) (skf

)+(

1LC

)]

= H(s)|s=s/kf

P 15.13 For the circuit in Fig. 15.31

H(s) =s2 +

(1

LC

)s2 + s

RC+(

1LC

)

Problems 15–11

It follows that

H ′(s) =s2 + 1

L′C′

s2 + sR′C′ + 1

L′C′

where R′ = kmR; L′ =km

kf

L;

C ′ =C

kmkf

·. .1

L′C ′ =k2

f

LC

1R′C ′ =

kf

RC

H ′(s) =s2 +

(k2

f

LC

)

s2 +(

kf

RC

)s +

k2f

LC

=

(skf

)2+ 1

LC(skf

)2+(

1RC

) (skf

)+ 1

LC

= H(s)|s=s/kf

P 15.14 [a] For the circuit in Fig. P15.14(a)

H(s) =Vo

Vi

=s +

1s

1Q

+ s +1s

=s2 + 1

s2 +(

1Q

)s + 1

For the circuit in Fig. P15.14(b)

H(s) =Vo

Vi

=Qs + Q

s

1 + Qs + Qs

=Q(s2 + 1)

Qs2 + s + Q

H(s) =s2 + 1

s2 +(

1Q

)s + 1


[b] k=f

ω′o

ωo

= 104; Q = 8;

Replace s with s/kf .

H ′(s) =

(s

104

)2+ 1(

s104

)2+ 1

8

(s

104

)+ 1

=s2 + 108

s2 + 1250s + 108

P 15.15 For prototype circuit (a):

H(s) =Vo

Vi

=Q

Q + 1s+ 1

s

=Q

Q + ss2+1

H(s) =Q(s2 + 1)

Q(s2 + 1) + s=

s2 + 1

s2 +(

1Q

)s + 1

For prototype circuit (b):

H(s) =Vo

Vi

=1

1 + (s/Q)(s2+1)

=s2 + 1

s2 +(

1Q

)s + 1

P 15.16 From the solution to Problem 14.15, ωo = 100 krad/s and β = 12.5 krad/s. Computethe two scale factors:

kf =ω′

o

ωo

=2π(200 × 103)

100 × 103 = 4π

km =1kf

C

C ′ =14π

10 × 10−9

2.5 × 10−9 =1π

Thus,

R′ = kmR =8000

π= 2546.48 Ω L′ =

km

kf

L =1/π4π

(10 × 10−3) = 253.303 µH

Calculate the cutoff frequencies:

ω′c1 = kfωc1 = 4π(93.95 × 103) = 1180.6 krad/s

ω′c2 = kfωc2 = 4π(106.45 × 103) = 1337.7 krad/s

To check, calculate the bandwidth:

β′ = ω′c2 − ω′

c1 = 157.1 krad/s = 4πβ (Checks!)

Problems 15–13

P 15.17 From the solution to Problem 14.24, ωo = 106 rad/s and β = 2π(10.61) krad/s.Calculate the scale factors:

kf =ω′

o

ωo

=50 × 103

106 = 0.05

km =kfL

′

L=

0.05(200 × 10−6)50 × 10−6 = 0.2

Thus,

R′ = kmR = (0.2)(750) = 150 Ω C ′ =C

kmkf

=20 × 10−9

(0.2)(0.05)= 2 µF

Calculate the bandwidth:

β′ = kfβ = (0.05)[2π(10.61 × 103)] = 3333 rad/s

To check, calculate the quality factor:

Q =ωo

β=

106

2π(10.61 × 103)= 15

Q′ =ω′

o

β′ =50 × 103

3333= 15 (Checks)

P 15.18 [a] km =R′

R=

10001

= 1000; kf =C

kmC ′ =1

(1000)(200 × 10−9)= 5000

L′ =km

kf

(L) =10005000

(1) = 200 mH

[b]V − 10/s

1000+

V

0.2s+

V

1000 + (5 × 106/s)= 0

V( 1

1000+

5s

+s

1000s + 5 × 106

)=

1100s

V =10(s + 5000)

2s2 + 10,000s + 25 × 106 =5(s + 5000)

s2 + 5000s + 12.5 × 106


Io =V

0.2s=

25(s + 5000)s(s2 + 5000s + 12.5 × 106)

=K1

s+

K2

s + 2500 − j2500+

K∗2

s + 2500 + j2500

K1 = 0.01; K2 = −0.005

io(t) = 10 − 10e−2500t cos 2500t mA

Since km = 1000 and the source voltage didn’t change, the amplitude of thecurrent is reduced by a factor of 1000. Since kf = 5000 the coefficients of t aremultiplied by 5000.

P 15.19 km =R′

R=

500050

= 100; kf =ω′

o

ωo

= 5000

C ′ =C

kmkf

=4 × 10−3

(100)(5000)= 8 nF

50 Ω → 5 kΩ; 700 Ω → 70 kΩ

L′ =km

kf

L =1005000

(20) = 0.4 H

0.05vφ → 0.05100

vφ = 5 × 10−4vφ

The original expression for the current:

io(t) = 1728 + 2880e−20t cos(15t + 126.87) mA

The frequency components will be multiplied by kf = 5000:

20 → 20(5000) = 105; 15 → 15(5000) = 75,000

The magnitudes will be reduced by km = 100:

1728 → 1728/100 = 17.28; 2880 → 2880/100 = 28.80

The expression for the current in the scaled circuit is thus,

io(t) = 17.28 + 28.80e−105t cos(75,000t + 126.87) mA

Problems 15–15

P 15.20 [a] From Eq 15.1 we have

H(s) =−Kωc

s + ωc

where K =R2

R1, ωc =

1R2C

·. . H ′(s) =−K ′ω′

c

s + ω′c

where K ′ =R′

2

R′1

ω′c =

1R′

2C′

By hypothesis R′1 = kmR1; R′

2 = kmR2,

and C ′ =C

kfkm

. It follows that

K ′ = K and ω′c = kfωc, therefore

H ′(s) =−Kkfωc

s + kfωc

=−Kωc(skf

)+ ωc

[b] H(s) =−K

(s + 1)

[c] H ′(s) =−K(skf

)+ 1

=−Kkf

s + kf

P 15.21 [a] From Eq. 15.4

H(s) =−Ks

s + ωc

where K =R2

R1and

ωc =1

R1C

·. . H ′(s) =−K ′ss + ω′

c

where K ′ =R′

2

R′1

and ω′c =

1R′

1C′

By hypothesis

R′1 = kmR1; R′

2 = kmR2; C ′ =C

kmkf

It follows that

K ′ = K and ω′c = kfωc

·. . H ′(s) =−Ks

s + kfωc

=−K(s/kf )(

skf

)+ ωc


[b] H(s) =−Ks

(s + 1)

[c] H ′(s) =−K(s/kf )(

skf

+ 1) =

−Ks

(s + kf )

P 15.22 [a] Hhp =−s

s + 1; kf =

ω′o

ω=

1000(2π)1

= 2000π

·. . H ′hp =

−s

s + 2000π1

RHCH

= 2000π; ·. . RH =1

(2000π)(0.1 × 10−6)= 1.59 kΩ

Hlp =−1

s + 1; kf =

ω′o

ω=

5000(2π)1

= 10,000π

·. . H ′lp =

−10,000πs + 10,000π

1RLCL

= 10,000π; ·. . RL =1

(10,000π)(0.1 × 10−6)= 318.3 Ω

[b] H ′(s) =−s

s + 2000π· −10,000πs + 10,000π

=10,000πs

(s + 2000π)(s + 10,000π)

[c] ωo =√

ωc1ωc1 =√

(2000π)(10,000π) = 1000π√

20 rad/s

H ′(jωo) =(10,000π)(j1000π

√20)

(2000π + j1000π√

20)(10,000π + j1000π√

20)

=j10

√20

(2 + j√

20)(10 + j√

20)= 0.8333/0

Problems 15–17

[d] G = 20 log10(0.8333) = −1.58 dB

[e]

P 15.23 [a] For the high-pass section:

kf =ω′

o

ω=

4000(2π)1

= 8000π

H ′(s) =−s

s + 8000π

·. .1

R1(10 × 10−9)= 8000π; R1 = 3.98 kΩ ·. . R2 = 3.98 kΩ

For the low-pass section:

kf =ω′

o

ω=

400(2π)1

= 800π

H ′(s) =−800π

s + 800π

·. .1

R2(10 × 10−9)= 800π; R2 = 39.8 kΩ ·. . R1 = 39.8 kΩ

0 dB gain corresponds to K = 1. In the summing amplifier we are free tochoose Rf and Ri so long as Rf/Ri = 1. To keep from having many differentresistance values in the circuit we opt for Rf = Ri = 39.8 kΩ.


[b]

[c] H ′(s) =s

s + 8000π+

800πs + 800π

=s2 + 1600πs + 64 × 105π2

(s + 800π)(s + 8000π)

[d] ωo =√

(8000π)(800π) = 800π√

10

H ′(j800π√

10) =−(800π

√10)2 + 1600π(j800π

√10) + 64 × 105π2

(800π + j800π√

10)(8000π + j800π√

10)

=j128 × 104

√10π2

(800π)2(1 + j√

10)(10 + j√

10)

=j2

√10

(1 + j√

10)(10 + j√

10)

= 0.1818/0

[e] G = 20 log10 0.1818 = −14.81 dB

Problems 15–19

[f]

P 15.24 [a] H(s) =(1/sC)

R + (1/sC)=

(1/RC)s + (1/RC)

H(jω) =(1/RC)

jω + (1/RC)

|H(jω)| =(1/RC)√

ω2 + (1/RC)2

|H(jω)|2 =(1/RC)2

ω2 + (1/RC)2

[b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then

Va − Vi

R1+ sCVa +

Va

R2 + sL= 0

Solving for Va yields

Va =(R2 + sL)Vi

R1LCs2 + (R1R2C + L)s + (R1 + R2)

But

Vo =sLVa

R2 + sL

Therefore

Vo =sLVi

R1LCs2 + (L + R1R2C)s + (R1 + R2)

H(s) =sL

R1LCs2 + (L + R1R2C)s + (R1 + R2)

H(jω) =jωL

[(R1 + R2) − R1LCω2] + jω(L + R1R2C)


|H(jω)| =ωL√

[R1 + R2 − R1LCω2]2 + ω2(L + R1R2C)2

|H(jω)|2 =ω2L2

(R1 + R2 − R1LCω2)2 + ω2(L + R1R2C)2

=ω2L2

R21L

2C2ω4 + (L2 + R21R

22C

2 − 2R21LC + 2R1R2LC)ω2 + (R1 + R2)2

[c] Let Va be the voltage across R2 positive at the upper terminal. Then

Va − Vi

R1+

Va

R2+ VasC + VasC = 0

(0 − Va)sC + (0 − Va)sC +0 − Vo

R3= 0

·. . Va =R2Vi

2R1R2Cs + R1 + R2

and Va = − Vo

2R3Cs

It follows directly that

H(s) =Vo

Vi

=−2R2R3Cs

2R1R2Cs + (R1 + R2)

·. . H(jω) =−2R2R3C(jω)

(R1 + R2) + jω(2R1R2C)

|H(jω)| =2R2R3Cω√

(R1 + R2)2 + ω24R21R

22C

2

|H(jω)|2 =4R2

2R23C

2ω2

(R1 + R2)2 + 4R21R

22C

2ω2

P 15.25 ωo = 2πfo = 400π rad/s

β = 2π(1000) = 2000π rad/s

·. . ωc2 − ωc1 = 2000π

√ωc1ωc2 = ωo = 400π

Solve for the cutoff frequencies:

ωc1ωc2 = 16 × 104π2

Problems 15–21

ωc2 =16 × 104π2

ωc1

·. .16 × 104π2

ωc1

− ωc1 = 2000π

or ω2c1

+ 2000πωc1 − 16 × 104π2 = 0

ωc1 = −1000π ±√

106π2 + 0.16 × 106π2

ωc1 = 1000π(−1 ±√

1.16) = 242.01 rad/s

·. . ωc2 = 2000π + 242.01 = 6525.19 rad/s

Thus, fc1 = 38.52 Hz and fc2 = 1038.52 Hz

Check: β = fc2 − fc1 = 1000Hz

ωc2 =1

RLCL

= 6525.19

RL =1

(6525.19)(5 × 10−6)= 30.65 Ω

ωc1 =1

RHCH

= 242.01

RH =1

(242.01)(5 × 10−6)= 826.43 Ω

P 15.26 ωo = 1000 rad/s; GAIN = 6

β = 4000 rad/s; C = 0.2 µF

β = ωc2 − ωc1 = 4000

ωo =√

ωc1ωc2 = 1000

Solve for the cutoff frequencies:

·. . ω2c1

+ 4000ωc1 − 106 = 0

ωc1 = −2000 ± 1000√

5 = 236.07 rad/s


ωc2 = 4000 + ωc1 = 4236.07 rad/s

Check: β = ωc2 − ωc1 = 4000 rad/s

ωc1 =1

RLCL

·. . RL =1

(0.2 × 10−6)(236.07)= 21.81 kΩ

1RHCH

= 4236.07

RH =1

(0.2 × 10−6)(4236.07)= 1.18 kΩ

Rf

Ri

= 6

If Ri = 1 kΩ Rf = 6Ri = 6 kΩ

P 15.27 [a] y = 20 log101√

1 + ω2n= −10 log10(1 + ω2n)

From the laws of logarithms we have

y =(−10

ln 10

)ln(1 + ω2n)

Thus

dy

dω=(−10

ln 10

) 2nω2n−1

(1 + ω2n)

x = log10 ω =ln ω

ln 10·. . ln ω = x ln 10

1ω

dω

dx= ln 10,

dω

dx= ω ln 10

dy

dx=(

dy

dω

)(dω

dx

)=

−20nω2n

1 + ω2ndB/decade

at ω = ωc = 1 rad/s

dy

dx= −10n dB/decade.

Problems 15–23

[b] y = 20 log101

[√

1 + ω2]n= −10n log10(1 + ω2)

=−10nln 10

ln(1 + ω2)

dy

dω=

−10nln 10

( 11 + ω2

)2ω =

−20nω

(ln 10)(1 + ω2)

As before

dω

dx= ω(ln 10); ·. .

dy

dx=

−20nω2

(1 + ω2)

At the corner ωc =√

21/n − 1 ·. . ω2c = 21/n − 1

dy

dx=

−20n[21/n − 1]21/n

dB/decade.

[c] For the Butterworth Filter For the cascade of identical sections

n dy/dx (dB/decade) n dy/dx (dB/decade)

1 −10 1 −10

2 −20 2 −11.72

3 −30 3 −12.38

4 −40 4 −12.73

∞ −∞ ∞ −13.86

[d] It is apparent from the calculations in part (c) that as n increases the amplitudecharacteristic at the cutoff frequency decreases at a much faster rate for theButterworth filter.Hence the transition region of the Butterworth filter will be much narrowerthan that of the cascaded sections.

P 15.28 [a] n ∼= (−0.05)(−30)log10(7000/2000)

∼= 2.76

·. . n = 3

[b] Gain = 20 log101√

1 + (7000/2000)6= −32.65 dB

P 15.29 [a] For the scaled circuit

H ′(s) =1/(R′)2C ′

1C′2

s2 + 2R′C′

1s + 1

(R′)2C′1C′

2

where

R′ = kmR; C ′1 = C1/kfkm; C ′

2 = C2/kfkm


It follows that

1(R′)2C ′

1C′2

=k2

f

R2C1C2

2R′C ′

1=

2kf

RC1

·. . H ′(s) =k2

f/R2C1C2

s2 + 2kf

RC1s +

k2f

R2C1C2

=1/R2C1C2(

skf

)2+ 2

RC1

(skf

)+ 1

R2C1C2

P 15.30 [a] H(s) =1

(s + 1)(s2 + s + 1)[b] fc = 2000 Hz; ωc = 4000π rad/s; kf = 4000π

H ′(s) =1

( skf

+ 1)[( skf

)2 + skf

+ 1]

=k3

f

(s + kf )(s2 + kfs + k2f )

=(4000π)3

(s + 4000π)[s2 + 4000πs + (4000π)2]

[c] H ′(j14,000π) =64

(4 + j14)(−180 + j56)

= 0.02332/− 236.77

Gain = 20 log10(0.02332) = −32.65 dB

P 15.31 [a] In the first-order circuit R = 1 Ω and C = 1 F.

km =R′

R=

10001

= 1000; kf =ω′

o

ωo

=2π(2000)

1= 4000π

R′ = kmR = 1000 Ω; C ′ =C

kmkf

=1

(1000)(4000π)= 79.58 nF

In the second-order circuit R = 1 Ω, 2/C1 = 1 so C1 = 2 F, andC2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit

R′ = kmR = 1000 Ω; C ′1 =

C1

kmkf

=2

(1000)(4000π)= 159.15 nF

C ′2 =

C2

kmkf

=0.5

(1000)(4000π)= 39.79 nF

Problems 15–25

[b]

P 15.32 [a] n =(−0.05)(−48)

log10(2000/500)= 3.99 ·. . n = 4

From Table 15.1 the transfer function of the first section is

H1(s) =s2

s2 + 0.765s + 1For the prototype circuit

2R2

= 0.765; R2 = 2.61 Ω; R1 =1R2

= 0.383 Ω

The transfer function of the second section is

H2(s) =s2

s2 + 1.848s + 1For the prototype circuit

2R2

= 1.848; R2 = 1.082 Ω; R1 =1R2

= 0.9240 Ω

The scaling factors are:

kf =ω′

o

ωo

=2π(2000)

1= 4000π

C ′ =C

kmkf

·. . 10 × 10−9 =1

4000πkm

·. . km =1

4000π(10 × 10−9)= 7957.75

Therefore in the first section

R′1 = kmR1 = 3.04 kΩ; R′

2 = kmR2 = 20.80 kΩ

In the second section

R′1 = kmR1 = 7.35 kΩ; R′

2 = kmR2 = 8.61 kΩ


[b]

P 15.33 n = 5: 1 + (−1)5s10 = 0; s10 = 1

s10 = 1/0 + 36k

k sk+1

0 1/0

1 1/36

2 1/72

3 1/108

4 1/144

5 1/180

6 1/216

7 1/252

8 1/288

9 1/324

Group by conjugate pairs to form denominator polynomial.

(s + 1)[s − (cos 108 + j sin 108)][s − (cos 252 + j sin 252)]

· [s − (cos 144 + j sin 144)][s − (cos 216 + j sin 216)]

= (s + 1)(s + 0.309 − j0.951)(s + 0.309 + j0.951)·

(s + 0.809 − j0.588)(s + 0.809 + j0.588)

which reduces to

(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)

Problems 15–27

n = 6: 1 + (−1)6s12 = 0 s12 = −1

s12 = 1/15 + 36k

k sk+1

0 1/15

1 1/45

2 1/75

3 1/105

4 1/135

5 1/165

6 1/195

7 1/225

8 1/255

9 1/285

10 1/315

11 1/345

Grouping by conjugate pairs yields

(s + 0.2588 − j0.9659)(s + 0.2588 + j0.9659)×

(s + 0.7071 − j0.7071)(s + 0.7071 + j0.7071)×

(s + 0.9659 − j0.2588)(s + 0.9659 + j0.2588)

or (s2 + 0.518s + 1)(s2 + 1.414s + 1)(s2 + 1.932s + 1)

P 15.34 H ′(s) =s2

s2 + 2kmR2(C/kmkf )s + 1

kmR1kmR2(C2/k2mk2

f)

H ′(s) =s2

s2 + 2kf

R2Cs +

k2f

R1R2C2

=(s/kf )2

(s/kf )2 + 2R2C

(skf

)+ 1

R1R2C2


P 15.35 [a] n =(−0.05)(−48)log10(32/8)

= 3.99 ·. . n = 4

From Table 15.1 the transfer function is

H(s) =1

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

The capacitor values for the first stage prototype circuit are

2C1

= 0.765 ·. . C1 = 2.61 F

C2 =1C1

= 0.38 F

The values for the second stage prototype circuit are

2C1

= 1.848 ·. . C1 = 1.08 F

C2 =1C1

= 0.92 F

The scaling factors are

km =R′

R= 1000; kf =

ω′o

ωo

= 16,000π

Therefore the scaled values for the components in the first stage are

R1 = R2 = R = 1000 Ω

C1 =2.61

(16,000π)(1000)= 52.01 nF

C2 =0.38

(16,000π)(1000)= 7.61 nF

The scaled values for the second stage are

R1 = R2 = R = 1000 Ω

C1 =1.08

(16,000π)(1000)= 21.53 nF

C2 =0.92

(16,000π)(1000)= 18.38 nF

Problems 15–29

[b]

P 15.36 [a] The cascade connection is a bandpass filter.

[b] The cutoff frequencies are 2 kHz and 8 kHz.The center frequency is

√(2)(8) = 4 kHz.

The Q is 4/(8 − 2) = 2/3 = 0.67

[c] For the high pass section kf = 4000π. The prototype transfer function is

Hhp(s) =s4

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

·. . H ′hp(s) =

(s/4000π)4

[(s/4000π)2 + 0.765(s/4000π) + 1]

· 1[(s/4000π)2 + 1.848(s/4000π) + 1]

=s4

(s2 + 3060πs + 16 × 106π2)(s2 + 7392πs + 16 × 106π2)

For the low pass section kf = 16,000π

Hlp(s) =1

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

·. . H ′lp(s) =

1[(s/16,000π)2 + 0.765(s/16,000π) + 1]

· 1[(s/16,000π)2 + 1.848(s/16,000π) + 1]

=(16,000π)4

([s2 + 12,240πs + (16,000π)2)][s2 + 29,568πs + (16,000π)2]

The cascaded transfer function is

H ′(s) = H ′hp(s)H

′lp(s)


For convenience let

D1 = s2 + 3060πs + 16 × 106π2

D2 = s2 + 7392πs + 16 × 106π2

D3 = s2 + 12,240πs + 256 × 106π2

D4 = s2 + 29,568πs + 256 × 106π2

Then

H ′(s) =65,536 × 1012π4s4

D1D2D3D4

[d] ωo = 2π(4000) = 8000π rad/s

s = j8000π

s4 = 4096 × 1012π4

D1 = (16 × 106π2 − 64 × 106π2) + j(8000π)(3060π)

= 106π2(−48 + j24.48) = 106π2(53.88/152.98)

D2 = (16 × 106π2 − 64 × 106π2) + j(8000π)(7392π)

= 106π2(−48 + j59.136) = 106π2(76.16/129.07)

D1 = (256 × 106π2 − 64 × 106π2) + j(8000π)(12,240π)

= 106π2(192 + j97.92) = 106π2(215.53/27.02)

D1 = (256 × 106π2 − 64 × 106π2) + j(8000π)(29,568π)

= 106π2(192 + j236.544) = 106π2(304.66/50.93)

H ′(jωo) =(65,536)(4096)π8 × 1024

(π8 × 1024)[(53.88)(76.16)(215.53)(304.66)/360]

= 0.996/− 360 = 0.996/0

P 15.37 [a] From the statement of the problem, K = 10 ( = 20 dB). Therefore for theprototype bandpass circuit

R1 =Q

K=

1610

= 1.6 Ω

R2 =Q

2Q2 − K=

16502

Ω

R3 = 2Q = 32 Ω

Problems 15–31


kf =ω′

o

ωo

= 2π(6400) = 12,800π

km =C

C ′kf

=1

(20 × 10−9)(12,800π)= 1243.40

Therefore,

R′1 = kmR1 = (1.6)(1243.40) = 1.99 kΩ

R′2 = kmR2 = (16/502)(1243.40) = 39.63 Ω

R′3 = kmR3 = 32(1243.40) = 39.79 kΩ

[b]

P 15.38 From Eq 15.58 we can write

H(s) =−(

2R3C

) (R3C

2

) (1

R1C

)s

s2 + 2R3C

s + R1+R2R1R2R3C2

or

H(s) =−(

R32R1

) (2

R3Cs)

s2 + 2R3C

s + R1+R2R1R2R3C2

Therefore

2R3C

= β =ωo

Q;

R1 + R2

R1R2R3C2 = ω2o ;

and K =R3

2R1

By hypothesis C = 1 F and ωo = 1 rad/s

·. .2R3

=1Q

or R3 = 2Q


R1 =R3

2K=

Q

K

R1 + R2

R1R2R3= 1

Q

K+ R2 =

(Q

K

)(2Q)R2

·. . R2 =Q

2Q2 − K

P 15.39 [a] First we will design a unity gain filter and then provide the passband gain withan inverting amplifier. For the high pass section the cut-off frequency is 500Hz. The order of the Butterworth is

n =(−0.05)(−20)log10(500/200)

= 2.51

·. . n = 3

Hhp(s) =s3

(s + 1)(s2 + s + 1)

For the prototype first-order section

R1 = R2 = 1 Ω, C = 1 F

For the prototype second-order section

R1 = 0.5 Ω, R2 = 2 Ω, C = 1 F


kf =ω′

o

ωo

= 2π(500) = 1000π

km =C

C ′kf

=1

(15 × 10−9)(1000π)=

106

15π

In the scaled first-order section

R′1 = R′

2 = kmR1 =106

15π(1) = 21.22 kΩ

C ′ = 15 nF

In the scaled second-order section

R′1 = 0.5km = 10.61 kΩ

R′2 = 2km = 42.44 kΩ

Problems 15–33

C ′ = 15 nF

For the low-pass section the cut-off frequency is 4500 Hz. The order of theButterworth filter is

n =(−0.05)(−20)

log10(11,250/4500)= 2.51; ·. . n = 3

Hlp(s) =1

(s + 1)(s2 + s + 1)

For the prototype first-order section

R1 = R2 = 1 Ω, C = 1 F

For the prototype second-order section

R1 = R2 = 1 Ω; C1 = 2 F; C2 = 0.5 F

The low-pass scaling factors are

km =R′

R= 104; kf =

ω′o

ωo

= (4500)(2π) = 9000π

For the scaled first-order section

R′1 = R′

2 = 10 kΩ; C ′ =C

kfkm

=1

(9000π)(104)= 3.54 nF

For the scaled second-order section

R′1 = R′

2 = 10 kΩ

C ′1 =

C1

kfkm

=2

(9000π)(104)= 7.07 nF

C ′2 =

C2

kfkm

=0.5

(9000π)(104)= 1.77 nF

GAIN AMPLIFIER

20 log10 K = 20 dB, ·. . K = 10

Since we are using 10 kΩ resistors in the low-pass stage, we will useRf = 100 kΩ and Ri = 10 kΩ in the inverting amplifier stage.


[b]

P 15.40 [a] Unscaled high-pass stage

Hhp(s) =s3

(s + 1)(s2 + s + 1)

The frequency scaling factor is kf = (ω′o/ωo) = 1000π. Therefore the scaled

transfer function is

H ′hp(s) =

(s/1000π)3(s

1000π+ 1

) [(s

1000π

)2+ s

1000π+ 1

]

=s3

(s + 1000π)[s2 + 1000πs + 106π2]

Unscaled low-pass stage

Hlp(s) =1

(s + 1)(s2 + s + 1)

The frequency scaling factor is kf = (ω′o/ωo) = 9000π. Therefore the scaled

transfer function is

H ′lp(s) =

1(s

9000π+ 1

) [(s

9000π

)2+(

s9000π

)+ 1

]

=(9000π)3

(s + 9000π)(s2 + 9000πs + 81 × 106π2)

Thus the transfer function for the filter is

H ′(s) = 10H ′hp(s)H

′lp(s) =

729 × 1010π3s3

D1D2D3D4

Problems 15–35

where

D1 = s + 1000π

D2 = s + 9000π

D3 = s2 + 1000πs + 106π2

D4 = s2 + 9000πs + 81 × 106π2

[b] At f = 200 Hz ω = 400π rad/s

D1(j400π) = 400π(2.5 + j1)

D2(j400π) = 400π(22.5 + j1)

D3(j400π) = 4 × 105π2(2.1 + j1.0)

D4(j400π) = 4 × 105π2(202.1 + j9)

Therefore

D1D2D3D4(j400π) = 256π61014(28,534.82/52.36)

H ′(j400π) =(729π3 × 1010)(64 × 106π3)

256π6 × 1014(28,534.82/52.36)

= 0.639/− 52.36

·. . 20 log10 |H ′(j400π)| = 20 log10(0.639) = −3.89 dB

At f = 1500 Hz, ω = 3000π rad/s

Then

D1(j3000π) = 1000π(1 + j3)

D2(j3000π) = 3000π(3 + j1)

D3(j3000π) = 106π2(−8 + j3)

D4(j3000π) = 9 × 106π2(8 + j3)

H ′(j3000π) =(729 × π3 × 1010)(27 × 109π3)

27 × 1018π6(730/270 )

= 9.99/90

·. . 20 log10 |H ′(j3000π)| = 19.99 dB


[c] From the transfer function the gain is down 19.99 + 3.89 or 23.88 dB at 200 Hz.Because the upper cut-off frequency is nine times the lower cut-off frequencywe would expect the high-pass stage of the filter to predict the loss in gain at200 Hz. For a 3nd order Butterworth

GAIN = 20 log101√

1 + (500/200)6= −23.89 dB.

1500 Hz is in the passband for this bandpass filter, and is in fact the centerfrequency. Hence we expect the gain at 1500 Hz to equal, or nearly equal,20 dB as specified in Problem 15.39. Thus our scaled transfer functionconfirms that the filter meets the specifications.

P 15.41 [a] From Table 15.1

Hlp(s) =1

(s2 + 0.518s + 1)(s2 +√

2s + 1)(s2 + 1.932s + 1)

Hhp(s) =1(

1s2 + 0.518

(1s

)+ 1

) (1s2 +

√2(

1s

)+ 1

) (1s2 + 1.932

(1s

)+ 1

)

Hhp(s) =s6

(s2 + 0.518s + 1)(s2 +√

2s + 1)(s2 + 1.932s + 1)

P 15.42 [a] kf = 25,000

H ′hp(s) =

(s/25,000)6

[(s/25,000)2 + 0.518(s/25,000) + 1]

· 1[(s/25,000)2 + 1.414s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 + 1]

=s6

(s2 + 12,950s + 625 × 106)(s2 + 35,350s + 625 × 106)

· 1(s2 + 48,300s + 625 × 106)

[b] H ′(j25,000) =−(25,000)6

[12,950(j25,000)][35,350(j25,000)][48,300(j25,000)]

=−(25,000)3

(12,950)(25,350)(48,300)j3

= 0.7067/− 90

20 log10 |H ′(j25,000)| = −3.02 dB

Problems 15–37

P 15.43 [a] At very low frequencies the two capacitor branches are open and because the opamp is ideal the current in R3 is zero. Therefore at low frequencies the circuitbehaves as an inverting amplifier with a gain of R2/R1. At very highfrequencies the capacitor branches are short circuits and hence the outputvoltage is zero.

[b] Let the node where R1, R2, R3, and C2 join be denoted as a, then

(Va − Vi)G1 + VasC2 + (Va − Vo)G2 + VaG3 = 0

−VaG3 − VosC1 = 0

or

(G1 + G2 + G3 + sC2)Va − G2Vo = G1Vi

Va =−sC1

G3Vo

Solving for Vo/Vi yields

H(s) =−G1G3

(G1 + G2 + G3 + sC2)sC1 + G2G3

=−G1G3

s2C1C2 + (G1 + G2 + G3)C1s + G2G3

=−G1G3/C1C2

s2 +[

(G1+G2+G3)C2

]s + G2G3

C1C2

=−G1G2G3

G2C1C2

s2 +[

(G1+G2+G3)C2

]s + G2G3

C1C2

=−Kbo

s2 + b1s + bo

where K =G1

G2; bo =

G2G3

C1C2

and b1 =G1 + G2 + G3

C2

[c] Equating coefficients we see that

G1 = KG2

G3 =boC1C2

G2=

boC1

G2

since by hypothesis C2 = 1 F

b1 =G1 + G2 + G3

C2= G1 + G2 + G3


·. . b1 = KG2 + G2 +boC1

G2

b1 = G2(1 + K) +boC1

G2

Solving this quadratic equation for G2 we get

G2 =b1

2(1 + K)±√

b21 − boC14(1 + K)

4(1 + K)2

=b1 ±

√b21 − 4bo(1 + K)C1

2(1 + K)

For G2 to be realizable

C1 <b21

4bo(1 + K)

[d] 1. Select C2 = 1 F

2. Select C1 such that C1 <b21

4bo(1 + K)

3. Calculate G2 (R2)

4. Calculate G1 (R1); G1 = KG2

5. Calculate G3 (R3); G3 = boC1/G2

P 15.44 [a] In the second order section of a third order Butterworth filter bo = b1 = 1Therefore,

C1 ≤ b21

4bo(1 + K)=

1(4)(1)(5)

= 0.05 F

·. . C1 = 0.05 F (limiting value)

[b] G2 =1

2(1 + 4)= 0.1 S

G3 =1

0.1(0.05) = 0.5 S

G1 = 4(0.1) = 0.4 S

Therefore,

R1 =1

G1= 2.5 Ω; R2 =

1G2

= 10 Ω; R3 =1

G3= 2 Ω

Problems 15–39

[c] kf =ω′

o

ωo

= 2π(2500) = 5000π

km =C2

C ′2kf

=1

(10 × 10−9)kf

= 6366.2

C ′1 =

0.05kfkm

= 0.5 × 10−9 = 500 pF

R′1 = (2.5)(6366.2) = 15.92 kΩ

R′2 = (10)(6366.2) = 63.66 kΩ

R′3 = (2)(6366.2) = 12.73 kΩ

[d] R′1 = R′

2 = (6366.2)(1) = 6.37 kΩ

C ′ =C

kfkm

=1

108 = 10 nF

[e]

P 15.45 [a] By hypothesis the circuit becomes:

For very small frequencies the capacitors behave as open circuits and thereforevo is zero. As the frequency increases, the capacitive branch impedancesbecome small compared to the resistive branches. When this happens thecircuit becomes an inverting amplifier with the capacitor C2 dominating thefeedback path. Hence the gain of the amplifier approaches(1/jωC2)/(1/jωC1) or C1/C2. Therefore the circuit behaves like a high-passfilter with a passband gain of C1/C2.


[b] Summing the currents away from the upper terminal of R2 yields

VaG2 + (Va − Vi)sC1 + (Va − Vo)sC2 + VasC3 = 0

or

Va[G2 + s(C1 + C2 + C3)] − VosC2 = sC1Vi

Summing the currents away from the inverting input terminal gives

(0 − Va)sC3 + (0 − Vo)G1 = 0

or

sC3Va = −G1Vo; Va =−G1Vo

sC3

Therefore we can write

−G1Vo

sC3[G2 + s(C1 + C2 + C3)] − sC2Vo = sC1Vi

Solving for Vo/Vi gives

H(s) =Vo

Vi

=−C1C3s

2

[C2C3s2 + G1(C1 + C2 + C3)s + G1G2]

=−C1C2

s2[s2 + G1

C2C3(C1 + C2 + C3)s + G1G2

C2C3

]

=−Ks2

s2 + b1s + bo

Therefore the circuit implements a second-order high-pass filter with apassband gain of C1/C2.

[c] C1 = K:

b1 =G1

(1)(1)(K + 2) = G1(K + 2)

·. . G1 =b1

K + 2; R1 =

(K + 2

b1

)

bo =G1G2

(1)(1)= G1G2

·. . G2 =bo

G1=

bo

b1(K + 2)

·. . R2 =b1

bo(K + 2)

Problems 15–41

[d] From Table 15.1 the transfer function of the second-order section of athird-order high-pass Butterworth filter is

H(s) =Ks2

s2 + s + 1

Therefore b1 = bo = 1

Thus

C1 = K = 8 F

R1 =8 + 2

1= 10 Ω

R2 =1

1(8 + 2)= 0.10 Ω

P 15.46 [a] Low-pass filter:

n =(−0.05)(−30)

log10(1000/400)= 3.77; ·. . n = 4

In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F

C1 ≤ b21

4bo(1 + K)≤ (0.765)2

(4)(2)≤ 0.0732

choose C1 = 0.03 F

G2 =0.765 ±

√(0.765)2 − 4(2)(0.03)

4=

0.765 ± 0.5884

Arbitrarily select the larger value for G2, then

G2 = 0.338 S; ·. . R2 =1

G2= 2.96 Ω

G1 = KG2 = 0.338 S; ·. . R1 =1

G1= 2.96 Ω

G3 =boC1

G2=

(1)(0.03)0.338

= 0.089 ·. . R3 = 1/G3 = 11.3 Ω

Therefore in the first second-order prototype circuit

R1 = R2 = 2.96 Ω; R3 = 11.3 Ω

C1 = 0.03 F; C2 = 1 F

In the second second-order prototype circuit: b1 = 1.848, b0 = 1, C2 = 1 F

·. . C1 ≤ (1.848)2

8≤ 0.427


choose C1 = 0.30 F

G2 =1.848 ±

√(1.848)2 − 8(0.3)

4=

1.848 ± 1.0084

Arbitrarily select the larger value, then

G2 = 0.7139 S; ·. . R2 =1

G2= 1.4008 Ω

G1 = KG2 = 0.7139 S; ·. . R1 =1

G1= 1.4008 Ω

G3 =boC1

G2=

(1)(0.30)0.7139

= 0.4202 S ·. . R3 = 1/G3 = 2.3796 Ω

In the low-pass section of the filter

kf =ω′

o

ωo

= 2π(400) = 800π

km =C

C ′kf

=1

(10 × 10−9)kf

=125,000

π

Therefore in the first scaled second-order section

R′1 = R′

2 = 2.96km = 118 kΩ

R′3 = 11.3km = 450 kΩ

C ′1 =

0.03kfkm

= 300 pF

C ′2 = 10 nF

In the second scaled second-order section

R′1 = R′

2 = 1.4008km = 55.74 kΩ

R′3 = 2.3796km = 94.68 kΩ

C ′1 =

0.3kfkm

= 3 nF

C ′2 = 10 nF

High-pass filter section

n =(−0.05)(−30)

log10(6400/2560)= 3.77; n = 4.

In the first prototype second-order section: b1 = 0.765; bo = 1; C2 = C3 = 1 F

C1 = K = 1 F

Problems 15–43

R1 =K + 2

b1=

30.765

= 3.92 Ω

R2 =b1

bo(K + 2)=

0.7653

= 0.255 Ω

In the second prototype second-order section: b1 = 1.848; bo = 1;C2 = C3 = 1 F

C1 = K = 1 F

R1 =K + 2

b1=

31.848

= 1.623 Ω

R2 =b1

bo(K + 2)=

1.8483

= 0.616 Ω

In the high-pass section of the filter

kf =ω′

o

ωo

= 2π(6400) = 12,800π

km =C

C ′kf

=1

(10 × 10−9)(12,800π)=

7812.5π

In the first scaled second-order section

R′1 = 3.92km = 9.75 kΩ

R′2 = 0.255km = 634 Ω

C ′1 = C ′

2 = C ′3 = 10 nF

In the second scaled second-order section

R′1 = 1.623km = 4.04 kΩ

R′2 = 0.616km = 1.53 kΩ

C ′1 = C ′

2 = C ′3 = 10 nF

In the gain section, let Ri = 10 kΩ and Rf = 10 kΩ.


[b]

P 15.47 [a] The prototype low-pass transfer function is

Hlp(s) =1

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

The low-pass frequency scaling factor is

kflp= 2π(400) = 800π

The scaled transfer function for the low-pass filter is

H ′lp(s) =

1[(s

800π

)2+ 0.765s

800π+ 1

] [(s

800π

)2+ 1.848s

800π+ 1

]

=4096 × 108π4

[s2 + 612πs + (800π)2] [s2 + 1478.4πs + (800π)2]

The prototype high-pass transfer function is

Hhp(s) =s4

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

The high-pass frequency scaling factor is

kfhp= 2π(6400) = 12,800π

Problems 15–45

The scaled transfer function for the high-pass filter is

H ′hp(s) =

(s/12,800π)4[(s

12,800π

)2+ 0.765s

12,800π+ 1

] [(s

12,800π

)2+ 1.848s

12,800π+ 1

]

=s4

[s2 + 9792πs + (12,800π)2][s2 + 23,654.4πs + (12,800π)2]

The transfer function for the filter is

H ′(s) =[H ′

lp(s) + H ′hp(s)

]

[b] fo =√

fc1fc2 =√

(400)(6400) = 1600 Hz

ωo = 2πfo = 3200π rad/s

(jωo)2 = −1024 × 104π2

(jωo)4 = 1,048,576 × 108π4

H ′lp(jωo) =

4096 × 108π4

[−960 × 104π2 + j612(3200π2)]×

1[−960 × 104π2 + j1478.4(3200π2)]

=40,000

(−3000 + j612)(−3000 + j1478.4)

= 3906.2 × 10−6/37.76

H ′hp(jωo) =

1,048,576 × 108π4

[15,360 × 104π2 + j9792(3200π2)]

1[15,360 × 104π2 + j23,654.4(3200π2)]

=10.24 × 106

(48,000 + j9792)(48,000 + j23,654.4)

= 3906.2 × 10−6/− 37.76

·. . H ′(jωo) = −3906.2 × 10−6(1/37.76 + 1/− 37.76)

= −3906.2 × 10−6(1.58/0) = −6176.35 × 10−6/0

G = 20 log10 |H ′(jωo)| = 20 log10(6176.35 × 10−6) = −44.19 dB

P 15.48 [a] At low frequencies the capacitor branches are open; vo = vi. At highfrequencies the capacitor branches are short circuits and the output voltage iszero. Hence the circuit behaves like a unity-gain low-pass filter.


[b] Let va represent the voltage-to-ground at the right-hand terminal of R1. Observethis will also be the voltage at the left-hand terminal of R2. The s-domainequations are

(Va − Vi)G1 + (Va − Vo)sC1 = 0

(Vo − Va)G2 + sC2Vo = 0

or

(G1 + sC1)Va − sC1Vo = G1Vi

−G2Va + (G2 + sC2)Vo = 0

·. . Va =G2 + sC2Vo

G2

·. .

[(G1 + sC1)

(G2 + sC2)G2

− sC1

]Vo = G1Vi

·. .Vo

Vi

=G1G2

(G1 + sC1)(G2 + sC2) − C1G2s

which reduces to

Vo

Vi

=G1G2/C1C2

s2 + G1C1

s + G1G2C1C2

=bo

s2 + b1s + bo

[c] There are four circuit components and two restraints imposed by H(s);therefore there are two free choices.

[d] b1 =G1

C1

·. . G1 = b1C1

bo =G1G2

C1C2

·. . G2 =bo

b1C2

[e] No, all physically realizeable capacitors will yield physically realizeableresistors.

[f] From Table 15.1 we know the transfer function of the prototype 4th orderButterworth filter is

H(s) =1

(s2 + 0.765s + 1)(s2 + 1.848s + 1)

In the first section bo = 1, b1 = 0.765

·. . G1 = (0.765)(1) = 0.765 S

R1 = 1/G1 = 1.307 Ω

G2 =1

0.765(1) = 1.307 S

Problems 15–47

R2 = 1/G2 = 0.765 Ω

In the second section bo = 1, b1 = 1.848

·. . G1 = 1.848 S

R1 = 1/G1 = 0.541 Ω

G2 =( 1

1.848

)(1) = 0.541 S

R2 = 1/G2 = 1.848 Ω

P 15.49 [a] kf =ω′

o

ωo

= 2π(3000) = 6000π

km =C

C ′kf

=1

(4.7 × 10−9)(6000π)=

106

28.2π

In the first section

R′1 = 1.307km = 14.75 kΩ

R′2 = 0.765km = 8.64 kΩ

In the second section

R′1 = 0.541km = 6.11 kΩ

R′2 = 1.848km = 20.86 kΩ


[b]

P 15.50 [a] Interchanging the Rs and Cs yields the following circuit.

At low frequencies the capacitors appear as open circuits and hence the outputvoltage is zero. As the frequency increases the capacitor branches approachshort circuits and va = vi = vo. Thus the circuit is a unity-gain, high-pass filter.

[b] The s-domain equations are

(Va − Vi)sC1 + (Va − Vo)G1 = 0

(Vo − Va)sC2 + VoG2 = 0

It follows that

Va(G1 + sC1) − G1Vo = sC1Vi

and Va =(G2 + sC2)Vo

sC2

Thus[(G2 + sC2)

sC2

](G1 + sC1) − G1

Vo = sC1Vi

Vos2C1C2 + sC1G2 + G1G2 = s2C1C2Vi

Problems 15–49

H(s) =Vo

Vi

=s2(

s2 +G2

C2s +

G1G2

C1C2

)

=Vo

Vi

=s2

s2 + b1s + bo

[c] There are 4 circuit components: R1, R2, C1 and C2.There are two transfer function constraints: b1 and bo.Therefore there are two free choices.

[d] bo =G1G2

C1C2; b1 =

G2

C2

·. . G2 = b1C2; R2 =1

b1C2

G1 =bo

b1C1

·. . R1 =b1

boC1

[e] No, all realizeable capacitors will produce realizeable resistors.

[f] The second-order section in a 3rd-order Butterworth high-pass filter iss2/(s2 + s + 1). Therefore bo = b1 = 1 and

R1 =1

(1)(1)= 1 Ω.

R2 =1

(1)(1)= 1 Ω.

P 15.51 [a] kf =ω′

o

ωo

= 104π

km =C

C ′kf

=1

(75 × 10−9)(104π)=

105

75π

C1 = C2 = 75 nF; R′1 = R′

2 = kmR = 424.4 Ω

[b] R = 424.4 Ω; C = 75 nF


[c]

[d] Hhp(s) =s3

(s + 1)(s2 + s + 1)

H ′hp(s) =

(s/104π)3

[(s/104π) + 1][(s/104π)2 + (s/104π) + 1]

=s3

(s + 104π)(s2 + 104πs + 108π2)

[e] H ′hp(j104π) =

(j104π)3

(j104π + 104π)[(j104π)2 + 104π(j104π) + 108π2]= 0.7071/135

·. . |H ′hp| = 0.7071 = −3 dB

P 15.52 [a] It follows directly from Eq 15.64 that

H(s) =s2 + 1

s2 + 4(1 − σ)s + 1

Now note from Eq 15.69 that (1 − σ) equals 1/4Q, hence

H(s) =s2 + 1

s2 + 1Qs + 1

[b] For Example 15.13, ωo = 5000 rad/s and Q = 5. Therefore kf = 5000 and

H ′(s) =(s/5000)2 + 1

(s/5000)2 +15

(s

5000

)+ 1

=s2 + 25 × 106

s2 + 1000s + 25 × 106

P 15.53 [a] ωo = 2000π rad/s

·. . kf =ω′

o

ωo

= 2000π

Problems 15–51

km =C

C ′kf

=1

(15 × 10−9)(2000π)=

105

3π

R′ = kmR =105

3π(1) = 10,610 Ω

R′

2= 5,305 Ω

σ = 1 − 14Q

= 1 − 14(20)

= 0.9875

σR′ = 10,478 Ω; (1 − σ)R′ = 133 Ω

C ′ = 15 nF

2C ′ = 30 nF

[b]

[c] kf = 2000π

H(s) =(s/2000π)2 + 1

(s/2000π)2 + 120(s/2000π) + 1

=s2 + 4 × 106π2

s2 + 100πs + 4 × 106π2

P 15.54 To satisfy the gain specification of 20 dB at ω = 0 and α = 1 requires

R1 + R2

R1= 10 or R2 = 9R1

Choose a standard resistor of 11.1 kΩ for R1 and a 100 kΩ potentiometer for R2.Since (R1 + R2)/R1 1 the value of C1 is

C1 =1

2π(40)(105)= 39.79 nF


Choose a standard capacitor value of 39 nF. Using the selected values of R1 and R2

the maximum gain for α = 1 is

20 log10

(111.111.1

)α=1

= 20.01 dB

When C1 = 39 nF the frequency 1/R2C1 is

1R2C1

=109

105(39)= 256.41 rad/s = 40.81 Hz

The magnitude of the transfer function at 256.41 rad/s is

|H(j256.41)|α=1 =|111.1 × 103 + j256.41(11.1)(100)(39)10−3||11.1 × 103 + j256.41(11.1)(100)(39)10−3| = 7.11

Therefore the gain at 40.81 Hz is

20 log10(7.11)α=1 = 17.04 dB

P 15.55 20 log10

(R1 + R2

R1

)= 13.98

·. .R1 + R2

R1= 5; ·. . R2 = 4R1

Choose R1 = 100 kΩ. Then R2 = 400 kΩ

1R2C1

= 100π rad/s; ·. . C1 =1

(100π)(400 × 103)= 7.96 nF

P 15.56 [a] |H(j0)| =R1 + αR2

R1 + (1 − α)R2=

11.1 + α(100)11.1 + (1 − α)100

Problems 15–53

P 15.57 [a] Combine the impedances of the capacitors in series in Fig. P15.53(b) to get

Ceq =1 − α

sC1+

α

sC1=

1sC1

which is identical to the impedance of the capacitor in Fig. P15.53(a).

[b]

Vx =α/sC1

(1 − α)/sC1 + α/sC1V = α

Vy =αR2

(1 − α)R2 + αR2= α = Vx

[c] Since x and y are both at the same potential, they can be shorted together, andthe circuit in Fig. 15.34 can thus be drawn as shown in Fig. 15.53(c).

[d] The feedback path between Vo and Vs containing the resistance R4 + 2R3 has noeffect on the ratio Vo/Vs, as this feedback path is not involved in the nodalequation that defines the voltage ratio. Thus, the circuit in Fig. 15.53(c) can besimplified into the form of Fig. 15.2, where the input impedance is theequivalent impedance of R1 in series with the parallel combination of(1 − α)/sC1 and (1 − α)R2, and the feedback impedance is the equivalentimpedance of R1 in series with the parallel combination of α/sC1 and αR2:

Zi = R1 +(1−α)sC1

· (1 − α)R2

(1 − α)R2 + (1−α)sC1

=R1 + (1 − α)R2 + R1R2C1s

1 + R2C1s

Zf = R1 +α

sC1· αR2

αR2 + αsC1

=R1 + αR2 + R1R2C1s

1 + R2C1s

P 15.58 As ω → 0

|H(iω)| → 2R3 + R4

2R3 + R4= 1


Therefore the circuit would have no effect on low frequency signals. As ω → ∞

|H(jω)| → [(1 − β)R4 + Ro](βR4 + R3)[(1 − β)R4 + R3](βR4 + Ro)

When β = 1

|H(j∞)|β=1 =Ro(R4 + R3)R3(R4 + Ro)

If R4 Ro

|H(j∞)|β=1∼= Ro

R3> 1

Thus, when β = 1 we have amplification or “boost”. When β = 0

|H(j∞)|β=0 =R3(R4 + R3)Ro(R4 + Ro)

If R4 Ro

|H(j∞)|β=0∼= R3

Ro

< 1

Thus, when β = 0 we have attenuation or “cut”.Also note that when β = 0.5

|H(jω)|β=0.5 =(0.5R4 + Ro)(0.5R4 + R3)(0.5R4 + R3)(0.5R4 + Ro)

= 1

Thus, the transition from amplification to attenuation occurs at β = 0.5. If β > 0.5we have amplification, and if β < 0.5 we have attenuation.Also note the amplification an attenuation are symmetric about β = 0.5. i.e.

|H(jω)|β=0.6 =1

|H(jω)|β=0.4

Yes, the circuit can be used as a treble volume control because

• The circuit has no effect on low frequency signals

• Depending on β the circuit can either amplify (β > 0.5) or attenuate (β < 0.5)signals in the treble range

• The amplification (boost) and attenuation (cut) are symmetric around β = 0.5.When β = 0.5 the circuit has no effect on signals in the treble frequency range.

Problems 15–55

P 15.59 [a] |H(j∞)|β=1 =Ro(R4 + R3)R3(R4 + Ro)

=(65.9)(505.9)(5.9)(565.9)

= 9.99

·. . maximum boost = 20 log10 9.99 = 19.99 dB

[b] |H(j∞)|β=0 =R3(R4 + R3)Ro(R4 + Ro)

·. . maximum cut = −21.93 dB

[c] R4 = 500 kΩ; Ro = R1 + R3 + 2R5 = 65.9 kΩ

·. . R4 = 7.59Ro

Yes, R4 is significantly greater than Ro.

[d] |H(j/R3C2)|β=1 =

∣∣∣∣∣∣(2R3 + R4) + j Ro

R3(R4 + R3)

(2R3 + R4) + j(R4 + Ro)

∣∣∣∣∣∣

=∣∣∣∣∣511.8 + j 65.9

5.9 (505.9)511.8 + j565.9

∣∣∣∣∣= 7.44

20 log10 |H(j/R3C2)|β=1 = 20 log10 7.44 = 17.43 dB

[e] When β = 0

|H(j/R3C2)|β=0 =(2R3 + R4) + j(R4 + Ro)

(2R3 + R4) + jRo

R3(R4 + R3)

Note this is the reciprocal of |H(j/R3C2)|β=1.

·. . 20 log10 |H(j/R3C2)|β+0 = −17.43 dB

[f] The frequency 1/R3C2 is very nearly where the gain is 3 dB off from itsmaximum boost or cut. Therefore for frequencies higher than 1/R3C2 thecircuit designer knows that gain or cut will be within 3 dB of the maximum.


P 15.60 |H(j∞)| =[(1 − β)R4 + Ro][βR4 + R3][(1 − βR4 + R3][βR4 + Ro]

=[(1 − β)500 + 65.9][β500 + 5.9][(1 − β)500 + 5.9][β500 + 65.9]

16Fourier Series

Assessment Problems

AP 16.1 av =1T

∫ 2T/3

0Vm dt +

1T

∫ T

2T/3

(Vm

3

)dt =

79Vm = 7π V

ak =2T

[∫ 2T/3

0Vm cos kω0t dt +

∫ T

2T/3

(Vm

3

)cos kω0t dt

]

=( 4Vm

3kω0T

)sin

(4kπ

3

)=(6

k

)sin

(4kπ

3

)

bk =2T

[∫ 2T/3

0Vm sin kω0t dt +

∫ T

2T/3

(Vm

3

)sin kω0t dt

]

=( 4Vm

3kω0T

) [1 − cos

(4kπ

3

)]=(6

k

) [1 − cos

(4kπ

3

)]

AP 16.2 [a] av = 7π = 21.99 V

[b] a1 = −5.196 a2 = 2.598 a3 = 0 a4 = −1.299 a5 = 1.039

b1 = 9 b2 = 4.5 b3 = 0 b4 = 2.25 b5 = 1.8

[c] ω0 =(2π

T

)= 50 rad/s

[d] f3 = 3f0 = 23.87 Hz

[e] v(t) = 21.99 − 5.2 cos 50t + 9 sin 50t + 2.6 cos 100t + 4.5 sin 100t

−1.3 cos 200t + 2.25 sin 200t + 1.04 cos 250t + 1.8 sin 250t + · · · V

AP 16.3 Odd function with both half- and quarter-wave symmetry.

vg(t) =(6Vm

T

)t, 0 ≤ t ≤ T/6; av = 0, ak = 0 for all k

16–1

16–2 CHAPTER 16. Fourier Series

bk = 0 for k even

bk =8T

∫ T/4

0f(t) sin kω0t dt, k odd

=8T

∫ T/6

0

(6Vm

T

)t sin kω0t dt +

8T

∫ T/4

T/6Vm sin kω0t dt

=(12Vm

k2π2

)sin

(kπ

3

)

vg(t) =12Vm

π2

∞∑n=1,3,5

1n2 sin

nπ

3sin nω0t V

AP 16.4 [a] Using the results from AP 16.2, and Equation (16.39),

A1 = −5.2 − j9 = 10.4/− 120; A2 = 2.6 − j4.5 = 5.2/− 60

A3 = 0; A4 = −1.3 − j2.25 = 2.6/− 120

A5 = 1.04 − j1.8 = 2.1/− 60

θ1 = −120; θ2 = −60; θ3 not defined;

θ4 = −120; θ5 = −60

[b] v(t) = 21.99 + 10.4 cos(50t − 120) + 5.2 cos(100t − 60)

+ 2.6 cos(200t − 120) + 2.1 cos(250t − 60) + · · · V

AP 16.5 The Fourier series for the input voltage is

vi =8Aπ2

∞∑n=1,3,5

( 1n2 sin

nπ

2

)sin nω0(t + T/4)

=8Aπ2

∞∑n=1,3,5

( 1n2 sin2 nπ

2

)cos nω0t

=8Aπ2

∞∑n=1,3,5

1n2 cos nω0t

8Aπ2 =

8(281.25π2)π2 = 2250 mV

ω0 =2πT

=2π

200π× 103 = 10

Problems 16–3

·. . vi = 2250∞∑

n=1,3,5

1n2 cos 10nt mV

From the circuit we have

Vo =Vi

R + (1/jωC)· 1jωC

=Vi

1 + jωRC

Vo =1/RC

1/RC + jωVi =

100100 + jω

Vi

Vi1 = 2250/0 mV; ω0 = 10 rad/s

Vi3 =2250

9/0 = 250/0 mV; 3ω0 = 30 rad/s

Vi5 =225025

/0 = 90/0 mV; 5ω0 = 50 rad/s

Vo1 =100

100 + j10(2250/0) = 2238.83/− 5.71 mV

Vo3 =100

100 + j30(250/0) = 239.46/− 16.70 mV

Vo5 =100

100 + j50(90/0) = 80.50/− 26.57 mV

·. . vo = 2238.33 cos(10t − 5.71) + 239.46 cos(30t − 16.70)

+ 80.50 cos(50t − 26.57) + . . . mV

AP 16.6 [a] The Fourier series of the input voltage is

vg =4Aπ

∞∑n=1,3,5

1n

sin nω0(t + T/4)

= 42∞∑

n=1,3,5

[ 1n

sin(

nπ

2

)]cos 2000nt V

From the circuit we have

VosC +Vo

sL+

Vo − Vg

R= 0

·. .Vo

Vg

= H(s) =s/RC

s2 + (s/RC) + (1/LC)


Substituting in the numerical values yields

H(s) =500s

s2 + 500s + 108

Vg1 = 42/0 ω0 = 2000 rad/s

Vg3 = 14/180 3ω0 = 6000 rad/s

Vg5 = 8.4/0 5ω0 = 10,000 rad/s

Vg7 = 6/180 7ω0 = 14,000 rad/s

H(j2000) =500(j2000)

108 − 4 × 106 + 500(j2000)=

j196 + j1

= 0.01042/89.40

H(j6000) = 0.04682/87.32

H(j10,000) = 1/0

H(j14,000) = 0.07272/− 85.83

Thus,

Vo1 = (42/0)(0.01042/89.40) = 0.4375/89.40 V

Vo3 = 0.6555/− 92.68 V

Vo5 = 8.4/0 V

Vo7 = 0.4363/94.17 V

Therefore,

vo = 0.4375 cos(2000t + 89.40) + 0.6555 cos(6000t − 92.68)

+ 8.4 cos(10,000t) + 0.4363 cos(14,000t + 94.17) + . . . V

[b] The 5th harmonic, that is, the term at 10,000 rad/s, dominates the outputvoltage. The circuit is a bandpass filter with a center frequency of 10,000 rad/sand a bandwidth of 500 rad/s. Thus, Q is 20 and the filter is quite selective.This causes the attenuation of the fundamental, third, and seventh harmonicterms in the output signal.

AP 16.7 ω0 =2π × 103

2094.4= 3 rad/s

jω0k = j3k

Problems 16–5

VR =2

2 + s + 1/s(Vg) =

2sVg

s2 + 2s + 1

H(s) =(

VR

Vg

)=

2ss2 + 2s + 1

H(jω0k) = H(j3k) =j6k

(1 − 9k2) + j6k

vg1 = 25.98 sin ω0t V; Vg1 = 25.98/0 V

H(j3) =j6

−8 + j6= 0.6/− 53.13; VR1 = 15.588/− 53.13 V

P1 =(15.588/

√2)2

2= 60.75 W

vg3 = 0, therefore P3 = 0 W

vg5 = −1.04 sin 5ω0t V; Vg5 = 1.04/180

H(j15) =j30

−224 + j30= 0.1327/− 82.37

VR5 = (1.04/180)(0.1327/− 82.37) = 138/97.63 mV

P5 =(0.138/

√2)2

2= 4.76 mW; therefore P ∼= P1

∼= 60.75 W

AP 16.8 Odd function with half- and quarter-wave symmetry, therefore av = 0, ak = 0 for allk, bk = 0 for k even; for k odd we have

bk =8T

∫ T/8

02 sin kω0t dt +

8T

∫ T/4

T/88 sin kω0t dt

=( 8

πk

) [1 + 3 cos

(kπ

4

)], k odd

Therefore Cn =(−j4

nπ

) [1 + 3 cos

(nπ

4

)], n odd


AP 16.9 [a] Irms =

√2T

[(2)2

(T

8

)(2) + (8)2

(3T8

− T

8

)]=

√34 = 5.831 A

[b] C1 =−j12.5

π; C3 =

j1.5π

; C5 =j0.9π

;

C7 =−j1.8

π; C9 =

−j1.4π

; C11 =j0.4π

Irms =

√√√√I2dc + 2

∞∑n=1,3,5

|Cn|2 ∼=√

2π2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42 + 0.42)

∼= 5.777 A

[c] % Error =5.777 − 5.831

5.831× 100 = −0.93%

[d] Using just the terms C1 – C9,

Irms =

√√√√I2dc + 2

∞∑n=1,3,5

|Cn|2 ∼=√

2π2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42)

∼= 5.774 A

% Error =5.774 − 5.831

5.831× 100 = −0.98%

Thus, the % error is still less than 1%.

AP 16.10 T = 32 ms, therefore 8 ms requires shifting the function T/4 to the right.

i =∞∑

n=−∞n(odd)

− j4

nπ

(1 + 3 cos

nπ

4

)ejnω0(t−T/4)

=4π

∞∑n=−∞

n(odd)

1n

(1 + 3 cos

nπ

4

)e−j(n+1)(π/2)ejnω0t

Problems 16–7

Problems

P 16.1 [a] ωoa =2π

200 × 10−6 = 31, 415.93 rad/s

ωob =2π

40 × 10−6 = 157.080 krad/s

[b] foa =1T

=1

200 × 10−6 = 5000 Hz; fob =1

40 × 10−6 = 25,000 Hz

[c] ava = 0; avb =100(10 × 10−6)

40 × 10−6 = 25 V

[d] The periodic function in Fig. P16.1(a) has half-wave symmetry. Therefore,

ava = 0; aka = 0 for k even; bka = 0 for k even

For k odd,

aka =4T

∫ T/4

040 cos

2πkt

Tdt +

4T

∫ T/2

T/480 cos

2πkt

Tdt

=160T

T

2πksin

2πkt

T

∣∣∣∣T/4

0+

320T

T

2πksin

2πkt

T

∣∣∣∣T/2

T/4

=80πk

sinπk

2+

160πk

(sin πk − sin

πk

2

)

= − 80πk

sinπk

2, k odd

bka =4T

∫ T/4

040 sin

2πkt

Tdt +

4T

∫ T/2

T/480 sin

2πkt

Tdt

=−160

T

T

2πkcos

2πkt

T

∣∣∣∣T/4

0− 320

T

T

2πkcos

2πkt

T

∣∣∣∣T/2

T/4

=−80πk

(0 − 1) − 160πk

(−1 − 0)

=240πk

The periodic function in Fig. P16.1(b) is even; therefore, bk = 0 for all k. Also,

avb = 25 V

akb =4T

∫ T/8

0100 cos

2πkt

Tdt

=400T

T

2πksin

2πk

Tt∣∣∣∣T/8

0

=200πk

sinπk

4


[e] For the periodic function in Fig. P16.1(a),

v(t) =80π

∞∑n=1,3,5

(− 1

nsin

nπ

2cos nωot +

3n

sin nωot)

V

For the periodic function in Fig. P16.1(b),

v(t) = 25 +200π

∞∑n=1

( 1n

sinnπ

4cos nωot

)V

P 16.2 In studying the periodic function in Fig. P16.2 note that it can be visualized as thecombination of two half-wave rectified sine waves, as shown in the figure below.Hence we can use the Fourier series for a half-wave rectified sine wave which isgiven as the answer to Problem 16.3(c).

v1(t) =100π

+ 50 sin ωot − 200π

∞∑n=2,4,6

cos nωot

(n2 − 1)V

v2(t) =60π

+ 30 sin ωo(t − T/2) − 120π

∞∑n=2,4,6

cos nωo(t − T/2)(n2 − 1)

V

Observe the following, noting that n is even:

sin ωo(t − T/2) = sin(ωot − 2π

T

T

2

)= sin(ωot − π) = − sin ωot

Problems 16–9

cos nωo(t − T/2) = cos(nωot − 2πn

T

T

2

)= cos(nωot − nπ) = cos nωot

Using the observations above,

v2(t) =60π

− 30 sin ωot − 120π

∞∑n=2,4,6

cos(nωot)(n2 − 1)

V

Thus,

v(t) = v1(t) + v2(t) =160π

+ 20 sin ωot − 320π

∞∑n=2,4,6

cos(nωot)(n2 − 1)

V

P 16.3 [a] Odd function with half- and quarter-wave symmetry, av = 0, ak = 0 for all k,bk = 0 for even k; for k odd we have

bk =8T

∫ T/4

0Vm sin kω0t dt =

4Vm

kπ, k odd

and v(t) =4Vm

π

∞∑n=1,3,5

1n

sin nω0t V

[b] Even function: bk = 0 for k

av =2T

∫ T/2

0Vm sin

π

Tt dt =

2Vm

π

ak =4T

∫ T/2

0Vm sin

π

Tt cos kω0t dt =

2Vm

π

( 11 − 2k

+1

1 + 2k

)

=4Vm/π

1 − 4k2

and v(t) =2Vm

π

[1 + 2

∞∑n=1

11 − 4n2 cos nω0t

]V

[c] av =1T

∫ T/2

0Vm sin

(2πT

)t dt =

Vm

π

ak =2T

∫ T/2

0Vm sin

2πT

t cos kω0t dt =Vm

π

(1 + cos kπ

1 − k2

)

Note: ak = 0 for k-odd, ak =2Vm

π(1 − k2)for k even,

bk =2T

∫ T/2

0Vm sin

2πT

t sin kω0t dt = 0 for k = 2, 3, 4, . . .

For k = 1, we have b1 =Vm

2; therefore

v(t) =Vm

π+

Vm

2sin ω0t +

2Vm

π

∞∑n=2,4,6

11 − n2 cos nω0t V


P 16.4 Starting with Eq. (16.2),

f(t) sin kω0t = av sin kω0t +∞∑

n=1an cos nω0t sin kω0t +

∞∑n=1

bn sin nω0t sin kω0t

Now integrate both sides from to to to + T. All the integrals on the right-hand sidereduce to zero except in the last summation when n = k, therefore we have∫ to+T

tof(t) sin kω0t dt = 0 + 0 + bk

(T

2

)or bk =

2T

∫ to+T

tof(t) sin kω0t dt

P 16.5 [a] I6 =∫ to+T

tosin mω0t dt = − 1

mω0cos mω0t

∣∣∣∣to+T

to

=−1mω0

[cos mω0(to + T ) − cos mω0to]

=−1mω0

[cos mω0to cos mω0T − sin mω0to sin mω0T − cos mω0to]

=−1mω0

[cos mω0to − 0 − cos mω0to] = 0 for all m,

I7 =∫ to+T

tocos mω0to dt =

1mω0

[sin mω0t]∣∣∣∣to+T

to

=1

mω0[sin mω0(to + T ) − sin mω0to]

=1

mω0[sin mω0to − sin mω0to] = 0 for all m

[b] I8 =∫ to+T

tocos mω0t sin nω0t dt =

12

∫ to+T

to[sin(m + n)ω0t − sin(m − n)ω0t] dt

But (m + n) and (m − n) are integers, therefore from I6 above, I8 = 0 for allm, n.

[c] I9 =∫ to+T

tosin mω0t sin nω0t dt =

12

∫ to+T

to[cos(m − n)ω0t − cos(m + n)ω0t] dt

If m = n, both integrals are zero (I7 above). If m = n, we get

I9 =12

∫ to+T

todt − 1

2

∫ to+T

tocos 2mω0t dt =

T

2− 0 =

T

2

[d] I10 =∫ to+T

tocos mω0t cos nω0t dt

=12

∫ to+T

to[cos(m − n)ω0t + cos(m + n)ω0t] dt

If m = n, both integrals are zero (I7 above). If m = n, we have

I10 =12

∫ to+T

todt +

12

∫ to+T

tocos 2mω0t dt =

T

2+ 0 =

T

2

Problems 16–11

P 16.6 av =1T

∫ to+T

tof(t) dt =

1T

∫ 0

−T/2f(t) dt +

∫ T/2

0f(t) dt

Let t = −x, dt = −dx, x =T

2when t =

−T

2

and x = 0 when t = 0

Therefore1T

∫ 0

−T/2f(t) dt =

1T

∫ 0

T/2f(−x)(−dx) = − 1

T

∫ T/2

0f(x) dx

Therefore av = − 1T

∫ T/2

0f(t) dt +

1T

∫ T/2

0f(t) dt = 0

ak =2T

∫ 0

−T/2f(t) cos kω0t dt +

2T

∫ T/2

0f(t) cos kω0t dt

Again, let t = −x in the first integral and we get

2T

∫ 0

−T/2f(t) cos kω0t dt = − 2

T

∫ T/2

0f(x) cos kω0x dx

Therefore ak = 0 for all k.

bk =2T

∫ 0

−T/2f(t) sin kω0t +

2T

∫ T/2

0f(t) sin kω0t dt

Using the substitution t = −x, the first integral becomes

2T

∫ T/2

0f(x) sin kω0x dx

Therefore we have bk =4T

∫ T/2

0f(t) sin kω0t dt

P 16.7 bk =2T

∫ 0

−T/2f(t) sin kω0t dt +

2T

∫ T/2

0f(t) sin kω0t dt

Now let t = x − T/2 in the first integral, then dt = dx, x = 0 when t = −T/2 andx = T/2 when t = 0, also sin kω0(x − T/2) = sin(kω0x − kπ) = sin kω0x cos kπ.Therefore

2T

∫ 0

−T/2f(t) sin kω0t dt = − 2

T

∫ T/2

0f(x) sin kω0x cos kπ dx and

bk =2T

(1 − cos kπ)∫ T/2

0f(x) sin kω0t dt

Now note that 1 − cos kπ = 0 when k is even, and 1 − cos kπ = 2 when k is odd.Therefore bk = 0 when k is even, and

bk =4T

∫ T/2

0f(t) sin kω0t dt when k is odd


P 16.8 Because the function is even and has half-wave symmetry, we have av = 0, ak = 0for k even, bk = 0 for all k and

ak =4T

∫ T/2

0f(t) cos kω0t dt, k odd

The function also has quarter-wave symmetry;therefore f(t) = −f(T/2 − t) in the interval T/4 ≤ t ≤ T/2;thus we write

ak =4T

∫ T/4

0f(t) cos kω0t dt +

4T

∫ T/2

T/4f(t) cos kω0t dt

Now let t = (T/2 − x) in the second integral, then dt = −dx, x = T/4 whent = T/4 and x = 0 when t = T/2. Therefore we get

4T

∫ T/2

T/4f(t) cos kω0t dt = − 4

T

∫ T/4

0f(x) cos kπ cos kω0x dx

Therefore we have

ak =4T

(1 − cos kπ)∫ T/4

0f(t) cos kω0t dt

But k is odd, hence

ak =8T

∫ T/4


P 16.9 Because the function is odd and has half-wave symmetry, av = 0, ak = 0 for all k,and bk = 0 for k even. For k odd we have

bk =4T

∫ T/2

0f(t) sin kω0t dt

The function also has quarter-wave symmetry, therefore f(t) = f(T/2 − t) in theinterval T/4 ≤ t ≤ T/2. Thus we have

bk =4T

∫ T/4

0f(t) sin kω0t dt +

4T

∫ T/2

T/4f(t) sin kω0t dt

Now let t = (T/2 − x) in the second integral and note that dt = −dx, x = T/4when t = T/4 and x = 0 when t = T/2, thus

4T

∫ T/2

T/4f(t) sin kω0t dt = − 4

Tcos kπ

∫ T/4

0f(x)(sin kω0x) dx

But k is odd, therefore the expression becomes

bk =8T

∫ T/4

0f(t) sin kω0t dt

Problems 16–13

P 16.10 [a] f =1T

=1

16 × 10−3 = 62.5 Hz

[b] no, because f(3 ms) = 10 mA but f(−3 ms) = −10 mA.

[c] yes, because f(−t) = −f(t) for all t.

[d] yes

[e] yes

[f] av = 0, function is odd

ak = 0, for all k; the function is odd

bk = 0, for k even, the function has half-wave symmetry

bk =8T

∫ T/4

0f(t) sin kωot, k odd

=8T

∫ T/8

05t sin kωot dt +

∫ T/4

T/80.01 sin kωot dt

=8T

Int1 + Int2

Int1 = 5∫ T/8

0t sin kωot dt

= 5[

1k2ω2

o

sin kωot − t

kωo

cos kωot

∣∣∣∣T/8

0

]

=5

k2ω2o

sinkπ

4− 0.625T

kωo

coskπ

4

Int2 = 0.01∫ T/4

T/8sin kωot dt =

−0.01kωo

cos kωot

∣∣∣∣T/4

T/8=

0.01kωo

coskπ

4

Int1 + Int2 =5

k2ω2o

sinkπ

4+(0.01

kωo

− 0.625Tkωo

)cos

kπ

4

0.625T = 0.625(16 × 10−3) = 0.01

·. . Int1 + Int2 =5

k2ω2o

sinkπ

4

bk =[ 8T

· 54π2k2 · T 2

]sin

kπ

4=

0.16π2k2 sin

kπ

4, k odd

i(t) =160π2

∞∑n=1,3,5

sin(nπ/4)n2 sin nωot mA


P 16.11 [a] T = 1; ωo =2πT

= 2π rad/s

[b] yes

[c] no

[d] no

P 16.12 [a] v(t) is even and has both half- and quarter-wave symmetry, therefore av = 0,bk = 0 for all k, ak = 0 for k-even; for odd k we have

ak =8T

∫ T/4

0Vm cos kω0t dt =

4Vm

πksin

(kπ

2

)

v(t) =4Vm

π

∞∑n=1,3,5

[ 1n

sinnπ

2

]cos nω0t V

[b] v(t) is even and has both half- and quarter-wave symmetry, therefore av = 0,ak = 0 for k-even, bk = 0 for all k; for k-odd we have

ak =8T

∫ T/4

0

(4Vp

Tt − Vp

)cos kω0t dt = − 8Vp

π2k2

Therefore v(t) = −8Vp

π2

∞∑n=1,3,5

1n2 cos nω0t V

P 16.13 [a] i(t) is even, therefore bk = 0 for all k.

av =12

· T

4· Im · 2 · 1

T=

Im

4A

ak =4T

∫ T/4

0

(Im − 4Im

Tt)

cos kωot dt

=4Im

T

∫ T/4

0cos kωot dt − 16Im

T 2

∫ T/4

0t cos kωot dt

= Int1 − Int2

Int1 =4Im

T

∫ T/4

0cos kωot dt =

2Im

πksin

kπ

2

Int2 =16Im

T 2

∫ T/4

0t cos kωot dt

=16Im

T 2

1

k2ω2o

cos kωot +t

kωo

sin kωot

∣∣∣∣T/4

0

=4Im

π2k2

(cos

kπ

2− 1

)+

2Im

kπsin

kπ

2

Problems 16–15

·. . ak =4Im

π2k2

(1 − cos

kπ

2

)A

·. . i(t) =Im

4+

4Im

π2

∞∑n=1

1 − cos(nπ/2)n2 cos nωot A

[b] Shifting the reference axis to the left is equivalent to shifting the periodicfunction to the right:

cos nωo(t − T/2) = cos nπ cos nωot

Thus

i(t) =Im

4+

4Im

π2

∞∑n=1

(1 − cos(nπ/2)) cos nπ

n2 cos nωot A

P 16.14 [a]

[b] Even, since f(t) = f(−t)

[c] Yes, since f(t) = −f(T/2 − t) in the interval 0 < t < 4.

[d] av = 0, ak = 0, for k even (half-wave symmetry)

bk = 0, for all k (function is even)

Because of the quarter-wave symmetry, the expression for ak is

ak =8T

∫ T/4


=88

∫ 2

04t2 cos kω0t dt = 4

[2t

k2ω20

cos kω0t +k2ω2

0t2 − 2

k3ω30

sin kω0t

]2

0


kω0(2) = k(2π

8

)(2) =

kπ

2

cos(kπ/2) = 0, since k is odd

·. . ak = 4[0 +

4k2ω20 − 2

k3ω30

sin(kπ/2)]

=16k2ω2

0 − 8k3ω3

0sin(kπ/2)

ω0 =2π8

=π

4; ω2

0 =π2

16; ω3

0 =π3

64

ak =(

k2π2 − 8k3π3

)(64) sin(kπ/2)

f(t) = 64∞∑

n=1,3,5

[n2π2 − 8

π3n3

]sin(nπ/2) cos(nω0t)

[e] cos nω0(t − 2) = cos(nω0t − π/2) = sin nω0t sin(nπ/2)

f(t) = 64∞∑

n=1,3,5

[n2π2 − 8

π3n3

]sin2(nπ/2) sin(nω0t)

P 16.15 [a]

[b] Odd, since f(−t) = −f(t)

[c] f(t) has quarter-wave symmetry, since f(T/2 − t) = f(t) in the interval0 < t < 4.

[d] av = 0, (half-wave symmetry); ak = 0, for all k (function is odd)

bk = 0, for k even (half-wave symmetry)

bk =8T

∫ T/4

0f(t) sin kω0t dt, k odd

=88

∫ 2

0t3 sin kω0t dt

Problems 16–17

=[

3t2

k2ω20

sin kω0t − 6k4ω4

0sin kω0t − t3

kω0cos kω0t +

6tk3ω3

0cos kω0t

]2

0

kω0(2) = k(2π

8

)(2) =

kπ

2

cos(kπ/2) = 0, since k is odd

·. . bk =[

12k2ω2

0sin(kπ/2) − 6

k4ω40

sin(kπ/2)]

kω0 = k(2π

8

)=

kπ

4; k2ω2

0 =k2π2

16; k4ω4

0 =k4π4

256

·. . bk =192π2k2

[1 − 8

π2k2

]sin(kπ/2), k odd

f(t) =192π2

∞∑n=1,3,5

[ 1n2

(1 − 8

π2n2

)sin(nπ/2)

]sin nω0t

[e] sin nω0(t − 2) = sin(nω0t − π/2) = − cos nω0t sin(nπ/2)

f(t) =−192π2

∞∑n=1,3,5

[ 1n2

(1 − 8

π2n2

)sin2(nπ/2)

]cos nω0t

P 16.16 [a]

[b] av = 0; ak = 0, for k even; bk = 0, for all k

ak =8T

∫ T/4

0f(t) cos kω0t dt, for k odd

=8T

∫ T/8

0

120tT

cos kω0t dt +8T

∫ T/4

T/8

(10 +

40T

t)

cos kω0t dt

=960T 2

∫ T/8

0t cos kω0t dt +

80T

∫ T/4

T/8cos kω0t dt +

320T 2

∫ T/4

T/8t cos kω0t dt


=960T 2

[cos kω0t

k2ω20

+t sin kω0t

kω0

]T/8

0

+80T

sin kω0t

kω0

∣∣∣∣T/4

T/8

+320T 2

[cos kω0t

k2ω20

+t sin kω0t

kω0

]T/4

T/8

kω0T

4=

kπ

2; kω0

T

8=

kπ

4

bk =960T 2

[cos(kπ/4)

k2ω20

+T

8kω0sin(kπ/4) − 1

k2ω20

]+

80kω0T

[sin(kπ/2) − sin(kπ/4)]

+320T 2

[cos(kπ/2)

k2ω20

+T

4sin(kπ/2)

kω0− cos(kπ/4)

k2ω20

− T sin(kπ/4)8kω0

]

=640

(kω0T )2 cos(kπ/4) +160

kω0T 2 sin(kπ/2) − 960(kω0T )2

kω0T = 2kπ; (kω0T )2 = 4k2π2

ak =160π2k2 cos(kπ/4) +

80πk

sin(kπ/2) − 240π2k2

[c] ak =80

π2k2 [2 cos(kπ/4) + πk sin(kπ/2) − 3]

a1 =80π2 [2 cos(π/4) + πk sin(π/2) − 3] ∼= 12.61

a3 =809π2 [2 cos(3π/4) + πk sin(3π/2) − 3] ∼= −12.46

a5 =80

25π2 [2 cos(5π/4) + πk sin(5π/2) − 3] ∼= 3.66

f(t) = 12.61 cos(ω0t) − 12.46 cos(3ω0t) + 3.66 cos(5ω0t) + . . .

[d] t =T

4; ω0t =

2πT

· T

4=

π

2

f(T/4) ∼= 12.61 cos(π/2) − 12.46 cos(3π/2) + 3.66 cos(5π/2) = 0

The result would have been non-trivial for t = T/8 or if the function had beenspecified as odd.

Problems 16–19

P 16.17 Let f(t) = v2(t − T/6).

av = −(2Vm/3)(T/3)(1/T ) = −(2Vm/9) and bk = 0 since f(t) is even

ak =4T

∫ T/6

0

(−2Vm

3

)cos kωotdt = − 4

T

2Vm

31

kωo

sin kωot∣∣∣∣T/6

0

= − 8Vm

3k2πsin

(kπ

3

)= −4Vm

3kπsin

(kπ

3

)

Therefore, v2(t − T/6) = −2Vm

9− 4Vm

3π

∞∑n=1

1n

sin(

nπ

3

)cos nωot

and v2(t) = −2Vm

9− 4Vm

3π

∞∑n=1

1n

sin(

nπ

3

)cos nωo(t + T/6)

Then, v(t) = v1(t) + v2(t). Simplifying,

v(t) =7Vm

9− 4Vm

3π

∞∑n=1

1n

[sin

(nπ

3

)cos

(nπ

3

)]cos nωot

+4Vm

3π

∞∑n=1

1n

[sin2

(nπ

3

)]sin nωot V

If Vm = 9π then av = 7π = 21.99 (Checks)

ak = −(12

n

)sin

(nπ

3

)cos

(nπ

3

)= −

(12n

)(12

)sin

(2nπ

3

)=( 6

n

)sin

(4nπ

3

)

bk =(12

n

)sin2

(nπ

3

)=(12

n

)(12

) [1 − cos

(2nπ

3

)]=( 6

n

) [1 − cos

(4nπ

3

)]

a1 = 6 sin(4π/3) = −5.2; b1 = 6[1 − cos(4π/3)] = 9

a2 = 3 sin(8π/3) = 2.6; b2 = 3[1 − cos(8π/3)] = 4.5

a3 = 2 sin(12π/3) = 0; b3 = 2[1 − cos(12π/3)] = 0

a4 = 1.5 sin(16π/3) = −1.3; b4 = 1.5[1 − cos(16π/3)] = 2.25

a5 = 1.2 sin(20π/3) = 1.04; b5 = 1.2[1 − cos(20π/3)] = 1.8

All coefficients check!


P 16.18 [a] The voltage has half-wave symmetry. Therefore,

av = 0; ak = bk = 0, k even

For k odd,

ak =4T

∫ T/2

0

(Im − 2Im

Tt)

cos kωot dt

=4T

∫ T/2

0Im cos kω0t dt − 8Im

T 2

∫ T/2

0t cos kω0t dt

=4Im

T

sin kω0t

kω0

∣∣∣∣T/2

0−8Im

T 2

[cos kωot

k2ω20

+t

kω0sin kω0T

]T/2

0

= 0 − 8Im

T 2

[cos kπ

k2ω20

− 1k2ω2

0

]

=(8Im

T 2

)( 1k2ω2

0

)(1 − cos kπ)

=4Im

π2k2 =20k2 , for k odd

bk =4T

∫ T/2

0

(Im − 2Im

Tt)

sin kωot dt

=4Im

T

∫ T/2

0sin kω0t dt − 8Im

T 2

∫ T/2

0t sin kω0t dt

=4Im

T

[− cos kω0t

kω0

]T/2

0

− 8Im

T 2

[sin kωot

k2ω20

− t

kω0cos kω0t

]T/2

0

=4Im

T

[1 − cos kπ

kω0

]− 8Im

T 2

[−T cos kπ

2kω0

]

=8Im

kω0T

[1 +

12

cos kπ]

=2Im

πk=

10πk

, for k odd

ak − jbk =20k2 − j

10πk

=10k

(2k

− jπ)

=10k2

√π2k2 + 4/− θk

where tan θk =πk

2

i(t) = 10∞∑

n=1,3,5

√(nπ)2 + 4

n2 cos(nω0t − θn)

Problems 16–21

[b] A1 = 10√

4 + π2 ∼= 37.24 A tan θ1 =π

2θ1

∼= 57.52

A3 =109

√4 + 9π2 ∼= 10.71 A tan θ3 =

3π2

θ3∼= 78.02

A5 =1025

√4 + 25π2 ∼= 6.33 A tan θ5 =

5π2

θ5∼= 82.74

A7 =1049

√4 + 49π2 ∼= 4.51 A tan θ7 =

7π2

θ7∼= 84.80

A9 =1081

√4 + 81π2 ∼= 3.50 A tan θ9 =

9π2

θ9∼= 85.95

i(t) ∼= 37.24 cos(ωot − 57.52) + 10.71 cos(3ωot − 78.02)

+ 6.33 cos(5ωot − 82.74) + 4.51 cos(7ωot − 84.80)

+ 3.50 cos(9ωot − 85.95) + . . .

i(T/4) ∼= 37.24 cos(90 − 57.52) + 10.71 cos(270 − 78.02)

+ 6.33 cos(450 − 82.74) + 4.51 cos(630 − 84.80)

+ 3.50 cos(810 − 85.95) ∼= 26.22 A

Actual value:

i(

T

4

)=

12(5π2) ∼= 24.67 A

P 16.19 The function has half-wave symmetry, thus ak = bk = 0 for k-even, av = 0; fork-odd

ak =4T

∫ T/2

0Vm cos kω0t dt − 8Vm

ρT

∫ T/2

0e−t/RC cos kω0t dt

where ρ =[1 + e−T/2RC

].

Upon integrating we get

ak =4Vm

T

sin kω0t

kω0

∣∣∣∣T/2

0

−8Vm

ρT· e−t/RC

(1/RC)2 + (kω0)2 ·[− cos kω0t

RC+ kω0 sin kω0t

] ∣∣∣∣T/2

0

=−8VmRC

T [1 + (kω0RC)2]


bk =4T

∫ T/2

0Vm sin kω0t dt − 8Vm

ρT

∫ T/2

0e−t/RC sin kω0t dt

= −4Vm

T

cos kω0t

kω0

∣∣∣∣T/2

0

−8Vm

ρT· −e−t/RC

(1/RC)2 + (kω0)2 ·[sin kω0t

RC+ kω0 cos kω0t

] ∣∣∣∣T/2

0

=4Vm

πk− 8kω0VmR2C2

T [1 + (kω0RC)2]

P 16.20 [a] a2k + b2

k = a2k +

(4Vm

πk+ kω0RCak

)2

= a2k [1 + (kω0RC)2] + 8Vm

πk

[2Vm

πk+ kω0RCak

]

But ak =−8VmRC

T [1 + (kω0RC)2]

Therefore a2k =

64V 2mR2C2

T 2[1 + (kω0RC)2]2, thus we have

a2k + b2

k =64V 2

mR2C2

T 2[1 + (kω0RC)2]+

16V 2m

π2k2 − 64V 2mkω0R

2C2

πkT [1 + (kω0RC)2]

Now let α = kω0RC and note that T = 2π/ω0, thus the expression for a2k + b2

k

reduces to a2k + b2

k = 16V 2m/π2k2(1 + α2). It follows that

√a2

k + b2k =

4Vm

πk√

1 + (kω0RC)2

[b] bk = kω0RCak +4Vm

πk

Thusbk

ak

= kω0RC +4Vm

πkak

= α − 1 + α2

α= − 1

α

Thereforeak

bk

= −α = −kω0RC

Problems 16–23

P 16.21 Since av = 0 (half-wave symmetry), Eq. 16.38 gives us

vo(t) =∞∑

1,3,5

4Vm

nπ

1√1 + (nω0RC)2

cos(nω0t − θn) where tan θn =bn

an

But from Eq. 16.57, we have tan βk = kω0RC. It follows from Eq. 16.72 thattan βk = −ak/bk or tan θn = − cot βn. Therefore θn = 90 + βn andcos(nω0t − θn) = cos(nω0t − βn − 90) = sin(nω0t − βn), thus ourexpression for vo becomes

vo =4Vm

π

∞∑n=1,3,5

sin(nω0t − βn)

n√

1 + (nω0RC)2

P 16.22 [a] e−x ∼= 1 − x for small x; therefore

e−t/RC ∼=(1 − t

RC

)and e−T/2RC ∼=

(1 − T

2RC

)

vo∼= Vm − 2Vm[1 − (t/RC)]

2 − (T/2RC)=(

Vm

RC

) [ 2t − (T/2)2 − (T/2RC)

]

∼=(

Vm

RC

)(t − T

4

)=(

Vm

RC

)t − VmT

4RCfor 0 ≤ t ≤ T

2

[b] ak =( −8

π2k2

)Vp =

( −8π2k2

)(VmT

4RC

)=

−4Vm

πω0RCk2

P 16.23 [a] Express vg as a constant plus a symmetrical square wave. The constant is Vm/2and the square wave has an amplitude of Vm/2, is odd, and has half- andquarter-wave symmetry. Therefore the Fourier series for vg is

vg =Vm

2+

2Vm

π

∞∑n=1,3,5

1n

sin nω0t

The dc component of the current is Vm/2R, and withsin nω0t = cos(nω0t − 90) the kth harmonic phase current is

Ik =2Vm/kπ

R + jkω0L/− 90 =2Vm

kπ√

R2 + (kω0L)2/− 90 − θk

where θk = tan−1

(kω0L

R

)

Thus the Fourier series for the steady-state current is

i =Vm

2R+

2Vm

π

∞∑n=1,3,5

sin(nω0t − θn)

n√

R2 + (nω0L)2A


[b]

The steady-state current will alternate between I1 and I2 in exponential tracesas shown. Assuming t = 0 at the instant i increases toward (Vm/R), we have

i =Vm

R+(I1 − Vm

R

)e−t/τ for 0 ≤ t ≤ T

2

and i = I2e−[t−(T/2)]/τ for T/2 ≤ t ≤ T, where τ = L/R. Now we solve for I1

and I2 by noting that

I1 = I2e−T/2τ and I2 =

Vm

R+(I1 − Vm

R

)e−T/2τ

These two equations are now solved for I1. Letting x = T/2τ, we get

I1 =(Vm/R)e−x

1 + e−x

Therefore the equations for i become

i =Vm

R−[

Vm

R(1 + e−x)

]e−t/τ for 0 ≤ t ≤ T

2and

i =[

Vm

R(1 + e−x)

]e−[t−(T/2)]/τ for

T

2≤ t ≤ T

A check on the validity of these expressions shows they yield an average valueof (Vm/2R):

Iavg =1T

∫ T/2

0

[Vm

R+(I1 − Vm

R

)e−t/τ

]dt +

∫ T

T/2I2e

−[t−(T/2)]/τ dt

=1T

VmT

2R+ τ(1 − e−x)

(I1 − Vm

R+ I2

)

=Vm

2Rsince I1 + I2 =

Vm

R

Problems 16–25

P 16.24 vi =4Aπ

∞∑n=1,3,5

1n

sin nω0(t + T/4)

=4Aπ

∞∑n=1,3,5

( 1n

sinnπ

2

)cos nω0t

ω0 =2π4π

× 103 = 500 rad/s;4Aπ

= 60

vi = 60∞∑

n=1,3,5

( 1n

sinnπ

2

)cos 500nt V

From the circuit

Vo =Vi

R + jωL· jωL =

jω

R/L + jωVi =

jω

1000 + jωVi

Vi1 = 60/0 V; ω = 500 rad/s

Vi3 = −20/0 = 20/180 V; 3ω = 1500 rad/s

Vi5 = 12/0 V; 5ω = 2500 rad/s

Vo1 =j500

1000 + j500(60/0) = 26.83/63.43 V

Vo3 =j1500

1000 + j1500(20/180) = 16.64/− 146.31 V

Vo5 =j2500

1000 + j2500(12/0) = 11.14/21.80 V

·. . vo = 26.83 cos(500t + 63.43) + 16.64 cos(1500t − 146.31)

+ 11.14 cos(2500t + 21.80) + . . . V

P 16.25 [a] From the solution to Assessment Problem 16.6 the Fourier series for the inputvoltage is

vg = 42∞∑

n=1,3,5

[ 1n

sin(

nπ

2

)]cos 2000nt V

Also from the solution to Assessment Problem 16.6 we have

Vg1 = 42/0 ω0 = 2000 rad/s


Vg3 = 14/180 3ω0 = 6000 rad/s

Vg5 = 8.4/0 5ω0 = 10,000 rad/s

Vg7 = 6/180 7ω0 = 14,000 rad/s

From the circuit in Fig. P16.26 we have

Vo

R+

Vo − Vg

sL+ (Vo − Vg)sC = 0

·. .Vo

Vg

= H(s) =s2 + 1/LC

s2 + (s/RC) + (1/LC)

Substituting in the numerical values gives

H(s) =s2 + 108

s2 + 500s + 108

H(j2000) =96

96 + j1= 0.9999/− 0.60

H(j6000) =64

64 + j3= 0.9989/− 2.68

H(j10,000) = 0

H(j14,000) =96

96 − j7= 0.9974/4.17

Vo1 = (42/0)(0.9999/− 0.60) = 41.998/− 0.60 V

Vo3 = (14/180)(0.9989/− 2.68) = 13.985/177.32 V

Vo5 = 0 V

Vo7 = (6/180)(0.9974/4.17) = 5.984/184.17 V

vo = 41.998 cos(2000t − 0.60) + 13.985 cos(6000t + 177.32)

+ 5.984 cos(14,000t + 184.17) + . . . V

[b] The 5th harmonic at the frequency√

1/LC = 10,000 rad/s has been eliminatedfrom the output voltage by the circuit, which is a bandreject filter with a centerfrequency of 10,000 rad/s.

P 16.26 [a] Note – find io(t)

V0 − Vg

16s+ V0(12.5 × 10−6s) +

V0

1000= 0

V0

[ 116s

+ 12.5 × 10−6s +1

1000

]=

Vg

16s

Problems 16–27

V0(1000 + 0.2s2 + 16s) = 1000Vg

V0 =5000Vg

s2 + 80s + 5000

I0 =V0

1000=

5Vg

s2 + 80s + 5000

H(s) =I0

Vg

=5

s2 + 80s + 5000

H(njω0) =5

(5000 − n2ω20) + j80nω0

ω0 =2πT

= 240π; ω20 = 57,600π2; 80ω0 = 19,200π

H(jnω0) =5

(5000 − 57,600π2n2) + j19,200πn

H(0) = 10−3

H(jω0) = 8.82 × 10−6/− 173.89

H(j2ω0) = 2.20 × 10−6/− 176.96

H(j3ω0) = 9.78 × 10−7/− 177.97

H(j4ω0) = 5.5 × 10−7/− 178.48

vg =680π

− 1360π

[13

cos ω0t +115

cos 2ω0t +135

cos 3ω0t +163

cos 4ω0t + . . .]

i0 =680π

× 10−3 − 13603π

(8.82 × 10−6) cos(ω0t − 173.89)

− 136015π

(2.20 × 10−6) cos(2ω0t − 176.96)

− 136035π

(9.78 × 10−7) cos(3ω0t − 177.97)

− 136063π

(5.5 × 10−7) cos(4ω0t − 178.48) − . . .

= 216.45 × 10−3 − 1.27 × 10−3 cos(ω0t − 173.89)

− 6.35 × 10−5 cos(2ω0t − 176.96)

− 1.21 × 10−5 cos(3ω0t − 177.97)

− 3.8 × 10−6 cos(4ω0t − 178.48) − . . .


i0 ∼= 216.45 − 1.27 cos(ω0t − 173.89) mA

Note that the sinusoidal component is very small compared to the dccomponent, so

i0 ∼= 216.45 mA (a dc current)

[b] Yes, the solution makes sense. The circuit is a low-pass filter which nearlyeliminates all but the dc component.

P 16.27 The function is odd with half-wave and quarter-wave symmetry. Therefore,

ak = 0, for all k; the function is odd

bk = 0, for k even, the function has half-wave symmetry

bk =8T

∫ T/4

0f(t) sin kωot, k odd

=8T

∫ T/10

0500t sin kωot dt +

∫ T/4

T/10sin kωot dt

=8T

Int1 + Int2

Int1 = 500∫ T/10

0t sin kωot dt

= 500[

1k2ω2

o

sin kωot − t

kωo

cos kωot

∣∣∣∣T/10

0

]

=500k2ω2

o

sinkπ

5− 50T

kωo

coskπ

5

Int2 =∫ T/4

T/10sin kωot dt =

−1kωo

cos kωot

∣∣∣∣T/4

T/10=

1kωo

coskπ

5

Int1 + Int2 =500k2ω2

o

sinkπ

5+( 1

kωo

− 50Tkωo

)cos

kπ

5

50T = 50(20 × 10−3) = 1

·. . Int1 + Int2 =500k2ω2

o

sinkπ

5

bk =[ 8T

· 5004π2k2 · T 2

]sin

kπ

5=

20π2k2 sin

kπ

5, k odd

Problems 16–29

i(t) =20π2

∞∑n=1,3,5

sin(nπ/5)n2 sin nωot A

From the circuit,

H(s) =Vo

Ig

= Zeq

Yeq =1R1

+1

R2 + sL+ sC

Zeq =1/C(s + R2/L)

s2 + s(R1R2C + L)/R1LC + (R1 + R2)/R1LC

Therefore,

H(s) =320 × 104(s + 32 × 104)

s2 + 32.8 × 104s + 28.8 × 108

We want the output for the third harmonic:

ω0 =2πT

=2π

20 × 10−3 = 100π; 3ω0 = 300π

Ig3 =209π2 sin

3π5 sin 3ω0t

= 0.214/− 90

H(j300π) =320 × 104(j300π + 32 × 104)

(j300π)2 + 32.8 × 104(j300π) + 28.8 × 108 = 353.6/− 5.96

Therefore,

Vo3 = H(j300π)Ig3 = (353.6/− 5.96)(0.214/− 90) = 75.7/− 90 − 5.96 V

vo3 = 75.7 sin(300πt − 5.96) V

P 16.28 ωo =2πT

=2π10π

× 106 = 200 krad/s

·. . n =3 × 106

0.2 × 106 = 15; n =5 × 106

0.2 × 106 = 25

H(s) =Vo

Vg

=(1/RC)s

s2 + (1/RC)s + (1/LC)


1RC

=1012

(250 × 103)(4)= 106;

1LC

=(103)(1012)

(10)(4)= 25 × 1012

H(s) =106s

s2 + 106s + 25 × 1012

H(jω) =jω × 106

(25 × 1012 − ω2) + j106ω

15th harmonic input:

vg15 = (150)(1/15) sin(15π/2) cos 15ωot = −10 cos 3 × 106t V

·. . Vg15 = 10/− 180 V

H(j3 × 106) =j3

16 + j3= 0.1843/79.38

Vo15 = (10)(0.1843)/− 100.62 V

vo15 = 1.84 cos(3 × 106t − 100.62) V

25th harmonic input:

vg25 = (150)(1/25) sin(25π/2) cos 5 × 106t = 6 cos 5 × 106t V

·. . Vg25 = 6/0 V

H(j5 × 106) =j5

0 + j5= 1/0

Vo25 = 6/0 V

vo25 = 6 cos 5 × 106t V

P 16.29 [a] av =T

2

[12

(T

2

)Im +

T

2Im

]=

3Im

4

i(t) =2Im

Tt, 0 ≤ t ≤ T/2

i(t) = Im, T/2 ≤ t ≤ T

ak =2T

∫ T/2

0

2Im

Tt cos kωot dt +

2T

∫ T

T/2Im cos kωot dt

Problems 16–31

=Im

π2k2 (cos kπ − 1)

bk =2T

∫ T/2

0

2Im

Tt sin kωot dt +

2T

∫ T

T/2Im sin kωot dt

=−Im

πk

av =3Im

4, a1 =

−2Im

π2 , a2 = 0

a3 =−2Im

9π2

b1 =−Im

π, b2 =

−Im

2π

·. . Irms = Im

√916

+2π4 +

12π2 +

18π2 = 0.8040Im (Eq. 16.81)

Irms = 192.95 mA

P = (0.19295)2(1000) = 37.23 W

[b] Area under i2:

A =∫ T/2

0

4I2m

T 2 t dt + I2m

T

2

=4I2

m

T 2

t3

3

∣∣∣∣T/2

0+I2

m

T

2

= I2mT

[16

+36

]=

23TI2

m

Irms =

√1T

· 23TI2

m =

√23Im = 195.96 mA

P = (0.19596)21000 = 38.4 W

[c] Error =(37.23

38.40− 1

)(100) = −3.05%

P 16.30 vg = 10 +80π2

∞∑n=1,3,5

1n2 cos nωot V

ωo =2πT

=2π4π

× 103 = 500 rad/s

vg = 10 +80π2 cos 500t +

809π2 cos 1500t + . . .


Vo − Vg

sL+ sCVo +

Vo

R= 0

Vo(RLCs2 + Ls + R) = RVg

H(s) =Vo

Vg

=1/LC

s2 + s/RC + 1/LC

1LC

=106

(0.1)(10)= 106

1RC

=106

(50√

2)(10)= 1000

√2

H(s) =106

s2 + 1000√

2s + 106

H(jω) =106

106 − ω2 + j1000ω√

2

H(j0) = 1

H(j500) = 0.9701/− 43.31

H(j1500) = 0.4061/− 120.51

vo = 10(1) +80π2 (0.9701) cos(500t − 43.31)

+809π2 (0.4061) cos(1500t − 120.51) + . . .

vo = 10 + 7.86 cos(500t − 43.31) + 0.3658 cos(1500t − 120.51) + . . .

Problems 16–33

Vrms∼=√√√√102 +

(7.86√

2

)2

+(

0.3658√2

)2

= 11.44 V

P ∼= V 2rms

50√

2= 1.85 W

Note – the higher harmonics are severely attenuated and can be ignored. Forexample, the 5th harmonic component of vo is

vo5 = (0.1580)( 80

25π2

)cos(2500t − 146.04) = 0.0512 cos(2500t − 146.04) V

P 16.31 [a] av =2(

12

T4 Vm

)T

=Vm

4

ak =4T

∫ T/4

0

[Vm − 4Vm

Tt]cos kωot dt

=4Vm

π2k2

[1 − cos

kπ

2

]

bk = 0, all k

av =604

= 15 V

a1 =240π2

a2 =2404π2 (1 − cos π) =

120π2

Vrms =

√√√√(15)2 +12

[(240π2

)2

+(120

π2

)2]

= 24.38 V

P =(24.38)2

10= 59.46 W

[b] Area under v2; 0 ≤ t ≤ T/4

v2 = 3600 − 28,800T

t +57,600

T 2 t2

A = 2∫ T/4

0

[3600 − 28,800

Tt +

57,600T 2 t2

]dt = 600T

Vrms =

√1T

600T =√

600 = 24.49 V

P =√

6002/10 = 60 W


[c] Error =(59.46

60.00− 1

)100 = −0.9041%

P 16.32 [a] v = 15 + 400 cos 500t + 100 cos(1500t − 90) V

i = 2 + 5 cos(500t − 30) + 3 cos(1500t − 15) A

P = (15)(2) +12(400)(5) cos(30) +

12(100)(3) cos(−75) = 934.85 W

[b] Vrms =

√√√√(15)2 +(

400√2

)2

+(

100√2

)2

= 291.93 V

[c] Irms =

√√√√(2)2 +(

5√2

)2

+(

3√2

)2

= 4.58 A

P 16.33 [a] Area under v2 = A = 4∫ T/6

0

36V 2m

T 2 t2 dt + 2V 2m

(T

3− T

6

)

=2V 2

mT

9+

V 2mT

3

Therefore Vrms =

√√√√ 1T

(2V 2

mT

9+

V 2mT

3

)= Vm

√29

+13

= 74.5356 V

[b] vg = 105.30 sin ω0t − 4.21 sin 5ω0t + 2.15 sin 7ω0t + · · · V

Therefore Vrms∼=√

(105.30)2 + (4.21)2 + (2.15)2

2= 74.5306 V

P 16.34 [a] v(t) =480π

sin ωot +13

sin 3ωot +15

sin 5ωot +17

sin 7ωot +19

sin 9ωot + · · ·

Vrms∼= 480

π

√√√√( 1√2

)2

+(

13√

2

)2

+(

15√

2

)2

+(

17√

2

)2

+(

19√

2

)2

=480π√

2

√1 +

19

+125

+149

+181

∼= 117.55 V

[b] % error =(117.55

120− 1

)(100) = −2.04%

[c] v(t) =960π2

sin ωot +

19

sin 3ωot +125

sin 5ωot

+149

sin 7ωot +181

sin 9ωot − · · ·

Vrms∼= 960

π2√

2

√1 +

181

+1

625+

12401

+1

6561

∼= 69.2765 V

Problems 16–35

Vrms =120√

3= 69.2820 V

% error =(69.2765

69.2820− 1

)(100) = −0.0081%

P 16.35 [a] v(t) ≈ 340π

− 680π

13

cos ωot +115

cos 2ωot + · · ·

Vrms ≈√√√√√(340

π

)2

+(680

π

)2(

13√

2

)2

+(

115

√2

)2

=340π

√1 + 4

( 118

+1

450

)= 120.0819 V

[b] Vrms =170√

2= 120.2082

% error =(120.0819

120.2082− 1

)(100) = −0.11%

[c] v(t) ≈ 170π

+ 85 sin ωot − 3403π

cos 2ωot

Vrms ≈√√√√(170

π

)2

+(

85√2

)2

+(

3403√

2π

)2

≈ 84.8021 V

Vrms =1702

= 85 V

% error = −0.23%

P 16.36 [a] Half-wave symmetry av = 0, ak = bk = 0, even k. For k odd,

ak =4T

∫ T/4

0

4Im

Tt cos kω0t dt =

16Im

T 2

∫ T/4

0t cos kω0t dt

=16Im

T 2

cos kω0t

k2ω20

+t

kω0sin kω0t

∣∣∣∣T/4

0

=16Im

T 2

0 +

T

4kω0sin

kπ

2− 1

k2ω20

ak =2Im

πk

[sin

(kπ

2

)− 2

πk

],

bk =4T

∫ T/4

0

4Im

Tt sin kω0t dt =

16Im

T 2

∫ T/4

0t sin kω0t dt

=16Im

T 2

sin kω0t

k2ω20

− t

kω0cos kω0t

∣∣∣∣T/4

0=

4Im

π2k2 sin(

kπ

2

)


[b] ak − jbk =2Im

πk

[sin

(kπ

2

)− 2

πk

]−[j

2πk

sin(

kπ

2

)]

a1 − jb1 =2Im

π

(1 − 2

π

)− j

2π

= 0.47Im/− 60.28

a3 − jb3 =2Im

3π

(−1 − 2

3π

)+ j

( 23π

)= 0.26Im/170.07

a5 − jb5 =2Im

5π

(1 − 2

5π

)− j

( 25π

)= 0.11Im/− 8.30

a7 − jb7 =2Im

7π

(−1 − 2

7π

)+ j

( 27π

)= 0.10Im/175.23

ig = 0.47Im cos(ω0t − 60.28) + 0.26Im cos(3ω0t + 170.07)

+ 0.11Im cos(5ω0t − 8.30) + 0.10Im cos(7ω0t + 175.23) + · · ·

[c] Ig =

√√√√ ∞∑n=1,3,5

(A2

n

2

)

∼= Im

√(0.47)2 + (0.26)2 + (0.11)2 + (0.10)2

2= 0.39Im

[d] Area under i2g = 2∫ T/4

0

(4Im

Tt)2

dt =(

32I2m

T 2

)(t3

3

) ∣∣∣∣T/4

0=

I2mT

6

Ig =

√√√√ 1T

(I2mT

6

)=

Im√6

= 0.41Im

[e] % error =(

estimatedexact

− 1)

100 =(

0.3927Im

(Im/√

6)− 1

)100 = −3.8%

P 16.37 [a] v has half-wave symmetry, quarter-wave symmetry, and is odd

·. . av = 0, ak = 0 all k, bk = 0 k-even

bk =8T

∫ T/4

0f(t) sin kωot dt, k-odd

=8T

∫ T/8

0

Vm

4sin kωot dt +

∫ T/4

T/8Vm sin kωot dt

=8Vm

4T

[−cos kωot

kωo

∣∣∣∣T/8

0

]+

8Vm

T

[−cos kωot

kωo

∣∣∣∣T/4

T/8

]

=8Vm

4kωoT

[1 − cos

kπ

4

]+

8Vm

Tkωo

[cos

kπ

4− 0

]

Problems 16–37

=8Vm

kωoT

14

− 14

coskπ

4+ cos

kπ

4

=4Vm

πk

14

+ 0.75 coskπ

4

=

1k[10 + 30 cos(kπ/4)]

b1 = 10 + 30 cos(π/4) = 31.21

b3 =13[10 + 30 cos(3π/4)] = −3.74

b5 =15[10 + 30 cos(5π/4)] = −2.24

b7 =17[10 + 30 cos(7π/4)] = 4.46

V (rms) ≈ Vm

√31.212 + 3.742 + 2.242 + 4.462

2= 22.51

[b] Area under v2 = 2[2(2.5π)2

(T

8

)+ 100π2

(T

4

)]= 53.125π2T

V (rms) =

√1T

(53.125π2)T =√

53.125π = 22.90

[c] % Error =(22.51

22.90− 1

)(100) = −1.7%

P 16.38 [a] From Problem 16.16,

The area under v2:

A = 4[∫ T/8

0

14,400T 2 t2 dt +

∫ T/4

T/8

(10 +

40tT

)2

dt

]


=57,600

T 2

t3

3

∣∣∣∣T/8

0+ 400t

∣∣∣∣T/4

T/8+

3200T

t2

2

∣∣∣∣T/4

T/8+

6400T 2

t3

3

∣∣∣∣T/4

T/8

=57,6001536

T + 400T

8+ 1600

3T64

+ 64007T

1536=

5753

T

Vrms =

√1T

(5753

T)

=

√5753

= 13.84 V

[b] P =V 2

rms

15= 12.78 W

[c] From Problem 16.16,

b1 =80π2 (2 cos 45 + π sin 90 − 3) = 12.61 V

vg∼= 12.61 sin ω0t V

P =(19.57/

√2)2

15= 5.30 W

[d] % error =( 5.30

13.84− 1

)(100) = −61.71%

P 16.39 Figure P16.39(b): ta = 0.2 s; tb = 0.6 s

v = 50t 0 ≤ t ≤ 0.2

v = −50t + 20 0.2 ≤ t ≤ 0.6

v = 25t − 25 0.6 ≤ t ≤ 1.0

Area 1 under v2 = A1 =∫ 0.2

02500t2 dt =

203

Area 2 = A2 =∫ 0.6

0.2100(4 − 20t + 25t2) dt =

403

Area 3 = A3 =∫ 1.0

0.6625(t2 − 2t + 1) dt =

403

A1 + A2 + A3 =1003

Vrms =

√11

(1003

)=

10√3

V.

Problems 16–39

Figure P16.39(c): ta = tb = 0.4 s

v(t) = 25t 0 ≤ t ≤ 0.4

v(t) =503

(t − 1) 0.4 ≤ t ≤ 1

A1 =∫ 0.4

0625t2 dt =

403

A2 =∫ 1.0

0.4

25009

(t2 − 2t + 1) dt =603

A1 + A2 =1003

Vrms =

√1T

(A1 + A2) =

√11

(1003

)=

10√3

V.

Figure P16.39 (d): ta = tb = 1

v = 10t 0 ≤ t ≤ 1

A1 =∫ 1

0100t2 dt =

1003

Vrms =

√11

(1003

)=

10√3

V.

P 16.40 cn =1T

∫ T/4

0Vme−jnωot dt =

Vm

T

[e−jnωot

−jnωo

∣∣∣∣T/4

0

]

=Vm

Tnωo

[j(e−jnπ/2 − 1)] =Vm

2πnsin

nπ

2+ j

Vm

2πn

(cos

nπ

2− 1

)

=Vm

2πn

[sin

nπ

2− j

(1 − cos

nπ

2

)]

v(t) =∞∑

n=−∞cne

jnωot

co = av =1T

∫ T/4

0Vm dt =

Vm

4


or

co =Vm

2πlimn→0

[sin(nπ/2)

n− j

1 − cos(nπ/2)n

]

=Vm

2πlimn→0

[(π/2) cos(nπ/2)

1− j

(π/2) sin(nπ/2)1

]

=Vm

2π

[π

2− j0

]=

Vm

4

Note it is much easier to use co = av than to use L’Hopital’s rule to find the limit of0/0.

P 16.41 co = av =VmT

2· 1T

=Vm

2

cn =1T

∫ T

0

Vm

Tte−jnωot dt

=Vm

T 2

[e−jnω0t

−n2ω20(−jnω0t − 1)

]T

0

=Vm

T 2

[e−jn2πT/T

−n2ω20

(−jn

2πT

T − 1)

− 1−n2ω2

0(−1)

]

=Vm

T 2

[1

n2ω20(1 + jn2π) − 1

n2ω20

]

= jVm

2nπ, n = ±1, ±2, ±3, . . .

P 16.42 [a] Vrms =

√1T

∫ T

0v2 dt =

√1T

∫ T

0

(Vm

T

)2

t2 dt

=

√V 2

m

T 3

t3

3

∣∣∣∣T0

=

√V 2

m

3=

Vm√3

P =(120/

√3)2

10= 480 W

[b] From the solution to Problem 16.41

c0 =1202

= 60 V; c4 = j1208π

= j15π

c1 = j1202π

= j60π

; c5 = j12010π

= j12π

Problems 16–41

c2 = j1204π

= j30π

; c6 = j12012π

= j10π

c3 = j1206π

= j20π

; c7 = j12014π

= j8.57π

Vrms =

√√√√c2o + 2

∞∑n=1

|cn|2

=√

602 + 2π2 (602 + 302 + 202 + 152 + 122 + 102 + 8.572)

= 68.58 V

[c] P =(68.58)2

10= 470.29 W

% error =(470.29

480− 1

)(100) = −2.02%

P 16.43 [a] Co = av =(1/2)(T/2)Vm

T=

Vm

4

Cn =1T

∫ T/2

0

2Vm

Tte−jnωot dt

=2Vm

T 2

[e−jnωot

−n2ω2o

(−jnωot − 1)]T/2

0

=Vm

2n2π2 [e−jnπ(jnπ + 1) − 1]

Since e−jnπ = cos nπ we can write

Cn =Vm

2π2n2 (cos nπ − 1) + jVm

2nπcos nπ

[b] Co =544

= 13.5 V

C−1 =−54π2 + j

27π

= 10.19/122.48 V

C1 = 10.19/− 122.48 V

C−2 = −j13.5π

= 4.30/− 90 V

C2 = 4.30/90 V

C−3 =−6π2 + j

9π

= 2.93/101.98 V

C3 = 2.93/− 101.98 V


C−4 = −j6.75π

= 2.15/− 90 V

C4 = 2.15/90 V

[c]

Vo

250+

Vo

sL+ VosC +

Vo − Vg

62.5 × 103 = 0

·. . (250LCs2 + 1.004sL + 250)Vo = 0.004sLVg

Vo

Vg

= H(s) =(1/62, 500C)s

s2 + 1/249C + 1/LC

H(s) =16s

s2 + 1/249Cs + 4 × 1010

ωo =2πT

=2π10π

× 106 = 2 × 105 rad/s

H(j0) = 0

H(j2 × 105k) =jk

12, 500(1 − k2) + j251k

Therefore,

H−1 = 0.0398/0; H1 = 0.0398/0

H−2 =−j2

−37, 500 − j20= 5.33 × 10−5/86.23; H2 = 5.33 × 10−5/− 89.23

H−3 =−3j

−10−5 − j753= 3.00 × 10−5/89.57; H2 = 3.00 × 10−5/− 89.57

H−4 =−4j

−187, 500 − j1004= 2.13 × 10−5/89.69; H2 = 2.13 × 10−5/− 89.69

The output voltage coefficients:

C0 = 0

C−1 = (10.19/122.48)(0.00398/0) = 0.0406/122.48 V

Problems 16–43

C1 = 0.0406/− 122.48 V

C−2 = (4.30/− 90)(5.33 × 10−5/86.23) = 2.29 × 10−4/− 3.77 V

C2 = 2.29 × 10−4/3.77 V

C−3 = (2.93/101.98)(3.00 × 10−5/89.57) = 8.79 × 10−5/191.55 V

C3 = 8.79 × 10−5/− 191.55 V

C−4 = (2.15/− 90)(2.13 × 10−5/89.69) = 4.58 × 10−5/− 0.31 V

C4 = 4.58 × 10−5/0.31 V

[d] Vrms∼=√√√√C2

o + 24∑

n=1|Cn|2 ∼=

√√√√24∑

n=1|Cn|2

∼=√

2(0.04062 + (2.29 × 10−4)2 + (8.79 × 10−5)2 + (4.58 × 10−5)2 ∼= 0.0574 V

P =(0.0574)2

250= 13.2 µW

P 16.44 [a] Vrms =

√1T

∫ T/2

0

(2Vm

Tt)2

dt

=

√√√√ 1T

[4V 2

m

T 2

t3

3

]T/2

0

=

√√√√ 4V 2m

(3)(8)=

Vm√6

Vrms =54√

6= 22.05 V

[b] From the solution to Problem 16.43

C0 = 13.5; |C3| = 2.93

|C1| = 10.19; |C4| = 2.15

|C2| = 4.30

Vg(rms) ∼=√

13.52 + 2(10.192 + 4.302 + 2.932 + 2.152) ∼= 21.29 V

[c] % Error =(21.29

22.05− 1

)(100) = −3.44%


P 16.45 [a] From Example 16.3 we have:

av =404

= 10 V, ak =40πk

sin(

kπ

2

)

bk =40πk

[1 − cos

(kπ

2

)], Ak/− θ

k = ak − jbk

A1 = 18.01 V θ1 = −45, A2 = 12.73 V, θ2 = −90

A3 = 6 V, θ3 = −135, A4 = 0, A5 = 3.6 V, θ5 = −45

A6 = 4.24 V, θ6 = −90, A7 = 2.57 V, θ7 = −135

[b] Cn =an − jbn

2, C−n =

an + jbn

2= C∗

n

C0 = av = 10 V C3 = 3/135 V C6 = 2.12/90 V

C1 = 9/45 V C−3 = 3/− 135 V C−6 = 2.12/− 90 V

C−1 = 9/− 45 V C4 = C−4 = 0 C7 = 1.29/135 V

C2 = 6.37/90 V C5 = 1.8/45 V C−7 = 1.29/− 135 V

C−2 = 6.37/− 90 V C−5 = 1.8/− 45 V

Problems 16–45


Ak = ak − jbk =Im

π2k2 (cos kπ − 1) + jIm

πk

A0 = 0.75Im = 180 mA

A1 =240π2 (−2) + j

240π

= 90.56/122.48 mA

A2 = j2402π

= 38.20/90 mA

A3 =2409π2 (−2) + j

2403π

= 26.03/101.98 mA

A4 = j2404π

= 19.10/90 mA

A5 =24025π2 (−2) + j

2405π

= 15.40/97.26 mA

A6 = j2406π

= 12.73/90 mA

[b] C0 = A0 = 180 mA

C1 =12A1/− θ1 = 45.28/122.48 mA

C−1 = 45.28/− 122.48 mA

C2 =12A2/− θ2 = 19.1/90 mA

C−2 = 19.1/− 90 mA


C3 =12A3/− θ3 = 13.02/101.98 mA

C−3 = 13.02/− 101.98 mA

C4 =12A4/− θ4 = 9.55/90 mA

C−4 = 9.55/− 90 mA

C5 =12A5/− θ5 = 7.70/97.26 mA

C−5 = 7.70/− 97.26 mA

C6 =12A6/− θ6 = 6.37/90 mA

C−6 = 6.37/− 90 mA

P 16.47 [a] v = A1 cos(ωot + 90) + A3 cos(3ωot − 90)

+A5 cos(5ωot + 90) + A7 cos(7ωot − 90)

v = −A1 sin ωot + A3 sin 3ωot − A5 sin 5ωot + A7 sin 7ωot

[b] v(−t) = A1 sin ωot − A3 sin 3ωot + A5 sin 5ωot − A7 sin 7ωot

·. . v(−t) = −v(t); odd function

[c] v(t − T/2) = −A1 sin(ωot − π) + A3 sin(3ωot − 3π)

−A5 sin(5ωot − 5π) + A7 sin(7ωot − 7π)

= A1 sin ωot − A3 sin 3ωot + A5 sin 5ωot − A7 sin 7ωot

·. . v(t − T/2) = −v(t), yes, the function has half-wave symmetry

Problems 16–47

[d] Since the function is odd, with hws, we test to see if

f(T/2 − t) = f(t)

f(T/2 − t) = −A1 sin(π − ωot) + A3 sin(3π − 3ωot)

A5 sin(5π − 5ωot) + A7 sin(7π − 7ωot)

= −A1 sin ωot + A3 sin 3ωot − A5 sin 5ωot + A7 sin 7ωot

·. . f(T/2 − t) = f(t) and the voltage has quarter-wave symmetry

P 16.48 [a] i = 11,025 cos 10,000t + 1225 cos(30,000t − 180) + 441 cos(50,000t − 180)

+ 225 cos 70,000t µA

= 11,025 cos 10,000t − 1225 cos 30,000t − 441 cos 50,000t

+ 225 cos 70,000t µA

[b] i(t) = i(−t), Function is even

[c] Yes, A0 = 0, An = 0 for n even

[d] Irms =

√11,0252 + 12252 + 4412 + 2252

2= 7.85 mA

[e] A1 = 11,025/0 µA; C1 = 5512.50/0 µA

A3 = 1225/180 µA; C3 = 612.5/180 µA

A5 = 441/180 µA; C5 = 220.5/180 µA

A7 = 225/0 µA; C7 = 112.50/0 µA

C−1 = 5512.50/0 µA; C−3 = 612.5/− 180 µA

C−5 = 220.5/− 180 µA; C−7 = 112.50/0 µA

i = 112.5e−j70,000t + 220.5e−j180e−j50,000t + 612.5e−j180

e−j30,000t

+ 5512.5e−j10,000t + 5512.5ej10,000t + 612.5ej180ej30,000t

+ 220.5ej180ej50,000t + 112.5ej70,000t µA


[f]

θn

(krad/s)10 30 50 70

−70 −50 −30 −10

−180˚

180˚

P 16.49 From Table 15.1 we have

H(s) =1

(s + 1)(s2 + s + 1)

After scaling we get

H ′(s) =106

(s + 100)(s2 + 100s + 104)

ωo =2πT

=2π5π

× 103 = 400 rad/s

·. . H ′(jnωo) =1

(1 + j4n)[(1 − 16n2) + j4n]

It follows that

H(j0) = 1/0

Problems 16–49

H(jωo) =1

(1 + j4)(−15 + j4)= 0.0156/− 241.03

H(j2ωo) =1

(1 + j8)(−63 + j8)= 0.00195/− 255.64

vg(t) =A

π+

A

2sin ωot − 2A

π

∞∑n=2,4,6,

cos nωot

n2 − 1

= 54 + 27π sin ωot − 36 cos 2ωot − · · · V

·. . vo = 54 + 1.33 sin(400t − 241.03) − 0.07 cos(800t − 255.64) − · · · V

P 16.50 Using the technique outlined in Problem 16.17 we can derive the Fourier series forvg(t). We get

vg(t) = 100 +800π2

∞∑n=1,3,5,

1n2 cos nωot

The transfer function of the prototype second-order low pass Butterworth filter is

H(s) =1

s2 +√

2s + 1, where ωc = 1 rad/s

Now frequency scale using kf = 2000 to get ωc = 2 krad/s:

H(s) =4 × 106

s2 + 2000√

2s + 4 × 106

H(j0) = 1

H(j5000) =4 × 106

(j5000)2 + 2000√

2(j5000)2 + 4 × 106= 0.1580/− 146.04

H(j15,000) =4 × 106

(j15,000)2 + 2000√

2(j15,000)2 + 4 × 106= 0.0178/− 169.13

Vdc = 100 V

Vg1 =800π2 /0 V

Vg3 =8009π2 /0 V


Vodc = 100(1) = 100 V

Vo1 =800π2 (0.1580/− 146.04) = 12.81/− 146.04 V

Vo3 =8009π2 (0.0178/− 169.13) = 0.16/− 169.13 V

vo(t) = 100 + 12.81 cos(5000t − 146.04)

+ 0.16 cos(15,000t − 169.13) + · · · V

P 16.51 [a] Let Va represent the node voltage across R2, then the node-voltage equations are

Va − Vg

R1+

Va

R2+ VasC2 + (Va − Vo)sC1 = 0

(0 − Va)sC2 +0 − Vo

R3= 0

Solving for Vo in terms of Vg yields

Vo

Vg

= H(s) =−1

R1C1s

s2 + 1R3

(1

C1+ 1

C2

)s + R1+R2

R1R2R3C1C2

It follows that

ω2o =

R1 + R2

R1R2R3C1C2

β =1R3

( 1C1

+1C2

)

Ko =R3

R1

(C2

C1 + C2

)

Note that

H(s) =−R3

R1

(C2

C1+C2

)1

R3

(1

C1+ 1

C2

)s

s2 + 1R3

(1

C1+ 1

C2

)s +

(R1+R2

R1R2R3C1C2

)[b] For the given values of R1, R2, R3, C1, and C2 we have

−R3

R1

(C2

C1 + C2

)= − R3

2R1= −400

313

1R3

( 1C1

+1C2

)= 2000

R1 + R2

R1R2R3C1C2= 0.16 × 1010 = 16 × 108

Problems 16–51

H(s) =−(400/313)(2000)s

s2 + 2000s + 16 × 108

ωo =2πT

=2π50π

× 106 = 4 × 104 rad/s

H(jnωo) =−(400/313)(2000)jnωo

16 × 108 − n2ω2o + j2000nωo

=−j(20/313)n

(1 − n2) + j0.05n

H(jωo) =−j(20/313)

j(0.050)= −400

313= −1.28

H(j3ωo) =−j(20/313)(3)

−8 + j0.15= 0.0240/91.07

H(j5ωo) =−j(100/313)−24 + j0.25

= 0.0133/90.60

vg(t) =4Aπ

∞∑n=1,3,5

1n

sin(nπ/2) cos nωot

A = 15.65π V

vg(t) = 62.60 cos ωot − 20.87 cos 3ωot + 12.52 cos 5ωot − · · ·

vo(t) = −80 cos ωot − 0.50 cos(3ωot + 91.07)

+ 0.17 cos(5ωot + 90.60) − · · · V

17The Fourier Transform

Assessment Problems

AP 17.1 [a] F (ω) =∫ 0

−τ/2(−Ae−jωt) dt +

∫ τ/2

0Ae−jωt dt

=A

jω[2 − ejωτ/2 − e−jωτ/2]

=2Ajω

[1 − ejωτ/2 + e−jωτ/2

2

]

=−j2A

ω[1 − cos

ωτ

2]

[b] F (ω) =∫ ∞

0te−ate−jωt dt =

∫ ∞

0te−(a+jω)t dt =

1(a + jω)2

AP 17.2 f(t) =12π

∫ −2

−34ejtω dω +

∫ 2

−2ejtω dω +

∫ 3

24ejtω dω

=1

j2πt4e−j2t − 4e−j3t + ej2t − e−j2t + 4ej3t − 4ej2t

=1πt

[3e−j2t − 3ej2t

j2+

4ej3t − 4e−j3t

j2

]

=1πt

(4 sin 3t − 3 sin 2t)

AP 17.3 [a] F (ω) = F (s) |s=jω= Le−at sin ω0ts=jω

=ω0

(s + a)2 + ω20

∣∣∣∣s=jω

=ω0

(a + jω)2 + ω20

[b] F (ω) = Lf(−t)s=−jω =[

1(s + a)2

]s=−jω

=1

(a − jω)2

17–1

17–2 CHAPTER 17. The Fourier Transform

[c] f+(t) = te−at, f−(t) = teat

Lf+(t) =1

(s + a)2 , Lf−(−t) =−1

(s + a)2

Therefore F (ω) =1

(a + jω)2 − 1(a − jω)2 =

−j4aω

(a2 + ω2)2

AP 17.4 [a] f ′(t) =2Aτ

,−τ

2< t < 0; f ′(t) =

−2Aτ

, 0 < t <τ

2

·. . f ′(t) =2Aτ

[u(t + τ/2) − u(t)] − 2Aτ

[u(t) − u(t − τ/2)]

=2Aτ

u(t + τ/2) − 4Aτ

u(t) +2Aτ

u(t − τ/2)

·. . f ′′(t) =2Aτ

δ(t +

τ

2

)− 4A

τδ(t) +

2Aτ

δ(t − τ

2

)

[b] Ff ′′(t) =[2A

τejωτ/2 − 4A

τ+

2Aτ

e−jωτ/2]

=4Aτ

[ejωτ/2 + e−jωτ/2

2− 1

]=

4Aτ

[cos

(ωτ

2

)− 1

]

[c] Ff ′′(t) = (jω)2F (ω) = −ω2F (ω); therefore F (ω) = − 1ω2Ff ′′(t)

Thus we have F (ω) = − 1ω2

4Aτ

[cos

(ωτ

2

)− 1

]

AP 17.5 v(t) = Vm

[u(t +

τ

2

)− u

(t − τ

2

)]

Fu(t +

τ

2

)=[πδ(ω) +

1jω

]ejωτ/2

Fu(t − τ

2

)=[πδ(ω) +

1jω

]e−jωτ/2

Therefore V (ω) = Vm

[πδ(ω) +

1jω

] [ejωτ/2 − e−jωτ/2

]

= j2Vmπδ(ω) sin(

ωτ

2

)+

2Vm

ωsin

(ωτ

2

)

=(Vmτ) sin(ωτ/2)

ωτ/2

Problems 17–3

AP 17.6 [a] Ig(ω) = F10sgn t =20jω

[b] H(s) =Vo

Ig

Using current division and Ohm’s law,

Vo = −I2s = −[ 44 + 1 + s

](−Ig)s =

4s5 + s

Ig

H(s) =4s

s + 5, H(ω) =

j4ω5 + jω

[c] Vo(ω) = H(ω) · Ig(ω) =(

j4ω5 + jω

)(20jω

)=

805 + jω

[d] vo(t) = 80e−5tu(t) V

[e] Using current division,

i1(0−) =15ig =

15(−10) = −2 A

[f] i1(0+) = ig + i2(0+) = 10 + i2(0−) = 10 + 8 = 18 A

[g] Using current division,

i2(0−) =45(10) = 8 A

[h] Since the current in an inductor must be continuous,

i2(0+) = i2(0−) = 8 A

[i] Since the inductor behaves as a short circuit for t < 0,

vo(0−) = 0 V

[j] vo(0+) = 1i2(0+) + 4i1(0+) = 80 V

AP 17.7 [a] Vg(ω) =1

1 − jω+ πδ(ω) +

1jω

H(s) =Va

Vg

=0.5‖(1/s)

1 + 0.5‖(1/s)=

1s + 3

, H(ω) =1

3 + jω

Va(ω) = H(ω)Vg(ω)

=1

(1 − jω)(3 + jω)+

1jω(3 + jω)

+πδ(ω)3 + jω

=1/4

1 − jω+

1/43 + jω

+1/3jω

− 1/33 + jω

+πδ(ω)3 + jω

=1/4

1 − jω+

1/3jω

− 1/123 + jω

+πδ(ω)3 + jω

=1/4

1 − jω+

1/3jω

− 1/123 + jω

+ πδ(ω)


Therefore va(t) =[14etu(−t) +

16

sgn t − 112

e−3tu(t) +16

]V

[b] va(0−) =14

− 16

+ 0 +16

=14

V

va(0+) = 0 +16

− 112

+16

=14

V

va(∞) = 0 +16

+ 0 +16

=13

V

AP 17.8 v(t) = 4te−tu(t); V (ω) =4

(1 + jω)2

Therefore |V (ω)| =4

1 + ω2

W1Ω =1π

∫ √3

0

[4

(1 + ω2)

]2

dω

=16π

12

[ω

ω2 + 1+ tan−1 ω

1

]√3

0

= 16[√

38π

+16

]= 3.769 J

W1Ω(total) =8π

[ω

ω2 + 1+ tan−1 ω

1

]∞0

=8π

[0 +

π

2

]= 4 J

Therefore % =3.769

4(100) = 94.23%

AP 17.9 |V (ω)| = 6 −( 6

2000π

)ω, 0 ≤ ω ≤ 2000π

|V (ω)|2 = 36 −( 72

2000π

)ω +

( 364π2 × 106

)ω2

W1Ω =1π

∫ 2000π

0

[36 − 72ω

2000π+

36 × 10−6

4π2 ω2

]dω

=1π

[36ω − 72ω2

4000π+

36 × 10−6ω3

12π2

]2000π

0

=1π

[36(2000π) − 72

4000π(2000π)2 +

36 × 10−6(2000π)3

12π2

]

Problems 17–5

= 36(2000) − 72(2000)2

4000+

36 × 10−6(2000)3

12

= 24 kJ

W6kΩ =24 × 103

6 × 103 = 4 J


Problems

P 17.1 [a] F (ω) =∫ 2

−2

[A sin

(π

2

)t]e−jωt dt =

−j4πA

π2 − 4ω2 sin 2ω

[b] F (ω) =∫ 0

−τ/2

(2Aτ

t + A)

e−jωt dt +∫ τ/2

0

(−2Aτ

t + A)

e−jωt dt

=4Aω2τ

[1 − cos

(ωτ

2

)]

P 17.2 [a] F (ω) =∫ τ/2

−τ/2

2Aτ

te−jωt dt

=2Aτ

[e−jωt

−ω2 (−jωt − 1)]τ/2

−τ/2

=2Aω2τ

[e−jωτ/2

(jωτ

2+ 1

)− ejωτ/2

(−jωτ

2+ 1

)]

F (ω) =2Aω2τ

[e−jωτ/2 − ejωτ/2 + j

ωτ

2

(e−jωτ/2 + ejωτ/2

)]

F (ω) = j2Aτ

[ωτ cos(ωτ/2) − 2 sin(ωτ/2)

ω2

]

[b] Using L’Hopital’s rule,

F (0) = limω→0

j2A[ωτ(τ/2)(− sin ωτ/2) + τ cos ω(τ/2) − 2(τ/2) cos(ωτ/2)

2ωτ

]

= limω→0

j2A[−ωτ(τ/2) sin(ωτ/2)

2ωτ

]

= limω→0

j2A[−τ sin(ωτ/2)

4

]= 0

·. . F (0) = 0

[c] When A = 1 and τ = 1

F (ω) = j2[ω cos(ω/2) − 2 sin(ω/2)

ω2

]

|F (ω)| =∣∣∣∣∣2ω cos(ω/2) − 4 sin(ω/2)

ω2

∣∣∣∣∣F (0) = 0

Problems 17–7

|F (2)| =∣∣∣∣4 cos 1 − 4 sin 1

4

∣∣∣∣ = 0.30

|F (4)| =∣∣∣∣8 cos 2 − 4 sin 2

16

∣∣∣∣ = 0.44

|F (6)| =∣∣∣∣12 cos 3 − 4 sin 3

36

∣∣∣∣ = 0.35

|F (8)| =∣∣∣∣16 cos 4 − 4 sin 4

64

∣∣∣∣ = 0.12

|F (9)| =∣∣∣∣18 cos 4.5 − 4 sin 4.5

81

∣∣∣∣ ∼= 0

|F (10)| =∣∣∣∣20 cos 5 − 4 sin 5

100

∣∣∣∣ = 0.10

|F (12)| =∣∣∣∣24 cos 6 − 4 sin 6

144

∣∣∣∣ = 0.17

|F (14)| =∣∣∣∣28 cos 7 − 4 sin 7

196

∣∣∣∣ = 0.09

|F (15.5)| =∣∣∣∣31 cos 7.75 − 4 sin 7.75

240.25

∣∣∣∣ ∼= 0

P 17.3 [a] F (ω) = A +2Aωo

ω, −ωo/2 ≤ ω ≤ 0

F (ω) = A − 2Aωo

ω, 0 ≤ ω ≤ ωo/2

F (ω) = 0 elsewhere


f(t) =12π

∫ 0

−ωo/2

(A +

2Aωo

ω)

ejtω dω

+12π

∫ ωo/2

0

(A − 2A

ωo

ω)

ejtω dω

f(t) =12π

[ ∫ 0

−ωo/2Aejtω dω +

∫ 0

−ωo/2

2Aωo

ωejtω dω

+∫ ωo/2

0Aejtω dω −

∫ ωo/2

0

2Aωo

ωejtω dω]

=12π

[ Int1 + Int2 + Int3 − Int4 ]

Int1 =∫ 0

−ωo/2Aejtω dω =

A

jt(1 − e−jtωo/2)

Int2 =∫ 0

−ωo/2

2Aωo

ωejtω dω =2Aωot2

(1 − jtωo

2e−jtωo/2 − e−jtωo/2)

Int3 =∫ ωo/2

0Aejtω dω =

A

jt(ejtωo/2 − 1)

Int4 =∫ ωo/2

0

2Aωo

ωejtω dω =2Aωot2

(−jtωo

2ejtωo/2 + ejtωo/2 − 1)

Int1 + Int3 =2At

sin(ωot/2)

Int2 − Int4 =4Aωot2

[1 − cos(ωot/2)] − 2At

sin(ωot/2)

·. . f(t) =12π

[ 4Aωot2

(1 − cos(ωot/2))]

=2A

πωot2

[2 sin2(ωot/4)

]

=4ωoA

πω2ot

2 sin2(ωot/4)

=ωoA

4π

[sin(ωot/4)(ωot/4)

]2

[b] f(0) =ωoA

4π(1)2 = 79.58 × 10−3ωoA

Problems 17–9

[c] A = 20π; ωo = 2 rad/s

f(t) = 10[sin(t/2)(t/2)

]2

P 17.4 [a] F (s) = Lte−at =1

(s + a)2

F (ω) = F (s)∣∣∣∣s=jω

+ F (s)∣∣∣∣s=−jω

F (ω) =[

1(a + jω)2

]+[

1(a − jω)2

]

=2(a2 − ω2)

(a2 − ω2)2 + 4a2ω2 =2(a2 − ω2)(a2 + ω2)2

[b] F (s) = Lt3e−at =6

(s + a)4

F (ω) = F (s)∣∣∣∣s=jω

+ F (s)∣∣∣∣s=−jω

F (ω) =6

(a + jω)4 − 6(a − jω)4 = −j48aω

a2 − ω2

(a2 + ω2)4

[c] F (s) = Le−at cos ω0t =s + a

(s + a)2 + ω20

=0.5

(s + a) − jω0+

0.5(s + a) + jω0

F (ω) = F (s)∣∣∣∣s=jω

+ F (s)∣∣∣∣s=−jω

F (ω) =0.5

(a + jω) − jω0+

0.5(a + jω) + jω0

+0.5

(a − jω) − jω0+

0.5(a − jω) + jω0

=a

a2 + (ω − ω0)2 +a

a2 + (ω + ω0)2


[d] F (s) = Le−at sin ω0t =ω0

(s + a)2 + ω20

=−j0.5

(s + a) − jω0+

j0.5(s + a) + jω0

F (ω) = F (s)∣∣∣∣s=jω

− F (s)∣∣∣∣s=−jω

F (ω) =−ja

a2 + (ω − ω0)2 +ja

a2 + (ω + ω0)2

[e] F (ω) =∫ ∞

−∞δ(t − to)e−jωt dt = e−jωto

(Use the sifting property of the Dirac delta function.)

P 17.5 Fsin ω0t = F

ejω0t

2j

− F

e−jω0t

2j

=12j

[2πδ(ω − ω0) − 2πδ(ω + ω0)]

= jπ[δ(ω + ω0) − δ(ω − ω0)]

P 17.6 f(t) =12π

∫ ∞

−∞[A(ω) + jB(ω)][cos tω + j sin tω] dω

=12π

∫ ∞

−∞[A(ω) cos tω − B(ω) sin tω] dω

+j

2π

∫ ∞

−∞[A(ω) sin tω + B(ω) cos tω] dω

But f(t) is real, therefore the second integral in the sum is zero.

P 17.7 By hypothesis, f(t) = −f(−t). From Problem 17.6, we have

f(−t) =12π

∫ ∞

−∞[A(ω) cos tω + B(ω) sin tω] dω

For f(t) = −f(−t), the integral∫∞−∞ A(ω) cos tω dω must be zero. Therefore, if

f(t) is real and odd, we have

f(t) =−12π

∫ ∞

−∞B(ω) sin tω dω

P 17.8 F (ω) =−j2ω

; therefore B(ω) =−2ω

; thus we have

f(t) = − 12π

∫ ∞

−∞

(−2ω

)sin tω dω =

1π

∫ ∞

−∞sin tω

ωdω

Butsin tω

ωis even; therefore f(t) =

2π

∫ ∞

0

sin tω

ωdω

Problems 17–11

Therefore,

f(t) =2π

· π

2= 1 t > 0

f(t) =2π

·(−π

2

)= −1 t < 0

from a table of definite integrals

Therefore f(t) = sgn t

P 17.9 From Problem 17.4[c] we have

F (ω) =ε

ε2 + (ω − ω0)2 +ε

ε2 + (ω + ω0)2

Note that as ε → 0, F (ω) → 0 everywhere except at ω = ±ω0. At ω = ±ω0,F (ω) = 1/ε, therefore F (ω) → ∞ at ω = ±ω0 as ε → 0. The area under eachbell-shaped curve is independent of ε, that is∫ ∞

−∞εdω

ε2 + (ω − ω0)2 =∫ ∞

−∞εdω

ε2 + (ω + ω0)2 = π

Therefore as ε → 0, F (ω) → πδ(ω − ω0) + πδ(ω + ω0)

P 17.10 A(ω) =∫ ∞

−∞f(t) cos ωt dt

=∫ 0

−∞f(t) cos ωt dt +

∫ ∞

0f(t) cos ωt dt

= 2∫ ∞

0f(t) cos ωt dt, since f(t) cos ωt is also even.

B(ω) = 0, since f(t) sin ωt is an odd function and∫ 0

−∞f(t) sin ωt dt = −

∫ ∞

0f(t) sin ωt dt

P 17.11 A(ω) =∫ 0

−∞f(t) cos ωt dt +

∫ ∞

0f(t) cos ωt dt = 0

since f(t) cos ωt is an odd function.

B(ω) = −2∫ ∞

0f(t) sin ωt dt, since f(t) sin ωt is an even function.

P 17.12 [a] F

df(t)dt

=∫ ∞

−∞df(t)dt

e−jωt dt

Let u = e−jωt, then du = −jωe−jωt; let dv = [df(t)/dt] dt, then v = f(t).

Therefore F

df(t)dt

= f(t)e−jωt

∣∣∣∣∞−∞−∫ ∞

−∞f(t)[−jωe−jωt dt]

= 0 + jωF (ω)


[b] Fourier transform of f(t) exists, i.e., f(∞) = f(−∞) = 0.

[c] To find F

d2f(t)dt2

, let g(t) =

df(t)dt

Then F

d2f(t)dt2

= F

dg(t)dt

= jωG(ω)

But G(ω) = F

df(t)dt

= jωF (ω)

Therefore we have F

d2f(t)dt2

= (jω)2F (ω)

Repeated application of this thought process gives

F

dnf(t)dtn

= (jω)nF (ω).

P 17.13 [a] F∫ t

−∞f(x) dx

=∫ ∞

−∞

[∫ t

−∞f(x) dx

]e−jωt dt

Now let u =∫ t

−∞f(x) dx, then du = f(t)dt

Let dv = e−jωt dt, then v =e−jωt

−jω

Therefore,

F∫ t

−∞f(x) dx

=

e−jωt

−jω

∫ t

−∞f(x) dx

∣∣∣∣∞−∞−∫ ∞

−∞

[e−jωt

−jω

]f(t) dt

= 0 +F (ω)jω

[b] We require∫ ∞

−∞f(x) dx = 0

[c] No, because∫ ∞

−∞e−axu(x) dx =

1a

= 0

P 17.14 [a] Ff(at) =∫ ∞

−∞f(at)e−jωt dt

Let u = at, du = a dt, u = ±∞ when t = ±∞Therefore,

Ff(at) =∫ ∞

−∞f(u)e−jωu/a

(du

a

)=

1aF(

ω

a

), a > 0

Problems 17–13

[b] Fe−|t| =1

1 + jω+

11 − jω

=2

1 + ω2

Therefore Fe−a|t| =(1/a)2

(ω/a)2 + 1

Therefore Fe−0.5|t| =4

4ω2 + 1, Fe−|t| =

2ω2 + 1

Fe−2|t| = 1/[0.25ω2 + 1], yes as “a” increases, the sketches show that f(t)approaches zero faster and F (ω) flattens out over the frequency spectrum.

P 17.15 [a] Ff(t − a) =∫ ∞

−∞f(t − a)e−jωt dt

Let u = t − a, then du = dt, t = u + a, and u = ±∞ when t = ±∞.Therefore,

Ff(t − a) =∫ ∞

−∞f(u)e−jω(u+a) du

= e−jωa∫ ∞

−∞f(u)e−jωu du = e−jωaF (ω)

[b] Fejω0tf(t) =∫ ∞

−∞f(t)e−j(ω−ω0)t dt = F (ω − ω0)

[c] Ff(t) cos ω0t = F

f(t)[ejω0t + e−jω0t

2

]

=12F (ω − ω0) +

12F (ω + ω0)

P 17.16 Y (ω) =∫ ∞

−∞

[∫ ∞

−∞x(λ)h(t − λ) dλ

]e−jωt dt

=∫ ∞

−∞x(λ)

[∫ ∞

−∞h(t − λ)e−jωt dt

]dλ


Let u = t − λ, du = dt, and u = ±∞, when t = ±∞.

Therefore Y (ω) =∫ ∞

−∞x(λ)

[∫ ∞

−∞h(u)e−jω(u+λ) du

]dλ

=∫ ∞

−∞x(λ)

[e−jωλ

∫ ∞

−∞h(u)e−jωu du

]dλ

=∫ ∞

−∞x(λ)e−jωλH(ω) dλ = H(ω)X(ω)

P 17.17 Ff1(t)f2(t) =∫ ∞

−∞

[ 12π

∫ ∞

−∞F1(u)ejtudu

]f2(t)e−jωt dt

=12π

∫ ∞

−∞

[∫ ∞

−∞F1(u)f2(t)e−jωtejtu du

]dt

=12π

∫ ∞

−∞

[F1(u)

∫ ∞

−∞f2(t)e−j(ω−u)t dt

]du

=12π

∫ ∞

−∞F1(u)F2(ω − u) du

P 17.18 [a] F (ω) =∫ ∞

−∞f(t)e−jωt dt

dF

dω=∫ ∞

−∞d

dω

[f(t)e−jωt

]dt = −j

∫ ∞

−∞tf(t)e−jωt dt = −jFtf(t)

Therefore jdF (ω)

dω= Ftf(t)

d2F (ω)dω2 =

∫ ∞

−∞(−jt)(−jt)f(t)e−jωt dt = (−j)2Ft2f(t)

Note that (−j)n =1jn

Thus we have jn

[dnF (ω)

dωn

]= Ftnf(t)

[b] (i) Fe−atu(t) =1

a + jω= F (ω);

dF (ω)dω

=−j

(a + jω)2

Therefore j

[dF (ω)

dω

]=

1(a + jω)2

Therefore Fte−atu(t) =1

(a + jω)2

Problems 17–15

(ii) F|t|e−a|t| = Fte−atu(t) − Fteatu(−t)

=1

(a + jω)2 − jd

dω

(1

a − jω

)

=1

(a + jω)2 +1

(a − jω)2

(iii) Fte−a|t| = Fte−atu(t) + Fteatu(−t)

=1

(a + jω)2 + jd

dω

(1

a − jω

)

=1

(a + jω)2 − 1(a − jω)2

P 17.19 [a] f1(t) = cos ω0t, F1(u) = π[δ(u + ω0) + δ(u − ω0)]

f2(t) = 1, −τ/2 < t < τ/2, and f2(t) = 0 elsewhere

Thus F2(u) =τ sin(uτ/2)

uτ/2

Using convolution,

F (ω) =12π

∫ ∞

−∞F1(u)F2(ω − u) du

=12π

∫ ∞

−∞π[δ(u + ω0) + δ(u − ω0)]τ

sin[(ω − u)τ/2](ω − u)(τ/2)

du

=τ

2

∫ ∞

−∞δ(u + ω0)

sin[(ω − u)τ/2](ω − u)(τ/2)

du

+τ

2

∫ ∞

−∞δ(u − ω0)

sin[(ω − u)τ/2](ω − u)(τ/2)

du

=τ

2· sin[(ω + ω0)τ/2]

(ω + ω0)(τ/2)+

τ

2· sin[(ω − ω0)τ/2]

(ω − ω0)τ/2

[b] As τ increases, the amplitude of F (ω) increases at ω = ±ω0 and at the sametime the duration of F (ω) approaches zero as ω deviates from ±ω0.The area under the [sin x]/x function is independent of τ, that is

τ

2

∫ ∞

−∞sin[(ω − ω0)(τ/2)]

(ω − ω0)(τ/2)dω =

∫ ∞

−∞sin[(ω − ω0)(τ/2)]

(ω − ω0)(τ/2)[(τ/2) dω] = π

Therefore as t → ∞,

f1(t)f2(t) → cos ω0t and F (ω) → π[δ(ω − ω0) + δ(ω + ω0)]


P 17.20 [a] vg = 100u(t)

Vg(ω) = 100[πδ(ω) +

1jω

]

H(s) =10

5s + 10=

2s + 2

H(ω) =2

jω + 2

Vo(ω) = H(ω)Vg(ω) =200πδ(ω)jω + 2

+200

jω(jω + 2)

= V1(ω) + V2(ω)

v1(t) =12π

∫ ∞

−∞200πejtω

jω + 2δ(ω) dω =

12π

(200π2

)= 50 (sifting property)

V2(ω) =K1

jω+

K2

jω + 2=

100jω

− 100jω + 2

v2(t) = 50sgn(t) − 100e−2tu(t)

vo(t) = v1(t) + v2(t) = 50 + 50sgn(t) − 100e−2tu(t)

= 100u(t) − 100e−2tu(t)

vo(t) = 100(1 − e−2t)u(t) V

[b]

Problems 17–17

P 17.21 [a] From the solution to Problem 17.20

H(ω) =2

jω + 2

Now,

Vg(ω) =200jω

Then,

Vo(ω) = H(ω)Vg(ω) =400

jω(jω + 2)=

K1

jω+

K2

jω + 2=

200jω

− 200jω + 2

·. . vo(t) = 100sgn(t) − 200e−2tu(t) V

[b]

P 17.22 [a] Find the Thévenin equivalent with respect to the terminals of the capacitor:

vTh =56vg; RTh = 60‖12 = 10 kΩ

Io =VTh

10,000 + 106/2s=

2sVTh

20,000s + 106

H(s) =Io

VTh=

10−4s

s + 50; H(ω) =

jω × 10−4

jω + 50


vTh =56vg = 30 sgn(t); VTh =

60jω

Io = H(ω)VTh(ω) =(

60jω

)(jω × 10−4

jω + 50

)=

6 × 10−3

jω + 50

io(t) = 6e−50tu(t) mA

[b] At t = 0− the circuit is

At t = 0+ the circuit is

ig(0+) =30 + 36

12= 5.5 mA

i60k(0+) =3060

= 0.5 mA

io(0+) = 5.5 + 0.5 = 6 mA

which agrees with our solution.We also know io(∞) = 0, which agrees with our solution.The time constant with respect to the terminals of the capacitor is RThC Thus,

τ = (10,000)(2 × 10−6) = 20 ms; ·. .1τ

= 50,

which also agrees with our solution.Thus our solution makes sense in terms of known circuit behavior.

Problems 17–19

P 17.23 [a] From the solution of Problem 17.22 we have

Vo =VTh

104 + (106/2s)· 106

2s

H(s) =Vo

VTh=

50s + 50

H(jω) =50

jω + 50

VTh(ω) =60jω

Vo(ω) = H(jω)VTh(ω) =(

60jω

)50

jω + 50

=3000

(jω)(jω + 50)=

60jω

− 60jω + 50

vo(t) = 30sgn(t) − 60e−50tu(t) V

[b] vo(0−) = −30 V

vo(0+) = 30 − 60 = −30 V

This makes sense because there cannot be an instantaneous change in thevoltage across a capacitor.

vo(∞) = 30 V

This agrees with vTh(∞) = 30 V.As in Problem 17.22 we know the time constant is 20 ms.

P 17.24 [a]Vo

Vg

= H(s) =4/s

0.5 + 0.01s + 4/s

H(s) =400

s2 + 50s + 400=

400(s + 10)(s + 40)

H(jω) =400

(jω + 10)(jω + 40)

Vg(ω) =6jω


Vo(ω) = Vg(ω)H(jω) =2400

jω(jω + 10)(jω + 40)

Vo(ω) =K1

jω+

K2

jω + 10+

K3

jω + 40

K1 =2400400

= 6; K2 =2400

(−10)(30)= −8

K3 =2400

(−40)(−30)= 2

Vo(ω) =6jω

− 8jω + 10

+2

jω + 40

vo(t) = 3sgn(t) − 8e−10tu(t) + 2e−40tu(t) V

[b] vo(0−) = −3 V

[c] vo(0+) = 3 − 8 + 2 = −3 V

[d] For t ≥ 0+:

Vo − 3/s0.5 + 0.01s

+(Vo + 3/s)s

4= 0

Vo

[ 100s + 50

+s

4

]=

300s(s + 50)

− 0.75

Vo =1200 − 3s2 − 150ss(s + 10)(s + 40)

=K1

s+

K2

s + 10+

K3

s + 40

K1 =1200400

= 3; K2 =1200 − 300 + 1500

(−10)(30)= −8

K3 =1200 − 4800 + 6000

(−40)(−30)= 2

vo(t) = (3 − 8e−10t + 2e−40t)u(t) V

[e] Yes.

Problems 17–21

P 17.25 [a] Io =Vg

0.5 + 0.01s + 4/s

H(s) =Io

Vg

=100s

s2 + 50s + 400=

100s(s + 10)(s + 40)

H(ω) =100(jω)

(jω + 10)(jω + 40)

Vg(ω) =6jω

Io(ω) = H(ω)Vg(ω) =600

(jω + 10)(jω + 40)

=20

jω + 10− 20

jω + 40

io(t) = (20e−10t − 20e−40t)u(t) A

[b] io(0−) = 0

[c] io(0+) = 0

[d]

Io =6/s

0.5 + 0.01s + 4/s=

600s2 + 50s + 400

=600

(s + 10)(s + 40)=

20s + 10

− 20s + 40

io(t) = (20e−10t − 20e−40t)u(t) A

[e] Yes.

P 17.26 [a] Io =IgR

R + 1/sC=

RCsIg

RCs + 1; H(s) =

Io

Ig

=s

s + 1/RC

1RC

=106

25 × 103 = 40; H(ω) =jω

jω + 40

ig = 200sgn(t) µA; Ig = (200 × 10−6)(

2jω

)=

400 × 10−6

jω


Io = Ig[H(ω)] =400 × 10−6

jω· jω

jω + 40=

400 × 10−6

jω + 40

io(t) = 400e−40tu(t) µA

[b] Yes, at the time the source current jumps from −200 µA to +200 µA thecapacitor is charged to (200)(50) × 10−3 = 10 V, positive at the lowerterminal. The circuit at t = 0− is

At t = 0+ the circuit is

The time constant is (50 × 103)(0.5 × 10−6) = 25 ms.

·. .1τ

= 40 ·. . for t > 0, io = 400e−40t µA

P 17.27 [a] Vo =IgR(1/sC)R + (1/sC)

=IgR

RCs + 1

H(s) =Vo

Ig

=1/C

s + (1/RC)=

2 × 106

s + 40

H(ω) =2 × 106

40 + jω; Ig(ω) =

400 × 10−6

jω

Vo(ω) = H(ω)Ig(ω) =(

400 × 10−6

jω

)(2 × 106

40 + jω

)

=800

jω(40 + jω)=

20jω

− 2040 + jω

vo(t) = 10sgn(t) − 20e−40tu(t) V

[b] Yes, at the time the current source jumps from −200 to +200 µA the capacitoris charged to −10 V. That is, at t = 0−,vo(0−) = (50 × 103)(−200 × 10−6) = −10 V.

Problems 17–23

At t = ∞ the capacitor will be charged to +10 V. That is,vo(∞) = (50 × 103)(200 × 10−6) = 10 VThe time constant of the circuit is (50 × 103)(0.5 × 10−6) = 25 ms, so1/τ = 40. The function vo(t) is plotted below:

P 17.28 [a] ig = 3e−5|t|

·. . Ig(ω) =3

jω + 5+

3−jω + 5

=30

(jω + 5)(−jω + 5)

Vo

10+

Vos

10= Ig

·. .Vo

Ig

= H(s) =10

s + 1; H(ω) =

10jω + 1

Vo(ω) = Ig(ω)H(ω) =300

(jω + 1)(jω + 5)(−jω + 5)

=K1

jω + 1+

K2

jω + 5+

K3

−jω + 5

K1 =300

(4)(6)= 12.5

K2 =300

(−4)(10)= −7.5

K3 =300

(6)(10)= 5

Vo(ω) =12.5

jω + 1− 7.5

jω + 5+

5−jω + 5

vo(t) = [12.5e−t − 7.5e−5t]u(t) + 5e5tu(−t) V


[b] vo(0−) = 5 V

[c] vo(0+) = 12.5 − 7.5 = 5 V

[d] ig = 3e−5tu(t), t ≥ 0+

Ig =3

s + 5; H(s) =

10s + 1

vo(0+) = 5 V; γC = 0.5

Vo

10+

Vos

10= Ig + 0.5

Vo(s + 1) =30

s + 5+ 5

Vo =30

(s + 5)(s + 1)+

5s + 1

=−7.5s + 5

+7.5

s + 1+

5s + 1

=12.5s + 1

− 7.5s + 5

·. . vo(t) = (12.5e−t − 7.5e−5t)u(t) V

[e] Yes, for t ≥ 0+ the solution in part (a) is also

vo(t) = (12.5e−t − 7.5e−5t)u(t) V

P 17.29 [a]

Vo − Vg

sL1+

Vo

sL2+

Vo

R= 0

·. . Vo =RVg

L1

[s + R

( 1L1

+1L2

)]

Problems 17–25

Io =Vo

sL2

·. .Io

Vg

= H(s) =R/L1L2

s(s + R[(1/L1) + (1/L2)])

R

L1L2= 12 × 105

R( 1

L1+

1L2

)= 3 × 104

·. . H(s) =12 × 105

s(s + 3 × 104)

H(ω) =12 × 105

jω(jω + 3 × 104)

Vg(ω) = 125π[δ(ω + 4 × 104) + δ(ω − 4 × 104)]

Io(ω) = H(ω)Vg(ω) =1500π × 105[δ(ω + 4 × 104) + δ(ω − 4 × 104)]

jω(jω + 3 × 104)

io(t) =1500π × 105

2π

∫ ∞

−∞[δ(ω + 4 × 104) + δ(ω − 4 × 104)]ejtω

jω(jω + 3 × 104)dω

io(t) = 750 × 105

e−j40,000t

−j40,000(30,000 − j40,000)

+ej40,000t

j40,000(30,000 + j40,000)

=75 × 106

4 × 108

e−j40,000t

−j(3 − j4)+

ej40,000t

j(3 + j4)

=75400

e−j40,000t

5/− 143.13 +ej40,000t

5/143.13

= 0.075 cos(40,000t − 143.13) A

io(t) = 75 cos(40,000t − 143.13) mA


[b] In the phasor domain:

Vo − 125j200

+Vo

j800+

Vo

120= 0

12Vo − 1500 + 3Vo + j20Vo = 0

Vo =1500

15 + j20= 60/− 53.13 V

Io =Vo

j800= 75 × 10−3/− 143.13 A

io(t) = 75 cos(40,000t − 143.13) mA

P 17.30 [a]

Vo =Vgs

25 + (100/s) + s=

Vgs2

s2 + 25s + 100

H(s) =Vo

Vg

=s2

(s + 5)(s + 20); H(ω) =

(jω)2

(jω + 5)(jω + 20)

vg = 25ig = −450e10tu(−t) − 450e−10tu(t) V

Vg = − 450−jω + 10

− 450jω + 10

Vo(ω) = H(ω)Vg =−450(jω)2

(−jω + 10)(jω + 5)(jω + 20)

+−450(jω)2

(jω + 10)(jω + 5)(jω + 20)

=K1

−jω + 10+

K2

jω + 5+

K3

jω + 20+

K4

jω + 5+

K5

jω + 10+

K6

jω + 20

Problems 17–27

K1 =450(100)(15)(30)

= −100 K4 =−450(25)(5)(15)

= −150

K2 =450(25)(15)(15)

= −50 K5 =−450(100)(−5)(10)

= 900

K3 =450(400)(30)(−15)

= 400 K6 =−450(400)(−15)(−10)

= −1200

Vo(ω) =−100

−jω + 10+

−200jω + 5

+−800

jω + 20+

900jω + 10

vo = −100e10tu(−t) + [900e−10t − 200e−5t − 800e−20t]u(t) V

[b] vo(0−) = −100 V

[c] vo(0+) = 900 − 200 − 800 = −100 V

[d] At t = 0− the circuit is

Therefore, the solution predicts v1(0−) will be −350 V.Now v1(0+) = v1(0−) because the inductor will not let the current in the 25 Ωresistor change instantaneously, and the capacitor will not let the voltageacross the 0.01 F capacitor change instantaneously.At t = 0+ the circuit is

From the circuit at t = 0+ we see that vo must be −100 V, which is consistentwith the solution for vo obtained in part (c).


P 17.31

Vo − Vg

25+

100Vo

s+

Vos

100s + 125 × 104 = 0

·. . Vo =s(100s + 125 × 104)Vg

125(s2 + 12,000s + 25 × 106)

Io =sVo

100s + 125 × 104

H(s) =Io

Vg

=s2

125(s2 + 12,000s + 25 × 106)

H(ω) =−8 × 10−3ω2

(25 × 106 − ω2) + j12,000ω

Vg(ω) = 300π[δ(ω + 5000) + δ(ω − 5000)]

Io(ω) = H(ω)Vg(ω) =−2.4πω2[δ(ω + 5000) + δ(ω − 5000)]

(25 × 106 − ω2) + j12,000ω

io(t) =−2.4π

2π

∫ ∞

−∞ω2[δ(ω + 5000) + δ(ω − 5000)](25 × 106 − ω2) + j12,000ω

ejtω dω

= −1.2

25 × 106e−j5000t

−j(12,000)(5000)+

25 × 106ej5000t

j(12,000)(5000)

=612

e−j5000t

−j+

ej5000t

j

= 0.5[e−j(5000t+90) + ej(5000t+90)]

io(t) = 1 cos(5000t + 90) A

Problems 17–29

P 17.32 [a]

From the plot of vg note that vg is −10 V for an infinitely long time beforet = 0. Therefore

·. . vo(0−) = −10 V

There cannot be an instantaneous change in the voltage across a capacitor, so

vo(0+) = −10 V

[b] io(0−) = 0 AAt t = 0+ the circuit is

io(0+) =30 − (−10)

5=

405

= 8 A

[c] The s-domain circuit is

Vo =[

Vg

5 + (10/s)

] (10s

)=

2Vg

s + 2

Vo

Vg

= H(s) =2

s + 2


H(ω) =2

jω + 2

Vg(ω) = 5(

2jω

)− 5[2πδ(ω)] +

30jω + 5

=10jω

− 10πδ(ω) +30

jω + 5

Vo(ω) = H(ω)Vg(ω) =2

jω + 2

[10jω

− 10πδ(ω) +30

jω + 5

]

=20

jω(jω + 2)− 20πδ(ω)

jω + 2+

60(jω + 2)(jω + 5)

=K0

jω+

K1

jω + 2+

K2

jω + 2+

K3

jω + 5− 20πδ(ω)

jω + 2

K0 =202

= 10; K1 =20−2

= −10; K2 =603

= 20; K3 =60−3

= −20

Vo(ω) =10jω

+10

jω + 2− 20

jω + 5− 20πδ(ω)

jω + 2=

10jω

+10

jω + 2+

20jω + 5

− 10πδ(ω)

vo(t) = 5sgn(t) + [10e−2t − 20e−5t]u(t) − 5 V

P 17.33 [a]

(Vo − Vg)s106 +

Vo

4s+

Vo

800= 0

·. . Vo =s2Vg

s2 + 1250s + 25 × 104

Vo

Vg

= H(s) =s2

(s + 250)(s + 1000)

H(ω) =(jω)2

(jω + 250)(jω + 1000)

vg = 45e−500|t|; Vg(ω) =45,000

(jω + 500)(−jω + 500)

·. . Vo(ω) = H(ω)Vg(ω) =45,000(jω)2

(jω + 250)(jω + 500)(jω + 1000)(−jω + 500)

=K1

jω + 250+

K2

jω + 500+

K3

jω + 1000+

K4

−jω + 500

Problems 17–31

K1 =45,000(−250)2

(250)(750)(750)= 20

K2 =45,000(−500)2

(−250)(500)(1000)= −90

K3 =45,000(−1000)2

(−750)(−500)(1500)= 80

K4 =45,000(500)2

(750)(1000)(1500)= 10

·. . vo(t) = [20e−250t − 90e−500t + 80e−1000t]u(t) + 10e500tu(−t) V

[b] vo(0−) = 10 V; Vo(0+) = 20 − 90 + 80 = 10 V

vo(∞) = 0 V

[c] IL =Vo

4s=

0.25sVg

(s + 250)(s + 1000)

H(s) =IL

Vg

=0.25s

(s + 250)(s + 1000)

H(ω) =0.25(jω)

(jω + 250)(jω + 1000)

IL(ω) =0.25(jω)(45,000)

(jω + 250)(jω + 500)(jω + 1000)(−jω + 500)

=K1

jω + 250+

K2

jω + 500+

K3

jω + 1000+

K4

−jω + 500

K4 =(0.25)(500)(45,000)(750)(1000)(1500)

= 5 mA

iL(t) = 5e500tu(−t); ·. . iL(0−) = 5 mA

K1 =(0.25)(−250)(45,000)

(250)(750)(750)= −20 mA

K2 =(0.25)(−500)(45,000)(−250)(500)(1000)

= 45 mA

K3 =(0.25)(−1000)(45,000)(−750)(−500)(1500)

= −20 mA

·. . iL(0+) = K1 + K2 + K3 = −20 + 45 − 20 = 5 mA

Checks, i.e., iL(0+) = iL(0−) = 5 mA


At t = 0−:

vC(0−) = 45 − 10 = 35 V

At t = 0+:

vC(0+) = 45 − 10 = 35 V

[d] We can check the correctness of our solution for t ≥ 0+ by using the Laplacetransform. Our circuit becomes

Vo

800+

Vo

4s+

(Vo − Vg)s106 + 35 × 10−6 +

5 × 10−3

s= 0

·. . (s2 + 1250s + 25 × 104)Vo = s2Vg − (35s + 5000)

vg(t) = 45e−500tu(t) V; Vg =45

s + 500

·. . (s + 250)(s + 1000)Vo =45s2 − (35s + 5000)(s + 500)

(s + 500)

·. . Vo =10s2 − 22,500s − 250 × 104

(s + 250)(s + 500)(s + 1000)

=20

s + 250− 90

s + 500+

80s + 1000

·. . vo(t) = [20e−250t − 90e−500t + 80e−1000t]u(t) V

This agrees with our solution for vo(t) for t ≥ 0+.

P 17.34 [a]

Vg(ω) =36

4 − jω− 36

4 + jω=

72jω(4 − jω)(4 + jω)

Problems 17–33

Vo(s) =(16/s)

10 + s + (16/s)Vg(s)

H(s) =Vo(s)Vg(s)

=16

s2 + 10s + 16=

16(s + 2)(s + 8)

H(ω) =16

(jω + 2)(jω + 8)

Vo(ω) = H(ω) · Vg(ω) =1152jω

(4 − jω)(4 + jω)(2 + jω)(8 + jω)

=K1

4 − jω+

K2

4 + jω+

K3

2 + jω+

K4

8 + jω

K1 =1152(4)

(8)(6)(12)= 8

K2 =1152(−4)(8)(−2)(4)

= 72

K3 =1152(−2)(6)(2)(6)

= −32

K4 =1152(−8)

(12)(−4)(−6)= −32

·. . Vo(jω) =8

4 − jω+

724 + jω

− 322 + jω

− 328 + jω

·. . vo(t) = 8e4tu(−t) + [72e−4t − 32e−2t − 32e−8t]u(t)V

[b] vo(0−) = 8 V

[c] vo(0+) = 72 − 32 − 32 = 8 V

The voltages at 0− and 0+ must be the same since the voltage cannot changeinstantaneously across a capacitor.

P 17.35 Vo(s) =10s

+30

s + 20− 40

s + 30=

600(s + 10)s(s + 20)(s + 30)

Vo(s) = H(s) · 15s

·. . H(s) =40(s + 10)

(s + 20)(s + 30)

·. . H(ω) =40(jω + 10)

(jω + 20)(jω + 30)


·. . Vo(ω) =30jω

· 40(jω + 10)(jω + 20)(jω + 30)

=1200(jω + 10)

jω(jω + 20)(jω + 30)

vo(ω) =20jω

+60

jω + 20− 80

jω + 30

vo(t) = 10sgn(t) + [60e−20t − 80e−30t]u(t) V

P 17.36 [a] f(t) =12π

∫ 0

−∞eωejtω dω +

∫ ∞

0e−ωejtω dω

=

1/π1 + t2

[b] W = 2∫ ∞

0

(1/π)2

(1 + t2)2 dt =2π2

∫ ∞

0

dt

(1 + t2)2 =12π

J

[c] W =1π

∫ ∞

0e−2ω dω =

1π

e−2ω

−2

∣∣∣∣∞0

=12π

J

[d]1π

∫ ω1

0e−2ω dω =

0.92π

, 1 − e−2ω1 = 0.9, e2ω1 = 10

ω1 = (1/2) ln 10 ∼= 1.15 rad/s

P 17.37

Io =IgR

R + (1/sC)=

RCsIg

RCs + 1

H(s) =Io

Ig

=s

s + (1/RC)

RC = (100 × 103)(1.25 × 10−6) = 125 × 10−3;1

RC=

10.125

= 8

H(s) =s

s + 8; H(ω) =

jω

jω + 8

Ig(ω) =30 × 10−6

jω + 2

Io(ω) = H(ω)Ig(ω) =30 × 10−6jω

(jω + 2)(jω + 8)

Problems 17–35

|Io(ω)| =ω(30 × 10−6)

(√

ω2 + 4)(√

ω2 + 64)

|Io(ω)|2 =900 × 10−12ω2

(ω2 + 4)(ω2 + 64)=

K1

ω2 + 4+

K2

ω2 + 64

K1 =(900 × 10−12)(−4)

(60)= −60 × 10−12

K2 =(900 × 10−12)(−64)

(−60)= 960 × 10−12

|Io(ω)|2 =960 × 10−12

ω2 + 64− 60 × 10−12

ω2 + 4

W1Ω =1π

∫ ∞

0|Io(ω)|2 dω =

960 × 10−12

π

∫ ∞

0

dω

ω2 + 64− 60 × 10−12

π

∫ ∞

0

dω

ω2 + 4

=120 × 10−12

πtan−1 ω

8

∣∣∣∣∞0

−30 × 10−12

πtan−1 ω

2

∣∣∣∣∞0

=(120

π· π

2− 30

π· π

2

)× 10−12 = (60 − 15) × 10−12 = 45 pJ

Between 0 and 4 rad/s

W1Ω =[120

πtan−1 1

2− 30

πtan−1 2

]× 10−12 = 7.14 pJ

% =7.1445

(100) = 15.86%

P 17.38 [a] Vg(ω) =60

(jω + 1)(−jω + 1)

H(s) =Vo

Vg

=0.4

s + 0.5; H(ω) =

0.4(jω + 0.5)

Vo(ω) =24

(jω + 1)(jω + 0.5)(−jω + 1)

Vo(ω) =−24

jω + 1+

32jω + 0.5

+8

−jω + 1

vo(t) = [−24e−t + 32e−t/2]u(t) + 8etu(−t) V


[b] |Vg(ω)| =60

(ω2 + 1)

[c] |Vo(ω)| =24

(ω2 + 1)√

ω2 + 0.25

[d] Wi = 2∫ ∞

0900e−2t dt = 1800

e−2t

−2

∣∣∣∣∣∞

0

= 900 J

[e] Wo =∫ 0

−∞64e2t dt +

∫ ∞

0(−24e−t + 32e−t/2)2 dt

= 32 +∫∞0 [576e−2t − 1536e−3t/2 + 1024e−t] dt

= 32 + 288 − 1024 + 1024 = 320 J

Problems 17–37

[f] |Vg(ω)| =60

ω2 + 1, |V 2

g (ω)| =3600

(ω2 + 1)2

Wg =3600

π

∫ 2

0

dω

(ω2 + 1)2

=3600

π

12

(ω

ω2 + 1+ tan−1 ω

) ∣∣∣∣20

=1800

π

(25

+ tan−1 2)

= 863.53 J

·. . % =(863.53

900

)× 100 = 95.95%

[g] |Vo(ω)|2 =576

(ω2 + 1)2(ω2 + 0.25)

=1024

ω2 + 0.25− 768

(ω2 + 1)2 − 1024(ω2 + 1)

Wo =1π

1024 · 2 · tan−1 2ω

∣∣∣∣20

−768(1

2

)(ω

ω2 + 1+ tan−1 ω

)2

0

−1024 tan−1 ω

∣∣∣∣20

=2048

πtan−1 4 − 384

π

(25

+ tan−1 2)

− 1024π

tan−1 2

= 319.2 J

% =319.2320

× 100 = 99.75%

P 17.39 Io =0.5sIg

0.5s + 25=

sIg

s + 50

H(s) =Io

Ig

=s

s + 50

H(ω) =jω

jω + 50

I(ω) =12

jω + 10

Io(ω) = H(ω)I(ω) =12(jω)

(jω + 10)(jω + 50)


|Io(ω)| =12ω√

(ω2 + 100)(ω2 + 2500)

|Io(ω)|2 =144ω2

(ω2 + 100)(ω2 + 2500)

=−6

ω2 + 100+

150ω2 + 2500

Wo(total) =1π

∫ ∞

0

150dω

ω2 + 2500− 1

π

∫ ∞

0

6dω

ω2 + 100

=3π

tan−1(

ω

50

) ∣∣∣∣∞0

−0.6π

tan−1(

ω

10

) ∣∣∣∣∞0

= 1.5 − 0.3 = 1.2 J

Wo(0–100 rad/s) =3π

tan−1(2) − 0.6π

tan−1(10)

= 1.06 − 0.28 = 0.78 J

Therefore, the percent between 0 and 100 rad/s is

0.781.2

(100) = 64.69%

P 17.40 [a] |Vi(ω)|2 =4 × 104

ω2 ; |Vi(100)|2 =4 × 104

1002 = 4; |Vi(200)|2 =4 × 104

2002 = 1

Problems 17–39

[b] Vo =ViR

R + (1/sC)=

RCVi

RCs + 1

H(s) =Vo

Vi

=s

s + (1/RC);

1RC

=10610−3

(0.5)(20)=

100010

= 100

H(ω) =jω

jω + 100

|Vo(ω)| =200|ω| · |ω|√

ω2 + 104=

200√ω2 + 104

|Vo(ω)|2 =4 × 104

ω2 + 104 , 100 ≤ ω ≤ 200 rad/s; |Vo(ω)|2 = 0, elsewhere

|Vo(100)|2 =4 × 104

104 + 104 = 2; |Vo(200)|2 =4 × 104

5 × 104 = 0.8

[c] W1Ω =1π

∫ 200

100

4 × 104

ω2 dω =4 × 104

π

[− 1

ω

]200

100

=4 × 104

π

[ 1100

− 1200

]=

200π

∼= 63.66 J

[d] W1Ω =1π

∫ 200

100

4 × 104

ω2 + 104 dω =4 × 104

π· tan−1 ω

100

∣∣∣∣200

100

=400π

[tan−1 2 − tan−1 1] ∼= 40.97 J


P 17.41 [a] Vi(ω) =A

a + jω; |Vi(ω)| =

A√a2 + ω2

H(s) =s

s + α; H(ω) =

jω

α + jω; |H(ω)| =

ω√α2 + ω2

Therefore |Vo(ω)| =ωA√

(a2 + ω2)(α2 + ω2)

Therefore |Vo(ω)|2 =ω2A2

(a2 + ω2)(α2 + ω2)

WIN =∫ ∞

0A2e−2at dt =

A2

2a; when α = a we have

WOUT(a) =A2

π

∫ a

0

ω2 dω

(ω2 + a2)2 =A2

π

∫ a

0

dω

a2 + ω2 −∫ a

0

a2 dω

(a2 + ω2)2

=A2

4aπ

(π

2− 1

)

WOUT(total) =A2

π

∫ ∞

0

[ω2

(a2 + ω2)2

]dω =

A2

4a

ThereforeWOUT(a)

WOUT(total)= 0.5 − 1

π= 0.1817 or 18.17%

[b] When α = a we have

WOUT(α) =1π

∫ α

0

ω2A2dω

(a2 + ω2)(α2 + ω2)

=A2

π

∫ α

0

[K1

a2 + ω2 +K2

α2 + ω2

]dω

where K1 =a2

a2 − α2 and K2 =−α2

a2 − α2

Therefore

WOUT(α) =A2

π(a2 − α2)

[a tan−1

(α

a

)− απ

4

]

WOUT(total) =A2

π(a2 − α2)

[aπ

2− α

π

2

]=

A2

2(a + α)

ThereforeWOUT(α)

WOUT(total)=

2π(a − α)

·[a tan−1

(α

a

)− απ

4

]

For α = a√

3, this ratio is 0.2723, or 27.23% of the output energy lies in thefrequency band between 0 and a

√3.

[c] For α = a/√

3, the ratio is 0.1057, or 10.57% of the output energy lies in thefrequency band between 0 and a/

√3.

Technology

Circuits nilsson 7th solution manual