Prexams · Web viewSolution : From symmetry, nodes aa are at the same voltage, as are nodes bb and cc. These nodes can therefore be joined together, which connects four resistors

EECE 210 – Final ExamJanuary 20, 2012

3%

1. Determine Vy, assuming VSRC = 1 V.

A. 2 V

B. 4 V

C. 5 V

D. 1 V

E. 3 V

Solution: Vx = -2Vx1, which makes Vx = 0. It follows that Vy = VSRC.

Version 1: VSRC = 1 V, Vy = VSRC = 1 V




Version 5: VSRC = 5 V, Vy = VSRC = 5 V.

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2. Determine Vac, assuming all

resistances are 10 and VSRC = 2 V.

A. 3 V

B. 1 V

C. 5 V

D. 4 V

E. 2 V

Solution: From symmetry, nodes aa are at the same voltage, as are nodes bb and cc.

These nodes can therefore be joined

together, which connects four resistors in

parallel in the two sets of resistors in the

middle, and results in the circuit shown.

Adding the two 2.5 resistors and

splitting the resulting 5 resistor into two

10 resistors in parallel gives Vac =

VSRC/2.

Version 1: VSRC = 2 V, Vac = VSRC/2 = 1 V



1

–+

–

+

VY 2Vx mA

– +Vx

VSRC

1 k

–

+

–

+VSRC VSRC

a

c

–

+

–

+VSRC VSRC

a

c

b

a

b

c


Version 5: VSRC = 10 V, Vac = VSRC/2 = 5 V.

3%

3. Given a load impedance 0.1(4 + j3) . Determine the susceptance of the load.

A. 0.6 S

B. -0.6 S

C. -1.2 S

D. -0.3 S

E. 1.2 S

Solution: The load admittance is Y L=

1Z L

= 1 /k4+ j3

= 125 k

(4− j3 ). Hence,

B=− 325 k S.

Version 1: k = 0.1, B=− 3

25 k=− 3

2 .5=−1.2

S

Version 2: k = 0.2, B=− 3

25 k=−3

5=−0.6

S

Version 3: k = 0.3, B=− 3

25 k=− 3

7 .5=−0 . 4

S

Version 4: k = 0.4, B=− 3

25 k=− 3

10=−0 . 3

S

Version 5: k = 0.5, B=− 3

25 k=− 3

12 . 5=−0. 24

S.

3%

4. If vSRC = 3cost V, determine at which

the voltage across the 10 resistor is a

maximum and specify this voltage.

A. 2 V, 2 krad/s

B. 4 V, 2 krad/s

C. 8 + j6 V, 1 krad/s

D. 8 – j6 V, 2 krad/s

E. 4 V, 4 krad/s

Solution: vSRC = Acost V. The voltage across the 10 resistor is maximum when the

parallel impedance of L and C is infinite, that is, when XL = -XC, which makes

( jX L)×( jX C )jX L+ jXC

. This occurs when ωL=1/ωC , or ω=1/√ LC=1/√0 .5×10−3×0. 5×10−3=2×103 rad/s.

2

–+vSRC 0.5 mF 0.5 mH 10

5

The voltage is VR = 2A/3 V.

Version 1: A = 3, VR = 2A/3 = 2 V

Version 2: A = 6, VR = 2A/3 = 4 V

Version 3: A = 9, VR = 2A/3 = 6 V

Version 4: A = 12, VR = 2A/3 = 8 V

Version 5: A = 15, VR = 2A/3 = 10 V.

3%

5. Determine the total reactive power absorbed by L and C in Problem 4 at = 4 krad/s

and with the 5 resistor replaced by a short circuit.

A. -13 VAR

B. +6.75 VAR

C. -27 VAR

D. 13 VAR

E. -6.75 VAR

Solution: The total reactive power absorbed is Q= A2

2 ( 1ωL

−ωC)=

A2

2 ( 14×103×0 . 5×10−3−4×103×0 .5×10−3)= A2

2(0 .5−2 )=−0 .75 A2

VAR.

Version 1: A = 3, Q=−0 .75 A2= -6.75 VAR

Version 2: A = 6, Q=−0 .75 A2= -27 VAR

Version 3: A = 9, Q=−0 .75 A2= -60.75 VAR

Version 4: A = 12, Q=−0 .75 A2= -108 VAR

Version 5: A = 15, Q=−0 .75 A2= -168.75 VAR.

3%6. Determine X so that the current in the 5 resistor

is zero, given that a = 2, and for the linear

transformer, L1 = M = 10 and L2 = 20 .

A. 2.22

B. -2.22

C. 2.5

D. -2.5

E. 1.88

Solution: To have zero current in the resistor, the

3

1:a

L1 L2

M

5 jX

–

+

100 V

jX

–

+

V

–

+

V

V/jM

aV/jM

–

+

V

–

+

V/a

1:a

open-circuited output of the linear transformer should be V. The input current is V/jM, and

the input voltage is V

jωL1

jωM=V

. The primary voltage of the ideal transformer is V/a, and the

primary current is aV/jM. It follows that jX =

V (1−1 /a )aV / jωM =( 1

a−1a2 ) jωM

.

Version 1: a = 2, X =( 1

a−1a2 )ωM=(0 .5−0 .25 )10=2 .5

Version 2: a = 3, X =( 1

a−1a2 )ωM=(0 .333−0 .111)10=2. 22

Version 3: a = 4, X =( 1

a−1a2 )ωM=(0 .25−0 . 0625)10=1 . 88

Version 4: a = 5, X =( 1

a−1a2 )ωM=(0 .2−0 .04 )10=1 .6

Version 5: a = 6, X =( 1

a−1a2 )ωM=(0 .167−0 . 028)10=1 .39

.

3%

7. A load absorbs a real power of 12 kW and a capacitive, reactive power of 5 kVAR.

Determine the power factor of the load.

A. 0.8

B. 0.92

C. 0.86

D. 0.83

E. 0.89

Solution: The complex power is √Q2+122. The p.f. is 12/√Q2+122

leading.

Version 1: Q = 5 kVAR, p.f. = 12/√Q2+122 = 0.92 leading




Version 5: Q = 9 kVAR, p.f. = 12/√Q2+122 = 0.8 leading.

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4

8. The load of Problem 7 is supplied at 250 V rms, 50 Hz. Determine the inductance that

must be placed in series with the load to make the total power factor of the load and the

inductor equal to unity.

A. 7.22 mH

B. 7.65 mH

C. 6.63 mH

D. 7.96 mH

E. 5.89 mH

Solution: The complex power in the preceding problem is SL = 12,000 – j1,000Q VA.

Considering the load voltage to be VL0 V rms, the load current is IL =

12 ,000+ j 1, 000 QV L A.

|I L|2=

106 (144+Q2)V L

2 A2. Hence,

1000 Q=106 (144+Q2 )

V L2 ωL

, or L =

QV L2

105 π (144+Q2) .

Version 1: Q = 5 kVAR, L =

QV L2

100 π (144+Q2) = 5.89 mH


QV L2

100 π (144+Q2) = 6.63 mH


QV L2

100 π (144+Q2) = 7.22 mH


QV L2

100 π (144+Q2) = 7.65 mH


QV L2

100 π (144+Q2) = 7.96 mH.

3%

9. If Rsrc = 2 and RL and XL are

variable over an arbitrary

range, determine RL and XL

that will maximize power

transfer to RL.

A. 8 , -j10

B. 10 , j10

C. 12 , -j10

D. 8 , j10

E. 10 , -j10

5

–+

RLj10

1:2

VSRC

Rsrc

jXL

Solution: XLm = -j10 so as to cancel the +j10 , and RLm = 4Rsrc so as to match the source

and load resistances.

Version 1: Rsrc = 2 , RLm = 4Rsrc = 8

Version 2: Rsrc = 2.5 , RLm = 4Rsrc = 10

Version 3: Rsrc = 3 , RLm = 4Rsrc = 12

Version 4: Rsrc = 3.5 , RLm = 4Rsrc = 14

Version 5: Rsrc = 4 , RLm = 4Rsrc = 16 .

3%

10. In the circuit of Problem 9, Rsrc = 20 , RL = 3 , and XL = -j6 . Determine the ratio of

primary to secondary turns that will maximize power transfer to RL.

A. 4

B. 5

C. 3

D. 2

E. 6

Solution: Let the primary to secondary turns ratio be a. The secondary load impedance is ZL

= 3 + j10 – j6 = 3 + j4 . When reflected to the primary side, this becomes (3 + j4)a2, of

magnitude 5a2. For maximum power transfer, 5a2 = Rsrc, or a=√ Rsrc /5 .

Version 1: Rsrc = 20 , a=√20/5=2

Version 2: Rsrc = 45 , a=√45 /5=3

Version 3: Rsrc = 80 , a=√80/5=4

Version 4: Rsrc = 125 , a=√125/5=5

Version 5: Rsrc = 180 , a=√180/5=6 .

3%

11. If the primary to secondary turns ratio in problem 10 is 4 and vSRC = 16sin(100t) V,

determine the power dissipated in the 3 resistor.

A. 0.70 W

B. 0.18 W

C. 0.11 W

D. 0.48 W

E. 0.30 W

Solution: The magnitude of the source voltage referred to the secondary is 16/4 = 4 V, and

6

the source resistance is Rsrc /16 . The magnitude of the load current is

|I L|=4√2

1

√ (Rsrc /16+3)2+ (4 )2 A rms, and the power dissipated is PL=

8×3

(R src /16+3)2+(4 )2

=24

(Rsrc /16+3 )2+16 W.

Version 1: Rsrc = 20 , PL=

24(20/16+3 )2+16 = 0.70 W


24(45 /16+3 )2+16 = 0.48 W


24(80/16+3 )2+16 = 0.3 W


24(125/16+3 )2+16 = 0.18 W


24(180/16+3 )2+16 = 0.11 W.

3%

12. The exponential form of the Fourier series expansion of a periodic function is

¿ n≠0 ¿¿ ¿n=∞

e− jn ω0 t

¿. Determine the second harmonic in trigonometric form.

A. 0.5cos(n0t + 60)

B. 2cos(n0t + 60)

C. cos(n0t + 60)

D. 2.5cos(n0t + 60)

E. 1.5cos(n0t + 60)

Solution: The ac component is

¿ n≠0 ¿¿ ¿n=∞

e− jn ω0 t

¿. Cn=

An3

(1+ j√3 )=an− jbn

2 , hence

an=2 An3

and bn=−2 A √3

n3. The trigonometric form of the nth harmonic is

2 An3

cosnω0 t−

2 A√3n3 sin ω0 t

=

4 An3 [ 1

2cosnω0 t−√3

2sin nω0 t ]=4 A

n3 [cos60∘cosnω0 t−sin 60∘sin nω0 t ] =

4 An3

cos ( nω0 t +60∘). The second harmonic is

f 2( t )= A2

cos (2 ω0 t+60∘ ).

7

Version 1: A = 1, f 2( t )= A

2cos (2 ω0 t+60∘ )=0.5 cos (2 ω0 t+60∘)

Version 2: A = 2, f 2( t )= A

2cos (2 ω0 t+60∘ )=cos (2 ω0 t+60∘ )

Version 3: A = 3, f 2( t )= A

2cos (2 ω0 t+60∘ )=1 . 5cos (2 ω0 t +60∘)

Version 4: A = 4, f 2( t )= A

2cos (2 ω0 t+60∘ )=2 cos (2ω0 t +60∘)

Version 5: A = 5, f 2( t )= A

2cos (2 ω0 t+60∘ )=2 . 5 cos (2 ω0 t+60∘ )

.

3%

13. The function shown has a0 = 0 and for n = 1, 2, 3,

etc.:

A. an = 0, bn = 0 for even n only and can be

determined by integrating the function from

t = 0 to t = T/2

B. an = 0, bn = 0 for even n only and can be

determined by integrating the function from

t = 0 to t = T/4

C. an = 0, bn = 0 for odd n only and can be determined by integrating the function from

t = 0 to t = T/2

D. an = 0, bn = 0 for odd n only and can be determined by integrating the function from

t = 0 to t = T/4

E. an 0, bn 0, for all n, and both can be determined by integrating the function from t

= 0 to t = T/2

Solution: The waveform is half-wave symmetric, is not quarter-wave symmetric, and is

neither odd nor even. Hence, an = 0, bn = 0 for even n, and can be determined by integrating

the function from t = 0 to t = T/2.

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14. If Frms is the rms value of a half-wave symmetric, periodic waveform, determine the rms

value of the same function but with the negative half-waves missing.

A. F rms/3

B. F rms/√3

C. πFrms/√2

8

A

2A

-A

-2A

T/4T/2

3T/4T t

D. F rms/√2

E. F rms/2

Solution: F rms2 =( rms of positive half cycles )2 + (rms of negative half cycles )2 =

2 (rms of positive half cycles )2 . Hence, (rms of positive half cycles )=F rms /√2

3%

15. Specify how the magnitude of the nth harmonic varies

with n for the periodic function shown.

A. n

B. n2

C. 1/n

D. 1/n2

E. 1/n3

Solution: Since the function is discontinuous, the magnitude of the nth harmonic varies as

1/n.

11%16. Determine Thevenin’s equivalent circuit

between terminals ab, assuming VSRC =

50 V. Express VTh in polar coordinates

and ZTh in rectangular coordinates.

Solution: The current in the LC branch is VX/j5

= -j0.2Vx, and the current in the middle

branch is 0.2(1 – j)Vx. From KVL, VSRC –

50.2(1 – j)Vx = VX + (-j2.5)(-j0.2)Vx =

0.5Vx, or Vx = VSRC/(1.5 – j), VTh = 0.5VX =

V SRC

3− j2=

V SRC

13(3+ j 2)=

V SRC

√13∠ tan−1(2/3 )=

V SRC

√13∠33 .7∘

V. When

terminals ab are short circuited, the current source is zero, and ISC =

V SRC

5 A. It follows that

ZTh = 5

13(3+ j 2)=

(1.15 + j0.77) .

9

–+VSRC

5 0.2VX

j5

-j2.5

b

a

–

+VX

–+VSRC

5

j5

-j2.5

b–

+VX

b

a

0.2VX

-j0.2VX

0.2(1 – j)VX

Version 1: VSRC = 50 V, V Th=

5√13

∠33. 7∘=1 .39∠33 .7∘

V


10√13

∠33.7∘=2 .77∠33 .7∘

V


15√13

∠33. 7∘=4 .16∠33 .7∘

V


20√13

∠33.7∘=5 .55∠33 .7∘

V


25√13

∠33. 7∘=6 . 93∠33. 7∘

V.

11%17. Determine vO(t), assuming vSRC =

2cos1000t V.

Solution: The impedances of the reactive

elements are: mH j8 , 1 mH j , 2 mH

j2 , and 0.5 mF -j2 . The circuit in the

frequency domain becomes as shown, with the

linear transformer replaced by its T-

equivalent circuit. The parallel

impedance to the right of terminals ab is

(3+ j3 )(− j3 )3 =3 (1− j ) , and Vab =

3(1− j)3− j3+2+ j 8

V SRC=

3(1− j )5(1+ j )

V SRC= -

j0.6VSRC. It follows that

V O=− j0 .6×33(1+ j)

V SRC= -0.3(1 + j)VSRC =−0 .3√2V SRC∠ 45∘

V, and vO(t) =

−0 .3√2V m cos(1000 t+45∘) V.

Version 1: Vm = 2, vO(t) = −0 .3√2V m cos(1000 t+45∘) = -0.85cos(1000t + 45) V




Version 5: Vm = 6, vO(t) = −0 .3√2 V SRC cos(1000 t+45∘) = -2.55cos(1000t + 45) V.

10

1 mH

1 mH 2 mH

6 mH2

–+

3

0.5 mF–

+

vOvSRC

2

–+

3

–

+

VOVSRC

-j2

-j

j2 j3 j6

b

a

11%18. RL is variable in the range (1, 8 ).

Determine RL for maximum power

transfer and the maximum power

transferred to it, assuming B = 4.

Solution: Thevenin’s voltage for the

Bsin20t source, as seen by RL, is

Bsin20t V. From superposition, Thevenin’s voltage for the cos0t sources is (2 +

100.2)cos0t = 4cos0t V. RTh = 10 . Hence, for maximum power transfer RL = 8 . The

power dissipated in RL is PL = ( B√2×18 )

2×8+( 4

√2×18 )2×8=( B2+16 )/81

W.

Version 1: B = 4, PL = (B2 + 16)/81 = 32/81 = 0.396 W

Version 2: B = 6, PL = (B2 + 16)/81 = 52/81 = 0.642 W

Version 3: B = 8 PL = (B2 + 16)/81 = 80/81 = 0.988 W

Version 4: B = 10, PL = (B2 + 16)/81 = 116/81 = 1.432 W

Version 5: B = 12, PL = (B2 + 16)/81 = 160/81 = 1.975 W.

11%19. Derive the first four terms of the trigonometric Fourier

series expansion of the periodic function f(t) = Asint,

-/2 t /2, with A = 1. Note that:

2sin α sin β=cos (α−β )−cos( α+β )

2 cosα cos β=cos (α−β )+cos (α+β )

2sin α cos β=sin(α+ β )+sin(α−β )

2 cosα sin β=sin(α+ β )−sin (α−β )

Solution: The function is odd and of zero average. Hence a0 = 0, an =0 for all integer n, and

bn=4T ∫0

T /2A sin t sin nω0 tdt

, where T = and 0 = 2/T = 2. Using the trigonometric relation for

the product of two sine functions, bn=

2 Aπ ∫0

π /2[cos(2 n−1) t−cos(2 n+1 )t ] dt

=

2 Aπ [sin(2n−1) t

2 n−1−

sin(2n+1) t2 n+1 ]

0

π /2

=

2 Aπ [ 1

2n−1sin(2 n−1) π

2− 1

2 n+1sin(2 n+1) π

2 ]0π /2

=(−1 )n+1 8 Aπ

n4 n2−1 . Hence,

11

–+

2cos0t V

10

–+

5

RL

Bsin20t V

0.2cos0t A

A

-A

/2-/2 t

f(t)

f ( t )=8 Aπ [ 1

3sin 2 t− 2

15sin 4 t+ 3

35sin 6 t− 4

63sin 8 t +.. .. ]

Version 1: A = 1, f ( t )= 8

π [ 13

sin 2 t− 215

sin 4 t + 335

sin 6 t− 463

sin 8 t+. . ..]Version 2: A = 2,

f ( t )=16π [ 1

3sin2 t− 2

15sin 4 t+ 3

35sin 6 t− 4

63sin 8 t +.. . .]

Version 3: A = 3, f ( t )=24

π [ 13

sin 2t− 215

sin 4 t+ 335

sin 6t− 463

sin 8 t+. .. .]Version 4: A = 4,

f ( t )=32π [ 1

3sin 2 t− 2

15sin 4 t + 3

35sin 6 t− 4

63sin 8 t+. . ..]

Version 5: A = 5, f ( t )=40

π [ 13

sin 2t− 215

sin 4 t+ 335

sin 6 t− 463

sin 8 t+. .. .] .11%20. Determine the rms value of the periodic function shown,

assuming A = 1.

Solution: The function has an average value of 0.25 A,

because the areas under the positive and negative triangles

cancel out, and the areas under the rectangles are 0.5 A –

0.25A. Alternatively, it can be argued that if 0.25A is

subtracted from f(t), the function becomes symmetrical with

respect to the horizontal axis The ac component of f(t) = A(2t + 0.75), 0 t 0.25 s. The

area under the square of the function over this time interval is A2∫0

0 .25(2t +0 .75 )2 dt

=

A2 [ 43

t 3+ 32

t2+ 916

t ]0

0. 25

=A2 ( 4

192+ 3

32+ 9

64 )=

A2 (49192 )

The mean square is obtained by

dividing by 0.25, which gives 49 A2

48 . The square of the rms value of f(t) is 49 A2

48+ A2

16 =

52 A2

48=13 A2

12 and the rms value is A√13/12 .

Version 1: A = 1, rms value is A√13/12= 1.041




12

A

1.5A

-0.5A-A

0.25 0.750.5 1 t, s

f(t)

Version 5: A = 5, rms value is A√161/48= 5.204.

13

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Prexams · Web viewSolution : From symmetry, nodes aa are at the same voltage, as are nodes bb and cc. These nodes can therefore be joined together, which connects four resistors