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CHAPTER 4
Exercises
E4.1 The voltage across the circuit is given by Equation 4.8: )/exp()( RCtVtv iC −=
E4.3 (a) In dc steady state, the capacitances act as open circuits and the
inductances act as short circuits. Thus the steady-state (i.e., t approaching infinity) equivalent circuit is:
From this circuit, we see that Then ohm’s law gives the voltage
as
in which Vi is the initial voltage. At the time t for which the voltage reaches 1% of the initial value, we have
1%
Taking the natural logarithm of both sides of the equation, we obtain ln( Solving and substituting values, we find t1% = 4.605RC = 23.03 ms.
)/exp(01.0 %1 RCt−=
RCt /605.4)01.0 %1−=−=
E4.2 The exponential transient shown in Figure 4.4 is given by
Taking the derivative with respect to time, we have )/exp()( τtVVtv ssC −−=
)/exp()(τ
τtV
dttdv sC −=
Evaluating at t = 0, we find that the initial slope is Because this matches the slope of the straight line shown in Figure 4.4, we have shown that a line tangent to the exponential transient at the origin reaches the final value in one time constant.
./τSV
A. 2 =aiV. 50== aa Riv
101
(b) The dc steady-state equivalent circuit is:
Here the two 10-Ω resistances are in parallel with an equivalent
resistance of 1/(1/10 + 1/10) = 5 Ω. This equivalent resistance is in series with the 5-Ω resistance. Thus the equivalent resistance seen by the source is 10 Ω, and Using the current division principle, this current splits equally between the two 10-Ω resistances, so we have
E4.4 (a)
(b) Just before the switch opens, the circuit is in dc steady state with an inductor current of This current continues to flow in the inductor immediately after the switch opens so we have . This current must flow (upward) through R2 so the initial value of the voltage is
A. 210/201 ==i
A. 132 == ii
ms 1100/1.0/ 2 === RLτ
A. 5.1/ 1 =RVs
A 5.1)0( =+i
V. 150)0()0( 2 −=+−=+ iRv (c) We see that the initial magnitude of v(t) is ten times larger than the source voltage. (d) The voltage is given by
)1000exp(150)/exp()(1
ttR
LVtv s −−=−−= ττ
Let us denote the time at which the voltage reaches half of its initial magnitude as tH. Then we have
)1000exp(5.0 Ht−= Solving and substituting values we obtain
ms 6931.0)2ln(10)5.0ln(10 33 ==−= −−Ht
102
E4.5 First we write a KCL equation for
Taking the derivative of each term of this equation with respect to time and multiplying each term by R, we obtain:
The solution to this equation is of the form: )(tv
.0≥t
∫ =++t
dxxvLR
tv0
20)(1)(
0)()(=+ tv
LR
dttdv
)/exp( τtK −=
in which / == RLτ is the time constant and K is a constant that must be chosen to fit the initial conditions in the circuit. Since the initial (t = 0+) inductor current is specified to be zero, the initial current in the resistor must be 2 A and the initial voltage is 20 V:
Thus, we have
s 2.0
Kv ==+ 20)0(
(tv
(tiL
)/exp(20) τt−= )/exp(2/ τtRviR −==
[ ] )/exp(22)/exp(2021)(1)
00
τττ txdxxvL
tt
−−=−−== ∫
E4.6 Prior to t = 0, the circuit is in DC steady state and the equivalent circuit is
Thus we have i(0-) = 1 A. However the current through the inductor cannot change instantaneously so we also have i(0+) = 1 A. With the switch open, we can write the KVL equation:
100)(200)(=+ ti
dttdi
The solution to this equation is of the form )/exp()( 21 τtKKti −+=
in which the time constant is In steady state with the switch open, we have Then using the initial
ms. 5200/1 ==τA. )( 5.0200/1001 ===∞ Ki
103
current, we have , from which we determine that Thus we have
211)0( KKi +==+
.5.02 =K
for )/exp(5.00 for A−+
<
τt( ti
t for )/exp(0 for
)(
>−
<
τtt
dt( ttv
)0()( −++vdxxi C(Ri
200sin(400)( + Cti
400)()( titt
×−=+
200sin()200cos( Bt +
)200sin(cos(200)200sin(
6 t 105 3 Bt−
−=
−× +
610400 −× B=A
sin(200)200cos(( t −
0.t 5.0 0.1)
>=
=t
0.100
V 0
)
−=
=
=diL
E4.7 As in Example 4.4, the KVL equation is
Taking the derivative and multiplying by C, we obtain
RC
Substituting values and rearranging the equation becomes
105 ×
0)200cos(21)0
=+ ∫ tC
tt
0))(=+ t
dttdi
)200sin(10 63 td
di −−
The particular solution is of the form (tip Substituting this into the differential equation and rearranging terms
results in
)) tA=
Equating the coefficients of the cos and sin terms gives the following equations:
−A and
[ ]10400
)200sin()200cos()200200 tBtAtA−×
++
−=+B 0=+A from which we determine that and .
Furthermore, the complementary solution is , and the complete solution is of the form
)ti
610200 −× 610200 −×−=B)/exp()( τtKtiC −=
A )/exp()200200 µτtKt −+=
At t = 0+, the equivalent circuit is
104
from which we determine that Then evaluating our solution at t = 0+, we have from which we determine that Thus the complete solution is
E4.8 The KVL equation is
Taking the derivative and multiplying by C, we obtain
RC
A. 4005000/2)0( µ==+i,200400)0( Ki +==+
µA. 200=KA )/exp(200)200sin(200)200cos(200)( µτtttti −+−=
0)exp(10)0()(1)(0
=−−+++ ∫ tvdxxiC
tRi C
t
0)exp(10)()(=−++ tCti
dttdi
Substituting values and rearranging, the equation becomes
2 d
)exp(1020)()( 6 ttidt
ti−×−=+ −
The particular solution is of the form (tip Substituting this into the differential equation and rearranging terms
results in
)exp() tA −=
)exp(1020)exp()exp(2 6 ttAtA −×−=−+−− −
Equating the coefficients gives Furthermore, the complementary solution is , and the complete solution is of the form
.1020 6−×=A)2/exp()( tKtiC −=
At t = 0+, the equivalent circuit is
A )2/exp()exp(20)( µtKtti −+−=
105
from which we determine that Then evaluating our solution at t = 0+, we have from which we determine that Thus the complete solution is
A. 510/5)0( 6 µ==+i,205)0( Ki +==+
µA. 15−=KA )2/exp(15)exp(20)( µttti −−−=
E4.9 (a) 2
5730 10
101011
=×
==−−LC
ω 51022
1×==
RCα ==
0ωαζ
(b) At t = 0+, the KCL equation for the circuit is
0 (1)
However, (( = vv , because the voltage across the capacitor cannot change instantaneously. Furthermore, , because the current through the inductance cannot change value instantaneously. Solving Equation (1) for and substituting values, we find that
V/s.
)0()0()0(1. +′++++
= vCiR
vL
0)0)0 =−+0)0()0( =−=+ LL ii
)0( +′v610)0( =+′v
(c) To find the particular solution or forced response, we can solve the circuit in steady-state conditions. For a dc source, we treat the capacitance as an open and the inductance as a short. Because the inductance acts as a short
(d) Because the circuit is overdamped ( the homogeneous solution is the sum of two exponentials. The roots of the characteristic solution are given by Equations 4.72 and 4.73:
.0)( =tv p
),1>ζ
320
21 102.373 ×−=−−−= ωααs
320
22 1079.26 ×−=−+−= ωααs
Adding the particular solution to the homogeneous solution gives the general solution:
106
)exp()exp()( 2211 tsKtsKtv += Now using the initial conditions, we have
210)0( KKv +==+ 2211610)0( sKsKv +==+′
Solving we find Thus the solution is:
.887.2 and 887.2 21 =−= KK)]exp()[exp(887.2)( 12 tststv −=
E4.10 (a) 1
5730 10
101011
=×
==−−LC
ω 5102
1==
RCα ==
0ωαζ
(b) The solution for this part is the same as that for Exercise 4.9b in which we found that V/s.
(c) The solution for this part is the same as that for Exercise 4.9c in
which we found
(d) The roots of the characteristic solution are given by Equations 4.72 and 4.73:
Because the circuit is critically damped the roots are equal and the homogeneous solution is of the form:
610)0( =+′v
.0)( =tv p
520
21 10−=−−−= ωααs 52
02
2 10−=−+−= ωααs),1( =ζ
)exp()exp()( 1211 tstKtsKtv += Adding the particular solution to the homogeneous solution gives the general solution:
)exp()exp()( 1211 tstKtsKtv +=
Now using the initial conditions we have
Solving we find Thus the solution is: 10)0( Kv ==+ 211
610)0( KsKv +==+′6
2 10=K)10exp(10)( 56 tttv −=
E4.11 (a) 0.2
5730 10
101011
=×
==−−LC
ω 41022
1×==
RCα ==
0ωαζ
(b) The solution for this part is the same as that for Exercise 4.9b in which we found that V/s.
610)0( =+′v
107
(c) The solution for this part is the same as that for Exercise 4.9c in which we found
(d) Because we have ( this is the underdamped case and we have
Adding the particular solution to the homogeneous solution gives the general solution:
Now using the initial conditions we have
.0)( =tv p
),1<ζ322
0 1098.97 ×=−= αωωn
)sin()exp()cos()exp()( 21 ttKttKtv nn ωαωα −+−=
10)0( Kv ==+ 21610)0( KKv nωα +−==+′
Solving we find Thus the solution is:
21.102 =KV )1098.97sin()102exp(21.10)( 34 tttv ××−=
Problems
P4.1 The time constant is the interval required for the voltage to fall to exp(-1) ≅ 0.368 times its initial value. The time constant is given by
Thus to attain a long time constant, we need large values for both R and C.
τ
.RC=τ
P4.2 We have ( )v and in which Vi is the initial voltage. The initial stored energy is After two time constants, we have so that 13.53 percent of the initial voltage remains. The energy is so only 1.83% of the initial energy remains.
P4.3 The solution is of the form given in Equation 4.19:
)/exp( τtVt i −= )()( 22
1 tCvtw =
.22
1ii CVW =
( ) i iVVtv 1353.0)/2exp( ≅−= ττ
iii WWCVCvw 0183.0)4exp()4exp()2()2( 22
122
1 ≅−=−== ττ
( ) ( )ms 11001.010
exp65 =××=
−−=−RCRCtVVtv ssC
Thus, we have
( ) ( )310exp100100 −−−= ttvC
108
P4.4* The solution is of the form of Equation 4.17: ( )vC
in which is a constant to be determined. At , we have
( ) ( )322 10exp100exp −−+=−+= tKRCtKVt s
2K += 0t( ) 2100500 KvC +=−=+
Solving, we find that and the solution is
1502 −=K( ) ( )310exp150100 −−−= ttvC
P4.5* The voltage across the capacitor is given by Equation 4.8.
( ) )/exp( RCtVtv i −=
in which V is the initial voltage, C = 100 µF is the capacitance, and R is the leakage resistance. The energy stored in the capacitance is
100=iV
Since we require the energy to be 90% of the initial value after one minute, we can write
)/2exp(100105.0)(21 242 RCttCvw −××== −
)/120exp(100105.0100105.09.0 2424 RC−××≤××× −−
109
Solving we determine that must be greater than 1139 s. Thus the leakage resistance must be greater than 11.39 MΩ.
P4.6 The voltage across the capacitor is given by Equation 4.8.
RC
( ) )/exp( RCtVtv i −=
in which Vi is the initial voltage. Substituting values, we have (v ) )5exp()0004.0/002.0exp(10102 -3 −=−==× ii VV
P4.7 (a)
V 1484=iV
(b)
(c)
( )( ) ( )
( )( ) ( ) 0 for 50exp5002.0exp50
0 for 00 for 50exp5002.0exp50
0 for 50vms 20
C
>−=−=
<=
>−=−=
<=
=
tttttv
ttttt
RC
R
( ) ( )[ ] ( ) W 100exp25002
µtRtvtp R
R −==
( )
( )
( )100exp25
100exp2500
0
0
0
−−=
−=
=
∞
∞
∞
∫
∫
t
dtt
dttpW R
(d) The initial energy stored in the capacitance is
J 25 µ=
=
( )[ ]
J 25
501002.021
021
26
2
µ=
×××
=
−
CvCW
110
P4.8 Equation 4.8 gives the expression for the voltage across a capacitance
discharging through a resistance: ( )vC ( R )CtVt i −= exp
After one-half-life, we have and .
Dividing by and taking the natural logarithm of both sides, we have
( )2i
halfCVtv = ( )RCtVV
halfii −= exp
2iV
( ) RCthalf−=− 2lnSolving, we obtain
P4.9 Prior to , we have because the switch is closed. After
, we can write the following KCL equation at the top node of the circuit:
Multiplying both sides by R and substituting values, we have
(1)
( ) τ6931.06931.02ln === RCRCthalf
0=t ( ) 0=tv0=t
( ) ( ) mA 1=+dt
tdvCRtv
( ) ( ) 1001.0 =+ tvdt
tdv
The solution is of the form (2) Substituting Equation (2) into Equation (1), we eventually obtain
( ) ( ) ( )tKKRCtKKtv 100expexp 2121 −+=−+=
The voltage across the capacitance cannot change instantaneously, so we have
101 =K
( ) ( )( ) 2100
000KKv
vv+==+
=−=+
Thus, , and the solution is
1012 −=−= KK( ) ( ) 0 for 100exp1010 >−−= tttv
111
P4.10* The initial energy is
( ) J 5010001010021
21 262
1 =××== −iVCW
At , half of the energy remains, and we have 25 , which
yields . The voltage across the capacitance is given by
2tt = ( )[ 222
1 tvC=
( ) V 1.7072 =tv
]
( ) ( ) ( ) 0 for 10exp1000exp >−=−= ttRCtVtv iC
Substituting, we have . Solving, we obtain ( )210exp10001.707 t−=
seconds 03466.010)7071.0ln(
2
2
=
−=
tt
P4.11* Because the voltage does not change, the current is zero, and the
voltmeter must be an open circuit. Thus, the resistance of the voltmeter is infinite.
P4.12 During the charging interval, the time constant is s, and
the voltage across the capacitor is given by (vC At the end of the charging interval (t = 28 s), this yields (Cv V
4011 == CRτ
280)]/exp(1[2890) 1 ≤≤−−= ttt τ
1455)]7.0exp(1[2890)28 =−−=
The time constant during the discharge interval is s. Working in terms of the time variable the voltage during the discharge interval is
vC ( At 2840 −=′t this yields 1455 V
5022 == CRτ,28−=′ tt
ttt ′≤′−=′ 0)/exp(1455) 2τ,12=
1145)24.0exp( =−
P4.13 The initial current is A. No wonder one
jumps!!! The time constant is 2.206127/26185/ ==RVi
.ns5.15== RCτ P4.14 The voltage across the resistance and capacitance is
The initial charge stored on the capacitance is
( ) ( )RCtVtv iC −= exp
ii CVQ =
112
The current through the resistance is
The total charge passing through the resistance is
P4.15* for P4.16 The final voltage for each 1 s interval is the initial voltage for the
succeeding interval. We have s. For we have
which yields V.
( ) ( )RCtRVti i
R −= exp
( )
( )
( )[ ]
1
0
0
0
exp
exp
CV
RCtRCRV
dtRCtRV
dttiQ
i
i
R
=
−−=
−=
=
∞
∞
∞
∫
∫
]/)(exp[)( 11 RCttVtv −−= 1tt ≥
2== RCτ ,10 ≤≤t)211)( /exp(11 ttv −−= 328.4)1( =v
For 21 ≤≤t we have which yields V.
For 32 ≤≤t we have which yields V.
Finally, for 43 ≤≤t we have which yields V.
, ]2/)1(exp[328.4)( −−= ttv625.2) =2(v
, ]2/)2(exp[)625.211(11)( −−−−= ttv920.5) =3(v
, ]2/)3(exp[920.5)( −−= ttv591.3)4( =v
P4.17 (a) The voltages across the capacitors cannot change instantaneously.
Thus, . Then, we can write
i
( ) ( ) ( ) ( ) 000 andV 10000 2211 =−=+=−=+ vvvv
( ) ( ) ( ) mA 1101000100000 3
21 =×−
=+−+
=+R
vv
(b) Applying KVL, we have
Taking a derivative with respect to time and rearranging, we obtain
( ) ( ) ( )
( ) ( ) ( )∫ ∫ =+++−
=++−t t
dttiC
tRidttiC
tvtRitv
0 021
21
00110010
113
(1)
(c) The time constant is .
(d) The solution to Equation (1) is of the form However, , so we have .
( ) ( ) 011121
=
++ ti
CCRdttdi
ms 5021
21 =+
=CC
CCRτ
( ) ( )τtKti −= exp1
( ) mA 10 =+i ( ) ( )mA 20exp and mA 11 ttiK −==
(e) The final value of is
( )tv2
( ) ( ) ( )
( )
( ) ( )V 50
05.0exp05.010
005.0exp1010
01
03
0
36
02
22
=
−−=
+−=
++=∞
∞
−
∞
∫
∫
t
dtt
vdttiC
v
t
Thus, the initial charge on is eventually divided equally between .
P4.18 For a dc steady-state analysis:
1C
21 and CC
1. Replace capacitances with open circuits. 2. Replace inductances with short circuits. 3. Solve the resulting circuit, which consists of dc sources and resistances.
P4.19 In dc steady state conditions, the voltages across the capacitors are
constant. Therefore, the currents through the capacitances, which are given by , are zero. Open circuits also have zero current.
dt
dvCiC =
Similarly, the currents through the inductances are constant. Therefore, the voltages across inductances, which are given by
, are zero. Short circuits also have zero voltage. dtdiLv =
114
P4.20* In steady state, the equivalent circuit is:
Thus, we have
1
A 3
0
23 ==
=
iii
P4.21* After the switch opens and the circuit reaches steady state, the 10-mA current flows through the 1-k resistance, and the voltage is 10 V. The initial voltage is . The time constant of the circuit is
. As a function of time, we have Let denote the time at which the voltage reaches 99% is its final value. Then we have Solving, we find
Ω( ) 00 =+Cv
ms 10=RC( ) ( ) ( )tRCttvC 100exp1010exp1010 −−=−−=
99t
( ) ( )9999 100exp10109.9 ttvC −−==
ms 05.4699 =t
P4.22 In steady state with a dc source, the inductance acts as a short circuit and the capacitance acts as an open circuit. The equivalent circuit is:
115
( ) ( )
( ) ( )
V 201mA 5.301
mA 201k 1V 2010
mA 5.100k 2V 201
4321
2
3
4
=
=++=
=Ω=
=
=Ω=
Cviiii
iii
P4.23 mA 6=Li V 18=xv 17−=Cv V
P4.24 Prior to , the steady-state equivalent circuit is:
and we see that .
0=t
V 18=cvA long time after , the steady-state equivalent circuit is:
and we have .
P4.25 With the circuit in steady state prior to the capacitor behaves as
an open circuit, the two 2-kΩ resistors are in parallel, and Because there cannot be infinite current
in this circuit, we have V. After the switch opens and the circuit reaches steady state, we have For the Thévenin resistance seen by the capacitor is Rt = 2 kΩ and the time constant is s. The general form of the solution for is However, we know that
0=t
V 75.91113
1318 =+
=cv
,0=t
V. 20)k (1mA) 20()0( =Ω×=−Cv20)0( =+Cv
V. 40)k (2mA) (20 =Ω×)( =∞Cv,0≥t
2.0== CRtτ0≥t )./exp()( 21 τtKKtvC −+=
116
2120)0( KKvC +==+ 140)( KvC ==∞ .202 −=K
0)/ ≥τ− tt204020)(
−=
= ttv
25=
,0≥t1== CRt
0≥t )./exp(2)( τtKtvC −=
2150)0( KKvC +==+ 125)( KvC ==∞ .252 =K
0)/ ≥− tt252550)(
+=
= ttvC
τ
and . Solving we find that Thus, we have
C
exp(0≤
P4.26 With the circuit in steady state prior to the capacitor behaves as an open circuit, the current is zero, and V. Because there cannot be infinite currents in this circuit, we have V. After the switch closes and the circuit reaches steady state, the source divides equally between the two 1-MΩ resistances, and we have V. For the Thévenin resistance seen by the capacitor is Rt = 500 kΩ and the time constant is τ s. The general form of the solution for is However, we know that
and . Solving we find that Thus we have
,0=t50)0( =−Cv
50)0( =+Cv
)(∞Cv
1 K+
A sketch of the capacitor voltage is:
exp(0≤
117
P4.27
mA 2=Ri 48=Cv V
P4.28* With the switch in position A and the circuit in steady state prior to the capacitor behaves as an open circuit, and
Because there cannot be infinite
current in this circuit, we have V. After switching and the circuit reaches steady state, we have For the resistance across the capacitor is R = 200 kΩ and the time constant is
s. The general form of the solution for is However, we know that and
. Solving we find that Thus, we have
R
P4.29 With the switch closed and the circuit in steady state prior to the capacitor behaves as an open circuit, and Because there cannot be infinite current in this circuit when the switch opens, we have
V. After switching and the circuit reaches steady state, we
have For the Thévenin resistance
,0=t ,0)0( =−Rv
V. 10k 20k 10
k 1030 =Ω+Ω
Ω=)0( −Cv
10)0()0( =+=+ RC vvV. 0)( =∞Rv ,0≥t
2== RCτ 0≥t)./exp(21)( τ−+= tKKtvR 2110)0( KKvR +==+
10)( KvR ∞ == .102 =K
0)/exp(1000)(
≥−=
<=
ttttv
τ
,0=tV. 25)0( =−Cv
25)0( =+Cv
V. 15k 15k 10
k 1525 =Ω+Ω
)( Ω=∞Cv ,0≥t
118
seen by the capacitor is and the time constant
is τ ms. The general form of the solution for is However, we know that
and . Solving, we find that Thus, we have
C
,k 615/110/1
1Ω=
+=tR
60== CRt 0≥t)./exp()( 21 τtKKtvC += − 21)0( 25 KKvC +=+ =
115)( KvC ==∞ .102 =K
0)/ ≥− ttexp(1015025
+
≤)(=
= ttvτ
τ
./RL=τ
)/exp( τtBA −+
).0( +xBAx +=+)0(
CRt=τ
tRL /=τ
( ) 0 0 < for= tti
A( ) (exp21 −+= KKti
P4.30 The time constant is the interval required for the current to fall to exp(−1) ≅ 0.368 times its initial value. The time constant is given by
Thus to attain a long time constant, we need a large value for L and a small value for R.
P4.31 The general form of the solution is . A is the steady-state solution for We determine the value of the desired current or voltage immediately after t = 0, denoted by Then, we solve for the value of B. Finally, we determine the Thévenin resistance Rt from the perspective of the energy storage element (i.e., the resistance seen looking back into the circuit with the energy storage element removed) and compute the time constant: for a capacitance or for an inductance.
)(tx =
.0>>t
P4.32* In steady state with the switch closed, we have because the closed switch shorts the source. In steady state with the switch open, the inductance acts as a short circuit and the current becomes . The current is of the form ( ) 1=∞i ) 0 for ≥tLRt
119
in which , because that is the Thévenin resistance seen looking back from the terminals of the inductance with the switch open. Also, we have
Ω= 20R
( ) ( )( ) 1
21
1000
KiKKii
==∞
+==−=+
Thus, and the current (in amperes) is given by
12 −=K( )
( ) 0 for 20exp10 for 0
≥−−=
<=
tttti
P4.33 The general form of the solution is ( )iL
At , we have
( )LtRKKt −+= exp21
+= 0t( ) ( ) 2100(0 KKii LL +==−=+
At , the inductance behaves as a short circuit, and we have ∞=t ( )iL Thus, the solution for the current is
L
The voltage is
11.0 K==∞
( )( ) 0 for 10-0.1exp-0.10 for 0
6 >=
<=
tttti
( ) ( )
( ) 0 for 10exp1000 for 0
6 >−=
<=
=
ttt
dttdiLtv
P4.34* The solution is similar to that for Problem P4.33.
( ) ( )( ) ( ) 0 for 10exp300
0 for 10exp3.01.06
6
>−=
>−−=
tttvtttiL
120
P4.35 The solution is of the form i ( ) ( )LRtKKt −+= exp21
At , we have and at , we have
The time constant is Thus, the
solution is The voltage across the inductor is
P4.36 In steady state, the inductor acts as a short circuit. With the switch
open, the steady-state current is ( . With the switch closed, the current eventually approaches . For
, the current has the form
+= 0t( ) 2100 KKi +==+
∞=t( ) ( ) ( )
15.0 200V 100
Ki
==
Ω=∞
ms. 5/ == RLτ
( ) ( )tti 200exp5.05.0 −−=
( ) ( ) ( )tdt
tdiLtvL 200exp100 −==
) ( ) A 1 100V 100 =Ω
( ) ( ) ( ) A 4 25V 100 =Ω=∞i0>t
i where , because that is the resistance with the switch closed. Now, we have
Thus, we have . The current is
( ) ( )LRtKKt −+= exp21
Ω= 25R
( ) ( )( ) 1
21
4100
KiKKii
==∞
+==−=+
32 −=K
121
( )( ) closed) (switch 0 512exp34
open) (switch 0 1≥−−=
<=
tt.tti
P4.37 Before the switch closes, 1 A of current circulates through the source
and the two 10-Ω resistors. Immediately after the switch closes, the inductor current remains 0 A because infinite voltage is not possible in this circuit. (Because the inductor current is zero we can consider the inductor to be an open circuit at t = 0+.) Therefore, the current through the resistors is unchanged, and In steady state, the inductor acts as a short circuit, and we have The Thévenin resistance seen by the inductor is 5 Ω because the two 10-Ω resistors are in parallel when we zero the source and look back into the circuit from the inductor terminals. Thus, the time constant is The general form of the solution is Using the initial and final values, we have and which yields Thus, the current is
A. 1)0( =+iA. 2)( =∞i
ms. 200/ == RLτ)./exp(2)( 1 τtKKti −+=
211)0( KKi +==+ 12)( Ki ==∞ .12 −=K
0)/exp(201)(
≥τ−−=
≤=
tttti
122
P4.38 Prior to the current source is shorted, so we have for After the switch opens at the current increases from zero headed for 0.5 A. The inductance sees a Thévenin resistance of 8 Ω, and the time constant is s, so we have for
,0=t0)( =tiL 0<t
,0=t )(tiL
375.08/ == Lτ)375.0/exp(5.05.0)( ttiL −−= 10 <<t
At the current reaches 0.4653 A. Then, the switch closes, the source is shorted, and the current decays toward zero. Because the inductance sees a resistance of 4 Ω, the time constant is s. for 1
,1=t
75.04/ == Lτ
)75.0/exp(4653.0)( ttiL −= t<
P4.39 (a) (b)
( ) ( ) 0 for exp ≥−= tLRtIti i
( ) ( ) ( ) ( ) 0 for 2exp22 ≥−== tLRtIRtRitp iR
(c)
( )
( ) ( )
( ) ( )
( )20
2
0
2
0
21
2exp2
2exp
i
t
ti
i
R
IL
LRtRLIR
dtLRtIR
dttpW
=
−−=
−=
=
∞=
=
∞
∞
∫
∫
which is precisely the expression for the energy stored in the inductance at .
P4.40 With the circuit in steady state before the switch opens, the inductor acts as a short circuit, the current through the inductor is and Immediately after the switch opens, the inductor current remains 0.1 A because infinite voltage is not possible in this circuit. Then the return path for the inductor current is through the 1 kΩ resistor so
After the switch opens, the current and voltage decay exponentially with a time constant τ Thus, we have
0=t
A, 1.0)( =tiL
.0)( =tvR
100)0 −=+ V. (Rvms. 2.0/ == RL
123
0)/exp(100
00)(>−−=
<=
ttttvR
τ
P4.41 In steady state with the switch closed, the current is 2 A for t < 0. The resistance of a voltmeter is very high -- ideally infinite. Thus, there is no path for the current in the inductance when the switch opens, and
is very large in magnitude at . Consequently, the voltage induced in the inductance is very large in magnitude and an arc occurs across the switch. With an ideal meter and switch, the voltage would be infinite. The voltmeter could be damaged in this circuit.
( ) =ti
dtdi 0=t
P4.42* The current in a circuit consisting of an inductance L and series
resistance R is given by in which Ii is the initial current. The initial energy stored in the inductance is and the energy stored as a function of time is
)/exp()( LRtIti i −=2)2/1( ii LIw =
Thus we require
Solving we determine that we require In practice the only practical way to attain such a small resistance for a 10-H inductance is to use superconductors.
)/2exp()2/1()( 2 LRtLItw i −=
]10/)3600(2exp[)2/1()2/1(75.0 22 RLILI ii −≤×
. 6.399 Ωµ≤R
P4.43 1. Write the circuit equation and, if it includes an integral, reduce the
equation to a differential equation by differentiating. 2. Form the particular solution. Often this can be accomplished by adding terms like those found in the forcing function and its derivatives, including an unknown coefficient in each term. Next, solve for the unknown coefficients by substituting the trial solution into the differential equation and requiring the two sides of the equation to be identical.
124
3. Form the complete solution by adding the complementary solution to the particular solution.
4. Use initial conditions to determine the value of K.
)/exp()( τtKtxc −=
P4.44* Applying KVL, we obtain the differential equation:
(1)
We try a particular solution of the form: (2) in which A is a constant to be determined. Substituting Equation (2) into Equation (1), we have which yields
The complementary solution is of the form
( ) ( ) ( ) 0 for exp5 >−=+ tttRidt
tdiL
( ) ( )tAtip −= exp
( ) ( ) ( )ttRAtLA −=−+−− exp5expexp
15−=
−=
LRA
( ) ( )LRtKtic −= exp1
The complete solution is Before the switch closes, the current must be zero. Furthermore, the current cannot change instantaneously, so we have . Therefore, we have which yields . Finally, the current is given by .
( ) ( ) ( ) ( ) ( )LRtKttititi cp −+−−=+= expexp 1
( ) 00 =+i( ) 1100 Ki +−==+ 11 =K( ) ( ) ( )for exp 0 ≥exp −+−−= tLRttti
P4.45 The differential equation is obtained by applying KVL for the node at the top end of the capacitance:
( )vC
Rearranging this equation and substituting , we have
( ) ( ) 0=+−
dttdvC
Rtvt C
( ) ttv =
RC (1)
We try a particular solution of the form (2)
( ) ( ) 0 for >=+ tttvdt
tdvC
C
( ) BtAtvCp +=
in which A and B are constants to be determined. Substituting Equation (2) into Equation (1), we have Solving, we have
tBtARCB =++
125
RCAB
−=
= 1
Thus, the particular solution is
The complementary solution (to the homogeneous equation) is of the form
( ) tRCtvCp +−=
Thus, the complete solution is
( ) ( )RCtKtvCc −= exp1
( ) ( ) ( ) ( )RCtKtRCtvtvtv CcCpC −++−=+= exp1
However, the solution must meet the given initial condition: Thus, and we have
( ) 100 KRCvC +−==
RCK =1
( ) ( ) ( ) ( )RCtRCtRCtvtvtv CcCpC −++−=+= exp
P4.46* Write a current equation at the top node:
2
Substitute the particular solution suggested in the hint: dt
tdvCRtvt CC )()()3exp( +=−
2
Solving for A and substituting values of the circuit parameters, we find The time constant is s, and the general form of the
solution is (tvC
)3exp(3)3exp()3exp( tACtRAt −−−≡−
.106−=A 1==τ RC
)3exp(10)exp()()exp() 611 ttKtvtK p −−−=+−=
However because of the closed switch, we have Substituting this into the general solution we find . Thus
(tvC P4.47* Write a current equation at the top node:
.0)0( =+Cv6
1 10=K)3exp(10)exp(10) 66 >−−−= ttt 0
126
cos(5
Differentiate each term with respect to time to obtain a differential equation:
50−
∫ ++=t
LidttvLR
tvt0
)0()(1)()10
Ltv
dttdv
Rt )()(1)10sin( +=
Substitute the particular solution suggested in the hint:
)]10sin()10cos([1)]10cos(10)10sin(10[1)10sin(50 tBtAL
tBtAR
t +++−≡−
Equating coefficients of sine and cosine terms, we have
Solving for A and B and substituting values of the circuit parameters, we find and The time constant is τ s, and the general form of the solution is
)(tv However, because the current in the inductor is zero at t = 0+, the 5 A
supplied by the source must flow through the 10-Ω resistor and we have Substituting this into the general solution we find .
Thus
LA
RB
LB
RA
+=
+−
=−
100
1050
25=A .25−=B 1.0/ == RL
)10sin(25)10cos(25)/exp()()/exp( 11 tttKtvtK p −+τ−=+τ−=
.50)0( =+v 251 =K
)(tv
)10sin(25)10cos(25)/exp(25 ≥−+−= tttt τ 0
P4.48 Using KVL, we obtain the differential equation
We try a particular solution of the form
( ) ( ) ( )
( ) ( ) ( )ttidt
tdi
tvtRidt
tdiL
300sin10300 =+
=+
( ) ( ) ( )tBtAtip 300sin300cos +=
in which A and B are constants to be determined. Substituting the proposed solution into the differential equation yields:
Equating coefficients of cosines yields
( ) ( ) ( ) (( )t
)tBtAtBtA300sin10
300sin300300cos300300cos300300sin300 =+++−
300300 =+ AB 0
127
Equating coefficients of sines yields From these equations, we find that . The complementary solution is and the complete solution is
10300300 =+− BA601 and 601 −== AB
( ) ( ) ( )tKLRtKtic 300expexp 11 −=−=
( ) ( ) ( )( ) ( ) ( ) ( ) ( )tttK
tititi pc
300sin601300cos601300exp1 +−−=
==
Finally, we use the given initial condition to determine that . Thus, the solution for the current in amperes is
( ) 60100 1 −== Ki6011 =K
( ) ( ) ( ) ( ) ( ) ( ) ( )tttti 300sin601300cos601300exp601 +−−=
P4.49 Applying KVL to the circuit, we have
( ) ( ) ( )
( ) ( ) ttidt
tdi
tvtRidt
tdiL
10102 =+
=+
We try a particular solution of the form in which A and B are constants to be determined. Substituting the proposed solution into the differential equation yields 2 . From this, we find that . The complementary solution is and the complete solution is
( ) BtAtip +=
tBtAB 101010 =++2.0 and 1 −== AB
( ) ( ) ( )tKLRtKtic 5expexp 11 −=−=
Finally, we use the given initial condition: to determine that . Thus, the solution is
P4.50. Applying KVL, we obtain the differential equation:
( ) ( ) ( ) ( ) ttKtititi pc +−−=+= 2.05exp1
( ) 2.000 1 −== Ki2.01 =K
( ) ( ) ttti +−−= 2.05exp2.0
(1)
Because the derivative of the forcing function is
( ) ( ) ( ) 0 for )sin(exp52 >−=+ ttttidt
tdi
( ) )cos()exp(5 )sin(exp5 tttt −+−−
we try a particular solution of the form:
128
(2) in which A and B are constants to be determined. Substituting Equation (2) into Equation (1), we have
( ) ( ) )cos()exp()sin(exp ttBttAtip −+−=
( )( ) )sin(exp5)cos()exp()sin()exp(
)sin()exp(2)cos()exp(2)cos()exp(2)sin(exp2ttttBttA
ttBttAttBttA−≡−+−+
−−−+−−−−
which yields and 2 Solving we find and
52 =−− BA 0=−BA1−=A 2−=B
The complementary solution is of the form The complete solution is
( ) ( ) )2/exp(exp 11 tKLRtKtic −=−=
Before the switch closes, the current must be zero. Furthermore, the current cannot change instantaneously, so we have . Therefore, we have which yields . Thus, the current is given by for t ≥ 0
( ) ( ) ( ) )2/exp()cos()exp(2)sin()exp( 1 tKtttttititi cp −+−−−−=+=
( ) 00 =+i( ) 1200 Ki +−==+ 21 =K
( ) )2/exp(2)cos()exp(2)sin()exp( tttttti −+−−−−=
P4.51 Usually, the particular solution includes terms with the same functional forms as the terms found in the forcing function and its derivatives. In this case, there are four different types of terms in the forcing function and its derivatives, namely and Thus, we we are led to try a particular solution of the form
),sin(tt ),cos(tt ),sin(t ).cos(t
)cos()sin()cos()sin()( tDtCtBttAttvp +++=
Substituting into the differential equation, we have
We require the two sides of the equation to be identical. Equating coefficients of like terms, we have
[ ])sin(5)cos()sin()cos()sin(
)sin()cos()sin()cos()cos()sin(2tttDtCtBttAt
tDtCtBttBtAttA≡+++
+−+−++
022 =+− CDA022 =++ DCB
02 =+BA
129
52 =− BASolving these equations, we obtain , , and Thus, the particular solution is
1=A 2−=B ,5/6=C .5/8=D
)cos(58)sin(
56)cos(2)sin()( tttttttvp ++−=
P4.52 (a)
(b) s (c) A particular solution of the form does not work because the left-hand side of the differential equation is identically zero for this choice.
)2exp(32)( tidt
tdi−=+
5.0/ == RLτ 2exp()( tAtic −= ))2exp()( tKtip −=
(d) Subsitituting into the differential equation produces
from which we have (e) Adding the particular solution and the complementary solution we have However, the current must be zero in the inductor prior to because of the open switch, and the current cannot change instantaneously in this circuit, so we have This yields Thus, the solution is
P4.53 First, we write the differential equation for the system and put it in the form
)2exp()( tKttip −=
)2exp(3)2exp(2)2exp()2exp(2 ttKttKtKt −≡−+−+−−
.3=K
)2exp(3)2exp()( tttAti −+−=
0=t
.0)0( =+i .0=A)2exp(3)( ttti −=
( ) ( ) ( ) )(2 202
2
tftdt
tdxdt
txd=++ ωα
Then compute the damping ratio ζ . If we have the system is underdamped, and the complementary solution is of the form
0/ωα=
,1<ζ
( ) ( ) ( ) ( ) ( )ttKttKtx nnc ωαωα sinexpcosexp 21 −+−=
in which . If we have the system is critically damped, and the complementary solution is of the form
220 αωω −=n
,1=ζ
)exp()exp()( 1211 tstKtsKtxc +=
130
in which s1 is the root of the characteristic equation . 02 20
2 =++ ωαss If we have the system is overdamped, and the complementary solution is of the form in which s1 and s2 are the roots of the characteristic equation
.
,1>ζ
)exp()exp()( 2211 tsKtsKtxc +=
02 20
2 =++ ωαss
P4.54 One way to determine the particular solution is to assume that it is a constant [ ], substitute into the differential equation, and solve for K.
A second method is to replace the inductors by short circuits the
capacitances by open circuits and solve for the steady-state dc response.
Ktxp =)(
P4.55 We look at a circuit diagram and combine all of the inductors that are in
series or parallel . Then, we combine all of the capacitances that are in series or parallel. Next, we count the number of energy storage elements (inductances and capacitances) in the reduced circuit. If there is only one energy-storage element, we have a first-order circuit. If there are two, we have a second-order circuit and so forth.
P4.56 The unit step function is defined by
)(tu
The unit step function is illustrated in Figure 4.27 in the book.
0 for 10 for 0
≥=
<=
tt
P4.57 The sketch should resemble the response shown in Figure 4.29 for Second-order circuits that are severely underdamped ( ζ )
have step responses that display considerable overshoot and ringing. P4.58* Applying KVL to the circuit, we obtain
.1.0=ζ 1<<
(1)
For the capacitance, we have
(2)
( ) ( ) ( ) 50==++ sC vtvtRidt
tdiL
( ) ( )dt
tdvCti C=
131
Using Equation (2) to substitute into Equation (1) and rearranging, we have
(3)
We try a particular solution of the form , resulting in . Thus, . (An alternative method to find the particular solution is to solve the circuit in dc steady state. Since the capacitance acts as an open circuit, the steady-state voltage across it is 50 V.) Comparing Equation (3) with Equation 4.67 in the text, we find
( ) ( ) ( ) ( ) ( ) LCtvLCdt
tdvLRdt
tvdC
CC 5012
2
=++
( ) ( ) ( ) 8842
2
105010104 ×=+×+ tvdt
tdvdt
tvdC
CC
( ) AtvCp = 50=A50=tvCp ( )
Since we have , this is the overdamped case. The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text.
40
4
101
1022
==
×==
LC
LR
ω
α
0ωα >
420
22
420
21
10732.3
102679.0
×−=−−−=
×−=−+−=
ωαα
ωαα
s
s
The complementary solution is ( )vCc and the complete solution is ( )vC
( ) ( )tsKtsKt 2211 expexp +=
( ) ( )tsKtsKt 2211 expexp50 ++=
The initial conditions are
( )Cv ( ) ( )000 and 00 ==== t
C
dttdvCi
Thus, we have
dvC
Solving, we find . Finally, the solution is
( )( )
22110
21
0
5000
KsKsdt
tKKv
t
C
+==
++==
=
867.3 and 87.53 21 =−= KK
132
( )vC P4.59* As in the solution to P4.58, we have
( ) ( )tstst 21 exp867.3exp87.5350 +−=
( )
40
4
101
102
50
==
==
=
LC
LR
tvCp
ω
α
Since we have ,this is the critically damped case. The roots of the characteristic equation are equal:
The complementary solution is given in Equation 4.75 in the text: and the complete solution is The initial conditions are
Thus, we have
0ωα =
420
22
420
21
10
10
−=−−−=
−=−+−=
ωαα
ωαα
s
s
( ) ( ) ( )tstKtsKtvCc 1211 expexp +=
( ) ( ) ( )tstKtsKtvC 1211 expexp50 ++=
( ) ( ) ( )000i and 00 ==== t
CC dt
tdvCv
( )( )
2110
1
0
5000
KKsdt
tdvKv
tC
C
+==
+==
=
Solving, we find . Finally, the solution is
P4.60* As in the solution to P4.58, we have
421 1050 and 50 ×−=−= KK
( ) ( ) ( ) ( )tsttstvC 14
1 exp1050exp5050 ×−−=
Since we have , this is the under-damped case. The natural frequency is given by Equation 4.76 in the text:
( )
40
4
101
105.02
50
==
×==
=
LC
LR
tvCp
ω
α
0ωα <
3220 10660.8 ×=−= αωωn
The complementary solution is given in Equation 4.77 in the text:
133
and the complete solution is
( ) ( ) ( ) ( ) ( )ttKttKtv nnCc ωαωα sinexpcosexp 21 −+−=
The initial conditions are
Thus, we have
Solving, we find . Finally, the solution is
( ) ( ) ( ) ( ) ( )ttKttKtv nnC ωαωα sinexpcosexp50 21 −+−+=
( ) ( ) ( )000 and 00 ==== t
CC dt
tdvCiv
( )( )
210
1
0
5000
KKdt
tdvKv
ntC
C
ωα +−==
+==
=
86.28 and 50 21 −=−= KK
P4.61 (a) Using Equation 4.103 from the text, the damping coefficient is
( ) ( ) ( ) ( ) ( ) ( )tttttv nnC ωαωα sinexp86.28cosexp5050 −−−−=
Equation 4.104 gives the undamped resonant frequency
Equation 4.71 gives the damping ratio
610202
1×==
RCα
60 10101
×==LC
ω
Thus, we have an overdamped circuit.
20 == ωαζ
(b) Writing a current equation at , we have
Substituting , yields
+= 0t( ) ( ) ( ) 1000
=+′++++ vCi
Rv
L
( ) ( ) 00 and 00 =+=+ Liv (c) Under steady-state conditions, the inductance acts as a short circuit. Therefore, the particular solution for is:
( ) 91010 ==+′ Cv
( )tv( ) 0=tv p
(d) The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text.
134
62
02
2
620
21
1032.37
10679.2
×−=−−−=
×−=−+−=
ωαα
ωαα
s
s
The complementary solution is and (since the particular solution is zero) the complete solution is
( ) ( ) ( )tsKtsKtvc 2211 expexp +=
The initial conditions are Thus, we have
( ) ( ) ( )tsKtsKtv 2211 exp +=
( ) ( ) 9100 and 00 =+′=+ vv
Solving, we find . Finally, the solution is
( )( ) 2211
921
10000
KsKsvKKv
+==+′
+==+
87.28 and 87.28 21 −== KK
( ) ( ) ( )tststv 21 exp87.28exp87.28 −= P4.62 (a) Using Equation 4.103 from the text, the damping coefficient is
=α
Equation 4.104 gives the undamped resonant frequency:
610102
1×=
RC
60 10101
×==LC
ω
Equation 4.71 gives the damping ratio: Thus, we have a critically damped circuit. (b) Writing a current equation at , we have
Substituting , yields (c) Under steady-state conditions, the inductance acts as a short circuit. Therefore, the particular solution for is:
/ 0 == ωαζ 1
+= 0t( ) ( ) ( ) 1000
=+′++++ vCi
Rv
L
( ) ( ) 00 and 00 =+=+ Liv( ) 91010 ==+′ Cv
( )tv (d) The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text.
( ) 0=tv p
135
62
02
2
620
21
1010
1010
×−=−−−=
×−=−+−=
ωαα
ωαα
s
s
The complementary solution is given in Equation 4.75 in the text: and the complete solution is
( ) ( ) ( )tstKtsKtvC 1211 expexp +=
The initial conditions are
( ) ( ) ( ) ( ) ( )tstKtsKtvtvtv pC 1211 expexp +=+=
( ) ( ) 9100 and 00 =+=+ vvThus, we have
Solving, we find . Finally, the solution is
P4.63 (a) Using Equation 4.103 from the text, the damping coefficient is
( )( ) 211
91
10000
KKsvKv
+==+′
==+
921 10 and 0 == KK
( ) ( )tttvC79 10exp10 −=
=α
Equation 4.104 gives the undamped resonant frequency:
6102
1=
RC
60 10101
×==LC
ω
Equation 4.71 gives the damping ratio: Thus, we have an underdamped circuit.
1.00 == ωαζ
(b) Writing a current equation at , we have
Substituting , yields
+= 0t
( ) ( ) ( ) 1000=+′+++
+ vCiR
vL
( ) ( ) 00 and 00 =+=+ Liv
(c) Under steady-state conditions, the inductance acts as a short circuit.
Therefore, the particular solution for is:
( ) 91010 ==+′ Cv
( )tv ( )v p 0=t
136
(d) The natural frequency is given by Equation 4.76 in the text: =ωn 622
0 10950.9 ×=− αω
The complementary solution is given in Equation 4.77 in the text: ( )vC and the complete solution is
( ) ( ) ( ) ( )ttKttKt nn ωαωα sinexpcosexp 21 −+−=
( )v The initial conditions are ( )v
( ) ( ) ( ) ( )ttKttKt nn ωαωα sinexpcosexp 21 −+−=
( ) 9100 and 00 =+′=+ v Thus, we have
Solving, we find . Finally, the solution is ( )v
( )( ) 21
91
10000
KKvKv
nωα +−==+′
==+
5.100 and 0 21 == KK( ) ( )ttt nωα sinexp5.100 −=
P4.64 Write a KVL equation for the circuit:
cos(10 ∫ +++=t
CvdttiC
tRidt
tdiLt0
)0()(1)()()100
Differentiate each term with respect to time to obtain a differential equation:
1000−
Substitute the particular solution suggested in the hint to obtain: Cti
dttdiR
dttidLt )()()()10sin( 2
2
++=
Equating coefficients of sine and cosine terms, we have
)]100sin()100cos([1)]100cos(100)100sin(100[
)]100sin(10)100cos(10[)10sin(10 443
tBtAC
tBtAR
tBtALt
+++−+
−−≡−
CABRAL
CBARBL
++−=
+−−=−
100100
1001010
4
43
Solving for A and B and substituting values of the circuit parameters, we find and Thus, the particular solution is
(tip Using Equations 4.60 and 4.61 from the text, we have
2.0=A .0=B)100cos(2.0) t=
137
1001
252
0 ==ω
==α
LC
LR
Because we have α , this is the under-damped case. The natural frequency is given by Equation 4.76 in the text: The complementary solution is given in Equation 4.77 in the text: and the complete solution is
0ω<
82.96220 =α−ω=ωn
( ) ( ) ( ) ( ) ( )ttKttKti nnc ωα−+ωα−= sinexpcosexp 21
( ) ( ) ( ) ( ) ( )ttKttKtti nn ωα−+ωα−+= sinexpcosexp)100cos(2.0 21
However, because the current is zero at t = 0+, the voltage across the inductor must be 10 V which implies that Thus, we can write
i
di
Solving we find and Finally, the solution is:
.10/)0( =+ dtdi
12.00)0( K+==+
2110)0( KKdt nω+α−==
+
2.01 −=K .05164.02 =K P4.65 As in the solution to P4.64, we have
( ) ( ) ( ) ( ) ( )tttttti nn ωα−+ωα−−= sinexp05164.0cosexp2.0)100cos(2.0
Solving for A and B and substituting values of the circuit parameters, we find and Thus the particular solution is
(tip
CABRAL
CBARBL
++−=
+−−=−
100100
1001010
4
43
05.0=A .0=B)100cos(05.0) t=
Using Equations 4.60 and 4.61 from the text, we have
Since we have ,this is the critically damped case. The roots of the characteristic equation are equal:
1001
1002
0 ==ω
==α
LC
LR
0ωα =
100
10020
22
20
21
−=ω−α−α−=
−=ω−α+α−=
s
s
138
The complementary solution is given in Equation 4.75 in the text: and the complete solution is
( ) ( ) ( )tstKtsKtic 1211 expexp +=
As in the solution to P4.51, The initial conditions are
( ) ( ) ( )tstKtsKtti 1211 expexp)100cos(05.0 ++=
i
di
Solving we find and Finally, the solution is i
105.00)0( K+==+
21110)0( KKsdt
+==+
05.01 −=K .52 =K( ) ( ) ( )ttttt 100exp5100exp05.0)100cos(05.0 −+−−=
P4.66 As in the solution to P4.64, we have
CABRAL
CBARBL
++−=
+−−=−
100100
1001010
4
43
Solving for A and B and substituting values of the circuit parameters, we find and Thus, the particular solution is
(tip Using Equations 4.60 and 4.61 from the text, we have
025.0=A .0=B)100cos(025.0) t=
1001
2002
0 ==ω
==α
LC
LR
Since we have , this is the overdamped case. The roots of the characteristic equation are found from Equations 4.72 and 4.73 in the text.
The complementary solution is
0ωα >
2.373
79.2620
22
20
21
−=ω−α−α−=
−=ω−α+α−=
s
s
( )ic and the complete solution is
( ) ( )tsKtsKt 2211 expexp +=
( ) ( ) ( )tsKtsKtti 2211 expexp)100cos(025.0 ++=
As in the solution to P4.51, The initial conditions are
139
i 21025.00)0( KK ++==+
di
Solving, we find and Finally, the solution is
221110)0( KsKsdt
+==+
00193.01 =K .02693.02 −=K
( ) ( ) ( )tststti 21 exp02693.0exp00193.0)100cos(025.0 −+=
P4.67 (a) Applying KCL, we have:
1L
)10sin(2)( 4
0
tdtdvCdttv
t
=+∫ Taking the derivative, multiplying by L, and rearranging, we have
LC
(b) This is a parallel RLC circuit having Using Equation 4.103 from the text, the damping coefficient is
)10cos(102)( 442
2
tLtvdt
vd=+
.∞=R
=α
Equation 4.104 gives the undamped resonant frequency:
02
1=
RC
=ω
The natural frequency is given by Equation 4.76 in the text:
40 101
=LC
=ωn The complementary solution given in Equation 4.77 becomes:
C
(c)The usual form for the particular solution doesn't work because it has the same form as the complementary solution. In other words, if we substitute a trial particular solution of the form
, the left-hand side of the differential equation becomes zero and cannot match the forcing function. (d) When we substitute into the differential equation, we eventually obtain
from which we obtain and Thus, the particular solution is (e) The complete solution is
4220 10=−αω
( ) ( ) ( ) ( ) ( )( ) ( )tKtK
ttKttKtv nn4
24
1
21
10sin10cossinexpcosexp
+=
−+−= ωαωα
( ) ( ) ( )tBtAtvp44 10sin10cos +=
( ) ( ) ( )tBttAttvp44 10sin10cos +=
)10cos(200)10cos(102)10sin(102 44444 ttBtA ≡+− −−
610=B .0=A( ) ( )tttvp
46 10sin10=
( ) ( ) ( ) ( )tKtKtttvtvtv Cp4
24
146 10sin10cos10sin10)()( ++=+=
140
However, we have which yields . Also, by KCL, the current through the capacitance is zero at so we have
which yields so the complete solution is
( ) ,00 =v 01 =K,0+=t
24100)0( K
dtdv
==+
,02 =K
( ) ( )tttvtvtv Cp46 10sin10)()( =+=
141
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