chapter 7 circuit
DESCRIPTION
RC & RL CIRCUITTRANSCRIPT
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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 ashort circuit, effectively eliminating the 2 resistor from the circuit.
First combine the 30 and 6 resistors in parallel:306 = 5Use voltage division to find the voltage drop across the parallel resistors:
v =5
5 + 3(120) = 75V
Now find the current using Ohms law:
i(0) = v
6=
75
6= 12.5A
[b] w(0) =1
2Li2(0) =
1
2(8 103)(12.5)2 = 625mJ
[c] To find the time constant, we need to find the equivalent resistance seenby the inductor for t > 0. When the switch opens, only the 2 resistorremains connected to the inductor. Thus,
=L
R=
8 103
2= 4ms
[d] i(t) = i(0)et/ = 12.5et/0.004 = 12.5e250tA, t 0
[e] i(5ms) = 12.5e250(0.005) = 12.5e1.25 = 3.58ASo w (5ms) = 1
2Li2(5ms) = 1
2(8) 103(3.58)2 = 51.3mJ
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72 CHAPTER 7. Response of First-Order RL and RC Circuits
w (dis) = 625 51.3 = 573.7mJ
% 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:
Using current division,
i(0) =10
10 + 6(6.4) = 4A
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, . . =L
Req=
0.32
3.2= 0.1 s
Use the initial inductor current and the time constant to find the currentin the inductor:i(t) = i(0)et/ = 4et/0.1 = 4e10tA, t 0Use current division to find the current in the 10 resistor:
io(t) =4
4 + 10 + 6(i) =
4
20(4e10t) = 0.8e10tA, t 0+
Finally, use Ohms law to find the voltage drop across the 10 resistor:vo(t) = 10io = 10(0.8e
10t) = 8e10tV, t 0+
[b] The initial energy stored in the inductor is
w(0) =1
2Li2(0) =
1
2(0.32)(4)2 = 2.56 J
Find the energy dissipated in the 4 resistor by integrating the powerover all time:
v4(t) = Ldi
dt= 0.32(10)(4e10t) = 12.8e10tV, t 0+
p4(t) =v244
= 40.96e20tW, t 0+
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Problems 73
w4(t) =
040.96e20tdt = 2.048 J
Find the percentage of the initial energy in the inductor dissipated in the4 resistor:
% dissipated =(2.048
2.56
)100 = 80%
AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves likean open circuit.
Find the voltage drop across the open circuit by finding the voltage dropacross the 50 k resistor. First use current division to find the currentthrough the 50 k resistor:
i50k =80 103
80 103 + 20 103 + 50 103(7.5 103) = 4mA
Use Ohms law to find the voltage drop:v(0) = (50 103)i50k = (50 10
3)(0.004) = 200V
[b] To find the time constant, we need to find the equivalent resistance seenby the capacitor for t > 0. When the switch opens, only the 50 kresistor remains connected to the capacitor. Thus, = RC = (50 103)(0.4 106) = 20ms
[c] v(t) = v(0)et/ = 200et/0.02 = 200e50tV, t 0
[d] w(0) =1
2Cv2 =
1
2(0.4 106)(200)2 = 8mJ
[e] w(t) =1
2Cv2(t) =
1
2(0.4 106)(200e50t)2 = 8e100tmJ
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:
8 103e100t = 2 103, e100t = 4, t = (ln 4)/100 = 13.86ms
AP 7.4 [a] This circuit is actually two RC circuits in series, and the requestedvoltage, vo, is the sum of the voltage drops for the two RC circuits. Thecircuit for t < 0 is shown below:
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74 CHAPTER 7. Response of First-Order RL and RC Circuits
Find the current in the loop and use it to find the initial voltage dropsacross the two RC circuits:
i =15
75,000= 0.2mA, v5(0
) = 4V, v1(0) = 8V
There are two time constants in the circuit, one for each RC subcircuit.5 is the time constant for the 5F 20k subcircuit, and 1 is the timeconstant for the 1F 40k subcircuit:5 = (20 10
3)(5 106) = 100ms; 1 = (40 103)(1 106) = 40ms
Therefore,v5(t) = v5(0
)et/5 = 4et/0.1 = 4e10tV, t 0v1(t) = v1(0
)et/1 = 8et/0.04 = 8e25tV, t 0Finally,vo(t) = v1(t) + v5(t) = [8e
25t + 4e10t] V, t 0
[b] Find the value of the voltage at 60 ms for each subcircuit and use thevoltage to find the energy at 60 ms:v1(60ms) = 8e
25(0.06) = 1.79V, v5(60ms) = 4e10(0.06) = 2.20V
w1(60ms) =12Cv21(60ms) =
12(1 106)(1.79)2 = 1.59J
w5(60ms) =12Cv25(60ms) =
12(5 106)(2.20)2 = 12.05J
w(60ms) = 1.59 + 12.05 = 13.64JFind the initial energy from the initial voltage:w(0) = w1(0) + w2(0) =
12(1 106)(8)2 + 1
2(5 106)(4)2 = 72J
Now calculate the energy dissipated at 60 ms and compare it to theinitial energy:wdiss = w(0) w(60ms) = 72 13.64 = 58.36J
% dissipated = (58.36 106/72 106)(100) = 81.05%
AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current inthe inductor:
i(0) = 24/2 = 12A = 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 dropacross the inductor at 0+. Note that this is the same as the voltage dropacross the 10 resistor, which has current from two sources 8 A fromthe current source and 12 A from the initial current through the inductor.
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Problems 75
v(0+) = 10(8 + 12) = 200V
[c] To calculate the time constant we need the equivalent resistance seen bythe inductor for t > 0. Only the 10 resistor is connected to the inductorfor t > 0. Thus, = L/R = (200 103/10) = 20ms
[d] To find i(t), we need to find the final value of the current in the inductor.When the switch has been in position a for a long time, the circuitreduces to the one below:
Note that the inductor behaves as a short circuit and all of the currentfrom the 8 A source flows through the short circuit. Thus,if = 8ANow,i(t) = if + [i(0
+) if ]et/ = 8 + [12 (8)]et/0.02
= 8 + 20e50tA, t 0
[e] To find v(t), use the relationship between voltage and current for aninductor:
v(t) = Ldi(t)
dt= (200 103)(50)(20e50t) = 200e50tV, t 0+
AP 7.6 [a]
From Example 7.6,
vo(t) = 60 + 90e100tV
Write a KCL equation at the top node and use it to find the relationshipbetween vo and vA:
vA vo8000
+vA
160,000+vA + 75
40,000= 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 + 90e100t) 12 = 60 + 72e100tV, t 0+
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76 CHAPTER 7. Response of First-Order RL and RC Circuits
[b] t 0+, since there is no requirement that the voltage be continuous in aresistor.
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
)(10 103) = 3.2mA
vc(0) = (3.2 103)(25 103) = 80V so vc(0
+) = 80V
Now use the next circuit, valid for 0 t 10ms, to calculate vc(t) forthat interval:
For 0 t 100ms:
= RC = (25 103)(1 106) = 25ms
vc(t) = vc(0)et/ = 80e40t V 0 t 10ms
[b] Calculate the starting capacitor voltage in the interval t 10ms, usingthe capacitor voltage from the previous interval:vc(0.01) = 80e
40(0.01) = 53.63VNow use the next circuit, valid for t 10ms, to calculate vc(t) for thatinterval:
For t 10ms :
Req = 25k100 k = 20k
= ReqC = (20 103)(1 106) = 0.02 s
Therefore vc(t) = vc(0.01+)e(t0.01)/ = 53.63e50(t0.01)V, t 0.01 s
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Problems 77
[c] To calculate the energy dissipated in the 25 k resistor, integrate thepower absorbed by the resistor over all time. Use the expressionp = v2/R to calculate the power absorbed by the resistor.
w25k =
0.010
[80e40t]2
25,000dt+
0.01
[53.63e50(t0.01)]2
25,000dt = 2.91mJ
[d] Repeat the process in part (c), but recognize that the voltage across thisresistor is non-zero only for the second interval:
w100k =
0.01
[53.63e50(t0.01)]2
100,000dt = 0.29mJ
We can check our answers by calculating the initial energy stored in thecapacitor. All of this energy must eventually be dissipated by the 25 kresistor and the 100 k resistor.
Check: wstored = (1/2)(1 106)(80)2 = 3.2mJ
wdiss = 2.91 + 0.29 = 3.2mJ
AP 7.8 [a] 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 + (30.8). Thecurrent leaving the 10 V source is
10
2 + (30.8)= 3.8A
The final current in the inductor, which is equal to the current in the0.8 resistor is
IF =3
3 + 0.8(3.8) = 3A
The resistance seen by the inductor is calculated to find the timeconstant:
[(23) + 0.8]36 = 1 =L
R=
2
1= 2 s
Therefore,
i = iF + [i(0+) iF]e
t/ = 3 3e0.5tA, 0 t 1 s
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78 CHAPTER 7. Response of First-Order RL and RC Circuits
For part (b) we need the value of i(t) at t = 1 s:
i(1) = 3 3e0.5 = 1.18A
.
[b] For t > 1 s
Use current division to find the final value of the current:
i =9
9 + 6(8) = 4.8A
The equivalent resistance seen by the inductor is used to calculate thetime constant:
3(9 + 6) = 2.5 =L
R=
2
2.5= 0.8 s
Therefore,
i = iF + [i(1+) iF]e
(t1)/
= 4.8 + 5.98e1.25(t1)A, t 1 s
AP 7.9 0 t 32ms:
vo = 1
RCf
321030
10dt+ 0 = 1
RCf(10t)
32103
0=
1
RCf(320 103)
RCf = (200 103)(0.2 106) = 40 103 so
1
RCf= 25
vo = 25(320 103) = 8V
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Problems 79
t 32ms:
vo = 1
RCf
t32103
5dy + 8 = 1
RCf(5y)
t32103
+8 = 1
RCf5(t 32 103) + 8
RCf = (250 103)(0.2 106) = 50 103 so
1
RCf= 20
vo = 20(5)(t 32 103) + 8 = 100t + 11.2
The output will saturate at the negative power supply value:
15 = 100t + 11.2 . . t = 262ms
AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at thenon-inverting input:
vp = Vf + [Vo Vf ]et/ = 2 + (0 + 2)et/
= (160 103)(10 109) = 103; 1/ = 625
vp = 2 + 2e625tV; vn = vp
Write a KVL equation at the inverting input, and use it to determine vo:
vn10,000
+vn vo40,000
= 0
. . vo = 5vn = 5vp = 10 + 10e625tV
The output will saturate at the negative power supply value:
10 + 10e625t = 5; e625t = 1/2; t = ln 2/625 = 1.11ms
[b] Use RC circuit analysis to determine the expression for the voltage at thenon-inverting input:
vp = Vf + [Vo Vf ]et/ = 2 + (1 + 2)e625t = 2 + 3e625tV
The analysis for vo is the same as in part (a):
vo = 5vp = 10 + 15e625tV
The output will saturate at the negative power supply value:
10 + 15e625t = 5; e625t = 1/3; t = ln 3/625 = 1.76ms
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710 CHAPTER 7. Response of First-Order RL and RC Circuits
Problems
P 7.1 [a] R =v
i= 25
[b] =1
10= 100ms
[c] =L
R= 0.1
L = (0.1)(25) = 2.5H
[d] w(0) =1
2L[i(0)]2 =
1
2(2.5)(6.4)2 = 51.2 J
[e] wdiss = t01024e20x dx = 1024
e20x
20
t0= 51.2(1 e20t) J
%dissipated =51.2(1 e20t)
51.2(100) = 100(1 e20t)
. . 100(1 e20t) = 60 so e20t = 0.4
Therefore t =1
20ln 2.5 = 45.81ms
P 7.2 [a] Note that there are several different possible solutions to this problem,and the answer to part (c) depends on the value of inductance chosen.
R =L
Choose a 10 mH inductor from Appendix H. Then,
R =0.01
0.001= 10 which is a resistor value from Appendix H.
[b] i(t) = Ioet/ = 10e1000tmA, t 0
[c] w(0) =1
2LI2o =
1
2(0.01)(0.01)2 = 0.5J
w(t) =1
2(0.01)(0.01e1000t)2 = 0.5 106e2000t
So 0.5 106e2000t =1
2w(0) = 0.25 106
e2000t = 0.5 then e2000t = 2
. . t =ln 2
2000= 346.57s (for a 10 mH inductor)
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Problems 711
P 7.3 [a] iL(0) =125
50= 2.5A
io(0+) =
125
25 2.5 = 5 2.5 = 2.5A
io() =125
25= 5A
[b] iL = 2.5et/ ; =
L
R=
50 103
25= 2ms
iL = 2.5e500tA
io = 5 iL = 5 2.5e500tA, t 0+
[c] 5 2.5e500t = 3
2 = 2.5e500t
e500t = 1.25 . . t = 446.29s
P 7.4 [a] t < 0
2 k6 k = 1.5 k
Find the current from the voltage source by combining the resistors inseries and parallel and using Ohms law:
ig(0) =
40
(1500 + 500)= 20mA
Find the branch currents using current division:
i1(0) =
2000
8000(0.02) = 5mA
i2(0) =
6000
8000(0.02) = 15mA
[b] The current in an inductor is continuous. Therefore,
i1(0+) = i1(0
) = 5mA
i2(0+) = i1(0
+) = 5mA (when switch is open)
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712 CHAPTER 7. Response of First-Order RL and RC Circuits
[c] =L
R=
0.4 103
8 103= 5 105 s;
1
= 20,000
i1(t) = i1(0+)et/ = 5e20,000tmA, t 0
[d] i2(t) = i1(t) when t 0+
. . i2(t) = 5e20,000tmA, t 0+
[e] The current in a resistor can change instantaneously. The switchingoperation forces i2(0
) to equal 15mA and i2(0+) = 5mA.
P 7.5 [a] io(0) = 0 since the switch is open for t < 0.
[b] For t = 0 the circuit is:
12060 = 40
. . ig =12
10 + 40= 0.24A = 240mA
iL(0) =
(120
180
)ig = 160mA
[c] For t = 0+ the circuit is:
12040 = 30
. . ig =12
10 + 30= 0.30A = 300mA
ia =(120
160
)300 = 225mA
. . io(0+) = 225 160 = 65mA
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Problems 713
[d] iL(0+) = iL(0
) = 160mA
[e] io() = ia = 225mA
[f ] iL() = 0, since the switch short circuits the branch containing the 20 resistor and the 100 mH inductor.
[g] =L
R=
100 103
20= 5ms;
1
= 200
. . iL = 0 + (160 0)e200t = 160e200tmA, 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
+) = 3.2V
[j] vL() = 0, since the current in the inductor is a constant at t =.
[k] vL(t) = 0 + (3.2 0)e200t = 3.2e200tV, t 0+
[l] io(t) = ia iL = 225 160e200tmA, t 0+
P 7.6 For t < 0
ig =48
6 + (181.5)= 6.5A
iL(0) =
18
18 + 1.5(6.5) = 6A = iL(0
+)
For t > 0
iL(t) = iL(0+)et/ A, t 0
=L
R=
0.5
10 + 12.45 + (5426)= 0.0125 s;
1
= 80
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714 CHAPTER 7. Response of First-Order RL and RC Circuits
iL(t) = 6e80tA, t 0
io(t) =54
80(iL(t)) =
54
80(6e80t) = 4.05e80tV, t 0+
P 7.7 [a] i(0) =24
12= 2A
[b] =L
R=
1.6
80= 20ms
[c] i = 2e50tA, t 0
v1 = Ld
dt(2e50t) = 160e50t V t 0+
v2 = 72i = 144e50tV t 0
[d] w(0) =1
2(1.6)(2)2 = 3.2 J
w72 = t072(4e100x) dx = 288
e100x
100
t0= 2.88(1 e100t) J
w72(15ms) = 2.88(1 e1.5) = 2.24 J
% dissipated =2.24
3.2(100) = 69.92%
P 7.8 w(0) =1
2(10 103)(5)2 = 125mJ
0.9w(0) = 112.5mJ
w(t) =1
2(10 103)i(t)2, i(t) = 5et/ A
. . w(t) = 0.005(25e2t/ ) = 125e2t/ )mJ
w(10s) = 125e20106/ mJ
. . 125e20106/ = 112.5 so e2010
6/ =10
9
=20 106
ln(10/9)=
L
R
R =10 103 ln(10/9)
20 106= 52.68
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Problems 715
P 7.9 [a] w(0) =1
2LI2g
wdiss = to
0I2gRe
2t/ dt = I2gRe2t/
(2/ )
to0
=1
2I2gR (1 e
2to/) =1
2I2gL(1 e
2to/)
wdiss = w(0)
. .1
2LI2g (1 e
2to/ ) = (1
2LI2g
)
1 e2to/ = ; e2to/ =1
(1 )
2to
= ln
[1
(1 )
];
R(2to)
L= ln[1/(1 )]
R =L ln[1/(1 )]
2to
[b] R =(10 103) ln[1/0.9]
20 106
R = 52.68
P 7.10 [a] vo(t) = vo(0+)et/
. . vo(0+)e10
3/ = 0.5vo(0+)
. . e103/ = 2
. . =L
R=
103
ln 2
. . L =10 103
ln 2= 14.43mH
[b] vo(0+) = 10iL(0
+) = 10(1/10)(30 103) = 30mV
vo(t) = 0.03et/ V
p10 =v2o10
= 9 105e2t/
w10 = 103
09 105e2t/ dt = 4.5 105(1 e210
3/)
=1
1000 ln 2. . w10 = 48.69nJ
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-
716 CHAPTER 7. Response of First-Order RL and RC Circuits
wL(0) =1
2Li2L(0) =
1
2(14.43 103)(3 103)2 = 64.92nJ
% diss in 1 ms =48.69
64.92 100 = 75%
P 7.11 [a] t < 0
iL(0) =
150
180(12) = 10A
t 0
=1.6 103
8= 200 106; 1/ = 5000
io = 10e5000tA t 0
[b] wdel =1
2(1.6 103)(10)2 = 80mJ
[c] 0.95wdel = 76mJ
. . 76 103 = to0
8(100e10,000t) dt
. . 76 103 = 80 103e10,000tto0= 80 103(1 e10,000to)
. . e10,000to = 0.05 so to = 299.57s
. .to=
299.57 106
200 106= 1.498 so to 1.498
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-
Problems 717
P 7.12 t < 0:
iL(0+) =
240
16 + 8= 10A; iL(0
) = 1040
50= 8A
t > 0:
Re =(10)(40)
50+ 10 = 18
=L
Re=
72 103
18= 4ms;
1
= 250
. . iL = 8e250tA
vo = 8io = 64e250tV, t 0+
P 7.13 p40 =v2o40
=(64)2
40e500t = 102.4e500tW
w40 =
0102.4e500t dt = 102.4
e500t
500
0= 204.8mJ
w(0) =1
2(72 103)(8)2 = 2304mJ
% diss =204.8
2304(100) = 8.89%
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-
718 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.14 [a] t < 0 :
iL(0) = 72
24 + 6= 2.4A
t > 0:
i = 100
160iT =
5
8iT
vT = 20i + iT(100)(60)
160= 12.5iT + 37.5iT
vTiT
= RTh = 12.5 + 37.5 = 25
=L
R=
250 103
25
1
= 100
iL = 2.4e100tA, t 0
[b] vL = 250 103(240e100t) = 60e100tV, t 0+
[c] i = 0.625iL = 1.5e100tA t 0+
P 7.15 w(0) =1
2(250 103)(2.4)2 = 720mJ
p60 = 60(1.5e100t)2 = 135e200tW
w60 =
0135e200t dt = 135
e200t
200
0= 675mJ
% dissipated =675
720(100) = 93.75%
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-
Problems 719
P 7.16 t < 0
iL(0) = iL(0
+) = 4A
t > 0
Find Thevenin resistance seen by inductor:
iT = 4vT ;vTiT
= RTh =1
4= 0.25
=L
R=
5 103
0.25= 20ms; 1/ = 50
io = 4e50tA, t 0
vo = Ldiodt
= (5 103)(200e50t) = e50tV, t 0+
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-
720 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.17 [a] t < 0 :
t = 0+:
120 = iab + 18 + 12, iab = 90A, t = 0+
[b] At t =:
iab = 240/2 = 120A, t =
[c] i1(0) = 18, 1 =2 103
10= 0.2ms
i2(0) = 12, 2 =6 103
15= 0.4ms
i1(t) = 18e5000tA, t 0
i2(t) = 12e2500tA, t 0
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-
Problems 721
iab = 120 18e5000t 12e2500tA, t 0
120 18e5000t 12e2500t = 114
6 = 18e5000t + 12e2500t
Let x = e2500t so 6 = 18x2 + 12x
Solving x =1
3= e2500t
. . e2500t = 3 and t =ln 3
2500= 439.44s
P 7.18 [a] t < 0
1 k4 k = 0.8 k
20k80 k = 16k
(105 103)(0.8 103) = 84V
iL(0) =
84
16,800= 5mA
t > 0
=L
R=
6
24 103 = 250s;
1
= 4000
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-
722 CHAPTER 7. Response of First-Order RL and RC Circuits
iL(t) = 5e4000tmA, t 0
p4k = 25 106e8000t(4000) = 0.10e8000tW
wdiss = t00.10e8000x dx = 12.5 106[1 e8000t] J
w(0) =1
2(6)(25 106) = 75J
0.10w(0) = 7.5J
12.5(1 e8000t) = 7.5; . . e8000t = 2.5
t =ln 2.5
8000= 114.54s
[b] wdiss(total) = 75(1 e8000t)J
wdiss(114.54s) = 45J
% = (45/75)(100) = 60%
P 7.19 [a] t > 0:
Leq = 1.25 +60
16= 5H
iL(t) = iL(0)et/ mA; iL(0) = 2A;
1
=
R
L=
7500
5= 1500
iL(t) = 2e1500tA, t 0
vR(t) = RiL(t) = (7500)(2e1500t) = 15,000e1500tV, t 0+
vo = 3.75diLdt
= 11,250e1500tV, t 0+
[b] io =1
6
t011,250e1500x dx+ 0 = 1.25e1500t 1.25A
P 7.20 [a] From the solution to Problem 7.19,
w(0) =1
2Leq[iL(0)]
2 =1
2(5)(2)2 = 10J
[b] wtrapped =1
2(10)(1.25)2 +
1
2(6)(1.25)2 = 12.5 J
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-
Problems 723
P 7.21 [a] R =v
i= 8k
[b]1
=
1
RC= 500; C =
1
(500)(8000)= 0.25F
[c] =1
500= 2ms
[d] w(0) =1
2(0.25 106)(72)2 = 648J
[e] wdiss = to0
(72)2e1000t
(800)dt
= 0.648e1000t
1000
to0= 648(1 e1000to)J
%diss = 100(1 e1000to) = 68 so e1000to = 3.125
. . t =ln 3.125
1000= 1139s
P 7.22 [a] Note that there are many different possible correct solutions to thisproblem.
R =
C
Choose a 100F capacitor from Appendix H. Then,
R =0.05
100 106= 500
Construct a 500 resistor by combining two 1k resistors in parallel:
[b] v(t) = Voet/ = 50e20t V, t 0
[c] 50e20t = 10 so e20t = 5
. . t =ln 5
20= 80.47ms
P 7.23 [a] v1(0) = v1(0
+) = 40V v2(0+) = 0
Ceq = (1)(4)/5 = 0.8F
= (25 103)(0.8 106) = 20ms;1
= 50
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-
724 CHAPTER 7. Response of First-Order RL and RC Circuits
i =40
25,000e50t = 1.6e50tmA, t 0+
v1 =1
106
t01.6 103e50x dx+ 40 = 32e50t + 8V, t 0
v2 =1
4 106
t01.6 103e50x dx+ 0 = 8e50t + 8V, t 0
[b] w(0) =1
2(106)(40)2 = 800J
[c] wtrapped =1
2(106)(8)2 +
1
2(4 106)(8)2 = 160J.
The energy dissipated by the 25 k resistor is equal to the energydissipated by the two capacitors; it is easier to calculate the energydissipated by the capacitors:
wdiss =1
2(0.8 106)(40)2 = 640J.
Check: wtrapped+ wdiss = 160 + 640 = 800J; w(0) = 800J.
P 7.24 [a] t < 0:
i1(0) = i2(0
) =3
30= 100mA
[b] t > 0:
i1(0+) =
0.2
2= 100mA
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-
Problems 725
i2(0+) =
0.2
8= 25mA
[c] Capacitor voltage cannot change instantaneously, therefore,
i1(0) = i1(0
+) = 100mA
[d] Switching can cause an instantaneous change in the current in a resistivebranch. In this circuit
i2(0) = 100mA and i2(0
+) = 25mA
[e] vc = 0.2et/ V, t 0
= ReC = 1.6(2 106) = 3.2s;
1
= 312,500
vc = 0.2e312,000tV, t 0
i1 =vc2= 0.1e312,000tA, t 0
[f ] i2 =vc8
= 25e312,000tmA, t 0+
P 7.25 [a] t < 0:
Req = 12k8 k = 10.2 k
vo(0) =10,200
10,200 + 1800(120) = 102V
t > 0:
= [(10/3) 106)(12,000) = 40ms;1
= 25
vo = 102e25tV, t 0
p =v2o
12,000= 867 103e50tW
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-
726 CHAPTER 7. Response of First-Order RL and RC Circuits
wdiss = 121030
867 103e50t dt
= 17.34 103(1 e50(12103)) = 7824J
[b] w(0) =(1
2
)(10
3
)(102)2 106 = 17.34mJ
0.75w(0) = 13mJ to0
867 103e50x dx = 13 103
. . 1 e50to = 0.75; e50to = 4; so to = 27.73ms
P 7.26 [a] t < 0:
io(0) =
6000
6000 + 4000(40m) = 24mA
vo(0) = (3000)(24m) = 72V
i2(0) = 40 24 = 16mA
v2(0) = (6000)(16m) = 96V
t > 0
= RC = (1000)(0.2 106) = 200s;1
= 5000
io(t) =24
1 103et/ = 24e5000tmA, t 0+
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-
Problems 727
[b]
vo =1
0.6 106
t024 103e5000x dx+ 72
= (40,000)e5000x
5000
t0+72
= 8e5000t + 8 + 72
vo = [8e5000t+ 80]V, t 0
[c] wtrapped = (1/2)(0.3 106)(80)2 + (1/2)(0.6 106)(80)2
wtrapped = 2880J.
Check:
wdiss =1
2(0.2 106)(24)2 = 57.6J
w(0) =1
2(0.3 106)(96)2 +
1
2(0.6 106)(72)2 = 2937.6J.
wtrapped + wdiss = w(0)
2880 + 57.6 = 2937.6 OK.
P 7.27 [a] At t = 0 the voltage on each capacitor will be 150V(5 30), positive atthe upper terminal. Hence at t 0+ we have
. . isd(0+) = 5 +
150
0.2+150
0.5= 1055A
At t =, both capacitors will have completely discharged.
. . isd() = 5A
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728 CHAPTER 7. Response of First-Order RL and RC Circuits
[b] isd(t) = 5 + i1(t) + i2(t)
1 = 0.2(106) = 0.2s
2 = 0.5(100 106) = 50s
. . i1(t) = 750e5106tA, t 0+
i2(t) = 300e20,000tA, t 0
. . isd = 5 + 750e5106t + 300e20,000tA, t 0+
P 7.28 [a]
vT = 20 103(iT + v) + 5 10
3iT
v = 5 103iT
vT = 25 103iT + 20 10
3(5 103iT )
RTh = 25,000 + 100 106
= RThC = 40 103 = RTh(0.8 10
6)
RTh = 50k = 25,000 + 100 106
=25,000
100 106= 2.5 104 A/V
[b] vo(0) = (5 103)(3600) = 18V t < 0
t > 0:
vo = 18e25tV, t 0
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-
Problems 729
v5000
+v vo20,000
+ 2.5 104v = 0
4v + v vo + 5v = 0
. . v =vo10
= 1.8e25tV, t 0+
P 7.29 [a]
pds = (16.2e25t)(450 106e25t) = 7290 106e50tW
wds =
0pds dt = 145.8J.
. . dependent source is delivering 145.8J.
[b] w5k =
0(5000)(0.36 103e25t)2 dt = 648 106
0e50t dt = 12.96J
w20k =
0
(16.2e25t)2
20,000dt = 13,122 106
0e50t dt = 262.44J
wc(0) =1
2(0.8 106)(18)2 = 129.6J
wdiss = 12.96 + 262.44 = 275.4Jwdev = 145.8 + 129.6 = 275.4J.
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-
730 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.30 t < 0
t > 0
vT = 5io 15io = 20io = 20iT . . RTh =vTiT
= 20
= RC = 40s;1
= 25,000
vo = 15e25,000tV, t 0
io = vo20
= 0.75e25,000tA, t 0+
P 7.31 [a] The equivalent circuit for t > 0:
= 2ms; 1/ = 500
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-
Problems 731
vo = 10e500tV, t 0
io = e500tmA, t 0+
i24k = e500t
(16
40
)= 0.4e500tmA, t 0+
p24k = (0.16 106e1000t)(24,000) = 3.84e1000tmW
w24k =
03.84 103e1000t dt = 3.84 106(0 1) = 3.84J
w(0) =1
2(0.25 106)(40)2 +
1
2(1 106)(50)2 = 1.45mJ
% diss (24 k) =3.84 106
1.45 103 100 = 0.26%
[b] p400 = 400(1 103e500t)2 = 0.4 103e1000t
w400 =
0p400 dt = 0.40J
% diss (400) =0.4 106
1.45 103 100 = 0.03%
i16k = e500t
(24
40
)= 0.6e500tmA, t 0+
p16k = (0.6 103e500t)2(16,000) = 5.76 103e1000tW
w16k =
05.76 103e1000t dt = 5.76J
% diss (16 k) = 0.4%
[c]
wdiss = 3.84 + 5.76 + 0.4 = 10J
wtrapped = w(0)
wdiss = 1.45 103 10 106 = 1.44mJ
% trapped =1.44
1.45 100 = 99.31%
Check: 0.26 + 0.03 + 0.4 + 99.31 = 100%
P 7.32 [a] Ce =(2 + 1)6
2 + 1 + 6= 2F
vo(0) = 5 + 30 = 25V
= (2 106)(250 103) = 0.5 s;1
= 2
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732 CHAPTER 7. Response of First-Order RL and RC Circuits
vo = 25e2tV, t 0+
[b] wo =1
2(3 106)(30)2 +
1
2(6 106)(5)2 = 1425J
wdiss =1
2(2 106)(25)2 = 625J
% diss =625
1425 100 = 43.86%
[c] io =vo
250 103= 100e2t A
v1 = 1
6 106
t0100 106e2x dx 5 = 16.67
t0e2x dx 5
= 16.67e2x
2
t05 = 8.33e2t 13.33V t 0
[d] v1 + v2 = vo
v2 = vo v1 = 25e2t 8.33e2t + 13.33 = 16.67e2t + 13.33V t 0
[e] wtrapped =1
2(6 106)(13.33)2 +
1
2(3 106)(13.33)2 = 800J
wdiss + wtrapped = 625 + 800 = 1425J (check)
P 7.33 [a] From Eqs. (7.35) and (7.42)
i =VsR
+(Io
VsR
)e(R/L)t
v = (Vs IoR)e(R/L)t
. .VsR
= 4; Io VsR
= 4
Vs IoR = 80;R
L= 40
. . Io = 4 +VsR
= 8A
Now since Vs = 4R we have
4R 8R = 80; R = 20
Vs = 80V; L =R
40= 0.5H
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Problems 733
[b] i = 4 + 4e40t; i2 = 16 + 32e40t + 16e80t
w =1
2Li2 =
1
2(0.5)[16 + 32e40t + 16e80t] = 4 + 8e40t + 4e80t
. . 4 + 8e40t + 4e80t = 9 or e80t + 2e40t 1.25 = 0
Let x = e40t:
x2 + 2x 1.25 = 0; Solving, x = 0.5; x = 2.5
But x 0 for all t. Thus,
e40t = 0.5; e40t = 2; t = 25 ln 2 = 17.33ms
P 7.34 [a] Note that there are many different possible solutions to this problem.
R =L
Choose a 1 mH inductor from Appendix H. Then,
R =0.001
8 106= 125
Construct the resistance needed by combining 100, 10, and 15resistors in series:
[b] i(t) = If + (Io If)et/
Io = 0A; If =VfR
=25
125= 200mA
. . i(t) = 200 + (0 200)e125,000tmA = 200 200e125,000tmA, t 0
[c] i(t) = 0.2 0.2e125,000t = (0.75)(0.2) = 0.15
e125,000t = 0.25 so e125,000t = 4
. . t =ln 4
125,000= 11.09s
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-
734 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.35 [a] t < 0
iL(0) = 5A
t > 0
iL() =40 80
4 + 16= 2A
=L
R=
4 103
4 + 16= 200s;
1
= 5000
iL = iL() + [iL(0+) iL()]e
t/
= 2 + (5 + 2)e5000t = 2 3e5000tA, t 0
vo = 16iL + 80 = 16(2 3e5000t) + 80 = 48 48e5000tV, t 0
[b] vL = LdiLdt
= 4 103(5000)[3e5000t] = 60e5000tV, t 0+
vL(0+) = 60V
From part (a) vo(0+) = 0V
Check: at t = 0+ the circuit is:
vL(0+) = 40 + (5A)(4) = 60V, vo(0
+) = 80 (16)(5A) = 0V
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Problems 735
P 7.36 [a] For t < 0, calculate the Thevenin equivalent for the circuit to the left andright of the 75 mH inductor. We get
i(0) =5 120
15 k + 8k= 5mA
i(0) = i(0+) = 5mA
[b] For t > 0, the circuit reduces to
Therefore i() = 5/15,000 = 0.333mA
[c] =L
R=
75 103
15,000= 5s
[d] i(t) = i() + [i(0+) i()]et/
= 0.333 + [5 0.333]e200,000t = 0.333 5.333e200,000tmA, t 0
P 7.37 [a] t < 0
KVL equation at the top node:
50 =vo8+
vo40
+vo10
Multiply by 40 and solve:
2000 = (5 + 1 + 4)vo; vo = 200V
. . io(0) =
vo10
= 200/10 = 20A
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-
736 CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0
Use voltage division to find the Thevenin voltage:
VTh = vo =40
40 + 120(800) = 200V
Remove the voltage source and make series and parallel combinations ofresistors to find the equivalent resistance:
RTh = 10 + 12040 = 10 + 30 = 40
The simplified circuit is:
=L
R=
40 103
40= 1ms;
1
= 1000
io() =200
40= 5A
. . io = io() + [io(0+) io()]e
t/
= 5 + (20 5)e1000t = 5 + 15e1000tA, t 0
[b] vo = 10io + Ldiodt
= 10(5 + 15e1000t) + 0.04(1000)(15e1000t)
= 50 + 150e1000t 600e1000t
vo = 50 450e1000tV, t 0+
P 7.38 [a]
VsR
+v
R+
1
L
t0v dt + Io = 0
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-
Problems 737
Differentiating both sides,
1
R
dv
dt+
1
Lv = 0
. .dv
dt+R
Lv = 0
[b]dv
dt=
R
Lv
dv
dtdt =
R
Lv dt so dv =
R
Lv dt
dv
v=
R
Ldt
v(t)Vo
dx
x=
R
L
t0dy
lnv(t)
Vo=
R
Lt
. . v(t) = Voe(R/L)t = (Vs RIo)e
(R/L)t
P 7.39 [a] vo(0+) = IgR2; =
L
R1 +R2
vo() = 0
vo(t) = IgR2e[(R1+R2)/L]tV, t 0+
[b] vo(0+), and the duration of vo(t) zero
[c] vsw = R2io; =L
R1 +R2
io(0+) = Ig; io() = Ig
R1R1 +R2
Therefore io(t) =IgR1
R1+R2+[Ig
IgR1R1+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+
[d] |vsw(0+)| ; duration 0
P 7.40 Opening the inductive circuit causes a very large voltage to be induced acrossthe inductor L. This voltage also appears across the switch (part [d] ofProblem 7.39), causing the switch to arc over. At the same time, the largevoltage across L damages the meter movement.
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-
738 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.41 t > 0; calculate vo(0+)
va15
+va vo(0
+)
5= 20 103
. . va = 0.75vo(0+) + 75 103
15 103 +vo(0
+) va5
+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+ 5 103
. . 360i = 4.5vo(0+) + 1800 103
8va = 6vo(0+) + 600 103
. . 13vo(0+) 6vo(0
+) 600 103 4.5vo(0+)
1800 103 = 2600 103
2.5vo(0+) = 200 103; vo(0
+) = 80mV
vo() = 0
Find the Thevenin resistance seen by the 4 mH inductor:
iT =vT20
+vT8 9i
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-
Problems 739
i =vT8 9i . . 10i =
vT8; i =
vT80
iT =vT20
+10vT80
9vT80
iTvT
=1
20+
1
80=
5
80=
1
16S
. . RTh = 16
=4 103
16= 0.25ms; 1/ = 4000
. . vo = 0 + (80 0)e4000t = 80e4000tmV, t 0+
P 7.42 For t < 0
vx15 0.8v +
vx 480
21= 0
v =vx 480
21
vx15 0.8
(vx 480
21
)+(vx 480
21
)
=vx15
+ 0.2(vs 480
21
)= 21vx + 3(vx 480) = 0
. . 24vx = 1440 so vx = 60V io(0) =
vx15
= 4A
t > 0
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-
740 CHAPTER 7. Response of First-Order RL and RC Circuits
Find Thevenin equivalent with respect to a, b
VTh 320
5 0.8
(VTh 320
5
)= 0 VTh = 320V
vT = (iT + 0.8v)(5) =(iT + 0.8
vT5
)(5)
vT = 5iT + 0.8vT . . 0.2vT = 5iT
vTiT
= RTh = 25
io() = 320/40 = 8A
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-
Problems 741
=80 103
40= 2ms; 1/ = 500
io = 8 + (4 8)e500t = 8 4e500tA, t 0
P 7.43 For t < 0, i80mH(0) = 50V/10 = 5AFor t > 0, after making a Thevenin equivalent we have
i =VsR
+(Io
VsR
)et/
1
=
R
L=
8
100 103= 80
Io = 5A; If =VsR
=80
8= 10A
i = 10 + (5 + 10)e80t = 10 + 15e80t A, t 0
vo = 0.08di
dt= 0.08(1200e80t) = 96e80tV, t 0+
P 7.44 [a] Let v be the voltage drop across the parallel branches, positive at the topnode, then
Ig +v
Rg+
1
L1
t0v dx+
1
L2
t0v dx = 0
v
Rg+(1
L1+
1
L2
) t0v dx = Ig
v
Rg+
1
Le
t0v dx = Ig
1
Rg
dv
dt+
v
Le= 0
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-
742 CHAPTER 7. Response of First-Order RL and RC Circuits
dv
dt+RgLe
v = 0
Therefore v = IgRget/ ; = Le/Rg
Thus
i1 =1
L1
t0IgRge
x/ dx =IgRgL1
ex/
(1/ )
t0=
IgLeL1
(1 et/)
i1 =IgL2
L1 + L2(1 et/) and i2 =
IgL1L1 + L2
(1 et/)
[b] i1() =L2
L1 + L2Ig; i2() =
L1L1 + L2
Ig
P 7.45 [a] t < 0
t > 0
iL(0) = iL(0
+) = 25mA; =24 103
120= 0.2ms;
1
= 5000
iL() = 50mA
iL = 50 + (25 + 50)e5000t = 50 + 75e5000tmA, t 0
vo = 120[75 103e5000t] = 9e5000tV, t 0+
[b] i1 =1
60 103
t09e5000x dx+ 10 103 = (30e5000t 20)mA, t 0
[c] i2 =1
40 103
t09e5000x dx+ 15 103 = (45e5000t 30)mA, t 0
P 7.46 t > 0
=1
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Problems 743
io = 5e40tA, t 0
vo = 40io = 200e40t V, t > 0+
200e40t = 100; e40t = 2
. . t =1
40ln 2 = 17.33ms
P 7.47 [a] wdiss =1
2Lei
2(0) =1
2(1)(5)2 = 12.5 J
[b] i3H =1
3
t0(200)e40x dx 5
= 1.67(1 e40t) 5 = 1.67e40t 3.33A
i1.5H =1
1.5
t0(200)e40x dx + 0
= 3.33e40t + 3.33A
wtrapped =1
2(4.5)(3.33)2 = 25J
[c] w(0) =1
2(3)(5)2 = 37.5 J
P 7.48 [a] v = IsR+ (Vo IsR)et/RC i =
(Is
VoR
)et/RC
. . IsR = 40, Vo IsR = 24
. . Vo = 16V
Is VoR
= 3 103; Is 16
R= 3 103; R =
40
Is
. . Is 0.4Is = 3 103; Is = 5mA
R =40
5 103 = 8k
1
RC= 2500; C =
1
2500R=
103
20 103= 50nF; = RC =
1
2500= 400s
[b] v() = 40V
w() =1
2(50 109)(1600) = 40J
0.81w() = 32.4J
v2(to) =32.4 106
25 109= 1296; v(to) = 36V
40 24e2500to = 36; e2500to = 6; . . to = 716.70s
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744 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.49 [a] Note that there are many different possible solutions to this problem.
R =
C
Choose a 10H capacitor from Appendix H. Then,
R =0.25
10 106= 25k
Construct the resistance needed by combining 10 k and 15k resistorsin series:
[b] v(t) = Vf + (Vo Vf )et/
Vo = 100V; Vf = (If)(R) = (1 103)(25 103) = 25V
. . v(t) = 25 + (100 25)e4t V = 25 + 75e4tV, t 0
[c] v(t) = 25 + 75e4t = 50 so e4t =1
3
. . t =ln 3
4= 274.65ms
P 7.50 [a]
io(0+) =
36
5000= 7.2mA
[b] io() = 0
[c] = RC = (5000)(0.8 106) = 4ms
[d] io = 0 + (7.2)e250t = 7.2e250tmA, t 0+
[e] vo = [36 + 1800(7.2 103e250t)] = 36 + 12.96e250t V, t 0+
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-
Problems 745
P 7.51 [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
+) = 0V
From the simplified circuit,
vo() = 60V
= RC = (20 103)(0.5 106) = 10ms 1/ = 100
vo = vo() + [vo(0+) vo()]e
t/ = (60 60e100t)V, t 0
[b] ic = Cdvodt
ic = 0.5 106(100)(60e100t) = 3e100tmA
v1 = 8000ic + vo = (8000)(3 103)e100t + (60 60e100t) = 60 36e100tV
io =v1
60 103= 1 0.6e100tmA, t 0+
[c] i1(t) = io + ic = 1 + 2.4e100tmA, t 0+
[d] i2(t) =v1
15 103= 4 2.4e100tmA, t 0+
[e] i1(0+) = 1 + 2.4 = 3.4mA
At t = 0+:
Re = 15k60 k8 k = 4800
v1(0+) = (5 103)(4800) = 24V
i1(0+) =
v1(0+)
60,000+v1(0
+)
8000= 0.4m + 3m = 3.4mA (checks)
P 7.52 [a] vo(0) = vo(0
+) = 120V
vo() = 150V; = 2ms;1
= 500
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-
746 CHAPTER 7. Response of First-Order RL and RC Circuits
vo = 150 + (120 (150))e500t
vo = 150 + 270e500tV, t 0
[b] io = 0.04 106(500)[270e500t] = 5.4e500tmA, t 0+
[c] vg = vo 12.5 103io = 150 + 202.5e
500tV
[d] vg(0+) = 150 + 202.5 = 52.5V
Checks:
vg(0+) = io(0
+)[37.5 103] 150 = 202.5 150 = 52.5V
i50k =vg50k
= 3 + 4.05e500tmA
i150k =vg150k
= 1 + 1.35e500tmA
-io + i50k + i150k + 4 = 0 (ok)
P 7.53 For t < 0
Simplify the circuit:
80/10,000 = 8mA, 10 k40 k24 k = 6k
8mA 3mA = 5mA
5mA 6 k = 30V
Thus, for t < 0
. . vo(0) = vo(0
+) = 30V
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-
Problems 747
t > 0
Simplify the circuit:
8mA + 2mA = 10mA
10k40 k24 k = 6k
(10mA)(6 k) = 60V
Thus, for t > 0
vo() = 10 103(6 103) = 60V
= RC = (10k)(0.05) = 0.5ms;1
= 2000
vo = vo() + [vo(0+) vo()]e
t/ = 60 + [30 (60)]e2000t
= 60 + 90e2000tV t 0
P 7.54 t < 0:
io(0) =
20
100(10 103) = 2mA; vo(0
) = (2 103)(50,000) = 100V
t =:
io() = 5 103(20
100
)= 1mA; vo() = io()(50,000) = 50V
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-
748 CHAPTER 7. Response of First-Order RL and RC Circuits
RTh = 50k50 k = 25k; C = 16nF
= (25,000)(16 109) = 0.4ms;1
= 2500
. . vo(t) = 50 + 150e2500tV, t 0
ic = Cdvodt
= 6e2500tmA, t 0+
i50k =vo
50,000= 1 + 3e2500tmA, t 0+
io = ic + i50k = (1 + 3e2500t)mA, t 0+
P 7.55 [a] vc(0+) = 50V
[b] Use voltage division to find the final value of voltage:
vc() =20
20 + 5(30) = 24V
[c] Find the Thevenin equivalent with respect to the terminals of thecapacitor:
VTh = 24V, RTh = 205 = 4,
Therefore = ReqC = 4(25 109) = 0.1s
The simplified circuit for t > 0 is:
[d] i(0+) =24 50
4= 18.5A
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-
Problems 749
[e] vc = vc() + [vc(0+) vc()]e
t/
= 24 + [50 (24)]et/ = 24 + 74e107tV, t 0
[f ] i = Cdvcdt
= (25 109)(107)(74e107t) = 18.5e10
7tA, t 0+
P 7.56 [a] Use voltage division to find the initial value of the voltage:
vc(0+) = v9k =
9k
9k + 3k(120) = 90V
[b] Use Ohms law to find the final value of voltage:
vc() = v40k = (1.5 103)(40 103) = 60V
[c] Find the Thevenin equivalent with respect to the terminals of thecapacitor:
VTh = 60V, RTh = 10k + 40k = 50k
= RThC = 1ms = 1000s
[d] vc = vc() + [vc(0+) vc()]e
t/
= 60 + (90 + 60)e1000t = 60 + 150e1000tV, t 0
We want vc = 60 + 150e1000t = 0:
Therefore t =ln(150/60)
1000= 916.3s
P 7.57 Use voltage division to find the initial voltage:
vo(0) =60
40 + 60(50) = 30V
Use Ohms law to find the final value of voltage:
vo() = (5mA)(20 k) = 100V
= RC = (20 103)(250 109) = 5ms;1
= 200
vo = vo() + [vo(0+) vo()]e
t/
= 100 + (30 + 100)e200t = 100 + 130e200tV, t 0
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-
750 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.58 For t < 0, vo(0) = 80Vt > 0:
vTh = 30 103i + 0.8(100) = 30 10
3(
100
100 103
)+ 80 = 50V
vT = 30 103i + 16 10
3iT = 30 103(0.8)iT + 16 10
3iT = 40 103iT
RTh =vTiT
= 40k
t > 0
vo = 50 + (80 50)et/
= RC = (40 103)(5 109) = 200 106;1
= 5000
vo = 50 + 30e5000tV, t 0
P 7.59 vo(0) = 50V; vo() = 80V
RTh = 16k
= (16)(5 106) = 80 106;1
= 12,500
v = 80 + (50 80)e12,500t = 80 30e12,500tV, t 0
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-
Problems 751
P 7.60 For t > 0
VTh = (25)(16,000)ib = 400 103ib
ib =33,000
80,000(120 106) = 49.5A
VTh = 400 103(49.5 106) = 19.8V
RTh = 16k
vo() = 19.8V; vo(0+) = 0
= (16, 000)(0.25 106) = 4ms; 1/ = 250
vo = 19.8 + 19.8e250tV, t 0
w(t) =1
2(0.25 106)v2o = w()(1 e
250t)2 J
(1 e250t)2 =0.36w()
w()= 0.36
1 e250t = 0.6
e250t = 0.4 . . t = 3.67ms
P 7.61 [a]
IsR = Ri +1
C
t0+i dx+ Vo
0 = Rdi
dt+
i
C+ 0
. .di
dt+
i
RC= 0
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-
752 CHAPTER 7. Response of First-Order RL and RC Circuits
[b]di
dt=
i
RC;
di
i=
dt
RC i(t)i(0+)
dy
y=
1
RC
t0+dx
lni(t)
i(0+)=t
RC
i(t) = i(0+)et/RC; i(0+) =IsR Vo
R=(Is
VoR
)
. . i(t) =(Is
VoR
)et/RC
P 7.62 [a] Let i be the current in the clockwise direction around the circuit. Then
Vg = iRg +1
C1
t0i dx+
1
C2
t0i dx
= iRg +(1
C1+
1
C2
) t0i dx = iRg +
1
Ce
t0i dx
Now differentiate the equation
0 = Rgdi
dt+
i
Ceor
di
dt+
1
RgCei = 0
Therefore i =VgRg
et/RgCe =VgRg
et/ ; = RgCe
v1(t) =1
C1
t0
VgRg
ex/ dx =Vg
RgC1
ex/
1/
t0=
VgCeC1
(et/ 1)
v1(t) =VgC2
C1 + C2(1 et/); = RgCe
v2(t) =VgC1
C1 + C2(1 et/); = RgCe
[b] v1() =C2
C1 + C2Vg; v2() =
C1C1 + C2
Vg
P 7.63 [a] For t > 0:
= RC = 250 103 8 109 = 2ms;1
= 500
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-
Problems 753
vo = 50e500tV, t 0+
[b] io =vo
250,000=
50e500t
250,000= 200e500t A
v1 =1
40 109 200 106
t0e500x dx+ 50 = 10e500t + 40V, t 0
P 7.64 [a] t < 0
t > 0
vo(0) = vo(0
+) = 40V
vo() = 80V
= (0.16 106)(6.25 103) = 1ms; 1/ = 1000
vo = 80 40e1000tV, t 0
[b] io = Cdvodt
= 0.16 106[40,000e1000t]
= 6.4e1000tmA; t 0+
[c] v1 =1
0.2 106
t06.4 103e1000x dx+ 32
= 64 32e1000tV, t 0
[d] v2 =1
0.8 106
t06.4 103e1000x dx+ 8
= 16 8e1000tV, t 0
[e] wtrapped =1
2(0.2 106)(64)2 +
1
2(0.8 106)(16)2 = 512J.
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-
754 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.65 [a] Leq =(3)(15)
3 + 15= 2.5H
=LeqR
=2.5
7.5=
1
3s
io(0) = 0; io() =120
7.5= 16A
. . io = 16 16e3tA, t 0
vo = 120 7.5io = 120e3t V, t 0+
i1 =1
3
t0120e3x dx =
40
3
40
3e3tA, t 0
i2 = io i1 =8
3
8
3e3tA, 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
= 120e3t V, t 0+
or
vo = 15di2dt
= 120e3t V, t 0+
The voltage solution is consistent with the current solutions.
1 = 3i1 = 40 40e3t Wb-turns
2 = 15i2 = 40 40e3t Wb-turns
. . 1 = 2 as it must, since
vo =d1dt
=d2dt
1() = 2() = 40 Wb-turns
1() = 3i1() = 3(40/3) = 40 Wb-turns
2() = 15i2() = 15(8/3) = 40 Wb-turns
. . i1() and i2() are consistent with 1() and 2().
P 7.66 [a] Leq = 5 + 10 2.5(2) = 10H
=L
R=
10
40=
1
4;
1
= 4
i = 2 2e4tA, t 0
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Problems 755
[b] v1(t) = 5di1dt 2.5
di
dt= 2.5
di
dt= 2.5(8e4t) = 20e4t V, t 0+
[c] v2(t) = 10di1dt 2.5
di
dt= 7.5
di
dt= 7.5(8e4t) = 60e4tV, t 0+
[d] i(0) = 2 2 = 0, which agrees with initial conditions.
80 = 40i1 + v1 + v2 = 40(2 2e4t) + 20e4t + 60e4t = 80V
Therefore, Kirchhoffs voltage law is satisfied for all values of t 0.Thus, the answers make sense in terms of known circuit behavior.
P 7.67 [a] Leq = 5 + 10 + 2.5(2) = 20H
=L
R=
20
40=
1
2;
1
= 2
i = 2 2e2tA, t 0
[b] v1(t) = 5di1dt
+ 2.5di
dt= 7.5
di
dt= 7.5(4e2t) = 30e2tV, t 0+
[c] v2(t) = 10di1dt
+ 2.5di
dt= 12.5
di
dt= 12.5(4e2t) = 50e2tV, t 0+
[d] i(0) = 0, which agrees with initial conditions.
80 = 40i1 + v1 + v2 = 40(2 2e2t) + 30e2t + 50e2t = 80V
Therefore, Kirchhoffs voltage law is satisfied for all values of t 0.Thus, the answers make sense in terms of known circuit behavior.
P 7.68 [a] From Example 7.10,
Leq =L1L2 M
2
L1 + L2 + 2M=
50 25
15 + 10= 1H
=L
R=
1
20;
1
= 20
. . io(t) = 4 4e20tA, t 0
[b] vo = 80 20io = 80 80 + 80e20t = 80e20tV, t 0+
[c] vo = 5di1dt 5
di2dt
= 80e20t V
io = i1 + i2
diodt
=di1dt
+di2dt
= 80e20t A/s
. .di2dt
= 80e20t di1dt
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756 CHAPTER 7. Response of First-Order RL and RC Circuits
. . 80e20t = 5di1dt 400e20t + 5
di1dt
. . 10di1dt
= 480e20t; di1 = 48e20t dt
t10
dx = t048e20y dy
i1 =48
20e20y
t0= 2.4 2.4e20tA, t 0
[d] i2 = io i1 = 4 4e20t 2.4 + 2.4e20t
= 1.6 1.6e20tA, t 0
[e] io(0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.
vo = Leqdiodt
= 1(80)e20t = 80e20tV, t 0+ (checks)
Also,
vo = 5di1dt 5
di2dt
= 80e20tV, t 0+ (checks)
vo = 10di2dt 5
di1dt
= 80e20tV, t 0+ (checks)
vo(0+) = 80V, which agrees with io(0
+) = 0A
io() = 4A; 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 withconservation of flux linkage. Hence, the answers make sense in terms ofknown circuit behavior.
P 7.69 [a] From Example 7.10,
Leq =L1L2 M
2
L1 + L2 + 2M=
0.125 0.0625
0.75 + 0.5= 50mH
=L
R=
1
5000;
1
= 5000
. . io(t) = 40 40e5000tmA, t 0
[b] vo = 10 250io = 10 250(0.04 + 0.04e5000t = 10e5000tV, t 0+
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Problems 757
[c] vo = 0.5di1dt 0.25
di2dt
= 10e5000tV
io = i1 + i2
diodt
=di1dt
+di2dt
= 200e5000t A/s
. .di2dt
= 200e5000t di1dt
. . 10e5000t = 0.5di1dt 50e5000t + 0.25
di1dt
. . 0.75di1dt
= 60e5000t; di1 = 80e5000t dt
t10
dx = t080e5000y dy
i1 =80
5000e5000y
t0= 16 16e5000tmA, t 0
[d] i2 = io i1 = 40 40e5000t 16 + 16e5000t
= 24 24e5000tmA, t 0
[e] io(0) = i1(0) = i2(0) = 0, consistent with zero initial stored energy.
vo = Leqdiodt
= (0.05)(200)e5000t = 10e5000tV, t 0+ (checks)
Also,
vo = 0.5di1dt 0.25
di2dt
= 10e5000tV, t 0+ (checks)
vo = 0.25di2dt 0.25
di1dt
= 10e5000tV, t 0+ (checks)
vo(0+) = 10V, which agrees with io(0
+) = 0A
io() = 40mA; io()Leq = (0.04)(0.05) = 2 mWb-turns
i1()L1 + i2()M = (16m)(500) + (24m)(250) = 2 mWb-turns (ok)
i2()L2 + i1()M = (24m)(250) + (16m)(250) = 2 mWb-turns (ok)
Therefore, the final values of io, i1, and i2 are consistent withconservation of flux linkage. Hence, the answers make sense in terms ofknown circuit behavior.
2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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758 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.70 t < 0:
iL(0) = 10V/5 = 2A = iL(0
+)
0 t 5:
= 5/0 =
iL(t) = 2et/ = 2e0 = 2
iL(t) = 2A 0 t 5 s
5 t
-
Problems 759
0 t 10ms:
i = 10e100tA
i(10ms) = 10e1 = 3.68A
10ms t 20ms:
Req =(5)(20)
25= 4
1
=
R
L=
4
50 103= 80
i = 3.68e80(t0.01)A
20ms t 20+ ms
vo(25ms) = 8.26e100(0.0250.02) = 5.013V
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760 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.72 From the solution to Problem 7.71, the initial energy is
w(0) =1
2(50mH)(10A)2 = 2.5 J
0.04w(0) = 0.1 J
. .1
2(50 103)i2L = 0.1 so iL = 2A
Again, from the solution to Problem 7.73, t must be between 10 ms and 20 mssince
i(10ms) = 3.68A and i(20ms) = 1.65A
For 10ms t 20ms:
i = 3.68e80(t0.01) = 2
e80(t0.01) =3.68
2so t 0.01 = 0.0076 . . t = 17.6ms
P 7.73 [a] t < 0:
Using Ohms law,
ig =800
40 + 6040= 12.5A
Using current division,
i(0) =60
60 + 40(12.5) = 7.5A = i(0+)
[b] 0 t 1ms:
i = i(0+)et/ = 7.5et/
1
=
R
L=
40 + 12060
80 103= 1000
i = 7.5e1000t
i(200s) = 7.5e103(200106) = 7.5e0.2 = 6.14A
[c] i(1ms) = 7.5e1 = 2.7591A
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Problems 761
1ms t
-
762 CHAPTER 7. Response of First-Order RL and RC Circuits
10s t
-
Problems 763
200s t 0, vo = (4/6)vc, where vc is the voltage across the 0.5Fcapacitor. Thus we will find vc first.t < 0
vc(0) =3
15(75) = 15V
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764 CHAPTER 7. Response of First-Order RL and RC Circuits
0 t 800s:
= ReC, Re =(6000)(3000)
9000= 2k
= (2 103)(0.5 106) = 1ms,1
= 1000
vc = 15e1000tV, t 0
vc(800s) = 15e0.8 = 6.74V
800s t 1.1ms:
= (6 103)(0.5 106) = 3ms,1
= 333.33
vc = 6.74e333.33(t800106)V
1.1ms t
-
Problems 765
P 7.77 w(0) =1
2(0.5 106)(15)2 = 56.25J
0 t 800s:
vc = 15e1000t; v2c = 225e
2000t
p3k = 75e2000tmW
w3k = 800106
075 103e2000t dt
= 75 103e2000t
2000
800106
0
= 37.5 106(e1.6 1) = 29.93J
1.1ms t :
vc = 6.1e1000(t1.1103)V; v2c = 37.19e
2000(t1.1103)
p3k = 12.4e2000(t1.1103)mW
w3k =
1.110312.4 103e2000(t1.110
3) dt
= 12.4 103e2000(t1.110
3)
2000
1.1103
= 6.2 106(0 1) = 6.2J
w3k = 29.93 + 6.2 = 36.13J
% =36.13
56.25(100) = 64.23%
P 7.78 t < 0:
vc(0) = (5)(1000) 103 = 5V = vc(0
+)
2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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766 CHAPTER 7. Response of First-Order RL and RC Circuits
0 t 5 s:
=; 1/ = 0; vo = 5e0 = 5V
5 s t
-
Problems 767
[b]
[c] vo(5ms) = 16.35V
io =+16.35
20= 817.68mA
P 7.80 [a] io(0) = 0; io() = 25mA
1
=
R
L=
2000
250 103 = 8000
io = (25 25e8000t)mA, 0 t 75s
vo = 0.25diodt
= 50e8000tV, 0 t 75s
75s t
-
768 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.81 [a] 0 t 1ms:
vc(0+) = 0; vc() = 50V;
RC = 400 103(0.01 106) = 4ms; 1/RC = 250
vc = 50 50e250t
vo = 50 50 + 50e250t = 50e250tV, 0 t 1ms
1ms t
-
Problems 769
[b]
[c] t 0 : vo = 00 t 4ms:
= (50 103)(0.025 106) = 1.25ms 1/ = 800
vo = 100 100e800tV, 0 t 4ms
vo(4ms) = 100 100e3.2 = 95.92V
4ms t 8ms:
vo = 100 + 195.92e800(t0.004)V, 4ms t 8ms
vo(8ms) = 100 + 195.92e3.2 = 92.01V
t 8ms:
vo = 92.01e800(t0.008)V, t 8ms
P 7.83 [a] = RC = (20,000)(0.2 106) = 4ms; 1/ = 250
io = vo = 0 t < 0
io(0+) = 20
(16
20
)= 16mA, io() = 0
. . io = 16e250tmA 0+ t 2ms
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770 CHAPTER 7. Response of First-Order RL and RC Circuits
i16k = 20 16e250tmA
. . vo = 320 256e250tV 0+ t 2ms
vc = vo 4 103io = 320 320e
250tV 0 t 2ms
vc(2ms) = 320 320e0.5 = 125.91V
. . io(2+ ms) = 16e0.5 = 9.7mA
io() = 0
vc = 125.91e250(t0.002), t 2ms
io = Cdvcdt
= (0.2 106)(250)(125.91)e250(t0.002)
= 6.3e250(t0.002)mA, t 2+ms
vo = 4000io + vc = 100.73e250(t0.002)V t 2+ ms
Summary part (a)
io = 0 t < 0
io = 16e250tmA (0+ t 2ms)
io = 6.3e250(t0.002)mA t 2+ms
vo = 0 t < 0
vo = 320 256e250tV, 0+ t 2ms
vo = 100.73e250(t0.002)V, t 2+ms
[b] io(0) = 0
io(0+) = 16mA
io(2ms) = 16e0.5 = 9.7mA
io(2+ms) = 6.3mA
[c] vo(0) = 0
vo(0+) = 64V
vo(2ms) = 320 256e0.5 = 164.73V
vo(2+ms) = 100.73
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Problems 771
[d]
[e]
P 7.84 t > 0:
vT = 12 104i + 16 10
3iT
i = 20
100iT = 0.2iT
. . vT = 24 103iT + 16 10
3iT
RTh =vTiT
= 8 k
= RC = (8 103)(2.5 106) = 0.02 1/ = 50
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772 CHAPTER 7. Response of First-Order RL and RC Circuits
vc = 20e50tV; 20e50t = 20,000
50t = ln 1000 . . t = 138.16ms
P 7.85 Find the Thevenin equivalent with respect to the terminals of the capacitor.RTh calculation:
iT =vT2000
+vT5000
4vT5000
. .iTvT
=5 + 2 8
10,000=
1
10,000
vTiT
= 10,000
1= 10 k
Open circuit voltage calculation:
The node voltage equations:
voc2000
+voc v11000
4i = 0
v1 voc1000
+v14000
5 103 = 0
The constraint equation:
i =v14000
Solving, voc = 80V, v1 = 60V
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Problems 773
vc(0) = 0; vc() = 80V
= RC = (10,000)(1.6 106) = 16ms;1
= 62.5
vc = vc() + [vc(0+) vc()]e
t/ = 80 + 80e62.5t = 14,400
Solve for the time of the maximum voltage rating:
e62.5t = 181; 62.5t = ln 181; t = 83.09ms
P 7.86
vT = 2000iT + 4000(iT 2 103v) = 6000iT 8v
= 6000iT 8(2000iT )
vTiT
= 10,000
=10
10,000= 1ms; 1/ = 1000
i = 25e1000tmA
. . 25e1000t 103 = 5; t =ln 200
1000= 5.3ms
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