Chapter 7 Circuit

7Response of First-Order RL and

RC Circuits

Assessment Problems

AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like 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

71 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.

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+

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.

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:



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.


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+



[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


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



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


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


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


10 + 15e625t = 5; e625t = 1/3; t = ln 3/625 = 1.76ms



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)


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)



[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


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



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


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



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


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%



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%


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+



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


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



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


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



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


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



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+


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



[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


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.



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


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



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


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



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


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



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


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.



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


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



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


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



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



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+


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



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


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



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


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



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


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



[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


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.



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


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



. . 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+


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.



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



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


Problems 761

1ms t


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



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

+)



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


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



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


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



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


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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