chapter 9 homework solution

7/24/2019 Chapter 9 Homework Solution

1/27

Chapter 9 Homework Solution

Chapter 9(1):P9.2-1, 5, 9

P9.3-2, 4

P9.4-1, 3P9.5-2, 3, 4

P9.6-1, 2, 5

Chapter 9(2):P9.7-1, 2

P9.8-2, 4, 9, 15, 17

P9.9-1

P 9.2-1 Find the differential equation for the

circuit shown in Figure P 9.2-1 using the directmethod.

Figure P 9.2-1


2/27

P 9.2-5 The input to the circuit shown in Figure P 9.2-5 is the voltage of the voltage source, vs. The

output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that showshow the output of this circuit is related to the input, for t> 0.

Hint: Use the direct method.

Figure P 9.2-5

Solution:

After the switch closes, use KCL to get

( ) ( ) ( )2

v t di t C v t

R dt= +

Use KVL to get

( ) ( ) ( )s 1d

v R i t L i t v t dt

= + +

Substitute to get

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

21

s 1 2

2 2

21 2

12

2 2

R d L d d v v t R C v t v t CL v t v t

R dt R dt dt

R Rd L dCL v t R C v t v t

dt R dt R CL

= + + + +

+= + + +

Finally,

( ) ( ) ( )2

s 1 1 2

2

2 2

1v R R Rd dv t v t v t

CL dt L R C dt R CL

+= + + +

P 9.2-9 The input to the circuit shown inFigure P 9.2-9 is the voltage of the voltage source,

vs. The output is the capacitor voltage v(t).Represent the circuit by a second-order differential

equation that shows how the output of this circuitis related to the input, for t> 0.

Hint: Use the direct method.

Figure P 9.2-9


3/27

Solution:

After the switch closes

( ) ( )d

i t C v t

dt

=

KCL and KVL give

( ) ( ) ( ) ( ) ( )s 21

1 d dv R i t L i t v t L i t v t

R dt dt

= + + + +

Substituting gives

( ) ( ) ( )

( ) ( ) ( )

22 2

s 22

1 1

22 2

22

1 1

1 1

1 1

R Rd dv LC v t R C v t v t

R dt dt R

R Rd dLC v t R C v t v t

R dt dt R

= + + + +

= + + + +

Finally

( ) ( )

( ) ( ) ( )

21 s 1 2

1 2 1 2

1R v R Rd dv t v t v t

dt dt LC LC R R L R R= + +

+ +

P 9.3-2 Find the characteristic equation and its

roots for the circuit of Figure P 9.3-2.

Answer: s2+ 400s+ 3 10

4= 0

roots:s= 300, 100

Figure P 9.3-2


4/27

Solution:

P 9.3-3 Find the characteristic equation and

its roots for the circuit shown in Figure P 9.3-3.

Figure P 9.3-3

Solution:

3 Ls L c

3 cL c

KVL: 40( ) = 100 10

1 103

dii i v

dt

dvi idt

+

= =

3 3 6L LL

2

L L

2 L

2

1 2

s

s

i

240 40 10010 10 10

23 3 3

400 30000 400

400 30000 0 ( 100)( 300) 0 100, 300

di di d i

dt dt dt

did i dii

dt dt dt

s s s s s s

=

+ + =

+ + = + + = = =

6sL

3 LL

v dvKCL: i 10 10 0

1 dt

diKVL: v = 2i +10

dt

v

+ + =

23 6 6 3

28

2

28 8

2 8

1 2

0 2 10 10 10 2 10 10 10

3 .00102 1 10

102000 3 10 1 10

102000 3 10 0, 3031, 98969

L L LL s L

L Ls L

L LL s

di di d ii v i

dt dt dt

di d iv idt dt

d i dii v

dt dt

s s s s

= + + + +

= + +

+ + =

+ + = = =


5/27

P 9.3-4 German automaker Volkswagen, in

its bid to make more efficient cars, has come

up with an auto whose engine saves energy byshutting itself off at stoplights. The stop

start system springs from a campaign to

develop cars in all its world markets that use

less fuel and pollute less than vehicles now onthe road. The stopstart transmission control

has a mechanism that senses when the car doesFigure P 9.3-4

not need fuel: coasting downhill and idling at an intersection. The engine shuts off, but a small starterflywheel keeps turning so that power can be quickly restored when the driver touches the accelerator.

A model of the stopstart circuit is shown in Figure P 9.3-4. Determine the characteristic equation

and the natural frequencies for the circuit.Answer:

s

2

+ 20s+ 400 = 0s= 10 j17.3

Solution:

Assume zero initial conditions

1 21

1 22

2

1 1loop 1 : 10 10 7

2 2

1 1loop 2 : 200 7

2 2

1 110 2 2

determinant :1 1 200

2 2

20 400 0, 10 17.3

di dii

dt dt

di dii dt

dt dt

s s

s ss

s s s j

+ =

+ + =

+

+

+ + = =


6/27

P 9.4-1 Determine v(t) for the circuit of

Figure P 9.4-1 whenL= 1 H and vs= 0 for t

0. The initial conditions are v(0) = 6 V anddv/dt(0) = 3000 V/s.

Answer: v(t) = 2e100t+ 8e400tVFigure P 9.4-1

Solution:

( ) ( )0

0 6, 3000dv

vdt

= =

( ) 2Using operators, the node equation is: 0 1v vv Ls

Csv or LCs s v vsR sL R

+ + = + + =

2

2

1,2

1 1So the characteristic equation is: 0

250 250 40,000 100, 400

s sRC LC

s

+ + =

= = ( )

( )

( )

( )

100 400

100 400

So

0 6

20 3000 100 400

8

2 8 t>0

t t

t t

v t Ae Be

v A B

AdvA B

Bdt

v t e e

= +

= = +

= = =

=

= +


7/27

P 9.4-2 AnRLCcircuit is shown in

Figure P 9.4-2, where v(0) = 2 V. The switch

has been open for a long time before closing att= 0. Determine and plot v(t).

Figure P 9.4-2

Solution:

( ) ( )

( )

( ) ( )

( )

( ) ( )

( ) ( )

2 2

3

1 2

0 2, 0 0

1 1Characteristic equation 0 4 3 0 1, 3

0 0

Use eq. 9.5 12

21 3 0 8 1

14

also have 0 2 2

From 1 & 2 get 1

t t

v i

s s s s sRC LC

v t Ae Be

v i

s A s B RC C

A B

v A B

A

= =

+ + = + + = =

= +

+ =

= =

= = +

=

( ) 3, 3

3 Vt t

B

v t e e

=

= +


8/27

P 9.4-3 Determine i1(t) and i2(t) for the circuit of

Figure P 9.4-3 when i1(0) = i2(0) = 11 A.

Figure P 9.4-3

Solution:

( ) ( )

( ) ( )( ) ( )

1 2 22

1 2

in operator form

1 5 3 0 1 thus 1 5 3 2 9 26 13 2 0 6,3 3 2 0

s i s is s s ss s

s i s i

+ + = = + + = + = + =

+ + =

( ) ( )

( ) ( )

( ) ( )

1 2

1 2

/6 2 /6 2

1 2

Now i 0 11 ; 0 11

from 1 & 2 get

0 033 143 2 ; 20

2 6 6 6

which yields 3, 8, 1, 0 12

( ) 3 8 & ( ) 12t t t t

A B i C D

di diA CB

dt dt

A B C

i t e e A i t e e A

= = + = = +

= = = =

= = = =

= + = +

( )

( )

1 21

1 22

KVL : 5 3 0 1

KVL : 3 3 2 0 2

di dii

dt dt

di dii

dt dt

+ =

+ + =

( )

( )

261

t- -2t62

Thus

i t = Ce + De

tt

i t Ae Be = +


9/27

P 9.5-2 Find vc(t) for t> 0 for the circuit of

Figure P 9.5-2. Assume steady-state

conditions exist at t= 0

.Answer: vc(t) = 8te

2tV

Figure P 9.5-2

Solution:

0t> ( ) cc c c

2

c

1KCL at v : 04

4 4 0

t

c c

dvv dt v

dt

d v dvv

dt dt

+ + =

+ + =

( )2 2 2c 1 24 4 0, 2, 2t t

s s s v t A e A t e + + = = = +

0 (Steady-State)t = ( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( )

200 0 0 & 0 2 0

10

Since 0 0 then 0 0 2

0 0V 8

S14

c c L L

c c L

c c

Vv v i A i

v i i A

dv i

dt

+ +

+ + +

+ +

= = = = =

= = =

= =

( )( )

( )

1

2

2

So 0 0

0 8

8

c

c

t

c

v A

dvA

dt

v t te V

+

+

= =

= =

=


10/27

P 9.5-3 Police often use stun guns toincapacitate potentially dangerous felons. The

hand-held device provides a series of high-

voltage, low-current pulses. The power of thepulses is far below lethal levels, but it is

enough to cause muscles to contract and put

the person out of action. The device provides a

pulse of up to 50,000 V, and a current of 1 mA

Figure P 9.5-3

flows through an arc. A model of the circuit for one period is shown in Figure P 9.5-3. Find v(t) for 0 < t ( )

( )

6

26

2

KVL a : .01 10 0 1

Also : .01 10 2

Lc L

L LcL

div i

dt

d i didvi C C

dt dt dt

+ + =

= = +

( ) ( )

( )

( )

26

2

26 6

2 6

12 2

7 7

75 10 5

1 2

0.01 10 0

10 10 4 .01Characteristic eq. 0.01 10 1 0

2 .01for critically damped: 10 C .04C = 0

0.04 pF 5 10 , 5 10

So

L LL

t

L

d i diC C i

dt dt

C C CC s s s

C

C s

i t A e A te

+ + =

+ + = =

= =

= +

( ) ( ) ( )

( ) ( )

( )

( ) ( )

7

7

710

6 6

6 6 5 10

L 1 2

6 12 5 10

Now from (1) 0 100 0 10 0 10

0So i 0 0 and 10 10 A

Now 10 10 V

t

Lc L

L t

L

t

L

di Av isdt

diA A i t te

dt

v t i t te

+ + +

= =

= = = = =

= =


11/27

P 9.5-4 Reconsider Problem P 9.4-1 whenL= 640

mH and the other parameters and conditions remain thesame.

Answer: v(t) = (6 1500t)e 250t V

Figure P 9.4-1

Solution:

2 31 1 1 10 with 500 and 62.5 10 yields 250, 250s s sRC LC RC LC

+ + = = = =

( )

( )

( )

( )

250 250

250 250

0 6

0 3000 250 1500

6 1500

t t

t t

v t Ae Bte

v A

dv A B Bdt

v t e te

= +

= =

= = + =

=

P 9.6-1 A communication system from a

space station uses short pulses to control a

robot operating in space. The transmittercircuit is modeled in Figure P 9.6-1. Find the

output voltage vc(t) for t> 0. Assume steady-

state conditions at t= 0

.

Answer:

vc(t) = e 400t[3 cos 300t+ 4 sin 300t] V

Figure P 9.6-1


12/27

Solution:

P 9.6-2 The switch of the circuit shown in Figure P 9.6-2 is opened at t= 0. Determine and plot v(t) whenC= 1/4 F. Assume steady state at t= 0

.

Answer: v(t) = 4e2t

sin 2tV

Figure P 9.6-2

t> 0

( )

( )

6KCL at : 5 10 0 1250

also : 0.8 2

ccc L

Lc

dvvv i

dt

divdt

+ + =

=

( ) ( )

( )

2

5 2

2

1 2

Solving for i in 1 & plugging into 2L

400 cos300 sin300

800 2.5 10 0 800 250,000 0, 400 300c c c

tv t e A t A t

c

d v dvv s s s j

dt dt = +

+ + = + + = =

t Steady State= 0 ( )

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )

( )

( )

( ) [ ]

5

1

1 2 2

400

6 V 60 A 0500500

60 250 6 3 V 0500

0Now from 1 : 2 10 0 800 0 0

So 0 3

0 0 400 300 4

3cos300 4sin 300

L L

c c

c

L c

c

c

tc

i i

v v

dvi v

dt

v A

dvA A A

dt

v t e t t V

+

+

+

+ +

+

+

= = =

= + = =

= =

= =

= = + =

= +


13/27

Solution:0t

=

( )

( )

0 2 A

0 0

i

v

=

=

0t =

( ) ( )0

KCL at node a:

1 0 0 1

1

tv dv

C vdt idt L

+ + + =

( )

2

1 1 12in operator form have 0 0 or 0

with 4 8 0 2 2

v Csv v i s s vLs C LC

s s s j

+ + + = + + =

+ + = = ( ) [ ]

( )

( ) ( )

( ) ( ) [ ]

( )

2

1 2

1

2 2

2

cos2 sin 2

0 0

0 1From 1 , 0 0 4 2 8 2 4

So 4 sin 2 V

t

t

v t e B t B t

v B

dvi v B or B

dt C

v t e t

= +

= =

= = = = =

=


14/27

P 9.6-5 The photovoltaic cells of theproposed space station shown in

Figure P 9.6-5aprovide the voltage v(t) of the

circuit shown in Figure P 9.6-5b. The spacestation passes behind the shadow of earth (at

t= 0) with v(0) = 2 V and i(0) = 1/10 A.

Determine and sketch v(t) for t> 0.

Figure P 9.6-5


15/27

Solution:

( ) ( ) 10 2 V and 0 A10

v i= =

2 21 1Char. eq. 0 or 2 5 0 thus the roots are 1 2s s s s s jRC LC

+ + = + + = =

2 11

So have ( ) cos 2 sin 2 now (0 ) 2tv t e B t B t v B + = + = =

( )( ) ( )

( )( )

( )1 2 2

0 01 1 VNeed 0 . KCL yields 0 0

5 2 s

0 1 3So 10 222

c c

dv vi i i

dt C

dvB B B

dt

+ +

+ + +

+

= = =

= = + =

( )3

Finally, we have 2 cos2 sin 2 V 02

t tv t e t e t t

= >


16/27

P 9.7-1 Determine the forced response for

the inductor current ifwhen (a) is= 1 A, (b) is=0.5tA, and (c) is= 2e

250tA for the circuit of

Figure P 9.7-1.

Figure P 9.7-1

Solution:

KCL :

KVL :

s L

L

v dvi i C

R dt

div L

dt

= + +

=2

2L Ls L

di d iLi i LC

R dt dt= + +

( ) ( )

2

2

5

3

( ) assume

1 1Let in

1to get: 0 0 1 1 10

.01 1 10

s f

L LL f L s

f

i l u t i A

d i dii i A i i

dt RC dt LC

A A i

= =

= = + + =

+ + = = =

( ) ( ) ( )

( )( )

0.5 ( ) assume

65 10 0.5

100 .001 .01 .001

650 100000 0 and 100000 0.5

s fi t u t i At B

A At B t

A B At t

= = +

+ + + =

+ = =

(a)

(b)

6

8

6 8

5 10

3.25 10

5 10 3.25 10f

A

B

i t A

=

=

=

250 250

250

2 Assumming does not work

because cannot have the same form as we choose

t t

s f

t

f s f

i e i Ae

i i i Bte

= =

=

250 250 250250

250

250 2

150 2

0.0133

0.0133 A

t t tt

t

f

Be Bte Btee

RC RC LC

B

B

i te

+ + =

=

=

=

(c)


17/27

P 9.7-2 Determine the forced response for the

capacitor voltage, vf, for the circuit of Figure P 9.7-2

when

(a) vs= 2 V, (b) v

s= 0.2tV, and (c) v

s= 1e

30tV.

Figure P 9.7-2

Solution:

Represent the circuit by the differential equation:2 1

s

d v R dvv v

dt L dt LC + + =

(a) 2 assume

1Then 0 0 12000 2 so6000

s f

f

v v A

A A v

= =

+ + = = =

0.2 assume

70 12000 12000 0.2 70 12000 0 and 12000 0.2s f

v t v At B

A At B t A B At t

= = +

+ + = + = =

(b)

1 70, 350

60000 12000

350 V60000

f

AA B B

tv

= = =

= +

(c)30 30

30 30 30 30 30

30

assume

1900 2100 12000 10800

10800

V10800

t t

s f

t t t t t

t

f

v e v Ae

A Ae Ae e Ae e A

ev

= =

+ = = =

=


18/27

P 9.8-2 Determine i(t) for t> 0 for the circuit shown in Figure P 9.8-2.

Hint: Show that 1 =2

2( ) 5 ( ) 5 ( ) for 0

d di t i t i t t

dt dt + + >

Figure P 9.8-2

Answer: i(t) = 0.2 + 0.246 e3.62t

0.646 e1.38t

A for t> 0.

Solution:

First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open

circuit at steady state, and the inductor will act like a short circuit.

( )1

0 0.2 A1 4

i = =

+

and

( ) ( )4

0 1 0.8 V1 4

v = = +

For t> 0

Apply KCL at node a:

s

10

v V dC v iR dt

+ + =

Apply KVL to the right mesh:

2 2 L0d d

R i L i v v R i L idt dt

+ = = +

After some algebra:2 2

1 2 1 2 s

2 2

1 1 1

5 5 1L R R C R R Vd d d d

i i i i i idt R L C dt R L C R L C dt dt

+ ++ + = + + =

The forced response will be a constant, if = B so2

21 5 5 0.2 A

d dB B B B

dt dt = + + = .

To find the natural response, consider the characteristic equation:

( )( )20 5 5 3.62 1.38s s s s= + + = + +


19/27

The natural response is

n 1 2

3.62 1.38t ti A e A e

= +

so

( ) 1 23.62 1.38

0.2

t t

i t A e A e

= + +

Then

( ) ( ) ( ) 1 23.62 1.38

4 4 10.48 1.52 0.8t td

v t i t i t A e A edt

= + = +

At t=0+

( ) 1 20.2 0 0.2i A A = + = + +

( ) 1 20.8 0 10.48 1.52 0.8v A A = + = +

soA1 = 0.246 andA2 = -0.646. Finally

( ) 3.62 1.380.2 0.246 0.646 At ti t e e = +


20/27

P 9.8-4 Find v(t) for t> 0 for the circuit shown inFigure P 9.8-4 when v(0) = 1 V and iL(0) = 0.

Answer:

3 4125 429 21cos 33 sin V17

t tv e e t t = +

Figure P 9.8-4

Solution:

t 0> ( )

( )

1KCL at top node : 0.5 5cos 0 112

1KVL at right loop : 0.5 212

LL

L

di dvt idt

dtdi dv

vdt dt

+ + =

= +

( )

( ) ( )

2 2

2 2

2 2

2 2

1of 1 0.5 5sin (3)

12

1of 2 0.5 412

L L

L

d i di d vd tdt dt dt dt

d i d v dvddt dt dt dt

+ + =

= +

( ) ( ) ( )2

2

22

2

Solving for in 4 and in 2 & plugging into 3

7 12 30sin 7 12 0 3, 4

L Ld i di

dt dt

d v dvv t s s s

dt dt + + = + + = =

3 4

1 2 1 2

1 2

so ( ) Try cos sin & plug

into D.E., equating like terms

3321yields ,17 17

t t

f fv t A e A e v v B t B t

B B

= + + = +

= =

0t +=

( )5 1 (0 ) V20 2 24

1 s1 1 12

c

dvi A

dt

++ = = = =

( )

1 21

21 2

3 4

21So (0 ) 1 2517

429(0 ) 33 24 3 4 1717

1 ( ) 25 429 21cos 33sin V17

t t

v A A A

dv AA Adt

v t e e t t

+

+

= = + + =

= = =

= +


21/27

P 9.8-9 In Figure P 9.8-9, determine the

inductor current i(t) when is= 5u(t) A. Assume

that i(0) = 0, vc(0) = 0.

Answer: i(t) = 5 + e2t

[ 5 cos 5t 2 sin 5t] A

Figure P 9.8-9

Solution:

P 9.8-15 The circuit shown in Figure P 9.8-15 is at steady state before the switch closes. Determine thecapacitor voltage, v(t), for t> 0.

Figure P 9.8-15

Solution:First, we find the initial conditions;

For t< 0, the switch is open and the circuit is at steady

( )

( )

( )

2

2

2

KCL:2

52

1 4 5

29 29

4 145

dv vC i i

sdt

d i L d iL C i u t

dt d t

d i d ii u t

dt d t

d i d ii u t

dt d t

+ + =

+ + =

+ + =

+ + =

[ ]

[ ]

2

2

2

Characteristic eqn: 4 29 0 roots : s = 2 j5

145 cos5 sin 5 and 529

So ( ) 5 cos5 sin 5

Now (0) 0 5 5

(0)0 2 5 2

n f

t

t

s s

i e A t B t i

i t e A t B t

i A A

diA B B

dt

+ + =

= + = =

= + +

= = + =

= = + =


22/27

state. At steady state, the capacitor acts like an open

circuit and the inductor acts like a short circuit.

( )0 0 Vv = and ( )0 0 Ai =

also

( ) ( ) ( )0 0

0 00.005 50 0.005

i vdv

dt= =

Next, represent the circuit after the switch closes by a differential equation.

After the switch closes, use KCL to get

( ) ( )

( )2

v t di t C v t

R dt= +

Use KVL to get

( ) ( ) ( )s 1d

v R i t L i t v t dt

= + +

Substitute to get

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

21

s 1 2

2 2

21 2

12

2 2

R d L d d v v t R C v t v t CL v t v t

R dt R dt dt

R Rd L dCL v t R C v t v t

dt R dt R

= + + + +

+= + + +

Finally,

( ) ( ) ( )2

s 1 1 2

2

2 2

1v R R Rd dv t v t v t

CL dt L R C dt R CL

+= + + +

Compare to

( ) ( ) ( )2

202

2 ( )d d

i t i t i t f t dtdt

+ + =

to get

1 1 2 s2

02 2

12 , and ( )

R R R vf t

L R C R CL CL

+= + = =

With the given element values, we have 14.5= and 20 200 = . Consequently, the roots of the

characteristic equation are 1 11.3s = and 2 17.7s = so the circuit is overdamped. The natural response

is


23/27

( ) 11.3 17.7n 1 2t t

v t A e A e = +

Next, determine the forced response.

The steady state response after the switch opens will beused as the forced response. At steady state, the

capacitor acts like an open circuit and the inductor actslike a short circuit.

f s

110 V

2v v= =

So

( ) 11.3 17.7n 1 210 t t

v t A e A e = + +

It remains to evaluateA1andA2using the initial conditions. At t= 0 we have

( ) 1 20 0 10v A A= = + +

and

( ) 1 20 0 11.3 17.7d

v A Adt

= =

Solving these equations gives

1 227.6 and 17.6A A= =

Finally,

( ) 11.3 17.710 27.6 17.6t tv t e e = +


24/27

P 9.8-17 The circuit shown in

Figure P 9.8-17 is at steady state before the

switch opens. Determine the inductor current,i2(t), for t> 0.

Figure P 9.8-17

Solution:First, we find the initial conditions;

For t< 0, the switch is closed and the

circuit is at steady state. At steady state, theinductors act like short circuits.

( )120

0 1.333 A15

i = =

and

( )2 0 0 Ai =

Next, represent the circuit by a differential equation.

After the switch opens, KVL gives

( ) ( ) ( )1 1 2 2 2 2d d

L i t R i t L i tdt dt

= +

KVL and KCL give

( ) ( ) ( )( )1 1 1 1 2 0d

L i t R i t i tdt

+ + =

Use the operator method to get

( )1 1 2 2 2 2

1 1 1 1 2 0

L s i R i L s i

L s i R i i

= +

+ + =


25/27

( ) ( )

2

1 1 1 1 1 2

1

2 2 2 2 2 2 2 2 1 2

1

2 1 22

2 2 2 1 1 2 2

1 1

2 1 1 1 22

2 2 2

2 2 1 1 2

0

0

0

0

L s i R s i R s i

Rs R i L s i R i L s i R s i

L

L R RL s i R R R s i i

L L

R R R R Rs i s i i

L L L L L

+ + =

+ + + + =

+ + + + =

+ + + + =

so

( ) ( ) ( )2

2 1 1 1 2

2 2 22

2 2 1 1 2

0R R R R Rd d

i t i t i t dt L L L dt L L

+ + + + =

Compare to

( ) ( ) ( )2 2

022 ( )d di t i t i t f t

dtdt + + =

to get

2 1 1 1 220

2 2 1 1 2

2 , and ( ) 0R R R R R

f tL L L L L

= + + = =

With the given element values, we have 33.9= and 20 281.25 = . Consequently, the roots of the

characteristic equation are2 2

1,2 0 4.4, 63.4s = = so the circuit is overdamped. The natural

response is

( ) 4.4 63.4n 1 2t t

i t A e A e = +

Next, determine the forced response.

The steady state response after the switch opens willbe used as the forced response. At steady state the

inductors act like short circuits.

f 0 Ai =

So

( ) ( ) ( ) 4.4 63.42 n f 1 2t t

i t i t i t A e A e = + = +

It remains to evaluateA1andA2using the initial conditions. At t= 0 we have


26/27

( )2 1 20 0i A A= = +

( ) ( ) ( ) ( ) ( )2 2 2 2 1 1 1 2 20 0 0 0 0 20d d

L i R i R i R i idt dt

+ + + =

and

( ) 1 220 0 4.4 63.4d

i A Adt

= =

Solving these equations givesA1= 0.339 andA2= 0.339 so

( ) 4.4 63.42 0.339 0.339 for 0t t

i t e e t = +


27/27

P 9.9-1 Find v(t) for t> 0 using the state variable method ofSection 9.9 when C= 1/5 F in the circuit of Figure P 9.9-1.

Sketch the response for v(t) for 0 < t< 10 s.

Answer: v(t) = 25et+ e 5t+ 24 V

Figure P 9.9-1

Solution:

2

2Solving for i in (1) & plugging into (2) 6 5 120

1 2

The characteristic equation is: 6 5 0,

The roots of the characteristic equation are 1, 5

Tthe natural response is:

d v dvv

dtdt

s s

s

+ + =

+ + =

=

5n 1 2

f f

L

1 21 2

5

1 2

( )

Try & plug into D.E. 24

(0) VFrom (1) 20 5 (0) 20s

So (0) 0 2425, 1

(0) ( ) 25 24 V20 5

t t

t t

v t A e A e

v B B v

dvi

dt

v A AA A

dvv t e eA A

dt

= +

= = =

= =

= = + + = = = + += =

0 circuit is source free (0) 0 & (0) 0Lt i v= = =

t > 0

( )

L

LL

(1)

1KCL at top node: 45

KVL at right loop: 1 6 0

dvi dt

div i

dt

+ =

=

chapter 9 homework solution

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