Solution manual fundamentals of electric circuits 3rd edition chapter07

No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior a Find the values of R and C.. No part of this Manual may be displayed,

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(a) Find the values of R and C

(b) Calculate the time constant τ

(c) Determine the time required for the voltage to decay half its initial value at

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4 ) t ( v

25

2 10 x x 10 x 40 RC ,

e v ) t ( v

V 4 ) 24 ( 2 10

2 ) 0 ( v v

t5.12

36

/oo

−

−τ

=

= +

=

=

Chapter 7, Problem 7

Assuming that the switch in Fig 7.87 has been in position A for a long time and is moved

to position B at t =0, find v0(t) for t ≥ 0

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Chapter 7, Problem 8

For the circuit in Fig 7.88, if

v = 10e−4t V and i = 0.2e − 4 t A, t > 0

(a) Find R and C

(b) Determine the time constant

(c) Calculate the initial energy in the capacitor

(d) -Obtain the time it takes to dissipate 50 percent of the initial energy

dt

dv C i

=

= C 4

1 ) 0 (

2

1 CV 2

1 2

1 w

2

1 e

e 1 5

0 = − - 8t 0 ⎯ ⎯→ - 8t 0 =

or 2 e8t 0 =

=

= ln ( 2 ) 8 1

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Chapter 7, Problem 12

The switch in the circuit of Fig 7.92 has been closed for a long time At t = 0 the switch

is opened Calculate i(t) for t > 0

12 )

2 R

Chapter 7, Problem 13

In the circuit of Fig 7.93,

v(t) = 20e−103t V, t > 0

i(t) = 4e−103t mA, t > 0

(a) Find R, L, and τ

(b) Calculate the energy dissipated in the resistance for 0 < t < 0.5 ms

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Th

(b) = 40 // 160 + 8 = 40 Ω , = = ( 20 10− 3) / 40 = 0 5 ms

x R

L R

Th

3 1 3 1 2 3 1

3 1 2

R R ) R R ( R R R

R R R R

+

+ +

= + +

=

= τ

3 1 3 1 2

3 1

R R ) R R ( R

) R R ( L

+ + +

(b) where

2 1

2 1

L L L

+

2 1

2 1 2 1 3 2 1

2 1 3

R R ) R R ( R R R

R R R R

+

+ +

= + +

=

= τ

) R R ) R R ( R ( L L (

) R R ( L L

2 1 2 1 3 2 1

2 1 2 1

+ + +

+

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4 1 R

-16te 2

-o 6 e ( 1 4 )( - 16 ) 2 e

dt

di L i 3 )

5 5

2 R

e ) 0 ( i

di -L )

t

(

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i 20 1

6 R

(a) Find L and R

(b) Determine the time constant

(c) Calculate the initial energy in the inductor

(d) What fraction of the initial energy is dissipated in 10 ms?

dt

di L v

L

ms 20

(c) = 2 = ( 0 08 )( 30 )2 =

2

1 ) 0 ( i L 2

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80 R 80

||

40

R 3 80

40 R

V ) ( i ) 0 ( i I

40 ) 2 ( 2

1 I L 2

1 w

=

=

3

40 R 1

3 80 R

= +

= ) t (

Using current division, the current through the 20 ohm resistor is

2.5t -

5

i - -i) ( 20 5

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1 1 3

2 2

2 i

+ +

= The Thevenin resistance R at the inductor’s terminals is th

3

4 ) 1 3 (

3 1 R

Lth

=

=

= τ 0 t , e -1.5 e

) 0 ( i ) t (

4t - o

dt

di L v

=o

v 2 e-4t V , t > 0

= +

1 3 1

v 0 5 e-4t V , t > 0

, 10

, 10

, 0

5

5 3

3 1 1

Otherwise

43

32

21

,

2

1

1 0 0

+

= ( t − 1 ) u ( t − 1 ) − ( t − 2 ) u ( t − 2 ) − ( t − 3 ) u ( t − 3 ) + ( t − 4 ) u ( t − 4 ) = r t − 1 ) − r t − 2 ) − r t − 3 ) + r t − 4 )

(d) y ( t ) = 2 u - t ) − 5 [ u ( t ) − u ( t − 1 ) ]

= 2 u - ) − 5 u ( ) + 5 u t − 1 )

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Chapter 7, Problem 25

Sketch each of the following waveforms

(a) i(t) = u(t -2) + u(t + 2)

1

v1 u t + 1 ) − 2 u ( ) + u t − 1 )

(b) v2( t ) = ( 4 − t ) [ u ( t − 2 ) − u ( t − 4 ) ]

) 4 t ( u ) 4 t ( ) 2 t ( u ) 4 t ( ) t (

= ) t (

v2 2 u(t − 2) − r(t − 2) + r(t − 4)

(c) v3( t ) = 2 [ u(t − 2) − u(t − 4) ] [ + 4 u(t − 4) − u(t − 6) ]

= ) t (

v3 2 u(t − 2) + 2 u(t − 4) − 4 u(t − 6)

(d) v4( t ) = -t [ u ( t − 1 ) − u ( t − 2 ) ] = -t u(t − 1) + t u ( t − 2 )

) 2 t ( u ) 2 2 t ( ) 1 t ( u ) 1 1 t - ) t (

= ) t (

v4 - r(t − 1) − u(t − 1) + r(t − 2) + 2 u(t − 2)

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t

λ λ λ

λ

2 ) 1 ( 0 )

1

1 4

1

2 1

The voltage across a 10-mH inductor is 20δ (t -2) mV Find the inductor current,

assuming that the inductor is initially uncharged

Chapter 7, Solution 33

) 0 ( i dt ) t ( v L

1 )

10 )

t

(

0 3 -

-3

+

− δ

) 1 t ( u ) 1 t ( ) 1 t ( u ) 1 t ( u dt d

− δ

= + δ

• +

•

− δ

= + δ

−

+ +

− δ

= +

−

) 6 t ( u ) 2 t ( 0 1 ) 6 t ( u ) 2 t ( ) 6 t (

) 2 t ( u ) 6 t ( u ) 2 t ( u ) 6 t ( dt d

−

=

− δ

• +

•

−

=

− δ

) 3 t ( 3 4 sin ) 3 t ( u t 4 cos 4

) 3 t ( t 4 sin ) 3 t ( u t 4 cos 4 ) 3 t ( u t 4 sin dt d

− δ

−

−

=

− δ +

−

=

− δ +

i t = e u(t)A

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(a) What is the time constant of the circuit?

(b) What is v( ∞ ) the final value of v?

(c) If v(0) = 2 find v(t) for t ≥ 0

Chapter 7, Solution 37

Let v = vh + vp, vp =10

4 /0

tAe

v = 10 + −0.25

8 10

2 ) 0

v

te

v = 10 − 8 −0.25(a) τ = 4 s

(b) v ( ∞ ) = 10 V

e

v = 10 − 8 −0.25 u(t)V

t u Ae i i

i•h+ h = ⎯ ⎯→ h = − t

Let

3

2 )

( 2 ) ( 3 , 0 ),

) ( 3

2

t u

ip =

) ( ) 3

2

t u Ae

If i(0) =0, then A + 2/3 = 0, i.e A=-2/3 Thus

) ( ) 1 ( 3

t u e

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Chapter 7, Problem 39

Calculate the capacitor voltage for t < 0 and t > 0 for each of the circuits in Fig 7.106

Figure 7.106

For Prob 7.39

Chapter 7, Solution 39

(a) Before t = 0,

= +

= ( 20 ) 1 4

1 ) t (

t ( v

8 ) 2 )(

4 (

=

τ , 4 v ( 0 ) = , 20 v ( ∞ ) =

8t-e ) 20 4 ( 20 ) t (

= ) t (

v 20 − 16 e-t 8 V (b) Before t = 0, v = v1+ v2, where v1 is due to the 12-V source and v2 is

due to the 2-A source

V 12

v1 =

To get v2, transform the current source as shown in Fig (a)

V -8

t ( v

12 ) (

v ∞ = , 4 v ( 0 ) = , 6 τ = RC = ( 2 )( 3 ) =

6 te ) 12 4 ( 12 ) t (

= ) t (

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v ∞ = , v ( 0 ) = 12 , τ = RC = ( 2 )( 3 ) = 6

6 te ) 4 12 ( 4 ) t (

= ) t (

v 4 + 8 et 6 V

(b) Before t = 0, v = 12 V

After t = 0, v ( t ) = v ( ∞ ) + [ v ( 0 ) − v ( ∞ ) ] et τAfter transforming the current source, the circuit is shown below

12 ) 0 (

5 36

) 30 )(

6 ( ) 1 )(

30

||

6 ( C

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o

o( t ) v ( ) v ( 0 ) v ( ) e v

0 ) 0 (

2 4

4 ) (

=

3

4 4

||

2

Req = = 4

) 3 ( 3

4

=

= τ

4 t

o( t ) 8 8 e

= ) t (

8 ) 12 ( 2 4

4 ) 0 (

+

= ) t (

vo 8 et 12 V

v 2 i

40

v 2 80

v

2

1

o o

(a)

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After t = 0, the circuit is as shown in Fig (b)

v

80

5 0 i 5 0

io = = Ω

=

=

5 0

80 i

1 R

o

V 64 )

0

(

vC =

480 t -

C( t ) 64 e

480 t C

480

1 -3 dt

dv -C i - i

i 0 8 e-t 480 u ( t ) A

0.5i 80 Ω

i vC

(b) 0.5i

3 mF

Chapter 7, Problem 44

The switch in Fig 7.111 has been in position a for a long time At t = 0 it moves to

position b Calculate i(t) for all t > 0

t ( v Using voltage division,

V 10 ) 30 ( 6 3

3 ) 0 (

+

6 3

3 ) (

e ) 4 10 ( 4 ) t (

2 ( dt

dv C ) t (

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2 25 0 ) 6 2

( 30 )]

( ) 0 ( [

v v v

t

v = ∞ + − ∞ − τ = − − V

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t ( v

( 1 e t)

24 ) t (

For t > 1, v ( 1 ) = 24 ( 1 − e- 1) = 15 17

30 ) ( v 0

24 - v(

6

1) - -(te ) 30 17 15 ( 30 ) t (

1) - -(te 83 14 30 ) t (

Thus,

= ) t (

1 t 0 ,

V e 1 24

-1) (t - t

t ( v

= ) t (

v 10 et 3 V

3 te 10 3

1 - ) 1 0 ( dt

dv C ) t (

3 1

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t ( v

( 1 e ) V 8

) t (

t ( v

V e

45 1 ) t (

v = t− 1 ) 5Thus,

= ) t (

1 t 0 , V e 1 8

5 ) 1 t 5 t

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t ( v 0 ) 0 (

For t > 0, we transform the current source to a voltage source as shown in Fig (a)

V 15 ) 30 ( 1 1 2

2 )

(

+ +

=

∞

Ω

= +

= ( 1 1 ) || 2 1 k

Rth

4

1 10 4

1 10 C

= τ

( 1 e ) , t 0 15

) t (

v i

1 5 7 ) t (

( 1 e ) mA 5

7 ) t (

Thus,

mA e 5 7 5 7 30 ) t (

= ) t (

Chapter 7, Problem 51

Rather than applying the short-cut technique used in Section 7.6, use KVL to obtain Eq (7.60)

Chapter 7, Solution 51.

Consider the circuit below

After the switch is closed, applying KVL gives

dt

di L Ri

di

dt L

R - R V i

diS

=

− Integrating both sides,

t L

R - R

V i

R V i ln

S 0 S

−

S 0

R V I

R V i

V ) ( i

which is the same as Eq (7.60)

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20 )

After t = 0, i ( t ) = i ( 0 ) et τ

2 2

4 R

i 5 e-t 2 u ( t ) A

(b) Before t = 0, the inductor acts as a short circuit so that the 2 Ω and 4 Ω

resistors are short-circuited

= ) t (

After t = 0, we have an RL circuit

τ

= i ( 0 ) et)

t (

2

3 R

L

=

= τ

= ) t (

i 6 e-2t 3 u ( t ) A

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Chapter 7, Problem 54

Obtain the inductor current for both t < 0 and t > 0 in each of the circuits in Fig 7.120

Figure 7.120

For Prob 7.54

Chapter 7, Solution 54.

(a) Before t = 0, i is obtained by current division or

= +

= ( 2 ) 4 4

4 ) t (

t ( i

5 3

=

= τ 1 ) 0 (

i = ,

7

6 ) 2 ( 3 4

3 ) 2 ( 12

||

4 4

12

||

4 ) (

+

= +

=

∞

t 2 -e 7

6 1 7

6 ) t (

=

= ) t (

10 ) t (

After t = 0, Req = 3 + 6 || 2 = 4 5

9

4 5 4

2 R

2 ) 0 (

To find i ( ∞ , consider the circuit below, at t = when the inductor ) becomes a short circuit,

9 v 3

v 6

v 24 2

v 10

−

A 3 3

v ) (

i ∞ = =

4 t 9 -e ) 3 2 ( 3 ) t (

= ) t (

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i 4 24

5 0 R

Lth

=

=

= τ-4te ) 48 ( )

∞

= i ( ) i ( 0 ) i ( ) et)

t ( i i(0) is found by applying nodal analysis to the following circuit

12 v 6

v 20

v 12

v 5

v 20

A 2 6

v ) 0 (

Since 4 20 || 5 = ,

6 1 ) 4 ( 6 4

4 ) (

t 1 6 0 4 e e

) 6 1 2 ( 6 1 ) t (

20t -e -20) ( 4 0 ( 2

1 dt

di L ) t (

= ) t (

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30 20

||

5 6

4 2 ) 0 (

For t > 0, the switch is closed so that the energies in L1 and L2 flow through the closed switch and become dissipated in the 5 Ω and 20 Ω resistors

1 t 1

1( t ) i ( 0 ) e

2

1 5

5 2 R

L1

1

τ

= ) t (

i1 2 4 e-2t u ( t ) A

2 t 2

2( t ) i ( 0 ) e

5

1 20

4 R

L2

2

τ

= ) t (

20 ) (

+

=

∞ Ω

= +

= 1 3 4

16

1 4

4 1 R

Lth

=

=

= τ

+

∞

= i ( ) i ( 0 ) i ( ) et)

t ( i

= ) t (

1 15 dt

di L i 3 ) t (

= ) t (

1 ) 3 ( 4 2

2 ) (

t ( i

-4te 1 ) t (

) -4)(-e )(

5 1 ( dt

di L ) t (

= ) t (

v 6 e-4tu ( t ) V

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4 ) (

t ( i

( 1 et 2)

4 ) t (

2 te 2

1 - ) 4 - )(

8 ( dt

di L ) t (

2 1 R

t ( i

= ) t (

i 10 − 5 e-8t u ( t ) A

8t -e ) 8 - )(

5 - 2

1 dt

di L ) t (

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||

3

2 R

i = ,

6

1 ) (

i ∞ =

( 1 et)

6

1 ) t (

For t > 1, ( 1 e ) 0 1054

6

1 ) 1 (

i = − - 1 =

2

1 6

1 3

1 ) (

i ∞ = + =

1) - -(te ) 5 0 1054 0 ( 5 0 ) t (

1) - -(te 3946 0 5 0 ) t (

Thus,

= ) t (

1 t 0 A

e 1 6

1

-1) (t - t

5 0 R

Lth

=

=

= τ

+

∞

= i ( ) i ( 0 ) i ( ) et)

t ( i

= ) t (

i 2 e-8t u ( t ) A

8t -e ) 2 )(

8 - 2

1 dt

di L ) t (

v - 8 e-8tu ( t ) V