Solution manual fundamentals of electric circuits 3rd edition chapter07

Chapter 7, Problem 1.
In the circuit shown in Fig. 7.81
v(t ) = 56e −200t V, t > 0
i(t ) = 8e −200t mA,

t>0

(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
t = 0.

Figure 7.81
For Prob. 7.1
Chapter 7, Solution 1.
τ=RC = 1/200

(a)

For the resistor, V=iR= 56e−200 t = 8Re −200t x10 −3
C=

⎯⎯

R=

56
= 7 kΩ
8

1
1
=
= 0.7143 µ F
200R 200 X7 X103

(b)
(c)

τ =1/200= 5 ms
If value of the voltage at = 0 is 56 .

1
x56 = 56e−200t
2
200to = ln2

⎯⎯

⎯⎯

to =

e200 t = 2
1
ln2 = 3.466 ms
200

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Chapter 7, Problem 2.
Find the time constant for the RC circuit in Fig. 7.82.

Figure 7.82
For Prob. 7.2.
Chapter 7, Solution 2.
τ = R th C
where R th is the Thevenin equivalent at the capacitor terminals.
R th = 120 || 80 + 12 = 60 Ω

τ = 60 × 0.5 × 10 -3 = 30 ms
Chapter 7, Problem 3.
Determine the time constant for the circuit in Fig. 7.83.

Figure 7.83
For Prob. 7.3.
Chapter 7, Solution 3.
R = 10 +20//(20+30) =10 + 40x50/(40 + 50)=32.22 kΩ

τ = RC = 32.22 X103 X100 X10 −12 = 3.222 µ S
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Chapter 7, Problem 4.
The switch in Fig. 7.84 moves instantaneously from A to B at t = 0. Find v for t > 0.

Figure 7.84
For Prob. 7.4.

Chapter 7, Solution 4.
For t<0, v(0-)=40 V.
For t >0. we have a source-free RC circuit.

τ = RC = 2 x103 x10 x10−6 = 0.02
v(t) = v(0)e− t / τ = 40e−50t V

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Chapter 7, Problem 5.
For the circuit shown in Fig. 7.85, find i(t), t > 0.

Figure 7.85
For Prob. 7.5.
Chapter 7, Solution 5.
Let v be the voltage across the capacitor.
For t <0,
4
v(0 − ) =
(24) = 16 V
2+4
For t >0, we have a source-free RC circuit as shown below.

i
5Ω
+
v

4Ω

1/3 F

1
3
= 16e− t / 3

τ = RC = (4 + 5) = 3 s

v(t) = v(0)e− t / τ
dv
1 1
i(t) = −C
= − (− )16e− t / 3 = 1.778e− t / 3 A
dt
3 3

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Chapter 7, Problem 6.
The switch in Fig. 7.86 has been closed for a long time, and it opens at t = 0. Find v(t) for
t ≥ 0.

Figure 7.86
For Prob. 7.6.

Chapter 7, Solution 6.

v o = v ( 0) =

2
(24) = 4 V
10 + 2

v( t ) = voe − t / τ , τ = RC = 40 x10−6 x 2 x103 =

2
25

v( t ) = 4e −12.5t V

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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 v 0 (t) for t ≥ 0.

Figure 7.87
For Prob. 7.7.

Chapter 7, Solution 7.
When the switch is at position A, the circuit reaches steady state. By voltage
division,
40
(12V ) = 8V
vo(0) =
40 + 20
When the switch is at position B, the circuit reaches steady state. By voltage
division,
30
(12V ) = 7.2V
vo(∞) =
30 + 20
20 x30
RTh = 20k // 30k =
= 12kΩ
50
τ = RThC = 12 x103 x2 x10 −3 = 24 s
vo(t) = vo(∞) + [vo(0) − vo(∞)]e−t /τ = 7.2 + (8 − 7.2)e− t / 24 = 7.2 + 0.8e− t / 24 V

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Chapter 7, Problem 8.
For the circuit in Fig. 7.88, if
v = 10e −4t V and

i = 0.2e − 4t 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.

Figure 7.88
For Prob. 7.8.
Chapter 7, Solution 8.
(a)

τ = RC =

1
4

dv
dt
-4t
- 0.2 e = C (10)(-4) e-4t
-i = C

⎯→ C = 5 mF

1
= 50 Ω
4C
1
τ = RC = = 0.25 s
4
1
1
w C (0) = CV02 = (5 × 10 -3 )(100) = 250 mJ
2
2
1 1
1
w R = × CV02 = CV02 (1 − e -2t 0 τ )
2 2
2
1
0.5 = 1 − e -8t 0 ⎯
⎯→ e -8t 0 =
2
8t 0
or
e =2
1
t 0 = ln (2) = 86.6 ms
8
R=

(b)
(c)
(d)

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Chapter 7, Problem 9.
The switch in Fig. 7.89 opens at t = 0. Find v 0 for t > 0

Figure 7.89
For Prob. 7.9.
Chapter 7, Solution 9.
For t < 0, the switch is closed so that
4
(6) = 4 V
vo(0) =
2+4
For t >0, we have a source-free RC circuit.
τ = RC = 3 x10−3 x4 x103 = 12 s

vo(t) = vo(0)e− t /τ = 4e− t /12 V

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Chapter 7, Problem 10.
For the circuit in Fig. 7.90, find v 0 (t) for t > 0. Determine the time necessary for the
capacitor voltage to decay to one-third of its value at t = 0.

Figure 7.90
For Prob. 7.10.
Chapter 7, Solution 10.
3
(36V ) = 9 V
3+9
For t>0, we have a source-free RC circuit
τ = RC = 3 x103 x20 x10−6 = 0.06 s

For t<0,

v(0 − ) =

vo(t) = 9e–16.667t V
Let the time be to.
3 = 9e–16.667to or e16.667to = 9/3 = 3
to = ln(3)/16.667 = 65.92 ms.

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Chapter 7, Problem 11.
For the circuit in Fig. 7.91, find i 0 for t > 0.

Figure 7.91
For Prob. 7.11.
Chapter 7, Solution 11.
For t<0, we have the circuit shown below.
3Ω

24 V

4H

4Ω

+

8Ω

4H
io
4Ω
8A

3Ω

8Ω

3//4= 4x3/7=1.7143
1.7143
(8) = 1.4118 A
io(0 − ) =
1.7143 + 8
For t >0, we have a source-free RL circuit.
L
4
τ= =
= 1/ 3
R 4+8
io(t) = io(0)e−t / τ = 1.4118e−3t A
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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.

Figure 7.92
For Prob. 7.12.
Chapter 7, Solution 12.
When t < 0, the switch is closed and the inductor acts like a short circuit to dc. The 4 Ω
resistor is short-circuited so that the resulting circuit is as shown in Fig. (a).

3Ω

12 V

i(0-)

+

4Ω

(a)

(b)

12
=4A
3
Since the current through an inductor cannot change abruptly,
i(0) = i(0 − ) = i(0 + ) = 4 A
i (0 − ) =

When t > 0, the voltage source is cut off and we have the RL circuit in Fig. (b).
L 2
τ = = = 0.5
R 4
Hence,
i( t ) = i(0) e - t τ = 4 e -2t A

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Chapter 7, Problem 13.
In the circuit of Fig. 7.93,
3

v(t) = 20e −10 t V,

t>0

3

i(t) = 4e −10 t mA,

t>0

(a) Find R, L, and τ .
(b) Calculate the energy dissipated in the resistance for 0 < t < 0.5 ms.

Figure 7.93
For Prob. 7.13.
Chapter 7, Solution 13.
(a) τ =

1
= 1ms
103

v = iR ⎯⎯
→ 20e−1000t = Rx4e−1000 t x10 −3
From this, R = 20/4 kΩ= 5 kΩ
5 x1000
L
⎯⎯

= 5H
But τ = = 1 3
L=
10
R
1000
(b) The energy dissipated in the resistor is
t

t

0

0

w = ∫ pdt = ∫ 80 x10 −3 e−2 x10 dt = −
3

−3

3
80 x10
e−2 x10 t
3
2 x10

0.5 x10 −3
0

= 40(1− e−1)µ J = 25.28 µ J

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Chapter 7, Problem 14.
Calculate the time constant of the circuit in Fig. 7.94.

Figure 7.94
For Prob. 7.14.
Chapter 7, Solution 14.
RTh = (40 + 20)//(10 + 30) =

60 x40
= 24kΩ
100

5 x10 −3
τ = L/R =
= 0.2083 µ s
24 x103

Chapter 7, Problem 15.
Find the time constant for each of the circuits in Fig. 7.95.

Figure 7.95
For Prob. 7.15.
Chapter 7, Solution 15
(a) RTh = 12 + 10 // 40 = 20Ω,
(b) RTh = 40 // 160 + 8 = 40Ω,

L
= 5 / 20 = 0.25s
RTh
L
τ=
= (20 x10 −3 ) / 40 = 0.5 ms
RTh

τ=

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Chapter 7, Problem 16.
Determine the time constant for each of the circuits in Fig. 7.96.

Figure 7.96
For Prob. 7.16.

Chapter 7, Solution 16.

τ=
(a)

L eq
R eq
L eq = L and R eq = R 2 +

τ=

(b)

R 1R 3
R 2 (R 1 + R 3 ) + R 1 R 3
=
R1 + R 3
R1 + R 3

L( R 1 + R 3 )
R 2 (R 1 + R 3 ) + R 1 R 3
R 3 (R 1 + R 2 ) + R 1 R 2
L1 L 2
R 1R 2
=
and R eq = R 3 +
L1 + L 2
R1 + R 2
R1 + R 2
L1L 2 (R 1 + R 2 )
τ=
(L 1 + L 2 ) ( R 3 ( R 1 + R 2 ) + R 1 R 2 )

where L eq =

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Chapter 7, Problem 17.
Consider the circuit of Fig. 7.97. Find v 0 (t) if i(0) = 2 A and v(t) = 0.

Figure 7.97
For Prob. 7.17.
Chapter 7, Solution 17.
i( t ) = i(0) e - t τ ,

τ=

14 1
L
=
=
R eq
4 16

i( t ) = 2 e -16t
v o ( t ) = 3i + L

di
= 6 e-16t + (1 4)(-16) 2 e-16t
dt

v o ( t ) = - 2 e -16t u ( t )V

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Chapter 7, Problem 18.
For the circuit in Fig. 7.98, determine v 0 (t) when i(0) = 1 A and v(t) = 0.

Figure 7.98
For Prob. 7.18.
Chapter 7, Solution 18.
If v( t ) = 0 , the circuit can be redrawn as shown below.

6
L 2 5 1
,
τ= = × =
5
R 5 6 3
-t τ
-3t
i( t ) = i(0) e = e
di - 2
(-3) e -3t = 1.2 e -3t V
v o ( t ) = -L =
dt
5
R eq = 2 || 3 =

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Chapter 7, Problem 19.
In the circuit of Fig. 7.99, find i(t) for t > 0 if i(0) = 2 A.

Figure 7.99
For Prob. 7.19.
Chapter 7, Solution 19.
i

1V
− +

10 Ω

i1

i1

i2

i/2

i2

40 Ω

To find R th we replace the inductor by a 1-V voltage source as shown above.
10 i1 − 1 + 40 i 2 = 0
But
i = i2 + i 2
and
i = i1
i.e.
i1 = 2 i 2 = i
1
⎯→ i =
10 i − 1 + 20 i = 0 ⎯
30
1
R th = = 30 Ω
i
L
6
τ=
=
= 0.2 s
R th 30

i( t ) = 2 e -5t u ( t ) A

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Chapter 7, Problem 20.

For the circuit in Fig. 7.100,
v = 120e −50t V

and
i = 30e −50t A, t > 0

(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?

Figure 7.100
For Prob. 7.20.
Chapter 7, Solution 20.

(a)

L
1
=
R 50
di
-v= L
dt
τ=

⎯→ R = 50L

- 120 e - 50t = L(30)(-50) e - 50t ⎯
⎯→ L = 80 mH
R = 50L = 4 Ω
L
1
τ= =
= 20 ms
(b)
R 50
1
1
w = L i 2 (0) = (0.08)(30) 2 = 36J
(c)
2
2
The value of the energy remaining at 10 ms is given by:
w10 = 0.04(30e–0.5)2 = 0.04(18.196)2 = 13.24J.
So, the fraction of the energy dissipated in the first 10 ms is given by:
(36–13.24)/36 = 0.6322 or 63.2%.

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Chapter 7, Problem 21.
In the circuit of Fig. 7.101, find the value of R
stored in the inductor will be 1 J.

Figure 7.101
For Prob. 7.21.
Chapter 7, Solution 21.
The circuit can be replaced by its Thevenin equivalent shown below.

Rth

Vth

+

2H

80
(60) = 40 V
80 + 40
80
R th = 40 || 80 + R =
+R
3
Vth
40
I = i(0) = i(∞) =
=
R th 80 3 + R
Vth =

2

⎟ =1
3⎠
40
40
=1 ⎯
⎯→ R =
R + 80 3
3
R = 13.333 Ω
1
1 ⎛ 40
w = L I 2 = (2)⎜
2
2 ⎝ R + 80

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Chapter 7, Problem 22.
Find i(t) and v(t) for t > 0 in the circuit of Fig. 7.102 if i(0) = 10 A.

Figure 7.102
For Prob. 7.22.
Chapter 7, Solution 22.
i( t ) = i(0) e - t τ ,

τ=

L
R eq

R eq = 5 || 20 + 1 = 5 Ω ,

τ=

2
5

i( t ) = 10 e -2.5t A
Using current division, the current through the 20 ohm resistor is
5
-i
io =
(-i) = = -2 e -2.5t
5 + 20
5
v( t ) = 20 i o = - 40 e -2.5t V

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Chapter 7, Problem 23.
Consider the circuit in Fig. 7.103. Given that v 0 (0) = 2 V, find v 0 and v x for t > 0.

Figure 7.103
For Prob. 7.23.
Chapter 7, Solution 23.
Since the 2 Ω resistor, 1/3 H inductor, and the (3+1) Ω resistor are in parallel,
they always have the same voltage.
2
2
+
= 1.5 ⎯
⎯→ i(0) = -1.5
2 3 +1
The Thevenin resistance R th at the inductor’s terminals is
13 1
4
L
R th = 2 || (3 + 1) = ,
τ=
=
=
3
R th 4 3 4
-i =

i( t ) = i(0) e - t τ = -1.5 e -4t , t > 0
di
v L = v o = L = -1.5(-4)(1/3) e -4t
dt
-4t
v o = 2 e V, t > 0
vx =

1
v = 0.5 e -4t V , t > 0
3 +1 L

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Chapter 7, Problem 24.
Express the following signals in terms of singularity functions.
⎧ 0,

(a) v(t) = ⎨
⎩− 5,
⎧ 0,
⎪− 10,

(b) i(t) = ⎨
⎪ 10,
⎪⎩ 0,

⎧t −1
⎪ 1,

(c) x(t) = ⎨
⎪4 − t
⎪⎩ 0,
⎧ 2,

(d) y(t) = ⎨− 5,
⎪ 0,

t<0
t>0
t <1
1< t < 3
3t <5

1< t < 2
23Otherwise

t<0
0 < t <1
t <1

Chapter 7, Solution 24.
(a) v( t ) = - 5 u(t)
(b) i( t ) = -10 [ u ( t ) − u ( t − 3)] + 10[ u ( t − 3) − u ( t − 5)]
= - 10 u(t ) + 20 u(t − 3) − 10 u(t − 5)
(c) x ( t ) = ( t − 1) [ u ( t − 1) − u ( t − 2)] + [ u ( t − 2) − u ( t − 3)]
+ (4 − t ) [ u ( t − 3) − u ( t − 4)]
= ( 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(-t ) − 5 u(t ) + 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)
(b) v(t) = r(t) – r(t - 3) + 4u(t - 5) – 8u(t - 8)

Chapter 7, Solution 25.
The waveforms are sketched below.
(a)

i(t)
2
1

-2 -1 0

1

2

3

4

t

(b)
v(t)
7

3

–1

0

1

2

3

4

5

6

7

8

t

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Chapter 7, Problem 26.
Express the signals in Fig. 7.104 in terms of singularity functions.

Figure 7.104
For Prob. 7.26.
Chapter 7, Solution 26.
(a)

(b)

(c)

(d)

v1 ( t ) = u ( t + 1) − u ( t ) + [ u ( t − 1) − u ( t )]
v1 ( t ) = u(t + 1) − 2 u(t ) + u(t − 1)
v 2 ( t ) = ( 4 − t ) [ u ( t − 2) − u ( t − 4) ]
v 2 ( t ) = -( t − 4) u ( t − 2) + ( t − 4) u ( t − 4)
v 2 ( t ) = 2 u(t − 2) − r(t − 2) + r(t − 4)

v 3 ( t ) = 2 [ u(t − 2) − u(t − 4)] + 4 [ u(t − 4) − u(t − 6)]
v 3 ( t ) = 2 u(t − 2) + 2 u(t − 4) − 4 u(t − 6)
v 4 ( t ) = -t [ u ( t − 1) − u ( t − 2)] = -t u(t − 1) + t u ( t − 2)
v 4 ( t ) = (-t + 1 − 1) u ( t − 1) + ( t − 2 + 2) u ( t − 2)
v 4 ( t ) = - r(t − 1) − u(t − 1) + r(t − 2) + 2 u(t − 2)

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Chapter 7, Problem 27.
Express v(t) in Fig. 7.105 in terms of step functions.

Figure 7.105
For Prob. 7.27.
Chapter 7, Solution 27.
v(t)= 5u(t+1)+10u(t)–25u(t–1)+15u(t-2)V

Chapter 7, Problem 28.
Sketch the waveform represented by
i(t) = r(t) – r(t -1) – u(t - 2) – r(t - 2)
+ r(t -3) + u(t - 4)

Chapter 7, Solution 28.
i(t) is sketched below.
i(t)
1

0

1

2

3

4

t

-1

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