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sadiuku chap 7 sol

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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.
for which the steady-state energy
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
3<t <5
t <5
1< t < 2
2<t <3
3<t < 4
Otherwise
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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Chapter 7, Problem 29.
Sketch the following functions:
(a) x(t) = 10e − t u(t-1)
(b) y(t) = 10e − ( t −1) u(t)
(c) z(t) = cos 4t δ (t - 1)
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Chapter 7, Solution 29
x(t)
(a)
3.679
0
(b)
1
t
y(t)
27.18
t
0
(c)
z (t ) = cos 4tδ (t − 1) = cos 4δ (t − 1) = −0.6536δ (t − 1) , which is sketched below.
z(t)
0
1
t
-0.653 δ (t )
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Chapter 7, Problem 30.
Evaluate the following integrals involving the impulse functions:
(a)
(b)
∫
∞
∫
∞
4t 2δ (t − 1)dt
−∞
−∞
4t 2 cos 2π t δ (t − 0.5)dt
Chapter 7, Solution 30.
∞
(a)
∫− ∞ 4t
(b)
∫-∞ 4t
∞
2
2
δ( t − 1) dt = 4t 2 t =1 = 4
cos(2πt ) δ( t − 0.5) dt = 4t 2 cos(2πt ) t =0.5 = cos π = - 1
Chapter 7, Problem 31.
Evaluate the following integrals:
(a)
∫
∞
(b)
∫
∞
2
e −4 r δ (t − 2)dt
−∞
−∞
[5δ (t ) + e −t δ (t ) + cos 2π t δ (t )] dt
Chapter 7, Solution 31.
(a)
(b)
= e = 112 × 10
∫ [ e δ(t − 2)] dt = e
∫ [ 5 δ(t ) + e δ(t ) + cos 2πt δ(t )] dt = ( 5 + e + cos(2πt ))
∞
-∞
∞
-∞
- 4t 2
- 4t 2
-t
t=2
-9
-16
-t
t =0
= 5 +1+1 = 7
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Chapter 7, Problem 32.
Evaluate the following integrals:
t
(a) ∫ u (λ ) dλ
l
4
(b) ∫ r (t − 1)dt
0
5
(c) ∫ (t − 6) 2 δ (t − 2)dt
1
Chapter 7, Solution 32.
(a)
(b)
t
t
t
1
4
1
1
∫ u (λ )dλ = ∫ 1dλ = λ
4
0
1
∫ r (t − 1)dt = ∫ 0dt + ∫ (t − 1)dt =
0
5
(c )
1
= t −1
∫ (t − 6)
2
δ (t − 2)dt = (t − 6) 2
t2
− t 14 = 4.5
2
t =2
= 16
1
Chapter 7, Problem 33.
The voltage across a 10-mH inductor is 20 δ (t -2) mV. Find the inductor current,
assuming that the inductor is initially uncharged.
Chapter 7, Solution 33.
i( t ) =
1 t
∫ v(t ) dt + i(0)
L 0
i( t ) =
10 -3
10 × 10 -3
∫ 20 δ(t − 2) dt + 0
t
0
i ( t ) = 2 u( t − 2 ) A
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Chapter 7, Problem 34.
Evaluate the following derivatives:
d
[u(t - 1) u(t + 1)]
dt
d
[r(t - 6) u(t - 2)]
(b)
dt
d
[sin 4tu(t - 31)]
(c)
dt
(a)
Chapter 7, Solution 34.
d
[u ( t − 1) u ( t + 1)] = δ( t − 1)u ( t + 1) +
dt
(a)
u ( t − 1)δ( t + 1) = δ( t − 1) • 1 + 0 • δ( t + 1) = δ( t − 1)
d
[r ( t − 6) u ( t − 2)] = u ( t − 6)u ( t − 2) +
dt
r ( t − 6)δ( t − 2) = u ( t − 6) • 1 + 0 • δ( t − 2) = u ( t − 6)
(b)
d
[sin 4t u (t − 3)] = 4 cos 4t u ( t − 3) + sin 4tδ( t − 3)
dt
= 4 cos 4t u ( t − 3) + sin 4x3δ( t − 3)
= 4 cos 4t u ( t − 3) − 0.5366δ( t − 3)
(c)
Chapter 7, Problem 35.
Find the solution to the following differential equations:
dv
+ 2v = 0,
dt
di
+ 3i = 0,
(b) 2
dt
(a)
v(0) = -1 V
i(0) = 2
Chapter 7, Solution 35.
(a)
v = Ae−2t ,
v(0) = A = −1
v = −e−2t u(t)V
i = Ae3 t / 2 ,
i(0) = A = 2
(b)
i(t) = 2e1.5t u(t)A
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Chapter 7, Problem 36.
Solve for v in the following differential equations, subject to the stated initial condition.
v(0) = 0
(a) dv / dt + v = u(t),
(b) 2 dv / dt – v =3u(t), v(0) = -6
Chapter 7, Solution 36.
(a)
v( t ) = A + B e-t , t > 0
A = 1,
v(0) = 0 = 1 + B
v( t ) =
1 − e -t V , t > 0
(b)
v( t ) = A + B e t 2 , t > 0
v(0) = -6 = -3 + B
A = -3 ,
v( t ) = - 3 ( 1 + e t 2 ) V , t > 0
or
B = -1
or
B = -3
Chapter 7, Problem 37.
A circuit is described by
4
dv
+ v = 10
dt
(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.
•
1
vh + 4 v
h
=0
⎯
⎯→
v h = Ae −t / 4
v = 10 + Ae −0.25t
(a) τ = 4 s
v(0) = 2 = 10 + A
v = 10 − 8e −0.25t
⎯
⎯→
A = −8
(b) v(∞) = 10 V
(c ) v = 10 − 8e −0.25t u(t)V
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Chapter 7, Problem 38.
A circuit is described by
di
+ 3i = 2u(t)
dt
Find i(t) for t > 0 given that i(0) = 0.
Chapter 7, Solution 38
Let i = ip +ih
•
i h + 3ih = 0
Let i p = ku (t ),
•
ip = 0,
3ku (t ) = 2u (t )
ip =
ih = Ae −3t u (t )
⎯
⎯→
⎯
⎯→
k=
2
3
2
u (t )
3
2
i = ( Ae −3t + )u (t )
3
If i(0) =0, then A + 2/3 = 0, i.e. A=-2/3. Thus
i=
2
(1 − e −3t )u (t )
3
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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.
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Chapter 7, Solution 39.
Before t = 0,
(a)
v( t ) =
1
(20) = 4 V
4 +1
After t = 0,
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
τ = RC = (4)(2) = 8 , v(0) = 4 ,
v(∞) = 20
v( t ) = 20 + (4 − 20) e -t 8
v( t ) = 20 − 16 e - t 8 V
Before t = 0, v = v1 + v 2 , where v1 is due to the 12-V source and v 2 is
due to the 2-A source.
v1 = 12 V
To get v 2 , transform the current source as shown in Fig. (a).
v 2 = -8 V
Thus,
v = 12 − 8 = 4 V
(b)
After t = 0, the circuit becomes that shown in Fig. (b).
2F
+
v2
2F
4Ω
−
+
−
8V
12 V
+
−
3Ω
3Ω
(a)
(b)
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v(∞) = 12 ,
v(0) = 4 ,
τ = RC = (2)(3) = 6
-t 6
v( t ) = 12 + (4 − 12) e
v( t ) = 12 − 8 e -t 6 V
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Chapter 7, Problem 40.
Find the capacitor voltage for t < 0 and t > 0 for each of the circuits in Fig. 7.107.
Figure 7.107
For Prob. 7.40.
Chapter 7, Solution 40.
Before t = 0, v = 12 V .
(a)
After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v(∞) = 4 ,
v(0) = 12 ,
τ = RC = (2)(3) = 6
v( t ) = 4 + (12 − 4) e - t 6
v( t ) = 4 + 8 e - t 6 V
(b)
Before t = 0, v = 12 V .
After t = 0, v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
After transforming the current source, the circuit is shown below.
t=0
2Ω
12 V
v(0) = 12 ,
v = 12 V
+
−
v(∞) = 12 ,
4Ω
5F
τ = RC = (2)(5) = 10
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Chapter 7, Problem 41.
For the circuit in Fig. 7.108, find v(t) for t > 0.
Figure 7.108
For Prob. 7.41.
Chapter 7, Solution 41.
v(0) = 0 ,
v(∞ ) =
R eq C = (6 || 30)(1) =
30
(12) = 10
36
(6)(30)
=5
36
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 10 + (0 − 10) e - t 5
v( t ) = 10 (1 − e -0.2t ) u ( t )V
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Chapter 7, Problem 42.
(a) If the switch in Fig. 7.109 has been open for a long time and is closed at t = 0, find
v o (t).
(b) Suppose that the switch has been closed for a long time and is opened at t = 0. Find
v o (t).
Figure 7.109
For Prob. 7.42.
Chapter 7, Solution 42.
(a)
v o ( t ) = v o (∞) + [ v o (0) − v o (∞)] e -t τ
4
v o (0) = 0 ,
v o (∞) =
(12) = 8
4+2
4
τ = R eq C eq , R eq = 2 || 4 =
3
4
τ = (3) = 4
3
v o (t ) = 8 − 8 e -t 4
v o ( t ) = 8 ( 1 − e -0.25t ) V
(b)
For this case, v o (∞) = 0 so that
v o ( t ) = v o (0) e - t τ
4
v o (0) =
(12) = 8 ,
4+2
v o ( t ) = 8 e -t 12 V
τ = RC = (4)(3) = 12
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Chapter 7, Problem 43.
Consider the circuit in Fig. 7.110. Find i(t) for t < 0 and t > 0.
Figure 7.110
For Prob. 7.43.
Chapter 7, Solution 43.
Before t = 0, the circuit has reached steady state so that the capacitor acts like an open
circuit. The circuit is equivalent to that shown in Fig. (a) after transforming the voltage
source.
0.5i
vo
i
40 Ω
2A
0.5i
80 Ω
(a)
0.5i = 2 −
Hence,
vo
,
40
vo
1 vo
= 2−
2 80
40
i=
i=
vo
80
⎯
⎯→ v o =
320
= 64
5
vo
= 0.8 A
80
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After t = 0, the circuit is as shown in Fig. (b).
0.5i
vC
i
3 mF
80 Ω
0.5i
(b)
v C ( t ) = v C (0) e - t τ ,
τ = R th C
To find R th , we replace the capacitor with a 1-V voltage source as shown in Fig. (c).
0.5i
vC
i
1V
+
−
0.5i
80 Ω
(c)
vC
1
0.5
=
,
i o = 0.5 i =
80 80
80
1 80
R th = =
= 160 Ω ,
τ = R th C = 480
i o 0.5
v C (0) = 64 V
i=
v C ( t ) = 64 e - t 480
dv C
⎛ 1 ⎞
⎟ 64 e - t 480
0.5 i = -i C = -C
= -3 ⎜
⎝ 480 ⎠
dt
i( t ) = 0.8 e - t 480 u ( t )A
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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.
Figure 7.111
For Prob. 7.44.
Chapter 7, Solution 44.
R eq = 6 || 3 = 2 Ω ,
τ = RC = 4
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
Using voltage division,
3
v(0) =
(30) = 10 V ,
3+ 6
v(∞) =
3
(12) = 4 V
3+ 6
Thus,
v( t ) = 4 + (10 − 4) e - t 4 = 4 + 6 e - t 4
⎛ - 1⎞
dv
i( t ) = C
= (2)(6) ⎜ ⎟ e - t 4 = - 3 e -0.25t A
⎝4⎠
dt
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Chapter 7, Problem 45.
Find v o in the circuit of Fig. 7.112 when v s = 6u(t). Assume that v o (0) = 1 V.
Figure 7.112
For Prob. 7.45.
Chapter 7, Solution 45.
To find RTh, consider the circuit shown below.
20 kΩ
10 kΩ
40 kΩ
RTh = 10 + 20 // 40 = 10 +
τ = RThC =
RTh
20 x40 70
kΩ
=
60
3
70
x103 x3 x10 −6 = 0.07
3
To find vo(∞), consider the circuit below.
20 kΩ
10 kΩ
+
6V
+
_
40 kΩ
vo
–
vo(∞) =
40
(6V) = 4V
40 + 20
vo(t) = vo(∞) + [vo(0) − vo(∞)]e−t / τ = 4 + (1− 4)e− t / 0.07 = 4 − 3e−14.286t V u(t)
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Chapter 7, Problem 46.
For the circuit in Fig. 7.113, i s (t) = 5u(t) Find v(t).
Figure 7.113
For Prob. 7.46.
Chapter 7, Solution 46.
τ = RTh C = (2 + 6) x0.25 = 2s,
v(0) = 0,
v(∞) = 6i s = 6 x5 = 30
v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ = 30(1 − e − t / 2 ) V
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Chapter 7, Problem 47.
Determine v(t) for t > 0 in the circuit of Fig. 7.114 if v(0) = 0.
Figure 7.114
For Prob. 7.47.
Chapter 7, Solution 47.
For t < 0, u ( t ) = 0 ,
u ( t − 1) = 0 ,
v(0) = 0
For 0 < t < 1, τ = RC = (2 + 8)(0.1) = 1
v(0) = 0 ,
v(∞) = (8)(3) = 24
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 24( 1 − e - t )
For t > 1,
v(1) = 24( 1 − e -1 ) = 15.17
- 6 + v(∞) - 24 = 0 ⎯
⎯→ v(∞) = 30
v( t ) = 30 + (15.17 − 30) e -(t-1)
v( t ) = 30 − 14.83 e -(t-1)
Thus,
(
)
⎧ 24 1 − e - t V ,
0<t<1
v( t ) = ⎨
-(t -1)
V,
t >1
⎩ 30 − 14.83 e
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Chapter 7, Problem 48.
Find v(t) and i(t) in the circuit of Fig. 7.115.
Figure 7.115
For Prob. 7.48.
Chapter 7, Solution 48.
For t < 0,
u (-t) = 1 ,
For t > 0,
u (-t) = 0 ,
R th = 20 + 10 = 30 ,
v(0) = 10 V
v(∞) = 0
τ = R th C = (30)(0.1) = 3
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 10 e -t 3 V
⎛ - 1⎞
dv
= (0.1) ⎜ ⎟10 e - t 3
⎝3⎠
dt
-1
i( t ) = e - t 3 A
3
i( t ) = C
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Chapter 7, Problem 49.
If the waveform in Fig. 7.116(a) is applied to the circuit of Fig. 7.116(b), find v(t).
Assume v(0) = 0.
Figure 7.116
For Prob. 7.49 and Review Question 7.10.
Chapter 7, Solution 49.
For 0 < t < 1, v(0) = 0 ,
R eq = 4 + 6 = 10 ,
v(∞) = (2)(4) = 8
τ = R eq C = (10)(0.5) = 5
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t ) = 8 ( 1 − e - t 5 ) V
For t > 1,
v(1) = 8 ( 1 − e -0.2 ) = 1.45 ,
v( t ) = v(∞) + [ v(1) − v(∞)] e -( t −1) τ
v( t ) = 1.45 e -( t −1) 5 V
Thus,
(
v(∞) = 0
)
⎧ 8 1 − e -t 5 V , 0 < t < 1
v( t ) = ⎨
- ( t −1 ) 5
V,
t >1
⎩ 1.45 e
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Chapter 7, Problem 50.
* In the circuit of Fig. 7.117, find i x for t > 0. Let R 1 = R 2 = 1k Ω , R 3 = 2k Ω , and C =
0.25 mF.
Figure 7.117
For Prob. 7.50.
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Chapter 7, Solution 50.
For the capacitor voltage,
v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ
v(0) = 0
For t > 0, we transform the current source to a voltage source as shown in Fig. (a).
1 kΩ
1 kΩ
+
30 V
+
−
2 kΩ
v
−
(a)
2
(30) = 15 V
2 +1+1
R th = (1 + 1) || 2 = 1 kΩ
1
1
τ = R th C = 10 3 × × 10 -3 =
4
4
v( t ) = 15 ( 1 − e -4t ) , t > 0
v(∞) =
We now obtain i x from v(t). Consider Fig. (b).
iT 1 kΩ
v
ix
30 mA
1 kΩ
1/4 mF
2 kΩ
(b)
But
i x = 30 mA − i T
v
dv
+C
iT =
R3
dt
i T ( t ) = 7.5 ( 1 − e -4t ) mA +
i T ( t ) = 7.5 ( 1 + e -4t ) mA
1
× 10 -3 (-15)(-4) e -4t A
4
Thus,
i x ( t ) = 30 − 7.5 − 7.5 e -4t mA
i x ( t ) = 7.5 ( 3 − e -4t ) mA , t > 0
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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.
t=0
R
+
VS
+
−
i
L
v
−
After the switch is closed, applying KVL gives
di
VS = Ri + L
dt
⎛
VS ⎞
di
⎟
L = -R ⎜ i −
or
⎝
dt
R⎠
di
-R
=
dt
i − VS R
L
Integrating both sides,
⎛ V ⎞ i( t ) - R
ln ⎜ i − S ⎟ I 0 =
t
⎝
R⎠
L
⎛ i − VS R ⎞ - t
⎟=
ln ⎜
⎝ I0 − VS R ⎠ τ
i − VS R
= e- t τ
or
I0 − VS R
i( t ) =
VS ⎛
VS ⎞ -t τ
⎟e
+ ⎜ I0 −
R ⎝
R⎠
which is the same as Eq. (7.60).
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Chapter 7, Problem 52.
For the circuit in Fig. 7.118, find i(t) for t > 0.
Figure 7.118
For Prob. 7.52.
Chapter 7, Solution 52.
20
= 2 A,
i(∞) = 2 A
10
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i(0) =
i( t ) = 2 A
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Chapter 7, Problem 53.
Determine the inductor current i(t) for both t < 0 and t > 0 for each of the circuits in Fig.
7.119.
Figure 7.119
For Prob. 7.53.
Chapter 7, Solution 53.
(a)
25
=5A
3+ 2
After t = 0,
i( t ) = i(0) e- t τ
L 4
τ = = = 2,
i(0) = 5
R 2
Before t = 0,
i=
i( t ) = 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.
i( t ) = 6 A
After t = 0, we have an RL circuit.
L 3
τ= =
i( t ) = i(0) e- t τ ,
R 2
i( t ) = 6 e - 2 t 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.
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Chapter 7, Solution 54.
(a)
Before t = 0, i is obtained by current division or
4
i( t ) =
(2) = 1 A
4+4
After t = 0,
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
L
,
R eq = 4 + 4 || 12 = 7 Ω
τ=
R eq
τ=
3.5 1
=
7
2
i(0) = 1 ,
i(∞) =
4 || 12
3
6
(2) =
(2) =
4 + 4 || 12
4+3
7
6 ⎛
6⎞
+ ⎜ 1 − ⎟ e -2 t
7 ⎝
7⎠
1
i( t ) = ( 6 − e - 2t ) A
7
10
Before t = 0, i( t ) =
=2A
2+3
After t = 0,
R eq = 3 + 6 || 2 = 4.5
i( t ) =
(b)
L
2
4
=
=
R eq 4.5 9
i(0) = 2
To find i(∞) , consider the circuit below, at t = when the inductor
becomes a short circuit,
v
τ=
i
10 V
2Ω
+
−
24 V
+
−
6Ω
2H
3Ω
10 − v 24 − v v
+
=
⎯
⎯→ v = 9
2
6
3
v
i(∞) = = 3 A
3
i( t ) = 3 + (2 − 3) e -9 t 4
i( t ) = 3 − e - 9 t 4 A
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Chapter 7, Problem 55.
Find v(t) for t < 0 and t > 0 in the circuit of Fig. 7.121.
Figure 7.121
For Prob. 7.55.
Chapter 7, Solution 55.
For t < 0, consider the circuit shown in Fig. (a).
0.5 H
io
3Ω
24 V
io
+
−
0.5 H
+
+
4io
v
i
2Ω
−
(a)
8Ω
20 V
+
v
+
−
2Ω
−
(b)
3i o + 24 − 4i o = 0 ⎯
⎯→ i o = 24
v
v( t ) = 4i o = 96 V
i = = 48 A
2
For t > 0, consider the circuit in Fig. (b).
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i(0) = 48 ,
i (∞ ) = 0
L
0.5 1
=
=
R th = 2 Ω , τ =
R th
2
4
i( t ) = (48) e -4t
v( t ) = 2 i( t ) = 96 e -4t u ( t )V
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Chapter 7, Problem 56.
For the network shown in Fig. 7.122, find v(t) for t > 0.
Figure 7.122
For Prob. 7.56.
Chapter 7, Solution 56.
R eq = 6 + 20 || 5 = 10 Ω ,
τ=
L
= 0.05
R
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i(0) is found by applying nodal analysis to the following circuit.
5Ω
vx
2A
12 Ω
20 Ω
i
6Ω
+
0.5 H
+
−
20 V
v
−
2+
20 − v x v x v x v x
=
+
+
5
12 20 6
vx
i ( 0) =
=2A
6
⎯
⎯→ v x = 12
Since 20 || 5 = 4 ,
4
i(∞) =
(4) = 1.6
4+6
i( t ) = 1.6 + (2 − 1.6) e- t 0.05 = 1.6 + 0.4 e-20t
di 1
v( t ) = L = (0.4) (-20) e -20t
dt 2
v( t ) = - 4 e -20t V
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Chapter 7, Problem 57.
*
Find i 1 (t) and i 2 (t) for t > 0 in the circuit of Fig. 7.123.
Figure 7.123
For Prob. 7.57.
* An asterisk indicates a challenging problem.
Chapter 7, Solution 57.
At t = 0 − , the circuit has reached steady state so that the inductors act like short
circuits.
6Ω
30 V
i
+
−
i1
i2
5Ω
20 Ω
20
30
30
=
= 3,
i1 =
(3) = 2.4 ,
25
6 + 5 || 20 10
i 1 ( 0 ) = 2 .4 A ,
i 2 ( 0 ) = 0 .6 A
i=
i 2 = 0 .6
For t > 0, the switch is closed so that the energies in L1 and L 2 flow through the
closed switch and become dissipated in the 5 Ω and 20 Ω resistors.
L
2.5 1
i1 ( t ) = i1 (0) e - t τ1 ,
τ1 = 1 =
=
R1
5
2
i1 ( t ) = 2.4 e -2t u ( t )A
i 2 ( t ) = i 2 (0) e- t τ 2 ,
τ2 =
L2
4 1
=
=
R 2 20 5
i 2 ( t ) = 0.6 e -5t u ( t )A
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Chapter 7, Problem 58.
Rework Prob. 7.17 if i(0) = 10 A and v(t) = 20u (t) V.
Chapter 7, Solution 58.
For t < 0,
v o (t) = 0
For t > 0,
i(0) = 10 ,
R th = 1 + 3 = 4 Ω ,
20
=5
1+ 3
L 14 1
τ=
=
=
R th
4 16
i(∞) =
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 5 ( 1 + e-16t ) A
di
1
v o ( t ) = 3 i + L = 15 ( 1 + e-16t ) + (-16)(5) e-16t
dt
4
-16t
v o ( t ) = 15 − 5 e V
Chapter 7, Problem 59.
Determine the step response v 0 (t) to v s in the circuit of Fig. 7.124.
Figure 7.124
For Prob. 7.59.
Chapter 7, Solution 59.
Let I be the current through the inductor.
i(0) = 0
For t < 0,
vs = 0 ,
For t > 0,
R eq = 4 + 6 || 3 = 6 ,
τ=
L 1 .5
=
= 0.25
R eq
6
2
(3) = 1
2+ 4
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 1 − e-4t
di
v o ( t ) = L = (1.5)(-4)(-e- 4t )
dt
i(∞) =
v o ( t ) = 6 e -4t u ( t ) V
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Chapter 7, Problem 60.
Find v(t) for t > 0 in the circuit of Fig. 7.125 if the initial current in the inductor is zero.
Figure 7.125
For Prob. 7.60.
Chapter 7, Solution 60.
Let I be the inductor current.
For t < 0,
u(t) = 0 ⎯
⎯→ i(0) = 0
For t > 0,
R eq = 5 || 20 = 4 Ω ,
τ=
L
8
= =2
R eq 4
i(∞) = 4
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 4 ( 1 − e - t 2 )
⎛ - 1⎞
di
= (8)(-4)⎜ ⎟ e- t 2
⎝2⎠
dt
-0.5t
V
v( t ) = 16 e
v( t ) = L
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Chapter 7, Problem 61.
In the circuit of Fig. 7.126, i s changes from 5 A to 10 A at t = 0 that is, i s = 5u (-t) +
10u(t) Find v and i.
Figure 7.126
For Prob. 7.61.
Chapter 7, Solution 61.
The current source is transformed as shown below.
4Ω
20u(-t) + 40u(t)
+
−
0.5 H
L 12 1
=
= ,
i(0) = 5 ,
R
4
8
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
τ=
i(∞) = 10
i( t ) = 10 − 5 e -8t u ( t )A
v( t ) = L
di ⎛ 1 ⎞
= ⎜ ⎟(-5)(-8) e -8t
dt ⎝ 2 ⎠
v( t ) = 20 e -8t u ( t ) V
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Chapter 7, Problem 62.
For the circuit in Fig. 7.127, calculate i(t) if i(0) = 0.
Figure 7.127
For Prob. 7.62.
Chapter 7, Solution 62.
L
2
=
=1
R eq 3 || 6
For 0 < t < 1, u ( t − 1) = 0 so that
1
i(0) = 0 ,
i(∞) =
6
1
i( t ) = ( 1 − e - t )
6
τ=
1
( 1 − e -1 ) = 0.1054
6
1 1 1
i(∞) = + =
3 6 2
i( t ) = 0.5 + (0.1054 − 0.5) e-(t -1)
i( t ) = 0.5 − 0.3946 e-(t -1)
For t > 1,
Thus,
i(1) =
⎧⎪
1
( 1 − e -t ) A
0<t<1
i( t ) = ⎨
6
⎪⎩ 0.5 − 0.3946 e -(t -1) A
t>1
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Chapter 7, Problem 63.
Obtain v(t) and i(t) in the circuit of Fig. 7.128.
Figure 7.128
For Prob. 7.63.
Chapter 7, Solution 63.
10
=2
5
For t < 0,
u (- t ) = 1 ,
i(0) =
For t > 0,
u (-t) = 0 ,
i(∞) = 0
L
0.5 1
τ=
=
=
R th
4 8
R th = 5 || 20 = 4 Ω ,
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 2 e -8t u ( t )A
v( t ) = L
di ⎛ 1 ⎞
= ⎜ ⎟(-8)(2) e-8t
dt ⎝ 2 ⎠
v( t ) = - 8 e -8t u ( t ) V
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Chapter 7, Problem 64.
Find v 0 (t) for t > 0 in the circuit of Fig. 7.129.
Figure 7.129
For Prob. 7.64.
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Chapter 7, Solution 64.
Let i be the inductor current.
For t < 0, the inductor acts like a short circuit and the 3 Ω resistor is shortcircuited so that the equivalent circuit is shown in Fig. (a).
6Ω
10 Ω
+
−
6Ω
i
3Ω
+
−
10 Ω
(a)
i = i(0) =
For t > 0,
io
v
i
3Ω
2Ω
(b)
10
= 1.667 A
6
R th = 2 + 3 || 6 = 4 Ω ,
τ=
L
4
= =1
R th 4
To find i(∞) , consider the circuit in Fig. (b).
10 − v v v
10
= +
⎯
⎯→ v =
6
3 2
6
v 5
i = i(∞) = =
2 6
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
5 ⎛ 10 5 ⎞
5
i( t ) = + ⎜ − ⎟ e - t = 1 + e - t A
6 ⎝ 6 6⎠
6
(
)
v o is the voltage across the 4 H inductor and the 2 Ω resistor
v o (t) = 2 i + L
⎛5⎞
di 10 10 - t
10 10
= + e + (4)⎜ ⎟(-1) e - t = − e - t
⎝6⎠
dt 6 6
6 6
(
)
v o ( t ) = 1.6667 1 − e - t V
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Chapter 7, Problem 65.
If the input pulse in Fig. 7.130(a) is applied to the circuit in Fig. 7.130(b), determine the
response i(t).
Figure 7.130
For Prob. 7.65.
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Chapter 7, Solution 65.
Since v s = 10 [ u ( t ) − u ( t − 1)] , this is the same as saying that a 10 V source is
turned on at t = 0 and a -10 V source is turned on later at t = 1. This is shown in
the figure below.
vs
10
1
t
-10
For 0 < t < 1, i(0) = 0 ,
R th = 5 || 20 = 4 ,
10
=2
5
L
2 1
τ=
= =
R th 4 2
i(∞) =
i( t ) = i(∞) + [ i(0) − i(∞)] e- t τ
i( t ) = 2 ( 1 − e -2t ) A
i(1) = 2 ( 1 − e-2 ) = 1.729
For t > 1,
i(∞) = 0
since vs = 0
i( t ) = i(1) e- ( t −1) τ
i( t ) = 1.729 e-2( t −1) A
Thus,
⎧ 2 ( 1 − e - 2t ) A 0 < t < 1
i( t ) = ⎨
t>1
⎩ 1.729 e - 2( t −1) A
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Chapter 7, Problem 66.
For the op amp circuit of Fig. 7.131, find v 0 . Assume that v s changes abruptly from 0 to
1 V at t = 0.
Figure 7.131
For Prob. 7.66.
Chapter 7, Solution 66.
For t<0-, vs =0 so that vo(0)=0
Let v be the capacitor voltage
For t>0, vs =1. At steady state, the capacitor acts like an open circuit so that we
have an inverting amplifier
vo(∞) = –(50k/20k)(1V) = –2.5 V
τ = RC = 50x103x0.5x10–6 = 25 ms
vo(t) = vo(∞) + (vo(0) – vo(∞))e–t/0.025 = 2.5(e–40t – 1) V.
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Chapter 7, Problem 67.
If v(0) = 5 V, find v 0 (t) for t > 0 in the op amp circuit of Fig. 7.132. Let R = 10k Ω and
C = 1 µ F.
Figure 7.132
For Prob. 7.67.
Chapter 7, Solution 67.
The op amp is a voltage follower so that v o = v as shown below.
R
v1
R
−
+
vo
+
R
vo
vo
C
−
At node 1,
v o − v1 v1 − 0 v1 − v o
2
=
+
⎯
⎯→ v1 = v o
R
R
R
3
At the noninverting terminal,
dv
v − v1
=0
C o + o
R
dt
dv
2
1
− RC o = v o − v1 = v o − v o = v o
dt
3
3
dv o
v
=− o
dt
3RC
v o ( t ) = VT e - t 3RC
VT = vo (0) = 5 V ,
τ = 3RC = (3)(10 × 103 )(1 × 10- 6 ) =
3
100
v o ( t ) = 5 e -100t 3 u ( t )V
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Chapter 7, Problem 68.
Obtain v 0 for t > 0 in the circuit of Fig. 7.133.
Figure 7.133
For Prob. 7.68.
Chapter 7, Solution 68.
This is a very interesting problem and has both an important ideal solution as well as an
important practical solution. Let us look at the ideal solution first. Just before the switch
closes, the value of the voltage across the capacitor is zero which means that the voltage
at both terminals input of the op amp are each zero. As soon as the switch closes, the
output tries to go to a voltage such that the input to the op amp both go to 4 volts. The
ideal op amp puts out whatever current is necessary to reach this condition. An infinite
(impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero
time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal
of the op amp). So vo will be equal to 8 volts for all t > 0.
What happens in a real circuit? Essentially, the output of the amplifier portion of the op
amp goes to whatever its maximum value can be. Then this maximum voltage appears
across the output resistance of the op amp and the capacitor that is in series with it. This
results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts.
vC(t) = Vop amp max(1 – e-t/(RoutC)) volts, for all values of vC less than 8 V,
= 8 V when t is large enough so that the 8 V is reached.
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Chapter 7, Problem 69.
For the op amp circuit in Fig. 7.134, find v 0 (t) for t > 0.
Figure 7.134
For Prob. 7.69.
Chapter 7, Solution 69.
Let v x be the capacitor voltage.
v x ( 0) = 0
For t < 0,
For t > 0, the 20 kΩ and 100 kΩ resistors are in series and together, they are in
parallel with the capacitor since no current enters the op amp terminals.
As t → ∞ , the capacitor acts like an open circuit so that
−4
v o (∞ ) =
(20 + 100) = −48
10
R th = 20 + 100 = 120 kΩ ,
τ = R th C = (120 × 103 )(25 × 10-3 ) = 3000
v o ( t ) = v o (∞) + [ v o (0) − v o (∞)] e - t τ
(
)
v o ( t ) = −48 1 − e - t 3000 V = 48(e–t/3000–1)u(t)V
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Chapter 7, Problem 70.
Determine v 0 for t > 0 when v s = 20 mV in the op amp circuit of Fig. 7.135.
Figure 7.135
For Prob. 7.70.
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Chapter 7, Solution 70.
Let v = capacitor voltage.
For t < 0, the switch is open and v(0) = 0 .
For t > 0, the switch is closed and the circuit becomes as shown below.
1
+
−
2
vS
+
+
−
vo
v
−
C
R
v1 = v 2 = v s
0 − vs
dv
=C
R
dt
⎯→ v o = v s − v
where v = v s − v o ⎯
(1)
(2)
(3)
From (1),
dv v s
=
=0
dt RC
- t vs
-1
v s dt + v(0) =
v=
∫
RC
RC
Since v is constant,
RC = (20 × 10 3 )(5 × 10 -6 ) = 0.1
- 20 t
v=
mV = -200 t mV
0.1
From (3),
v o = v s − v = 20 + 200 t
v o = 20 ( 1 + 10t ) mV
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Chapter 7, Problem 71.
For the op amp circuit in Fig. 7.136, suppose v 0 = 0 and v s = 3 V. Find v(t) for t > 0.
Figure 7.136
For Prob. 7.71.
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Chapter 7, Solution 71.
We temporarily remove the capacitor and find the Thevenin equivalent at its
terminals. To find RTh, we consider the circuit below.
Ro
20 kΩ
RTh
Since we are assuming an ideal op amp, Ro = 0 and RTh=20kΩ . The op amp circuit
is a noninverting amplifier. Hence,
10
)vs = 2vs = 6V
10
The Thevenin equivalent is shown below.
VTh = (1+
20 kΩ
+
6V
+
_
v
10 µF
–
Thus,
v(t) = 6(1− e− t /τ ) , t > 0
where τ = RTH C = 20 x10 −3 x10 x10 −6 = 0.2
v(t) = 6(1− e−5t ), t > 0 V
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Chapter 7, Problem 72.
Find i 0 in the op amp circuit in Fig. 7.137. Assume that v(0) = -2 V, R = 10 k Ω , and C =
10 µF .
Figure 7.137
For Prob. 7.72.
Chapter 7, Solution 72.
The op amp acts as an emitter follower so that the Thevenin equivalent circuit is
shown below.
C
+
3u(t)
Hence,
+
−
v
−
io
R
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v(0) = -2 V , v(∞) = 3 V , τ = RC = (10 × 10 3 )(10 × 10 -6 ) = 0.1
v( t ) = 3 + (-2 - 3) e -10t = 3 − 5 e -10t
dv
= (10 × 10 -6 )(-5)(-10) e -10t
dt
i o = 0.5 e -10t mA , t > 0
io = C
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Chapter 7, Problem 73.
For the op amp circuit in Fig. 7.138, let R 1 = 10 k Ω , R f = 20 k Ω , C = 20 µ F, and v(0)
= 1 V. Find v 0 .
Figure 7.138
For Prob. 7.73.
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Chapter 7, Solution 73.
Consider the circuit below.
Rf
v1
R1
v2
+
v1
+
−
C
v
v3
−
−
+
+
vo
−
At node 2,
v1 − v 2
dv
=C
R1
dt
At node 3,
dv v 3 − v o
C
=
dt
Rf
But v 3 = 0 and v = v 2 − v 3 = v 2 . Hence, (1) becomes
v1 − v
dv
=C
R1
dt
dv
v1 − v = R 1C
dt
v1
dv
v
or
+
=
dt R 1C R 1C
which is similar to Eq. (7.42). Hence,
⎧
vT
t<0
v( t ) = ⎨
-t τ
t>0
⎩ v1 + ( v T − v1 ) e
where v T = v(0) = 1 and v1 = 4
τ = R 1C = (10 × 10 3 )(20 × 10 -6 ) = 0.2
⎧ 1
t<0
v( t ) = ⎨
⎩ 4 − 3 e -5t t > 0
From (2),
dv
v o = -R f C
= (20 × 10 3 )(20 × 10 -6 )(15 e -5t )
dt
v o = -6 e -5t , t > 0
(1)
(2)
v o = - 6 e -5t u(t ) V
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Chapter 7, Problem 74.
Determine v 0 (t) for t > 0 in the circuit of Fig. 7.139. Let i s = 10u(t) µ A and assume that
the capacitor is initially uncharged.
Figure 7.139
For Prob. 7.74.
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Chapter 7, Solution 74.
Let v = capacitor voltage. For t < 0, v(0) = 0
Rf
C
is
R
−
+
is
+
vo
−
i s = 10 µA . Consider the circuit below.
For t > 0,
dv v
+
dt R
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
is = C
(1)
(2)
It is evident from the circuit that
τ = RC = (2 × 10 −6 )(50 × 10 3 ) = 0.1
At steady state, the capacitor acts like an open circuit so that i s passes through R.
Hence,
v(∞) = i s R = (10 × 10 −6 )(50 × 10 3 ) = 0.5 V
Then,
v( t ) = 0.5 ( 1 − e -10t ) V
(3)
But
is =
0 − vo
Rf
⎯
⎯→ v o = -i s R f
(4)
Combining (1), (3), and (4), we obtain
- Rf
dv
vo =
v − RfC
R
dt
-1
dv
v o = v − (10 × 10 3 )(2 × 10 -6 )
5
dt
-10t
-2
v o = -0.1 + 0.1e − (2 × 10 )(0.5)( - 10 e -10t )
v o = 0.2 e -10t − 0.1
v o = 0.1 ( 2 e -10t − 1) V
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Chapter 7, Problem 75.
In the circuit of Fig. 7.140, find v 0 and i 0 , given that v s = 4u(t) V and v(0) = 1 V.
Figure 7.140
For Prob. 7.75.
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Chapter 7, Solution 75.
Let v1 = voltage at the noninverting terminal.
Let v 2 = voltage at the inverting terminal.
For t > 0,
v1 = v 2 = v s = 4
0 − vs
= i o , R 1 = 20 kΩ
R1
vo = -ioR
v
dv
R 2 = 10 kΩ , C = 2 µF
Also, i o =
+C ,
R2
dt
- vs
v
dv
i.e.
=
+C
R1
R2
dt
This is a step response.
v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ ,
where τ = R 2 C = (10 × 10 3 )(2 × 10 -6 ) =
(1)
(2)
v(0) = 1
1
50
At steady state, the capacitor acts like an open circuit so that i o passes through
R 2 . Hence, as t → ∞
- vs
v(∞)
= io =
R1
R2
- R2
- 10
i.e.
v(∞) =
vs =
(4) = -2
R1
20
v( t ) = -2 + (1 + 2) e -50t
v( t ) = -2 + 3 e -50t
But
v = vs − vo
or
v o = v s − v = 4 + 2 − 3 e -50 t
v o = 6 − 3 e -50 t u ( t )V
- vs
-4
=
= -0.2 mA
R 1 20k
v
dv
io =
+C
= - 0.2 mA
R2
dt
io =
or
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Chapter 7, Problem 76.
Repeat Prob. 7.49 using PSpice.
Chapter 7, Solution 76.
The schematic is shown below. For the pulse, we use IPWL and enter the corresponding
values as attributes as shown. By selecting Analysis/Setup/Transient, we let Print Step =
25 ms and Final Step = 2 s since the width of the input pulse is 1 s. After saving and
simulating the circuit, we select Trace/Add and display –V(C1:2). The plot of V(t) is
shown below.
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Chapter 7, Problem 77.
The switch in Fig. 7.141 opens at t = 0. Use PSpice to determine v(t) for t > 0.
Figure 7.141
For Prob. 7.77.
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Chapter 7, Solution 77.
The schematic is shown below. We click Marker and insert Mark Voltage Differential at
the terminals of the capacitor to display V after simulation. The plot of V is shown
below. Note from the plot that V(0) = 12 V and V(∞) = -24 V which are correct.
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Chapter 7, Problem 78.
The switch in Fig. 7.142 moves from position a to b at t = 0. Use PSpice to find i(t) for
t > 0.
Figure 7.142
For Prob. 7.78.
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Chapter 7, Solution 78.
(a)
When the switch is in position (a), the schematic is shown below. We insert
IPROBE to display i. After simulation, we obtain,
i(0) = 7.714 A
from the display of IPROBE.
(b)
When the switch is in position (b), the schematic is as shown below. For inductor
I1, we let IC = 7.714. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms
and Final Step = 2 s. After Simulation, we click Trace/Add in the probe menu and
display I(L1) as shown below. Note that i(∞) = 12A, which is correct.
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Chapter 7, Problem 79.
In the circuit of Fig. 7.143, the switch has been in position a for a long time but moves
instantaneously to position b at t = 0 Determine i 0 (t).
Figure 7.143
For Prob. 7.79.
Chapter 7, Solution 79.
When the switch is in position 1, io(0) = 12/3 = 4A. When the switch is in position 2,
4
L
i o (∞ ) = −
= −0.5 A,
R Th = (3 + 5) // 4 = 8 / 3,
τ=
= 3 / 80
5+3
R Th
i o ( t ) = i o (∞) + [i o (0) − i o (∞)]e − t / τ = − 0.5 + 4.5e −80 t / 3 u(t)A
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Chapter 7, Problem 80.
In the circuit of Fig. 7.144, assume that the switch has been in position a for a long time,
find:
(a) i 1 (0), i 2 (0), and v 0 (0)
(b) i L (t)
(c) i 1 ( ∞ ), i 2 ( ∞ ), and v 0 ( ∞ ).
Figure 7.144
For Prob. 7.80.
Chapter 7, Solution 80.
(a) When the switch is in position A, the 5-ohm and 6-ohm resistors are shortcircuited so that
i1 (0) = i2 (0) = vo (0) = 0
but the current through the 4-H inductor is iL(0) =30/10 = 3A.
(b) When the switch is in position B,
R Th = 3 // 6 = 2Ω,
τ=
L
= 4 / 2 = 2 sec
R Th
i L ( t ) = i L (∞) + [i L (0) − i L (∞)]e − t / τ = 0 + 3e − t / 2 = 3e − t / 2 A
(c) i1 (∞) =
30
= 2 A,
10 + 5
vo (t ) = L
3
i 2 (∞ ) = − i L (∞ ) = 0 A
9
di L
dt
⎯
⎯→
v o (∞ ) = 0 V
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Chapter 7, Problem 81.
Repeat Prob. 7.65 using PSpice.
Chapter 7, Solution 81.
The schematic is shown below. We use VPWL for the pulse and specify the attributes as
shown. In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final
Step = 3 S. By inserting a current marker at one terminal of LI, we automatically obtain
the plot of i after simulation as shown below.
2.0A
1.5A
1.0A
0.5A
0A
0s
0.5s
-I(L1)
1.0s
1.5s
2.0s
2.5s
3.0s
Time
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Chapter 7, Problem 82.
In designing a signal-switching circuit, it was found that a 100- µ F capacitor was needed
for a time constant of 3 ms. What value resistor is necessary for the circuit?
Chapter 7, Solution 82.
3 × 10 -3
τ
= 30 Ω
τ = RC ⎯
⎯→ R = =
C 100 × 10 -6
Chapter 7, Problem 83.
An RC circuit consists of a series connection of a 120-V source, a switch, a 34-M Ω
resistor, and a 15- µ F capacitor. The circuit is used in estimating the speed of a horse
running a 4-km racetrack. The switch closes when the horse begins and opens when the
horse crosses the finish line. Assuming that the capacitor charges to 85.6 V, calculate the
speed of the horse.
Chapter 7, Solution 83.
v(∞) = 120,
τ = RC = 34 x10 6 x15 x10 −6 = 510s
v(0) = 0,
v(t ) = v(∞) + [v(0) − v(∞)]e − t / τ
Solving for t gives
⎯
⎯→
85.6 = 120(1 − e − t / 510 )
t = 510 ln 3.488 = 637.16 s
speed = 4000m/637.16s = 6.278m/s
Chapter 7, Problem 84.
The resistance of a 160-mH coil is 8 Ω . Find the time required for the current to build up
to 60 percent of its final value when voltage is applied to the coil.
Chapter 7, Solution 84.
Let Io be the final value of the current. Then
i (t ) = I o (1 − e − t / τ ),
0.6 I o = I o (1 − e −50t )
τ = R / L = 0.16 / 8 = 1 / 50
⎯
⎯→
t=
1
1
ln
= 18.33 ms.
50 0.4
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Chapter 7, Problem 85.
A simple relaxation oscillator circuit is shown in Fig. 7.145. The neon lamp fires when its
voltage reaches 75 V and turns off when its voltage drops to 30 V. Its resistance is 120 Ω
when on and infinitely high when off.
(a) For how long is the lamp on each time the capacitor discharges?
(b) What is the time interval between light flashes?
Figure 7.145
For Prob. 7.85.
Chapter 7, Solution 85.
(a)
The light is on from 75 volts until 30 volts. During that time we
essentially have a 120-ohm resistor in parallel with a 6-µF capacitor.
v(0) = 75, v(∞) = 0, τ = 120x6x10-6 = 0.72 ms
v(t1) = 75 e − t1 / τ = 30 which leads to t1 = –0.72ln(0.4) ms = 659.7 µs of
lamp on time.
(b)
τ = RC = (4 × 106 )(6 × 10-6 ) = 24 s
Since v( t ) = v(∞) + [ v(0) − v(∞)] e - t τ
v( t 1 ) − v(∞) = [ v(0) − v(∞)] e - t1 τ
v( t 2 ) − v(∞) = [ v(0) − v(∞)] e- t 2 τ
(1)
(2)
Dividing (1) by (2),
v( t1 ) − v(∞)
= e( t 2 − t1 ) τ
v( t 2 ) − v(∞)
⎛ v( t ) − v(∞) ⎞
⎟
t 0 = t 2 − t1 = τ ln ⎜ 1
⎝ v( t 2 ) − v(∞) ⎠
⎛ 75 − 120 ⎞
⎟ = 24 ln (2) = 16.636 s
t 0 = 24 ln ⎜
⎝ 30 − 120 ⎠
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Chapter 7, Problem 86.
Figure 7.146 shows a circuit for setting the length of time voltage is applied to the
electrodes of a welding machine. The time is taken as how long it takes the capacitor to
charge from 0 to 8 V. What is the time range covered by the variable resistor?
Figure 7.146
For Prob. 7.86.
Chapter 7, Solution 86.
v( t ) = v(∞) + [ v(0) − v(∞)] e- t τ
v(∞) = 12 ,
v(0) = 0
-t τ
v( t ) = 12 ( 1 − e )
v( t 0 ) = 8 = 12 ( 1 − e- t 0 τ )
8
1
= 1 − e- t 0 τ ⎯
⎯→ e- t 0 τ =
12
3
t 0 = τ ln (3)
For R = 100 kΩ ,
τ = RC = (100 × 103 )(2 × 10-6 ) = 0.2 s
t 0 = 0.2 ln (3) = 0.2197 s
For R = 1 MΩ ,
τ = RC = (1 × 106 )(2 × 10-6 ) = 2 s
t 0 = 2 ln (3) = 2.197 s
Thus,
0.2197 s < t 0 < 2.197 s
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 7, Problem 87.
A 120-V dc generator energizes a motor whose coil has an inductance of 50 H and a
resistance of 100 Ω . A field discharge resistor of 400 Ω is connected in parallel with the
motor to avoid damage to the motor, as shown in Fig. 7.147. The system is at steady
state. Find the current through the discharge resistor 100 ms after the breaker is tripped.
Figure 7.147
For Prob. 7.87.
Chapter 7, Solution 87.
Let i be the inductor current.
For t < 0,
i (0 − ) =
120
= 1.2 A
100
For t > 0, we have an RL circuit
L
50
τ= =
= 0.1 ,
i(∞) = 0
R 100 + 400
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 1.2 e -10t
At t = 100 ms = 0.1 s,
i(0.1) = 1.2 e -1 = 441mA
which is the same as the current through the resistor.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 7, Problem 88.
The circuit in Fig. 7.148(a) can be designed as an approximate differentiator or an
integrator, depending on whether the output is taken across the resistor or the capacitor,
and also on the time constant τ = RC of the circuit and the width T of the input pulse in
Fig. 7.148(b). The circuit is a differentiator if τ << T, say τ < 0.1T, or an integrator if τ
>> T, say τ > 10T.
(a) What is the minimum pulse width that will allow a differentiator output to
appear across the capacitor?
(b) If the output is to be an integrated form of the input, what is the maximum
value the pulse width can assume?
Figure 7.148
For Prob. 7.88.
Chapter 7, Solution 88.
(a)
τ = RC = (300 × 10 3 )(200 × 10 -12 ) = 60 µs
As a differentiator,
T > 10 τ = 600 µs = 0.6 ms
Tmin = 0.6 ms
i.e.
(b)
τ = RC = 60 µs
As an integrator,
T < 0.1τ = 6 µs
i.e.
Tmax = 6 µs
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 7, Problem 89.
An RL circuit may be used as a differentiator if the output is taken across the inductor and
τ << T (say τ < 0.1T), where T is the width of the input pulse. If R is fixed at 200 k Ω
determine the maximum value of L required to differentiate a pulse with T = 10 µ s.
Chapter 7, Solution 89.
Since τ < 0.1 T = 1 µs
L
< 1 µs
R
L < R × 10 -6 = (200 × 10 3 )(1 × 10 -6 )
L < 200 mH
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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you are using it without permission.
Chapter 7, Problem 90.
An attenuator probe employed with oscilloscopes was designed to reduce the magnitude
of the input voltage v i by a factor of 10. As shown in Fig. 7.149, the oscilloscope has
internal resistance R s and capacitance C s while the probe has an internal resistance R p
If R p is fixed at 6 M Ω find R s and C s for the circuit to have a time constant of 15 µ s.
Figure 7.149
For Prob. 7.90.
Chapter 7, Solution 90.
We determine the Thevenin equivalent circuit for the capacitor C s .
Rs
v th =
v,
R th = R s || R p
Rs + Rp i
Rth
Vth
+
−
Cs
The Thevenin equivalent is an RC circuit. Since
Rs
1
1
v th = v i ⎯
⎯→
=
10
10 R s + R p
Rs =
1
6 2
R p = = MΩ
9
9 3
Also,
τ = R th C s = 15 µs
6 (2 3)
where R th = R p || R s =
= 0.6 MΩ
6+2 3
15 × 10 -6
τ
Cs =
= 25 pF
=
R th 0.6 × 10 6
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 7, Problem 91.
The circuit in Fig. 7.150 is used by a biology student to study “frog kick.” She noticed
that the frog kicked a little when the switch was closed but kicked violently for 5 s when
the switch was opened. Model the frog as a resistor and calculate its resistance. Assume
that it takes 10 mA for the frog to kick violently.
Figure 7.150
For Prob. 7.91.
Chapter 7, Solution 91.
12
= 240 mA ,
i(∞) = 0
50
i( t ) = i(∞) + [ i(0) − i(∞)] e - t τ
i( t ) = 240 e - t τ
i o (0) =
L 2
=
R R
i( t 0 ) = 10 = 240 e - t 0
τ=
τ
e t 0 τ = 24 ⎯
⎯→ t 0 = τ ln (24)
t0
5
2
τ=
=
= 1.573 =
ln (24) ln (24)
R
2
= 1.271 Ω
R=
1.573
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 7, Problem 92.
To move a spot of a cathode-ray tube across the screen requires a linear increase in the
voltage across the deflection plates, as shown in Fig. 7.151. Given that the capacitance of
the plates is 4 nF, sketch the current flowing through the plates.
Figure 7.151
For Prob. 7.92.
Chapter 7, Solution 92.
⎧ 10
⎪ 10 -3
dv
= 4 × 10 -9 ⋅ ⎨ 2 ×- 10
i=C
dt
⎪
⎩ 5 × 10 -6
0 < t < tR
tR < t < tD
⎧ 20 µA
0 < t < 2 ms
i( t ) = ⎨
⎩- 8 mA 2 ms < t < 2 ms + 5 µs
which is sketched below.
5 µs
20 µA
t
2 ms
-8 mA
(not to scale)
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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