Circuit Variables
1
Assessment Problems
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
per second to miles per second:
✓ ◆
2 3 ⇥ 108 m 100 cm
1 in
1 ft
1 mile
124,274.24 miles
·
·
·
·
=
.
3
1s
1m
2.54 cm 12 in 5280 feet
1s
Now set up a proportion to determine how long it takes this signal to travel
1100 miles:
1100 miles
124,274.24 miles
=
.
1s
xs
Therefore,
x=
1100
= 0.00885 = 8.85 ⇥ 10 3 s = 8.85 ms.
124,274.24
AP 1.2 To solve this problem we use a product of ratios to change units from
dollars/year to dollars/millisecond. We begin by expressing $10 billion in
scientific notation:
$100 billion = $100 ⇥ 109 .
Now we determine the number of milliseconds in one year, again using a
product of ratios:
1 year
1 day
1 hour 1 min
1 sec
1 year
·
·
·
·
=
.
365.25 days 24 hours 60 mins 60 secs 1000 ms
31.5576 ⇥ 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a
product of ratios:
100
$100 ⇥ 109
1 year
·
=
= $3.17/ms.
9
1 year
31.5576 ⇥ 10 ms
31.5576
1–1
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1–2
CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. 1.2, current is the time rate of change of charge, or i = dq
dt
In this problem, we are given the current and asked to find the total charge.
To do this, we must integrate Eq. 1.2 to find an expression for charge in terms
of current:
q(t) =
Z t
i(x) dx.
0
We are given the expression for current, i, which can be substituted into the
above expression. To find the total charge, we let t ! 1 in the integral. Thus
we have
qtotal =
Z 1
0
20
(0
5000
=
1
20
e 5000x =
5000
0
20e 5000x dx =
1) =
20
(e 1
5000
e0 )
20
= 0.004 C = 4000 µC.
5000
AP 1.4 Recall from Eq. 1.2 that current is the time rate of change of charge, or
i = dq
. In this problem we are given an expression for the charge, and asked to
dt
find the maximum current. First we will find an expression for the current
using Eq. 1.2:
dq
d 1
i=
=
dt
dt ↵2

✓
1
t
+ 2 e ↵t
↵ ↵
✓
✓
◆
=
=0
=
◆
d t ↵t
e
dt ↵
1 ↵t
e
↵
t
↵ e ↵t
↵
d 1
dt ↵2
✓
✓
◆
◆
◆
✓
d 1 ↵t
e
dt ↵2
✓
◆
1
↵ 2 e ↵t
↵
◆
1
1
+t+
e ↵t
↵
↵
= te ↵t .
Now that we have an expression for the current, we can find the maximum
value of the current by setting the first derivative of the current to zero and
solving for t:
di
d
= (te ↵t ) = e ↵t + t( ↵)e↵t = (1
dt
dt
↵t)e ↵t = 0.
Since e ↵t never equals 0 for a finite value of t, the expression equals 0 only
when (1 ↵t) = 0. Thus, t = 1/↵ will cause the current to be maximum. For
this value of t, the current is
i=
1 ↵/↵
1
e
= e 1.
↵
↵
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Problems
1–3
Remember in the problem statement, ↵ = 0.03679. Using this value for ↵,
i=
1
e 1⇠
= 10 A.
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four figures from Fig. 1.6:
[a] Now we have to match the voltage and current shown in the first figure
with the polarities shown in Fig. 1.6. Remember that 4A of current
entering Terminal 2 is the same as 4A of current leaving Terminal 1. We
get
(a) v =
20 V,
(c) v = 20 V,
i=
i=
4 A;
4 A;
(b) v =
20 V,
(d) v = 20 V,
i = 4 A;
i = 4 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = ( 20)( 4) = 80 W.
[c] Since the power is greater than 0, the box is absorbing power.
AP 1.6 [a] Applying the passive sign convention to the power equation using the
voltage and current polarities shown in Fig. 1.5, p = vi. To find the time
at which the power is maximum, find the first derivative of the power
with respect to time, set the resulting expression equal to zero, and solve
for time:
p = (80,000te 500t )(15te 500t ) = 120 ⇥ 104 t2 e 1000t ;
dp
= 240 ⇥ 104 te 1000t
dt
120 ⇥ 107 t2 e 1000t = 0.
Therefore,
240 ⇥ 104
120 ⇥ 107 t = 0.
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1–4
CHAPTER 1. Circuit Variables
Solving,
t=
240 ⇥ 104
= 2 ⇥ 10 3 = 2 ms.
120 ⇥ 107
[b] The maximum power occurs at 2 ms, so find the value of the power at 2
ms:
p(0.002) = 120 ⇥ 104 (0.002)2 e 2 = 649.6 mW.
[c] From Eq. 1.3, we know that power is the time rate of change of energy, or
p = dw/dt. If we know the power, we can find the energy by integrating
Eq. 1.3. To find the total energy, the upper limit of the integral is
infinity:
wtotal =
Z 1
0
120 ⇥ 104 x2 e 1000x dx
120 ⇥ 104 1000x
=
e
[( 1000)2 x2
( 1000)3
=0
120 ⇥ 104 0
e (0
( 1000)3
2( 1000)x + 2)
1
0
0 + 2) = 2.4 mJ.
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and
thus entering the lower terminal where the polarity marking of the voltage is
negative. Thus, using the passive sign convention, p = vi. Substituting the
values of voltage and current given in the figure,
p=
(800 ⇥ 103 )(1.8 ⇥ 103 ) =
1440 ⇥ 106 =
1440 MW.
Thus, because the power associated with the Oregon end of the line is
negative, power is being generated at the Oregon end of the line and
transmitted by the line to be delivered to the California end of the line.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
1–5
Chapter Problems
5280 ft 2526 lb 1 kg
·
·
= 20.5 ⇥ 106 kg.
1 mi
1000 ft 2.2 lb
P 1.1
(4 cond.) · (845 mi) ·
P 1.2
[a] To begin, we calculate the number of pixels that make up the display:
npixels = (3840)(2160) = 8,294,400 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one byte,
each pixel requires 3 bytes of information. We can calculate the number
of bytes of information required for the display by multiplying the
number of pixels in the display by 3 bytes per pixel:
nbytes =
8,294,400 pixels 3 bytes
·
= 24,883,200 bytes/display.
1 display
1 pixel
Finally, we use the fact that there are 106 bytes per MB:
1 MB
24,883,200 bytes
· 6
= 24.88 MB/display.
1 display
10 bytes
[b]
2 hr
24,883,200 bytes 30 images 60 s 60 min
·
·
·
·
1 image
1s
1 min
1 hr
1 video
= 5.375 ⇥ 1012 bytes/video = 5.375 TB/video.
[c]
24,883,200 bytes 8 bits 30 images
·
·
= 5,971,968,000 bits/s
1 image
1 byte
1 sec
= 5.972 Gb/s.
P 1.3
[a] We can set up a ratio to determine how long it takes the bamboo to grow
10 µm First, recall that 1 mm = 103 µm. Let’s also express the rate of
growth of bamboo using the units mm/s instead of mm/day. Use a
product of ratios to perform this conversion:
250 mm
1 day
1 hour 1 min
250
10
·
·
·
=
=
mm/s.
1 day 24 hours 60 min 60 sec
(24)(60)(60)
3456
Use a ratio to determine the time it takes for the bamboo to grow 10 µm:
10 ⇥ 10 6 m
10/3456 ⇥ 10 3 m
=
1s
xs
[b]
P 1.4
so
10 ⇥ 10 6
x=
= 3.456 s.
10/3456 ⇥ 10 3
1 cell length 3600 s (24)(7) hr
·
·
= 175,000 cell lengths/week.
3.456 s
1 hr
1 week
(480)(320) pixels 2 bytes 30 frames
·
·
= 9.216 ⇥ 106 bytes/sec;
1 frame
1 pixel
1 sec
(9.216 ⇥ 106 bytes/sec)(x secs) = 32 ⇥ 230 bytes;
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–6
P 1.5
CHAPTER 1. Circuit Variables
x=
32 ⇥ 230
= 3728 sec = 62 min ⇡ 1 hour of video.
9.216 ⇥ 106
[a]
x photos
20,000 photos
=
;
3
(11)(15)(1) mm
1 mm3
x=
[b]
(20,000)(1)
= 121 photos.
(11)(15)(1)
16 ⇥ 230 bytes
x bytes
=
;
3
(11)(15)(1) mm
(0.2)3 mm3
x=
(16 ⇥ 230 )(0.008)
= 832,963 bytes.
(11)(15)(1)
P 1.6
(260 ⇥ 106 )(540)
= 104.4 gigawatt-hours.
109
P 1.7
First we use Eq. 1.2 to relate current and charge:
i=
dq
= 24 cos 4000t.
dt
Therefore, dq = 24 cos 4000t dt.
To find the charge, we can integrate both sides of the last equation. Note that
we substitute x for q on the left side of the integral, and y for t on the right
side of the integral:
Z q(t)
q(0)
dx = 24
Z t
0
cos 4000y dy.
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
q(t)
q(0) = 24
sin 4000y
4000
t
=
0
24
sin 4000t
4000
24
24
sin 4000(0) =
sin 4000t.
4000
4000
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum
value at t = 0, so q(t) = 6 ⇥ 10 3 sin 4000t C = 6 sin 4000t mC.
P 1.8
w = qV = (1.6022 ⇥ 10 19 )(6) = 9.61 ⇥ 10 19 = 0.961 aJ.
P 1.9
n=
35 ⇥ 10 6 C/s
= 2.18 ⇥ 1014 elec/s.
1.6022 ⇥ 10 19 C/elec
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Problems
P 1.10
1–7
[a] First we use Eq. 1.2 to relate current and charge:
i=
dq
= 0.125e 2500t .
dt
Therefore, dq = 0.125e 2500t dt.
To find the charge, we can integrate both sides of the last equation. Note
that we substitute x for q on the left side of the integral, and y for t on
the right side of the integral:
Z q(t)
q(0)
dx = 0.125
Z t
0
e 2500y dy.
We solve the integral and make the substitutions for the limits of the
integral:
t
e 2500y
= 50 ⇥ 10 6 (1
q(0) = 0.125
2500 0
q(t)
e 2500t ).
But q(0) = 0 by hypothesis, so
q(t) = 50(1
e 2500t ) µC.
[b] As t ! 1, qT = 50 µC.
[c] q(0.5 ⇥ 10 3 ) = (50 ⇥ 10 6 )(1
P 1.11
e( 2500)(0.0005) ) = 35.675 µC.
[a] First we use Eq. (1.2) to relate current and charge:
i=
dq
= 40te 500t .
dt
Therefore, dq = 40te 500t dt.
To find the charge, we can integrate both sides of the last equation. Note
that we substitute x for q on the left side of the integral, and y for t on
the right side of the integral:
Z q(t)
q(0)
dx = 40
Z t
0
ye 500y dy.
We solve the integral and make the substitutions for the limits of the
integral:
q(t)
q(0) = 40
e 500y
( 500y
( 500)2
= 160 ⇥ 10 6 (1
t
1)
500te 500t
0
= 160 ⇥ 10 6 e 500t ( 500t
1) + 160 ⇥ 10 6
e 500t ).
But q(0) = 0 by hypothesis, so
q(t) = 160(1
500te 500t
[b] q(0.001) = (160)[1
e 500t ) µC.
500(0.001)e 500(0.001)
e 500(0.001) = 14.4 µC.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–8
P 1.12
CHAPTER 1. Circuit Variables
[a] In Car B, the current i is in the direction of the voltage drop across the
12 V battery(the current i flows into the + terminal of the battery of
Car B). Therefore using the passive sign convention,
p = vi = (40)(12) = 480 W.
Since the power is positive, the battery in Car B is absorbing power, so
Car B must have the “dead” battery.
Z t
60 s
[b] w(t) =
= 90 s;
p dx;
1.5 min = 1.5 ·
1 min
0
w(90) =
Z 90
0
480 dx;
w = 480(90
P 1.13
0) = 480(90) = 43,200 J = 43.2 kJ.
Assume we are standing at box A looking toward box B. Use the passive sign
convention to get p = vi, since the current i is flowing into the + terminal of
the voltage v. Now we just substitute the values for v and i into the equation
for power. Remember that if the power is positive, B is absorbing power, so
the power must be flowing from A to B. If the power is negative, B is
generating power so the power must be flowing from B to A.
[a] p = (30)(6) = 180 W
P 1.14
[b] p = ( 20)( 8) = 160 W
160 W from A to B;
[c] p = ( 60)(4) =
240 W
240 W from B to A;
[d] p = (40)( 9) =
360 W
360 W from B to A.
p = (12)(0.1) = 1.2 W;
w(t) =
P 1.15
180 W from A to B;
Z t
0
p dt;
4 hr ·
w(14,400) =
3600 s
= 14,400 s;
1 hr
Z 14,400
0
1.2 dt = 1.2(14,400) = 17.28 kJ.
[a]
p = vi = ( 20)(5) = 100 W.
Power is being delivered by the box.
[b] Entering.
[c] Gain.
P 1.16
[a] p = vi = ( 20)( 5) = 100 W, so power is being absorbed by the box.
[b] Leaving.
[c] Lose.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 1.17
p = vi;
w=
Z t
0
1–9
p dx.
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
Note that in constructing the plot above, we used the fact that 60 hr
= 216,000 s = 216 ks.
p(0) = (6)(15 ⇥ 10 3 ) = 90 ⇥ 10 3 W;
p(216 ks) = (4)(15 ⇥ 10 3 ) = 60 ⇥ 10 3 W;
1
w = (60 ⇥ 10 3 )(216 ⇥ 103 ) + (90 ⇥ 10 3
2
P 1.18
[a] p = vi = (0.05e 1000t )(75
dp
=
dt
60 ⇥ 10 3 )(216 ⇥ 103 ) = 16,200 J.
75e 1000t ) = (3.75e 1000t
3750e 1000t + 7500e 2000t = 0
2 = e1000t
so
ln 2 = 1000t
so
thus
3.75e 2000t ) W;
2e 2000t = e 1000t ;
p is maximum at t = 693.15 µs;
pmax = p(693.15 µs) = 937.5 mW.
[b] w =
Z 1
0
=
P 1.19
[3.75e
3.75
1000
1000t
3.75e
2000t
] dt =

3.75
e 1000t
1000
1
3.75
2000t
e
2000
0
3.75
= 1.875 mJ.
2000
[a] p = vi = (15e 250t )(0.04e 250t ) = 0.6e 500t W;
p(0.01) = 0.6e 500(0.01) = 0.6e 5 = 0.00404 = 4.04 mW.
[b] wtotal =
=
Z 1
0
p(x) dx =
0.0012(e 1
Z 1
0
0.6e 500x dx =
1
0.6
e 500x
500
0
e0 ) = 0.0012 = 1.2 mJ.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–10
P 1.20
CHAPTER 1. Circuit Variables
[a]
p = vi
dp
dt
=
[(1500t + 1)e 750t ](0.04e 750t )
=
(60t + 0.04)e 1500t ;
=
60e 1500t
1500e 1500t (60t + 0.04)
90,000te 1500t .
dp
Therefore,
= 0 when t = 0
dt
so pmax occurs at t = 0.
=
[b] pmax
[c] w
=
[(60)(0) + 0.04]e0 = 0.04
=
40 mW.
Z t
pdx
=
Z0t
=
0
60xe 1500x dx +
Z t
0
0.04e 1500x dx
t
t
e 1500x
60e 1500x
(
1500x
1)
+
0.04
.
( 1500)2
1500 0
0
When t = 1 all the upper limits evaluate to zero, hence
0.04
60
= 53.33 µJ.
+
w=
4
225 ⇥ 10
1500
=
P 1.21
[a] p = vi = 0.25e 3200t 0.5e 2000t + 0.25e 800t ;
p(625 µs) = 42.2 mW.
[b]
w(t)
=
=
w(625 µs)
=
Z t
0
(0.25e 3200t
140.625
0.5e 2000t + 0.25e 800t )
78.125e 3200t + 250e 2000t
312.5e 800t µJ;
12.14 µJ.
[c] wtotal = 140.625 µJ.
P 1.22
[a]
p = vi = [104 t + 5)e 400t ][(40t + 0.05)e 400t ]
=
dp
dt
400 ⇥ 103 t2 e 800t + 700te 800t + 0.25e 800t
=
e 800t [400,000t2 + 700t + 0.25];
=
{e 800t [800 ⇥ 103 t + 700]
800e 800t [400,000t2 + 700t + 0.25]}
[ 3,200,000t2 + 2400t + 5]100e 800t .
dp
Therefore,
= 0 when 3,200,000t2 2400t 5 = 0
dt
so pmax occurs at t = 1.68 ms.
=
[b] pmax
=
[400,000(.00168)2 + 700(.00168) + 0.25]e 800(.00168)
=
666.34 mW.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[c] w
Z t
=
Z0t
=
0
pdx
2
400,000x e
800x
dx +
Z t
0
700xe
800x
dx +
t
Z t
0
1–11
0.25e 800x dx
400,000e 800x
[64 ⇥ 104 x2 + 1600x + 2] +
=
512 ⇥ 106
0
t
t
700e 800x
e 800x
.
( 800x 1) + 0.25
64 ⇥ 104
800 0
0
When t ! 1 all the upper limits evaluate to zero, hence
(400,000)(2)
700
0.25
w=
= 2.97 mJ.
+
+
512 ⇥ 106
64 ⇥ 104
800
P 1.23
[a] We can find the time at which the power is a maximum by writing an
expression for p(t) = v(t)i(t), taking the first derivative of p(t)
and setting it to zero, then solving for t. The calculations are shown below:
p =
p
dp
dt
dp
dt
t1
0 t < 0,
= vi = t(1
p = 0 t > 40 s;
0.025t)(4
0.2t) = 4t
0.6t + 0.015t2 = 0.015(t2
=
4
=
0
=
8.453 s;
when t2
0.3t2 + 0.005t3 W,
0  t  40 s;
40t + 266.67);
40t + 266.67 = 0;
t2 = 31.547 s;
(using the polynomial solver on your calculator)
p(t1 )
= 4(8.453)
0.3(8.453)2 + 0.005(8.453)3 = 15.396 W;
p(t2 ) = 4(31.547) 0.3(31.547)2 + 0.005(31.547)3 = 15.396 W.
Therefore, maximum power is being delivered at t = 8.453 s.
[b] The maximum power was calculated in part (a) to determine the time at
which the power is maximum: pmax = 15.396 W (delivered).
[c] As we saw in part (a), the other “maximum” power is actually a
minimum, or the maximum negative power. As we calculated in part (a),
maximum power is being extracted at t = 31.547 s.
[d] This maximum extracted power was calculated in part (a) to determine
the time at which power is maximum: pmax = 15.396 W (extracted).
[e] w =
Z t
0
pdx =
Z t
0
(4x
0.3x2 + 0.005x3 )dx = 2t2
0.1t3 + 0.00125t4 .
w(0)
=
0 J;
w(30)
= 112.5 J;
w(10)
= 112.5 J;
w(40)
= 0 J;
w(20) = 200 J.
To give you a feel for the quantities of voltage, current, power, and energy
and their relationships among one another, they are plotted below:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–12
P 1.24
CHAPTER 1. Circuit Variables
[a] p = vi = 2000 cos(800⇡t) sin(800⇡t) = 1000 sin(1600⇡t) W.
Therefore, pmax = 1000 W.
[b] pmax (extracting) = 1000 W.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
=
[c] pavg
=
[d]
pavg
P 1.25
Z 2.5⇥10 3
1
1000 sin(1600⇡t) dt
2.5 ⇥ 10 3 0
3
cos 1600⇡t 2.5⇥10
250
5
[1
=
4 ⇥ 10
1600⇡
⇡
0
cos 4⇡] = 0 .
Z 15.625⇥10 3
1
=
1000 sin(1600⇡t) dt
15.625 ⇥10 3 0
3
40
cos 1600⇡t 15.625⇥10
3
= [1 cos 25⇡] = 25.46 W.
= 64 ⇥ 10
1600⇡
⇡
0
[a] v(20 ms) = 100e 1 sin 3 = 5.19 V;
i(20 ms) = 20e 1 sin 3 = 1.04 A;
p(20 ms) = vi = 5.39 W.
[b]
vi = 2000e 100t sin.2 150t

1 1
cos 300t
= 2000e 100t
2 2
= 1000e 100t 1000e 100t cos 300t;
p =
w
=
=
Z 1
0
1000e
100t
1
e 100t
1000
100 0
(
dt
Z 1
0
1000e 100t cos 300t dt
e 100t
1000
[ 100 cos 300t + 300 sin 300t]
2 + (300)2
(100)

100
= 10 1000
= 10 1
4
1 ⇥ 10 + 9 ⇥ 104
= 9 J.
P 1.26
1–13
)1
0
[a]
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1–14
CHAPTER 1. Circuit Variables
= 10 + 0.5 ⇥ 10 3 t mA,
0  t  10 ks;
i(t)
= 15 mA,
10 ks  t  20 ks;
i(t)
= 25
i(t)
= 0,
[b] i(t)
0.5 ⇥ 10 3 t mA,
20 ks  t  30 ks;
t > 30 ks.
p = vi = 120i so
p(t)
= 1200 + 0.06t mW,
0  t  10 ks;
p(t)
= 1800 mW,
10 ks  t  20 ks;
p(t)
= 3000
20 ks  t  30 ks;
p(t)
= 0,
0.06t mW,
t > 30 ks.
[c] To find the energy, calculate the area under the plot of the power:
1
w(10 ks) = (0.6)(10,000) + (1.2)(10,000) = 15 kJ;
2
w(20 ks) = w(10 ks) + (1.8)(10,000) = 33 kJ;
1
w(10 ks) = w(20 ks) + (0.6)(10,000) + (1.2)(10,000) = 48 kJ.
2
P 1.27
[a] q
[b] w
=
area under i vs. t plot
=
1
(8)(12,000) + (16)(12,000) + 21 (16)(4000)
2
=
48,000 + 192,000 + 32,000 = 272,000 C.
=
Z
p dt =
Z
vi dt;
v = 250 ⇥ 10 6 t + 8,
0  t  12,000s:
0  t  16 ks.
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Problems
i
=
p =
w1
=
666.67 ⇥ 10 6 t;
24
192 + 666.67 ⇥ 10 6 t
Z 12,000
0
1–15
166.67 ⇥ 10 9 t2 ;
(192 + 666.67 ⇥ 10 6 t
166.67 ⇥ 10 9 t2 ) dt
= (2304 + 48 96)103 = 2256 kJ.
12,000 s  t  16,000 s:
i =
p =
w2
wT
P 1.28
64
4 ⇥ 10 3 t;
512
16 ⇥ 10 3 t
10 6 t2 ;
=
Z 16,000
(512
=
(2048
896
=
w1 + w2 = 2256 + 362.667 = 2618.667 kJ.
12,000
16 ⇥ 10 3 t
10 6 t2 ) dt
789.33)103 = 362.667 kJ;
[a] 0 s  t < 4 s:
v = 2.5t V;
i = 1 µA;
p = 2.5t µW;
i = 0 A;
p = 0 W;
i=
p = 2.5t
4 s < t  8 s:
v = 10 V;
8 s  t < 16 s:
v=
2.5t + 30 V;
1 µA;
30 µW;
16 s < t  20 s:
v=
10 V;
i = 0 A;
p = 0 W;
i = 0.4 µA;
p = 0.4t
i = 0 A;
p = 0 W;
20 s  t < 36 s:
v=t
30 V;
12 µW;
36 s < t  46 s:
v = 6 V;
46 s  t < 50 s:
v=
1.5t + 75 V;
i=
0.6 µA; p = 0.9t
45 µW;
t > 50 s:
v = 0 V;
i = 0 A;
p = 0 W.
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1–16
CHAPTER 1. Circuit Variables
[b] Calculate the area under the curve from zero up to the desired time:
P 1.29
1
(4)(10) = 20 µJ;
2
w(4)
=
w(12)
= w(4)
w(36)
= w(12) + 12 (4)(10)
w(50)
= w(36)
1
(4)(10) = 0 J;
2
1
(10)(4) + 12 (6)(2.4) = 7.2 µJ;l
2
1
(4)(3.6) = 0 J.
2
We use the passive sign convention to determine whether the power equation
is p = vi or p = vi and substitute into the power equation the values for v
and i, as shown below:
pa
=
pb
=
vb ib = ( 18)(0.045) =
810 mW;
pc
=
vc ic = (2)( 0.006) =
12 mW;
pd
=
vd id =
(20)( 0.020) = 400 mW;
pe
=
ve ie =
(16)( 0.014) = 224 mW;
va ia =
( 18)( 0.051) =
918 mW;
pf = vf if = (36)(0.031) = 1116 mW.
Remember that if the power is positive, the circuit element is absorbing
power, whereas is the power is negative, the circuit element is developing
power. We can add the positive powers together and the negative powers
together
— if the power balances, these power sums should be equal:
X
Pdev = 918 + 810 + 12 = 1740 mW;
X
Pabs = 400 + 224 + 1116 = 1740 mW.
Thus, the power balances and the total power developed in the circuit is 1740
mW.
P 1.30
[a] Remember that if the circuit element is absorbing power, the power is
positive, whereas if the circuit element is supplying power, the power is
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Problems
1–17
negative. We can add the positive powers together and the negative
powers X
together — if the power balances, these power sums should be
equal:
Psup = 600 + 50 + 600 + 1250 = 2500 W;
X
Pabs = 400 + 100 + 2000 = 2500 W.
Thus, the power balances.
[b] The current can be calculated using i = p/v or i =
application of the passive sign convention:
P 1.31
ia
=
ib
=
pb /vb = ( 50)/( 100) = 0.5 A;
ic
=
pc /vc = (400)/(200) = 2.0 A;
id
=
pd /vd = ( 600)/(300) =
2.0 A;
ie
=
pe /ve = (100)/( 200) =
0.5 A;
if
=
ig
=
pa
=
va ia =
( 3000)( 0.250) =
pb
=
vb ib =
(4000)( 0.400) = 1600 W;
pc
=
vc ic =
(1000)(0.400) =
pd
=
vd id = (1000)(0.150) = 150 W;
pe
=
ve ie = ( 4000)(0.200) =
pa /va =
pf /vf =
p/v, with proper
( 600)/(400) = 1.5 A;
(2000)/(500) =
4.0 A;
pg /vg = ( 1250)/( 500) = 2.5 A.
750 W;
400 W;
800 W;
pf = vf if = (4000)(0.050) = 200 W.
Therefore,
X
X
Pabs = 1600 + 150 + 200 = 1950 W;
Pdel = 750 + 400 + 800 = 1950 W =
X
Pabs .
Thus, the interconnection does satisfy the power check.
P 1.32
[a] If the power balances, the sum of the power values should be zero:
ptotal = 0.175 + 0.375 + 0.150
0.320 + 0.160 + 0.120
0.660 = 0.
Thus, the power balances.
[b] When the power is positive, the element is absorbing power. Since
elements a, b, c, e, and f have positive power, these elements are
absorbing power.
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1–18
CHAPTER 1. Circuit Variables
[c] The voltage can be calculated using v = p/i or v =
application of the passive sign convention:
P 1.33
va
=
pa /ia = (0.175)/(0.025) = 7 V;
vb
=
pb /ib = (0.375)/(0.075) = 5 V;
vc
=
vd
=
ve
=
vf
=
pf /if = (0.120)/( 0.03) =
4 V;
vg
=
pg /ig = ( 0.66)/(0.055) =
12 V.
pc /ic =
(0.150)/( 0.05) = 3 V;
pd /id = ( 0.320)/(0.04) =
pe /ie =
p/i, with proper
8 V;
(0.160)/(0.02) =
8 V;
[a] From the diagram and the table we have
pa
=
va ia =
(900)( 22.5) = 20,250 W;
pb
=
vb ib =
(105)( 52.5) = 5512.5 W;
pc
=
vc ic =
( 600)( 30) =
pd
=
pe
=
pf
=
pg
=
vg ig =
(585)(82.5) =
ph
=
vh ih =
( 165)(82.5) = 13,612.5 W.
X
Pdel
=
18,000 + 30,712.5 + 48,262.5 = 96,975 W;
Pabs
=
20,250 + 5512.5 + 3600 + 18,000 + 13,612.5 = 60,975 W.
X
vd id = (585)( 52.5) =
ve ie =
18,000 W;
30,712.5 W;
( 120)(30) = 3600 W;
vf if = (300)(60) = 18,000 W;
Therefore,
X
Pdel 6=
X
48,262.5 W;
Pabs and the subordinate engineer is correct.
[b] The di↵erence between the power delivered to the circuit and the power
absorbed by the circuit is
96,975
60,975 = 36,000.
One-half of this di↵erence is 18,000 W, so it is likely that pc or pf is in
error. Either the voltage or the current probably has the wrong sign. (In
Chapter 2, we will discover that using KCL at the top node, the current
ic should be 30 A, not 30 A!) If the sign of pc is changed from negative
to positive, we can recalculate the power delivered and the power
absorbed as follows:
X
Pdel = 30,712.5 + 48,262.5 = 78,975 W;
X
Pabs = 20,250 + 5512.5 + 18,000 + 3600 + 18,000 + 13,612.5 = 78,975 W.
Now the power delivered equals the power absorbed and the power
balances for the circuit.
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Problems
P 1.34
pa
=
pb
=
pc
=
pd
=
pe
=
pf
=
pg
=
vg ig = (120)(4) = 480 W;
ph
=
vh ih = ( 220)( 5) = 1100 W.
X
va ia = (120)( 10) =
vb ib =
(120)(9) =
1–19
1200 W;
1080 W;
vc ic = (10)(10) = 100 W;
vd id =
(10)( 1) = 10 W;
ve ie = ( 10)( 9) = 90 W;
vf if =
( 100)(5) = 500 W;
Pdel = 1200 + 1080 = 2280 W;
Pabs = 100 + 10 + 90 + 500 + 480 + 1100 = 2280 W.
X
X
Pdel = Pabs = 2280 W.
Therefore,
X
Thus, the interconnection now satisfies the power check.
P 1.35
[a] The revised circuit model is shown below:
[b] The expression for the total power in this circuit is
va ia
v b ib
v f if + v g ig + v h ih
= (120)( 10)
(120)(10)
( 120)(3) + 120ig + ( 240)( 7) = 0.
Therefore,
120ig = 1200 + 1200
360
1680 = 360
so
360
= 3 A.
120
Thus, if the power in the modified circuit is balanced the current in
component g is 3 A.
ig =
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Circuit Elements
Assessment Problems
AP 2.1
[a] Note that the current ib is in the same circuit branch as the 8 A current
source; however, ib is defined in the opposite direction of the current
source. Therefore,
ib =
8 A.
Next, note that the dependent voltage source and the independent
voltage source are in parallel with the same polarity. Therefore, their
voltages are equal, and
8
ib
vg =
=
= 2 V.
4
4
[b] To find the power associated with the 8 A source, we need to find the
voltage drop across the source, vi . Note that the two independent sources
are in parallel, and that the voltages vg and v1 have the same polarities,
so these voltages are equal:
vi = v g =
2 V.
Using the passive sign convention,
ps = (8 A)(vi ) = (8 A)( 2 V) =
16 W.
Thus the current source generated 16 W of power.
2–1
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2–2
CHAPTER 2. Circuit Elements
AP 2.2
[a] Note from the circuit that vx = 25 V. To find ↵ note that the two
current sources are in the same branch of the circuit but their currents
flow in opposite directions. Therefore
↵vx =
15 A.
Solve the above equation for ↵ and substitute for vx ,
15 A
15 A
↵=
= 0.6 A/V.
=
vx
25 V
[b] To find the power associated with the voltage source we need to know the
current, iv . Note that this current is in the same branch of the circuit as
the dependent current source and these two currents flow in the same
direction. Therefore, the current iv is the same as the current of the
dependent source:
iv = ↵vx = (0.6)( 25) =
15 A.
Using the passive sign convention,
ps =
(iv )(25 V) =
( 15 A)(25 V) = 375 W.
Thus the voltage source dissipates 375 W.
AP 2.3
[a] The resistor and the voltage source are in parallel and the resistor voltage
and the voltage source have the same polarities. Therefore these two
voltages are the same:
vR = vg = 1 kV.
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Problems
2–3
Note from the circuit that the current through the resistor is ig = 5 mA.
Use Ohm’s law to calculate the value of the resistor:
vR
1 kV
R=
= 200 k⌦.
=
ig
5 mA
Using the passive sign convention to calculate the power in the resistor,
pR = (vR )(ig ) = (1 kV)(5 mA) = 5 W.
The resistor is dissipating 5 W of power.
[b] Note from part (a) the vR = vg and iR = ig . The power delivered by the
source is thus
psource
3W
psource = vg ig
= 40 V.
so
vg =
=
ig
75 mA
Since we now have the value of both the voltage and the current for the
resistor, we can use Ohm’s law to calculate the resistor value:
R=
vg
40 V
= 533.33 ⌦.
=
ig
75 mA
The power absorbed by the resistor must equal the power generated by
the source. Thus,
pR =
psource =
( 3 W) = 3 W.
[c] Again, note the iR = ig . The power dissipated by the resistor can be
determined from the resistor’s current:
pR = R(iR )2 = R(ig )2 .
Solving for ig ,
480 mW
pr
=
= 0.0016
R
300 ⌦
Then, since vR = vg
i2g =
so
ig =
vR = RiR = Rig = (300 ⌦)(40 mA) = 12 V
p
0.0016 = 0.04 A = 40 mA.
so
vg = 12 V.
AP 2.4
[a] Note from the circuit that the current through the conductance G is ig ,
flowing from top to bottom, because the current source and the
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2–4
CHAPTER 2. Circuit Elements
conductance are in the same branch of the circuit so must have the same
current. The voltage drop across the current source is vg , positive at the
top, because the current source and the conductance are also in parallel
so must have the same voltage. From a version of Ohm’s law,
0.5 A
ig
=
= 10 V.
G
50 mS
Now that we know the voltage drop across the current source, we can
find the power delivered by this source:
vg =
psource =
v g ig =
(10)(0.5) =
5 W.
Thus the current source delivers 5 W to the circuit.
[b] We can find the value of the conductance using the power, and the value
of the current using Ohm’s law and the conductance value:
pg = Gvg2
so
G=
pg
9
= 2 = 0.04 S = 40 mS;
2
vg
15
ig = Gvg = (40 mS)(15 V) = 0.6 A.
[c] We can find the voltage from the power and the conductance, and then
use the voltage value in Ohm’s law to find the current:
pg = Gvg2
Thus
vg2 =
so
vg =
8W
pg
=
= 40,000.
G
200 µS
q
40,000 = 200 V;
ig = Gvg = (200 µS)(200 V) = 0.04 A = 40 mA.
AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every
circuit element.
Write a KVL equation clockwise around the circuit, starting below the
voltage source:
24 V + v2 + v5
v1 = 0.
Next, use Ohm’s law to calculate the three unknown voltages from the
three currents:
v2 = 3i2 ;
v5 = 7i5 ;
v1 = 2i1 .
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Problems
2–5
A KCL equation at the upper right node gives i2 = i5 ; a KCL equation at
the bottom right node gives i5 = i1 ; a KCL equation at the upper left
node gives is = i2 . Now replace the currents i1 and i2 in the Ohm’s law
equations with i5 :
v2 = 3i2 = 3i5 ;
v5 = 7i5 ;
v1 = 2i1 =
2i5 .
Now substitute these expressions for the three voltages into the first
equation:
24 = v2 + v5
v1 = 3i5 + 7i5
( 2i5 ) = 12i5 .
Therefore i5 = 24/12 = 2 A.
[b] v1 =
2i5 =
2(2) =
4 V.
[c] v2 = 3i5 = 3(2) = 6 V.
[d] v5 = 7i5 = 7(2) = 14 V.
[e] A KCL equation at the lower left node gives is = i1 . Since i1 = i5 ,
is = 2 A. We can now compute the power associated with the voltage
source:
p24 = (24)is = (24)( 2) =
48 W.
Therefore 24 V source is delivering 48 W.
AP 2.6 Redraw the circuit labeling all voltages and currents:
We can find the value of the unknown resistor if we can find the value of its
voltage and its current. To start, write a KVL equation clockwise around the
right loop, starting below the 24 ⌦ resistor:
120 V + v3 = 0.
Use Ohm’s law to calculate the voltage across the 8 ⌦ resistor in terms of its
current:
v3 = 8i3 .
Substitute the expression for v3 into the first equation:
120 V + 8i3 = 0
so
i3 =
120
= 15 A.
8
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2–6
CHAPTER 2. Circuit Elements
Also use Ohm’s law to calculate the value of the current through the 24 ⌦
resistor:
i2 =
120 V
= 5 A.
24 ⌦
Now write a KCL equation at the top middle node, summing the currents
leaving:
i1 + i2 + i3 = 0
so
i1 = i2 + i3 = 5 + 15 = 20 A.
Write a KVL equation clockwise around the left loop, starting below the
voltage source:
200 V + v1 + 120 V = 0
so
v1 = 200
120 = 80 V.
Now that we know the values of both the voltage and the current for the
unknown resistor, we can use Ohm’s law to calculate the resistance:
R =
80
v1
= 4 ⌦.
=
i1
20
AP 2.7 [a] Plotting a graph of vt versus it gives
Note that when it = 0, vt = 25 V; therefore the voltage source must be
25 V. Since the plot is a straight line, its slope can be used to calculate
the value of resistance:
v
25 0
25
R=
=
=
= 100 ⌦
i
0.25 0
0.25
A circuit model having the same v i characteristic is a 25 V source in
series with a 100⌦ resistor, as shown below:
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Problems
2–7
[b] Draw the circuit model from part (a) and attach a 25 ⌦ resistor:
To find the power delivered to the 25 ⌦ resistor we must calculate the
current through the 25 ⌦ resistor. Do this by first using KCL to recognize
that the current in each of the components is it , flowing in a clockwise
direction. Write a KVL equation in the clockwise direction, starting
below the voltage source, and using Ohm’s law to express the voltage
drop across the resistors in the direction of the current it flowing through
the resistors:
25
25 V + 100it + 25it = 0
= 0.2 A.
so
125it = 25
so
it =
125
Thus, the power delivered to the 25 ⌦ resistor is
p25 = (25)i2t = (25)(0.2)2 = 1 W.
AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0,
it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot
is a straight line, its slope can be used to calculate the value of resistance:
v
25 0
25
=
=
= 100 ⌦.
i
0.25 0
0.25
A circuit model having the same v i characteristic is a 0.25 A current
source in parallel with a 100⌦ resistor, as shown below:
R=
[b] Draw the circuit model from part (a) and attach a 25 ⌦ resistor:
Note that by writing a KVL equation around the right loop we see that
the voltage drop across both resistors is vt . Write a KCL equation at the
top center node, summing the currents leaving the node. Use Ohm’s law
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2–8
CHAPTER 2. Circuit Elements
to specify the currents through the resistors in terms of the voltage drop
across the resistors and the value of the resistors.
vt
vt
0.25 +
+
= 0,
so
5vt = 25,
thus
vt = 5 V;
100 25
p25 =
vt2
= 1 W.
25
AP 2.9 First note that we know the current through all elements in the circuit except
the 6 k⌦ resistor (the current in the three elements to the left of the 6 k⌦
resistor is i1 ; the current in the three elements to the right of the 6 k⌦ resistor
is 30i1 ). To find the current in the 6 k⌦ resistor, write a KCL equation at the
top node:
i1 + 30i1 = i6k = 31i1 .
We can then use Ohm’s law to find the voltages across each resistor in terms
of i1 . The results are shown in the figure below:
[a] To find i1 , write a KVL equation around the left-hand loop, summing
voltages in a clockwise direction starting below the 5 V source:
5 V + 54,000i1
1 V + 186,000i1 = 0.
Solving for i1
54,000i1 + 186,000i1 = 6 V
so
240,000i1 = 6 V.
Thus,
i1 =
6
= 25 µA.
240,000
[b] Now that we have the value of i1 , we can calculate the voltage for each
component except the dependent source. Then we can write a KVL
equation for the right-hand loop to find the voltage v of the dependent
source. Sum the voltages in the clockwise direction, starting to the left of
the dependent source:
+v
54,000i1 + 8 V
186,000i1 = 0.
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Problems
2–9
Thus,
v = 240,000i1
8 V = 240,000(25 ⇥ 10 6 )
8V = 6V
8V =
2 V.
We now know the values of voltage and current for every circuit element.
Let’s construct a power table:
Element
Current Voltage
(µA)
(V)
5V
25
5
54 k⌦
25
1.35
1V
25
1
6 k⌦
775
4.65
Dep. source
750
2
1.8 k⌦
750
1.35
8V
750
8
Power
Power
Equation
(µW)
p=
vi
125
p = Ri2
33.75
p=
vi
25
p = Ri2
3603.75
p=
vi
1500
p = Ri2
1012.5
p=
vi
6000
[c] The total power generated in the circuit is the sum of the negative power
values in the power table:
125 µW + 25 µW + 6000 µW =
6150 µW.
Thus, the total power generated in the circuit is 6150 µW.
[d] The total power absorbed in the circuit is the sum of the positive power
values in the power table:
33.75 µW + 3603.75 µW + 1500 µW + 1012.5 µW = 6150 µW.
Thus, the total power absorbed in the circuit is 6150 µW.
AP 2.10 Given that i = 2 A, we know the current in the dependent source is
2i = 4 A. We can write a KCL equation at the left node to find the current in
the 10 ⌦ resistor. Summing the currents leaving the node,
5 A + 2 A + 4 A + i10⌦ = 0
so
i10⌦ = 5 A
2A
4A =
1 A.
Thus, the current in the 10 ⌦ resistor is 1 A, flowing right to left, as seen in
the circuit below.
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2–10
CHAPTER 2. Circuit Elements
[a] To find vs , write a KVL equation, summing the voltages counter-clockwise
around the lower right loop. Start below the voltage source.
vs + (1 A)(10 ⌦) + (2 A)(30 ⌦) = 0
so
vs = 10 V + 60 V = 70 V.
[b] The current in the voltage source can be found by writing a KCL equation
at the right-hand node. Sum the currents leaving the node
4 A + 1 A + iv = 0
so
iv = 4 A
1 A = 3 A.
The current in the voltage source is 3 A, flowing top to bottom. The
power associated with this source is
p = vi = (70 V)(3 A) = 210 W.
Thus, 210 W are absorbed by the voltage source.
[c] The voltage drop across the independent current source can be found by
writing a KVL equation around the left loop in a clockwise direction:
v5A + (2 A)(30 ⌦) = 0
so
v5A = 60 V.
The power associated with this source is
p=
v5A i =
(60 V)(5 A) =
300 W.
This source thus delivers 300 W of power to the circuit.
[d] The voltage across the controlled current source can be found by writing a
KVL equation around the upper right loop in a clockwise direction:
+v4A + (10 ⌦)(1 A) = 0
so
v4A =
10 V.
The power associated with this source is
p = v4A i = ( 10 V)(4 A) =
40 W.
This source thus delivers 40 W of power to the circuit.
[e] The total power dissipated by the resistors is given by
(i30⌦ )2 (30 ⌦) + (i10⌦ )2 (10 ⌦) = (2)2 (30 ⌦) + (1)2 (10 ⌦) = 120 + 10 = 130 W.
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Problems
2–11
Problems
P 2.1
The interconnection is valid. The 10 A current source has a voltage drop of
100 V, positive at the top, because the 100 V source supplies its voltage drop
across a pair of terminals shared by the 10 A current source. The right hand
branch of the circuit must also have a voltage drop of 100 V from the left
terminal of the 40 V source to the bottom terminal of the 5 A current source,
because this branch shares the same terminals as the 100 V source. This
means that the voltage drop across the 5 A current source is 140 V, positive at
the top. Also, the two voltage sources can carry the current required of the
interconnection. This is summarized in the figure below:
From the values of voltage and current in the figure, the power supplied by the
current sources is calculated as follows:
P10A =
(100)(10) = 1000 W (1000 W supplied);
P5A
P 2.2
X
=
(140)(5) =
700 W
(700 W supplied);
Pdev = 1700 W.
[a] Yes, independent voltage sources can carry whatever current is required by
the connection; independent current source can support any voltage
required by the connection.
[b] 18 V source:
absorbing;
5 mA source:
delivering;
7 V source:
absorbing.
[c]
P18V
=
P5mA
=
P7V
=
X
Pabs =
(5 ⇥ 10 3 )(18) = 90 mW (abs);
(5 ⇥ 10 3 )(25) =
125 mW (del);
(5 ⇥ 10 3 )(7) = 35 mW
X
(abs);
Pdel = 125 mW.
[d] Yes; 18 V source is delivering, the 5 mA source is absorbing, and the 7 V
source is absorbing
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2–12
CHAPTER 2. Circuit Elements
P18V
=
P5mA
=
P7V
=
X
Pabs =
(5 ⇥ 10 3 )(18) =
90 mW (del);
(5 ⇥ 10 3 )(11) = 55 mW (abs);
(5 ⇥ 10 3 )(7) = 35 mW (abs;)
X
Pdel = 90 mW.
P 2.3
The interconnection is not valid. Note that the 3 A and 4 A sources are both
connected in the same branch of the circuit. A valid interconnection would
require these two current sources to supply the same current in the same
direction, which they do not.
P 2.4
The interconnect is valid since the voltage sources can all carry 5 A of current
supplied by the current source, and the current source can carry the voltage
drop required by the interconnection. Note that the branch containing the 10
V, 40 V, and 5 A sources must have the same voltage drop as the branch
containing the 50 V source, so the 5 A current source must have a voltage
drop of 20 V, positive at the right. The voltages and currents are summarize
in the circuit below:
P 2.5
P50V
=
(50)(5) = 250 W (abs);
P10V
=
(10)(5) = 50 W
P40V
=
(40)(5) =
200 W (dev);
P5A
=
(20)(5) =
100 W
X
(abs);
(dev).
Pdev = 300 W.
First there is no violation of Kirchho↵’s laws, hence the interconnection is
valid.
Kirchho↵’s voltage law requires
20 + 60 + v1
v2 = 0
so
v1
v2 =
40 V.
The conservation of energy law requires
(5 ⇥ 10 3 )v2
(15 ⇥ 10 3 )v2
(20 ⇥ 10 3 )(20) + (20 ⇥ 10 3 )(60) + (20 ⇥ 10 3 )v1 = 0
or
v1
v2 =
40 V.
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Problems
Hence any combination of v1 and v2 such that v1
solution.
P 2.6
v2 =
2–13
40 V is a valid
[a] The voltage drop from the top node to the bottom node in this circuit
must be the same for every path from the top to the bottom. Therefore,
the voltages of the two voltage sources are equal:
↵i = 6.
Also, the current i is in the same branch as the 15 mA current source,
but in the opposite direction, so
i =
0.015A.
Substituting,
6
= 400.
0.015
The interconnection is valid if ↵ = 400 V/A.
[b] The voltage across the current source must equal the voltage across the 6
V source, since both are connected between the top and bottom nodes.
Using the passive sign convention,
↵( 0.015) = 6
!
↵=
p = vi = (6)(0.015) = 0.09 = 90 mW.
[c] Since the power is positive, the current source is absorbing power.
P 2.7
[a] Because both current sources are in the same branch of the circuit, their
values must be the same. Therefore,
v1
= 0.4 ! v1 = 0.4(50) = 20 V.
50
[b] p = v1 (0.4) = (20)(0.4) = 8 W (absorbed).
P 2.8
The interconnection is invalid. In the middle branch, the value of the current
ix must be 50 mA, since the 50 mA current source supplies current in this
branch in the same direction as the current ix . Therefore, the voltage supplied
by the dependent voltage source in the right hand branch is 1800(0.05) = 90
V. This gives a voltage drop from the top terminal to the bottom terminal in
the right hand branch of 90 + 60 = 150 V. But the voltage drop between these
same terminals in the left hand branch is 30 V, due to the voltage source in
that branch. Therefore, the interconnection is invalid.
P 2.9
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2–14
CHAPTER 2. Circuit Elements
First, 10va = 5 V, so va = 0.5 V. Then recognize that each of the three
branches is connected between the same two nodes, so each of these branches
must have the same voltage drop. The voltage drop across the middle branch
is 5 V, and since va = 0.5 V, vg = 0.5 5 = 4.5 V. Also, the voltage drop
across the left branch is 5 V, so 20 + v9A = 5 V, and v9A = 15 V, where v9A
is positive at the top. Note that the current through the 20 V source must be
9 A, flowing from top to bottom, and the current through the vg is 6 A flowing
from top to bottom. Let’s find the power associated with the left and middle
branches:
p9A = (9)( 15) = 135 W;
p20V = (9)(20) = 180 W;
pvg = (6)( 4.5) = 27 W;
p6A = (6)(0.5) = 3 W.
Since there is only one component left, we can find the total power:
ptotal = 135 + 180 + 27 + 3 + pds = 75 + pds = 0
so pds must equal 75 W.
Therefore,
P 2.10
X
Pdev =
X
Pabs = 210 W.
[a] Yes, Kirchho↵’s laws are not violated. (Note that i =
8 A.)
[b] No, because the voltages across the independent and dependent current
sources are indeterminate. For example, define v1 , v2 , and v3 as shown:
Kirchho↵’s voltage law requires
v1 + 20 = v3 ;
v2 + 100 = v3 .
Conservation of energy requires
8(20)
8v1
16v2
16(100) + 24v3 = 0
or
v1 + 2v2
3v3 =
220.
Now arbitrarily select a value of v3 and show the conservation of energy
will be satisfied. Examples:
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Problems
2–15
If v3 = 200 V then v1 = 180 V and v2 = 100 V. Then
180 + 200
If v3 =
120
P 2.11
600 =
220 (CHECKS).
100 V, then v1 =
400 + 300 =
120 V and v2 =
200 V. Then
220 (CHECKS).
[a] Using the passive sign convention and Ohm’s law,
i=
v
40
=
= 0.016 = 16 mA.
R
2500
[b] PR = Ri2 = (2500)(0.016)2 = 0.64 = 640 mW.
[c] Using the passive sign convention with the voltage polarity reversed,
i=
v
=
R
40
=
2500
0.016 =
16 mA;
PR = Ri2 = (2500)( 0.016)2 = 0.64 = 640 mW.
P 2.12
[a] Using the passive sign convention and Ohm’s law,
v = Ri = (3000)(0.015) = 45 V.
452
v2
=
= 0.675 = 675 mW.
R
3000
[c] Using the passive sign convention with the current direction reversed,
[b] PR =
P 2.13
v=
Ri =
(3000)(0.015) =
45 V;
PR =
452
v2
=
= 0.675 = 675 mW.
R
3000
[a]
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2–16
CHAPTER 2. Circuit Elements
[b]
P 2.14
Vbb
=
no-load voltage of battery;
Rbb
=
internal resistance of battery;
Rx
=
resistance of wire between battery and switch;
Ry
=
resistance of wire between switch and lamp A;
Ra
=
resistance of lamp A;
Rb
=
resistance of lamp B;
Rw
=
resistance of wire between lamp A and lamp B;
Rg1
=
resistance of frame between battery and lamp A;
Rg2
=
resistance of frame between lamp A and lamp B;
S
=
switch.
Since we know the device is a resistor, we can use Ohm’s law to calculate the
resistance. From Fig. P2.14(a),
v = Ri
so
v
R= .
i
Using the values in the table of Fig. P2.14(b),
120
=
0.01
R=
60
60
120
180
=
=
=
= 12 k⌦.
0.005
0.005
0.01
0.015
Note that this value is found in Appendix H.
P 2.15
The resistor value is the ratio of the power to the square of the current:
p
R = 2 . Using the values for power and current in Fig. P2.15(b),
i
33 ⇥ 10 3
74.25 ⇥ 10 3
132 ⇥ 10 3
8.25 ⇥ 10 3
=
=
=
(0.5 ⇥ 10 3 )2
(1 ⇥ 10 3 )2
(1.5 ⇥ 10 3 )2
(2 ⇥ 10 3 )2
=
297 ⇥ 10 3
206.25 ⇥ 10 3
=
= 33 k⌦.
(2.5 ⇥ 10 3 )2
(3 ⇥ 10 3 )2
Note that this is a value from Appendix H.
P 2.16
Since we know the device is a resistor, we can use the power equation. From
Fig. P2.16(a),
p = vi =
v2
R
so
R=
v2
.
p
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Problems
2–17
Using the values in the table of Fig. P2.16(b)
R=
( 4)2
(4)2
(8)2
( 8)2
=
=
=
640 ⇥ 10 3
160 ⇥ 10 3
160 ⇥ 10 3
640 ⇥ 10 3
(12)2
(16)2
=
=
= 100 ⌦.
1440 ⇥ 10 3
2560 ⇥ 10 3
Note that this value is found in Appendix H.
P 2.17
[a] Write a KCL equation at the top node:
1.5 + i1 + i2 = 0
so
i1 + i2 = 1.5.
Write a KVL equation around the right loop:
v1 + v2 + v3 = 0.
From Ohm’s law,
v1 = 100i1 ,
v2 = 150i2 ,
v3 = 250i2 .
Substituting,
100i1 + 150i2 + 250i2 = 0
so
100i1 + 400i2 = 0.
Solving the two equations for i1 and i2 simultaneously,
i1 = 1.2 A
and
i2 = 0.3 A.
[b] Write a KVL equation clockwise around the left loop:
vo + v1 = 0
So
but
v1 = 100i1 = 100(1.2) = 120 V.
vo = v1 = 120 V.
[c] Calculate power using p = vi for the source and p = Ri2 for the resistors:
psource =
vo (1.5) =
(120)(1.5) =
180 W;
p100⌦ = 1.22 (100) = 144 W;
p150⌦ = 0.32 (150) = 13.5 W;
p250⌦ = 0.32 (250) = 22.5 W.
X
Pdev = 180 W
X
Pabs = 144 + 13.5 + 22.5 = 180 W.
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2–18
CHAPTER 2. Circuit Elements
P 2.18
[a] Write a KVL equation clockwise aroud the right loop, starting below the
300 ⌦ resistor:
v a + vb =
0
so
va = v b .
Using Ohm’s law,
va = 300ia
and
vb = 75ib .
so
ib = 4ia .
Substituting,
300ia = 75ib
Write a KCL equation at the top middle node, summing the currents
leaving:
ig + ia + ib = 0
so
ig = ia + ib = ia + 4ia = 5ia .
Write a KVL equation clockwise around the left loop, starting below the
voltage source:
200 V + v40 + va = 0.
From Ohm’s law,
v40 = 40ig
and
va = 300ia .
Substituting,
200 V + 40ig + 300ia = 0
Substituting for ig :
200 V + 40(5ia ) + 300ia =
200 V + 200ia + 300ia =
200 V + 500ia = 0.
Thus,
200 V
= 0.4 A.
500
[b] From part (a), ib = 4ia = 4(0.4 A) = 1.6 A.
[c] From the circuit, vo = 75 ⌦(ib ) = 75 ⌦(1.6 A) = 120 V.
[d] Use the formula pR = Ri2R to calculate the power absorbed by each
resistor:
500ia = 200 V
so
ia =
p40⌦ = i2g (40 ⌦) = (5ia )2 (40 ⌦) = [5(0.4)]2 (40 ⌦) = (2)2 (40 ⌦) = 160 W;
p300⌦ = i2a (300 ⌦) = (0.4)2 (300 ⌦) = 48 W;
p75⌦ = i2b (75 ⌦) = (4ia )2 (75 ⌦) = [4(0.4)]2 (75 ⌦) = (1.6)2 (75 ⌦) = 192 W.
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Problems
2–19
[e] Using the passive sign convention,
psource =
=
(200 V)ig =
(200 V)(5ia ) =
(200 V)(2 A) =
(200 V)[5(0.4 A)]
400 W.
Thus the voltage source delivers 400 W of power to the circuit. Check:
X
X
P 2.19
[a]
Pdis = 160 + 48 + 192 = 400 W;
Pdel = 400 W.
vo
=
800
= 10io ;
io
=
8ia + 14ia + 18ia = 40(20) = 800 V;
800/10 = 80 A.
[b] ig = ia + io = 20 + 80 = 100 A.
[c] pg (delivered) = (100)(800) = 80,000 W = 80 kW.
P 2.20
Label the unknown resistor currents and voltages:
[a] KCL at the top node:
0.02 = i1 + i2 ;
KVL around the right loop:
vo + v2 5 = 0.
Use Ohm’s law to write the resistor voltages in the previous equation in
terms of the resistor currents:
5000i1 + 2000i2
5=0
!
Multiply the KCL equation by
eliminate i2 :
5000i1 + 2000i2 = 5.
2000 and add it to the KVL equation to
2000(i1 + i2 ) + ( 5000i1 + 2000i2 ) =
2000(0.02) + 5
!
7000i1 =
35.
Solving,
35
= 0.005 = 5 mA.
7000
Therefore,
i1 =
vo = Ri1 = (5000)(0.005) = 25 V.
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2–20
CHAPTER 2. Circuit Elements
[b] p20mA =
(0.02)vo =
i2 = 0.02
p5V =
(0.02)(25) =
i1 = 0.02
(5)i2 =
0.5 W;
0.005 = 0.015 A;
(5)(0.015) =
0.075 W;
p5k = 5000i21 = 5000(0.005)2 = 0.125 W;
p2k = 2000i22 = 2000(0.015)2 = 0.45 W;
ptotal = p20mA + p5V + p5k + p2k =
0.5
0.075 + 0.125 + 0.45 = 0.
Thus the power in the circuit balances.
P 2.21
Label the unknown resistor voltages and currents:
[a] ia =
3.5
= 0.02 A (Ohm’s law);
175
i1 = ia = 0.02 A (KCL).
[b] vb = 200i1 = 200(0.02) = 4 V (Ohm’s law);
v1 + vb + 3.5 = 0
so
v1 = 3.5 + vb = 3.5 + 4 = 7.5 V (KVL).
[c] va = 0.05(50) = 2.5 V (Ohm’s law);
vg + va + v1 = 0
so
vg = va + v1 = 2.5 + 7.5 = 10 V (KVL).
[d] pg = vg (0.05) = 10(0.05) = 0.5 W.
P 2.22
[a]
v2 = 2(20) = 40 V;
v8⌦ = 80 40 = 40 V;
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Problems
2–21
i2 = 40 V/8 ⌦ = 5 A;
i3 = io i2 = 2 5 = 3 A;
v4⌦ = ( 3)(4) = 12 V;
v1 = 4i3 + v2 = 12 + 40 = 28 V;
i1 = 28 V/4 ⌦ = 7 A.
[b] i4 = i1 + i3 = 7
[c]
3 = 4 A.
p13⌦
=
42 (13) = 208 W;
p8⌦
=
(5)2 (8) = 200 W;
p4⌦
=
72 (4) = 196 W;
p4⌦
=
( 3)2 (4) = 36 W;
p20⌦
=
22 (20) = 80 W.
X
Pdis = 208 + 200 + 196 + 36 + 80 = 720 W;
ig = i4 + i2 = 4 + 5 = 9 A;
Pdev = (9)(80) = 720 W.
P 2.23
[a] Start with the 22.5 ⌦ resistor. Since the voltage drop across this resistor is
90 V, we can use Ohm’s law to calculate the current:
90 V
= 4 A.
22.5 ⌦
Next we can calculate the voltage drop across the 15 ⌦ resistor by writing
a KVL equation around the outer loop of the circuit:
i22.5 ⌦ =
240 V + 90 V + v15 ⌦ = 0
so
v15 ⌦ = 240
90 = 150 V.
Now that we know the voltage drop across the 15 ⌦ resistor, we can use
Ohm’s law to find the current in this resistor:
150 V
i15 ⌦ =
= 10 A.
15 ⌦
Write a KCL equation at the middle right node to find the current
through the 5 ⌦ resistor. Sum the currents entering:
4A
10 A + i5 ⌦ = 0
so
i5 ⌦ = 10 A
4 A = 6 A.
Write a KVL equation clockwise around the upper right loop, starting
below the 4 ⌦ resistor. Use Ohm’s law to express the voltage drop across
the resistors in terms of the current through the resistors:
v4 ⌦ + 90 V + (5 ⌦)( 6 A) = 0
so
v4 ⌦ = 90 V
30 V = 60 V.
Using Ohm’s law we can find the current through the 4 ⌦ resistor:
i4 ⌦ =
60 V
= 15 A.
4⌦
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2–22
CHAPTER 2. Circuit Elements
Write a KCL equation at the middle node. Sum the currents entering:
15 A
6A
i20 ⌦ = 0
so
i20 ⌦ = 15 A
6 A = 9 A.
Use Ohm’s law to calculate the voltage drop across the 20 ⌦ resistor:
v20 ⌦ = (20 ⌦)(9 A) = 180 V.
All of the voltages and currents calculated above are shown in the figure
below:
Calculate the power dissipated by the resistors using the equation
pR = Ri2R :
p4⌦ = (4)(15)2 = 900 W
p20⌦ = (20)(9)2 = 1620 W;
p5⌦ = (5)(6)2 = 180 W
p22.5⌦ = (22.5)(4)2 = 360 W;
p15⌦ = (15)(10)2 = 1500 W.
[b] We can calculate the current in the voltage source, ig by writing a KCL
equation at the top middle node:
ig = 15 A + 4 A = 19 A.
Now that we have both the voltage and the current for the source, we can
calculate the power supplied by the source:
pg =
[c]
P 2.24
[a]
240(19) =
X
4560 W
thus
pg (supplied) = 4560 W.
Pdis = 900 + 1620 + 180 + 360 + 1500 = 4560 W.
Therefore,
X
Psupp =
X
Pdis .
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Problems
vo
=
20(8) + 16(15) = 400 V;
io
=
400/80 = 5 A;
ia
=
25 A.
2–23
P230 (supplied) = (230)(25) = 5750 W;
ib = 5 + 15 = 20 A;
P260 (supplied) = (260)(20) = 5200 W.
[b]
X
X
Pdis
Psup
=
(25)2 (2) + (20)2 (8) + (5)2 (4) + (15)2 16 + (20)2 2 + (5)2 (80)
=
1250 + 3200 + 100 + 3600 + 800 + 2000 = 10,950 W;
=
5750 + 5200 = 10,950 W.
Therefore,
P 2.25
[a]
X
Pdis =
X
Psup = 10,950 W.
v2 = 100 + 4(15) = 160 V;
i1 =
100
v1
=
= 5 A;
4 + 16
20
v1 = 160
i3 = i1
(9 + 11 + 10)(2) = 100 V;
2=5
2 = 3 A;
vg = v1 + 30i3 = 100 + 30(3) = 190 V;
i4 = 2 + 4 = 6 A;
ig =
i4
i3 =
6
3=
9 A.
[b] Calculate power using the formula p = Ri2 :
p9 ⌦ = (9)(2)2 = 36 W;
p11 ⌦ = (11)(2)2 = 44 W;
p10 ⌦ = (10)(2)2 = 40 W;
p5 ⌦ = (5)(6)2 = 180 W;
p30 ⌦ = (30)(3)2 = 270 W;
p4 ⌦ = (4)(5)2 = 100 W;
p16 ⌦ = (16)(5)2 = 400 W;
p15 ⌦ = (15)(4)2 = 240 W.
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2–24
CHAPTER 2. Circuit Elements
[c] vg = 190 V.
[d] Sum the power dissipated by the resistors:
X
pdiss = 36 + 44 + 40 + 180 + 270 + 100 + 400 + 240 = 1310 W.
The power associated with the sources is
pvolt source = (100)(4) = 400 W;
pcurr source = vg ig = (190)( 9) =
1710 W.
Thus the total power dissipated is 1310 + 400 = 1710 W and the total
power developed is 1710 W, so the power balances.
P 2.26
[a]
va = (5 + 10)(4) = 60 V;
240 + va + vb = 0
so
vb = 240
ie = vb /(14 + 6) = 180/20 = 9 A;
id = ie 4 = 9 4 = 5 A;
vc = 4id + vb = 4(5) + 180 = 200 V;
ic = vc /10 = 200/10 = 20 A;
vd = 240 vc = 240 200 = 40 V;
ia = id + ic = 5 + 20 = 25 A;
R = vd /ia = 40/25 = 1.6 ⌦.
va = 240
60 = 180 V;
[b] ig = ia + 4 = 25 + 4 = 29 A;
pg (supplied) = (240)(29) = 6960 W.
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Problems
P 2.27
2–25
Label all unknown resistor voltages and currents:
Ohms’ law for 5 k⌦ resistor:
v1 = (0.01)(5000) = 50 V.
KVL for lower left loop:
80 + v2 + 50 = 0 ! v2 = 80 50 = 30 V.
Ohm’s law for 1.5 k⌦ resistor:
i2 = v2 /1500 = 30/1500 = 20 mA.
KCL at center node:
i2 = i3 + 0.01 ! i3 = i2 0.01 = 0.02 0.01 = 0.01 = 10 mA.
Ohm’s law for 3 k⌦ resistor
v3 = 3000i3 = 3000(0.01) = 30 V.
KVL for lower right loop:
v1 + v3 + v4 = 0 ! v4 = v1 v3 = 50 30 = 20 V.
Ohm’s law for 500 ⌦ resistor:
i4 = v4 /500 = 20/500 = 0.04 = 40 mA.
KCL for right node:
i3 + iR = i4 ! iR = i4 i3 = 0.04 0.01 = 0.03 = 30 mA.
KVL for outer loop:
80 + vR + v4 = 0 ! vR = 80 v4 = 80 20 = 60 V.
Therefore,
R=
P 2.28
vR
60
= 2000 = 2 k⌦.
=
iR
0.03
[a] Plot the v—i characteristic:
From the plot:
R=
130 ( 30)
v
=
= 20 ⌦.
i
8 0
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2–26
CHAPTER 2. Circuit Elements
When it = 0, vt = 30 V; therefore the ideal voltage source has a voltage
of 30 V. Thus the device can be modeled as a 30 V source in series
with a 20 ⌦ resistor, as shown below:
[b] We attach a 40 ⌦ resistor to the device model developed in part (a):
Write a KVL equation clockwise around the circuit, using Ohm’s law to
express the voltage drop across the resistors in terms of the current it
through the resistors:
( 30 V)
Thus
20it
it =
40it = 0
so
60it =
30 V.
30 V
= 0.5 A.
60
Now calculate the power dissipated by the resistor:
p40 ⌦ = 40i2t = (40)(0.5)2 = 10 W.
P 2.29
[a] Plot the v
i characteristic
From the plot:
R=
v
(125
=
i
(15
50)
= 5 ⌦.
0)
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Problems
2–27
When it = 0, vt = 50 V; therefore the ideal current source has a current
of 10 A
[b]
10 + it = i1
and
5i1 =
So, 10 + it =
4it so it =
20it .
2 A. Thus,
p20 ⌦ = 20i2t = (20)( 2)2 = 80 W.
P 2.30
[a] Begin by constructing a plot of voltage versus current:
[b] Since the plot is linear for 0  is  24 mA amd since R =
calculate R from the plotted values as follows:
v/ i, we can
v
24 18
6
=
=
= 250 ⌦.
i
0.024 0
0.024
We can determine the value of the ideal voltage source by considering the
value of vs when is = 0. When there is no current, there is no voltage
drop across the resistor, so all of the voltage drop at the output is due to
the voltage source. Thus the value of the voltage source must be 24 V.
The model, valid for 0  is  24 mA, is shown below:
R=
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2–28
CHAPTER 2. Circuit Elements
[c] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the
voltage source. Use Ohm’s law to express the voltage drop across the
resistors in terms of the current i:
24 V + 250i + 1000i = 0
Thus,
i=
so
1250i = 24 V.
24 V
= 19.2 mA.
1250 ⌦
[d] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the
voltage source. Use Ohm’s law to express the voltage drop across the
resistors in terms of the current i:
24 V + 250i = 0
Thus,
i=
so
250i = 24 V.
24 V
= 96 mA.
250 ⌦
[e] The short circuit current can be found in the table of values (or from the
plot) as the value of the current is when the voltage vs = 0. Thus,
isc = 48 mA
(from table).
[f ] The plot of voltage versus current constructed in part (a) is not linear (it
is piecewise linear, but not linear for all values of is ). Since the proposed
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Problems
2–29
circuit model is a linear model, it cannot be used to predict the nonlinear
behavior exhibited by the plotted data.
P 2.31
[a]
[b]
v = 30V;
i = 15 mA;
[c] 2000i1 = 3000is ,
40 = i1 + is = 2.5is ,
R=
v
= 2 k⌦.
i
i1 = 1.5is ;
is = 16 mA.
[d] vs (open circuit) = (40 ⇥ 10 3 )(2 ⇥ 103 ) = 80 V.
[e] The open circuit voltage can be found in the table of values (or from the
plot) as the value of the voltage vs when the current is = 0. Thus,
vs (open circuit) = 55 V (from the table).
[f ] Linear model cannot predict the nonlinear behavior of the practical
current source.
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2–30
P 2.32
CHAPTER 2. Circuit Elements
[a] The circuit:
v1 = (4000)(0.01) = 40 V
(Ohm’s law);
v1
= 2000io + 6000io = 8000io
2
(KVL).
Thus,
40/2
v1 /2
=
= 0.0025 = 2.5 mA.
8000
8000
[b] Calculate the power for all components:
io =
p10mA =
pd.s. =
p4k =
(0.01)v1 =
(v1 /2)io =
(0.01)(40) =
0.4 W;
(40/2)(2.5 ⇥ 10 3 ) =
0.05 W;
402
v12
=
= 0.4 W;
4000
4000
p2k = 2000i2o = 2000(2.5 ⇥ 10 3 )2 = 0.0125 W;
p6k = 6000i2o = 6000(2.5 ⇥ 10 3 )2 = 0.0375 W.
Therefore,
ptotal =
0.4
0.05 + 0.4 + 0.0125 + 0.0375 = 0.
Thus the power in the circuit balances.
P 2.33
Label unknown current:
20 + 450i + 150i = 0
so
600i = 20
!
(KVL and Ohm’s law);
i = 33.33 mA.
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Problems
vx = 150i = 150(0.0333) = 5 V
vo = 300
✓
2–31
(Ohm’s law);
◆
vx
= 300(5/100) = 15 V
100
(Ohm’s law).
Calculate the power for all components:
p20V =
20i =
pd.s. =
vo
✓
20(0.0333) =
◆
vx
=
100
0.667 W;
(15)(5/100) =
0.75 W;
p450 = 450i2 = 450(0.033)2 = 0.5 W;
p150 = 150i2 = 150(0.033)2 = 0.1667 W;
p300 =
152
vo2
=
= 0.75 W.
300
300
Thus the total power absorbed is
pabs = 0.5 + 0.1667 + 0.75 = 1.4167 W.
P 2.34
Label unknown voltage and current:
vx + vo + 2ix = 0
vx = 6ix
(KVL);
(Ohm’s law).
Therefore
6ix + vo + 2ix = 0 so vo = 4ix .
Thus
ix =
vo
.
4
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2–32
CHAPTER 2. Circuit Elements
Also,
vo
2
(Ohm’s law);
45 = ix + i1
(KCL).
i1 =
Substituting for the currents ix and i1 :
45 =
3vo
vo vo
+
=
.
4
2
4
Thus
vo = 45
✓ ◆
4
= 60 V.
3
The only two circuit elements that could supply power are the two sources, so
calculate the power for each source:
vx = 6ix = 6
p45V =
vo
= 6(60/4) = 90 V;
4
45vx =
45(90) =
4050 W;
pd.s. = (2ix )i1 = 2(vo /4)(vo /2) = 2(60/4)(60/2) = 900 W.
Only the independent voltage source is supplying power, so the total power
supplied is 4050 W.
P 2.35
[a] io = 0 because no current can exist in a single conductor connecting two
parts of a circuit.
[b]
200 + 8000ig + 12,000ig = 0
so
ig = 200/20,000 = 10 mA;
3
3
v = (12 ⇥ 10 )(10 ⇥ 10 ) = 120 V;
5 ⇥ 10 3 v = 0.6 A;
9000i1 = 3000i2
so
i2 = 3i1 ;
0.6 + i1 + i2 = 0
so
0.6 + i1 + 3i1 = 0
thus
i1 = 0.15 A.
[c] i2 = 3i1 =
0.45 A.
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Problems
P 2.36
[a] 12
2–33
2i = 5i ;
5i = 8i + 2i = 10i .
Therefore, 12
2i = 10i , so i = 1 A.
5i = 10i = 10; so i = 2 A.
vo = 2i = 2 V.
[b] ig = current out of the positive terminal of the 12 V source;
vd = voltage drop across the 8i source;
ig = i + i + 8i = 9i + i = 19 A;
vd = 2 + 8 = 10 V.
X
X
Pgen
=
12ig + 8i (8) = 12(19) + 8(2)(8) = 356 W;
Pdiss
=
2i ig + 5i2 + 8i (i + 8i ) + 2i2 + 8i vd
=
2(1)(19) + 5(2)2 + 8(1)(17) + 2(1)2 + 8(2)(10)
X
P 2.37
40i2 +
Pgen
356 W; Therefore,
=
X
5
5
+
= 0;
40 10
v1 = 80i2 =
25i1 +
=
Pdiss = 356 W.
i2 =
15.625 mA;
1.25 V;
( 1.25)
+ ( 0.015625) = 0;
20
i1 = 3.125 mA;
vg = 60i1 + 260i1 = 320i1 .
Therefore, vg = 1 V.
P 2.38
iE
iB
iC = 0;
iC = i B
i2 =
therefore iE = (1 + )iB ;
iB + i1 ;
Vo + iE RE
i1 R1 + VCC
(i1
iB )R2 = 0;
(i1
iB )R2 = 0
or
i1 =
VCC + iB R2
;
R 1 + R2
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2–34
CHAPTER 2. Circuit Elements
Vo + iE RE + iB R2
VCC + iB R2
R2 = 0.
R 1 + R2
Now replace iE by (1 + )iB and solve for iB . Thus
iB =
P 2.39
[VCC R2 /(R1 + R2 )] Vo
.
(1 + )RE + R1 R2 /(R1 + R2 )
Here is Equation 2.21:
iB =
(VCC R2 )/(R1 + R2 ) V0
;
(R1 R2 )/(R1 + R2 ) + (1 + )RE
(10)(60,000)
VCC R2
= 6V;
=
R 1 + R2
100,000
(40,000)(60,000)
R1 R2
= 24 k⌦;
=
R 1 + R2
100,000
iB =
5.4
6 0.6
=
= 0.18 mA;
24,000 + 50(120)
30,000
iC = iB = (49)(0.18) = 8.82 mA;
iE = iC + iB = 8.82 + 0.18 = 9 mA;
v3d = (0.009)(120) = 1.08V;
vbd = Vo + v3d = 1.68V;
i2 =
1.68
vbd
= 28 µA;
=
R2
60,000
i1 = i2 + iB = 28 + 180 = 208 µA;
vab = 40,000(208 ⇥ 10 6 ) = 8.32 V;
iCC = iC + i1 = 8.82 + 0.208 = 9.028 mA;
v13 + (8.82 ⇥ 10 3 )(750) + 1.08 = 10 V;
v13 = 2.305 V.
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Problems
P 2.40
2–35
[a]
[b]
P 2.41
Each radiator is modeled as a 48 ⌦ resistor:
Write a KVL equation for each of the three loops:
240
= 5 A;
48
240 + 48i1 = 0
!
i1 =
48i1 + 48i2 = 0
!
i2 = i1 = 5 A;
48i2 + 48i3 = 0
!
i3 = i2 = 5 A.
Therefore, the current through each radiator is 5 A and the power for each
radiator is
prad = Ri2 = 48(5)2 = 1200 W.
There are three radiators, so the total power for this heating system is
ptotal = 3prad = 3(1200) = 3600 W.
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2–36
P 2.42
CHAPTER 2. Circuit Elements
Each radiator is modeled as a 48 ⌦ resistor:
Write a KVL equation for the left and right loops:
240 + 48i1 = 0
!
48i1 + 48i2 + 48i2 = 0
i1 =
240
= 5 A;
48
!
i2 =
5
i1
= = 2.5 A.
2
2
The power for the center radiator is
pcen = 48i21 = 48(5)2 = 1200 W.
The power for each of the radiators on the right is
pright = 48i22 = 48(2.5)2 = 300 W.
Thus the total power for this heating system is
ptotal = pcen + 2pright = 1200 + 2(300) = 1800 W.
The center radiator produces 1200 W, just like the three radiators in Problem
2.41. But the other two radiators produce only 300 W each, which is 1/4th of
the power of the radiators in Problem 2.41. The total power of this
configuration is 1/2 of the total power in Fig. P2.41.
P 2.43
Each radiator is modeled as a 48 ⌦ resistor:
Write a KVL equation for the left and right loops:
240 + 48i1 + 48i2 = 0;
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Problems
48i2 + 48i3 = 0
!
2–37
i2 = i3 .
Write a KCL equation at the top node:
i1 = i2 + i3
!
i1 = i2 + i2 = 2i2 .
Substituting into the first KVL equation gives
240 + 48(2i2 ) + 48i2 = 0
!
i2 =
240
= 1.67 A.
3(48)
Solve for the currents i1 and i3 :
i3 = i2 = 1.67 A;
i1 = 2i2 = 2(1.67) = 3.33 A.
Calculate the power for each radiator using the current for each radiator:
pleft = 48i21 = 48(3.33)2 = 533.33 W;
pmiddle = pright = 48i22 = 48(1.67)2 = 133.33 W.
Thus the total power for this heating system is
ptotal = pleft + pmiddle + pright = 533.33 + 133.33 + 133.33 = 800 W.
All radiators in this configuration have much less power than their
counterparts in Fig. P2.41. The total power for this configuration is only
22.2% of the total power for the heating system in Fig. P2.41.
P 2.44
Each radiator is modeled as a 48 ⌦ resistor:
Write a KVL equation for this loop:
240 + 48i + 48i + 48i = 0
!
i=
240
= 1.67 A.
3(48)
Calculate the power for each radiator:
prad = 48i2 = 48(1.67)2 = 133.33 W.
Calculate the total power for this heating system:
ptotal = 3prad = 3(133.33) = 400 W.
Each radiator has much less power than the radiators in Fig. P2.41, and the
total power of this configuration is just 1/9th of the total power in Fig. P2.41.
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Simple Resistive Circuits
3
Assessment Problems
AP 3.1
Start from the right hand side of the circuit and make series and parallel
combinations of the resistors until one equivalent resistor remains. Begin by
combining the 6 ⌦ resistor and the 10 ⌦ resistor in series:
6 ⌦ + 10 ⌦ = 16 ⌦.
Now combine this 16 ⌦ resistor in parallel with the 64 ⌦ resistor:
16 ⌦k64 ⌦ =
(16)(64)
1024
=
= 12.8 ⌦.
16 + 64
80
This equivalent 12.8 ⌦ resistor is in series with the 7.2 ⌦ resistor:
12.8 ⌦ + 7.2 ⌦ = 20 ⌦.
Finally, this equivalent 20 ⌦ resistor is in parallel with the 30 ⌦ resistor:
20 ⌦k30 ⌦ =
(20)(30)
600
=
= 12 ⌦.
20 + 30
50
Thus, the simplified circuit is as shown:
3–1
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3–2
CHAPTER 3. Simple Resistive Circuits
[a] With the simplified circuit we can use Ohm’s law to find the voltage across
both the current source and the 12 ⌦ equivalent resistor:
v = (12 ⌦)(5 A) = 60 V.
[b] Now that we know the value of the voltage drop across the current source,
we can use the formula p = vi to find the power associated with the
source:
p=
(60 V)(5 A) =
300 W.
Thus, the source delivers 300 W of power to the circuit.
[c] We now can return to the original circuit, shown in the first figure. In this
circuit, v = 60 V, as calculated in part (a). This is also the voltage drop
across the 30 ⌦ resistor, so we can use Ohm’s law to calculate the current
through this resistor:
60 V
= 2 A.
30 ⌦
Now write a KCL equation at the upper left node to find the current iB :
iA =
5 A + iA + iB = 0
so
iB = 5 A
iA = 5 A
2 A = 3 A.
Next, write a KVL equation around the outer loop of the circuit, using
Ohm’s law to express the voltage drop across the resistors in terms of the
current through the resistors:
v + 7.2iB + 6iC + 10iC = 0.
So
16iC = v
Thus
iC =
7.2iB = 60 V
(7.2)(3) = 38.4 V.
38.4
= 2.4 A.
16
Now that we have the current through the 10 ⌦ resistor we can use the
formula p = Ri2 to find the power:
p10 ⌦ = (10)(2.4)2 = 57.6 W.
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Problems
3–3
AP 3.2
[a] We can use voltage division to calculate the voltage vo across the 75 k⌦
resistor:
75,000
vo (no load) =
(200 V) = 150 V.
75,000 + 25,000
[b] When we have a load resistance of 150 k⌦ then the voltage vo is across the
parallel combination of the 75 k⌦ resistor and the 150 k⌦ resistor. First,
calculate the equivalent resistance of the parallel combination:
75 k⌦k150 k⌦ =
(75,000)(150,000)
= 50,000 ⌦ = 50 k⌦.
75,000 + 150,000
Now use voltage division to find vo across this equivalent resistance:
vo =
50,000
(200 V) = 133.3 V.
50,000 + 25,000
[c] If the load terminals are short-circuited, the 75 k⌦ resistor is e↵ectively
removed from the circuit, leaving only the voltage source and the 25 k⌦
resistor. We can calculate the current in the resistor using Ohm’s law:
i=
200 V
= 8 mA.
25 k⌦
Now we can use the formula p = Ri2 to find the power dissipated in the
25 k⌦ resistor:
p25k = (25,000)(0.008)2 = 1.6 W.
[d] The power dissipated in the 75 k⌦ resistor will be maximum at no load
since vo is maximum. In part (a) we determined that the no-load voltage
is 150 V, so be can use the formula p = v 2 /R to calculate the power:
p75k (max) =
(150)2
= 0.3 W.
75,000
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3–4
CHAPTER 3. Simple Resistive Circuits
AP 3.3
[a] We will write a current division equation for the current through the 80⌦
resistor and use this equation to solve for R:
i80⌦ =
Thus
R
(20 A) = 4 A
R + 40 ⌦ + 80 ⌦
16R = 480
and
so
R=
20R = 4(R + 120).
480
= 30 ⌦.
16
[b] With R = 30 ⌦ we can calculate the current through R using current
division, and then use this current to find the power dissipated by R,
using the formula p = Ri2 :
iR =
40 + 80
(20 A) = 16 A
40 + 80 + 30
so
pR = (30)(16)2 = 7680 W.
[c] Write a KVL equation around the outer loop to solve for the voltage v,
and then use the formula p = vi to calculate the power delivered by the
current source:
v + (60 ⌦)(20 A) + (30 ⌦)(16 A) = 0
Thus,
psource =
(1680 V)(20 A) =
so
v = 1200 + 480 = 1680 V.
33,600 W.
Thus, the current source generates 33,600 W of power.
AP 3.4
[a] First we need to determine the equivalent resistance to the right of the
40 ⌦ and 70 ⌦ resistors:
1
1
1
1
1
Req = 20 ⌦k30 ⌦k(50 ⌦ + 10 ⌦)
+
+
=
.
so
=
Req
20 ⌦ 30 ⌦ 60 ⌦
10 ⌦
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Problems
Thus,
3–5
Req = 10 ⌦.
Now we can use voltage division to find the voltage vo :
vo =
40
(60 V) = 20 V.
40 + 10 + 70
[b] The current through the 40 ⌦ resistor can be found using Ohm’s law:
vo
20 V
=
= 0.5 A.
40
40 ⌦
This current flows from left to right through the 40 ⌦ resistor. To use
current division, we need to find the equivalent resistance of the two
parallel branches containing the 20 ⌦ resistor and the 50 ⌦ and 10 ⌦
resistors:
(20)(60)
20 ⌦k(50 ⌦ + 10 ⌦) =
= 15 ⌦.
20 + 60
Now we use current division to find the current in the 30 ⌦ branch:
15
i30⌦ =
(0.5 A) = 0.16667 A = 166.67 mA.
15 + 30
i40⌦ =
[c] We can find the power dissipated by the 50 ⌦ resistor if we can find the
current in this resistor. We can use current division to find this current
from the current in the 40 ⌦ resistor, but first we need to calculate the
equivalent resistance of the 20 ⌦ branch and the 30 ⌦ branch:
(20)(30)
= 12 ⌦.
20 + 30
Current division gives:
20 ⌦k30 ⌦ =
i50⌦ =
Thus,
12
(0.5 A) = 0.08333 A.
12 + 50 + 10
p50⌦ = (50)(0.08333)2 = 0.34722 W = 347.22 mW.
AP 3.5 [a]
We can find the current i using Ohm’s law:
i=
1V
= 0.01 A = 10 mA.
100 ⌦
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3–6
CHAPTER 3. Simple Resistive Circuits
[b]
Rm = 50 ⌦k5.555 ⌦ = 5 ⌦.
We can use the meter resistance to find the current using Ohm’s law:
imeas =
1V
= 0.009524 = 9.524 mA.
100 ⌦ + 5 ⌦
AP 3.6 [a]
Use voltage division to find the voltage v:
v=
75,000
(60 V) = 50 V.
75,000 + 15,000
[b]
The meter resistance is a series combination of resistances:
Rm = 149,950 + 50 = 150,000 ⌦.
We can use voltage division to find v, but first we must calculate the
equivalent resistance of the parallel combination of the 75 k⌦ resistor and
the voltmeter:
(75,000)(150,000)
75,000 ⌦k150,000 ⌦ =
= 50 k⌦.
75,000 + 150,000
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Problems
Thus,
vmeas =
3–7
50,000
(60 V) = 46.15 V.
50,000 + 15,000
AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite
resistors must be equal. Therefore,
(1000)(150)
= 1500 ⌦ = 1.5 k⌦.
100
[b] When the bridge is balanced, there is no current flowing through the
meter, so the meter acts like an open circuit. This places the following
branches in parallel: The branch with the voltage source, the branch with
the series combination R1 and R3 and the branch with the series
combination of R2 and Rx . We can find the current in the latter two
branches using Ohm’s law:
5V
5V
iR1 ,R3 =
= 20 mA;
iR2 ,Rx =
= 2 mA.
100 ⌦ + 150 ⌦
1000 + 1500
We can calculate the power dissipated by each resistor using the formula
p = Ri2 :
100Rx = (1000)(150)
so
Rx =
p100⌦ = (100 ⌦)(0.02 A)2 = 40 mW;
p150⌦ = (150 ⌦)(0.02 A)2 = 60 mW;
p1000⌦ = (1000 ⌦)(0.002 A)2 = 4 mW;
p1500⌦ = (1500 ⌦)(0.002 A)2 = 6 mW.
Since none of the power dissipation values exceeds 250 mW, the bridge
can be left in the balanced state without exceeding the power-dissipating
capacity of the resistors.
AP 3.8 Convert the three Y-connected resistors, 20 ⌦, 10 ⌦, and 5 ⌦ to three
-connected resistors Ra , Rb , and Rc . To assist you the figure below has both
the Y-connected resistors and the -connected resistors
(5)(10) + (5)(20) + (10)(20)
= 17.5 ⌦;
20
(5)(10) + (5)(20) + (10)(20)
= 35 ⌦;
Rb =
10
(5)(10) + (5)(20) + (10)(20)
= 70 ⌦.
Rc =
5
Ra =
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3–8
CHAPTER 3. Simple Resistive Circuits
The circuit with these new
-connected resistors is shown below:
From this circuit we see that the 70 ⌦ resistor is parallel to the 28 ⌦ resistor:
70 ⌦k28 ⌦ =
(70)(28)
= 20 ⌦.
70 + 28
Also, the 17.5 ⌦ resistor is parallel to the 105 ⌦ resistor:
17.5 ⌦k105 ⌦ =
(17.5)(105)
= 15 ⌦.
17.5 + 105
Once the parallel combinations are made, we can see that the equivalent 20 ⌦
resistor is in series with the equivalent 15 ⌦ resistor, giving an equivalent
resistance of 20 ⌦ + 15 ⌦ = 35 ⌦. Finally, this equivalent 35 ⌦ resistor is in
parallel with the other 35 ⌦ resistor:
35 ⌦k35 ⌦ =
(35)(35)
= 17.5 ⌦.
35 + 35
Thus, the resistance seen by the 2 A source is 17.5 ⌦, and the voltage can be
calculated using Ohm’s law:
v = (17.5 ⌦)(2 A) = 35 V.
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Problems
3–9
Problems
P 3.1
[a] The 6 ⌦ and 12 ⌦ resistors are in series, as are the 9 ⌦ and 7 ⌦ resistors.
The simplified circuit is shown below:
[b] The 3 k⌦, 5 k⌦, and 7 k⌦ resistors are in series. The simplified circuit is
shown below:
[c] The 300 ⌦, 400 ⌦, and 500 ⌦ resistors are in series. The simplified circuit is
shown below:
[d] The 50 ⌦ and 90 ⌦ resistors are in series, as are the 80 ⌦ and 70 ⌦
resistors. The simplified circuit is shown below:
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3–10
P 3.2
CHAPTER 3. Simple Resistive Circuits
Always work from the side of the circuit furthest from the source. Remember
that the current in all series-connected circuits is the same, and that the
voltage drop across all parallel-connected resistors is the same.
[a] Circuit in Fig. P3.1(a):
Req = 6 + 12 + [4k(9 + 7)] = 18 + (4k16) = 18 + 3.2 = 21.2 ⌦.
Circuit in Fig. P3.1(b):
Req = 4000 + [10,000k(3000 + 5000 + 7000)] = 4000 + (10,000k15,000)
= 4000 + 6000 = 10 k⌦.
Circuit in Fig. P3.1(c):
Req = (300 + 400 + 500) + (600k1200) = 1200 + 400 = 1600 ⌦.
Circuit in Fig. P3.1(d):
Req = ([(70 + 80)k100] + 50 + 90)k300 = [(150k100) + 50 + 90]k300
= (60 + 50 + 90)k300 = 200k300 = 120 ⌦.
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.1(a):
P =
Vs2
102
= 4.717 W.
=
Req
21.2
For the circuit in Fig. P3.1(b):
P = Is2 Req = 0.0032 (10,000) = 0.09 = 90 mW.
For the circuit in Fig. P3.1(c):
Vs2
0.22
P =
= 2.5 ⇥ 10 5 = 25 µW.
=
Req
1600
For the circuit in Fig. P3.1(d):
P = Is2 (Req ) = (0.03)2 (120) = 0.108 = 108 mW.
P 3.3
[a] The 10 ⌦ and 40 ⌦ resistors are in parallel, as are the 100 ⌦ and 25 ⌦
resistors. The simplified circuit is shown below:
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Problems
3–11
[b] The 9 k⌦, 18 k⌦, and 6 k⌦ resistors are in parallel. The simplified circuit
is shown below:
[c] The 750 ⌦ and 500 ⌦ resistors are in parallel, as are the 1.5 k⌦ and 3 k⌦
resistors. The simplified circuit is shown below:
[d] The 600 ⌦, 200 ⌦, and 300 ⌦ resistors are in series. The simplified circuit
is shown below:
P 3.4
Always work from the side of the circuit furthest from the source. Remember
that the current in all series-connected circuits is the same, and that the
voltage drop across all parallel-connected resistors is the same.
[a] Circuit in Fig. P3.3(a):
Req = 18 + (100k25k(22 + (10k40))) = 18 + (20k(22 + 8) = 18 + 12 = 30 ⌦.
Circuit in Fig. P3.3(b):
Req = 10,000k(5000 + 2000 + (9000k18,000k6000)) = 10,000k(7000 + 3000)
= 10,000k10,000 = 5 k⌦.
Circuit in Fig. P3.3(c):
Req = (900 + 600)k750k500 + (1500k3000) + 2000 = 250 + 1000 + 2000 = 3250 ⌦.
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3–12
CHAPTER 3. Simple Resistive Circuits
Circuit in Fig. P3.3(d):
Req = 600k200k300k(250 + 150) = 600k200k300k400 = 80 ⌦.
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.3(a):
P =
Vs2
52
= 0.833 W.
=
Req
30
For the circuit in Fig. P3.3(b):
P = Is2 (Req ) = (0.05)2 (5000) = 12.5 W.
For the circuit in Fig. P3.3(c):
Vs2
652
P =
= 1.3 W.
=
Req
3250
For the circuit in Fig. P3.3(d):
P = Is2 (Req ) = 0.22 (80) = 3.2 W.
P 3.5
[a] Rab = 12 + (24k(30 + 18)) + 10 = 12 + (24k48) + 10 = 12 + 16 + 10 = 38 ⌦.
[b] Rab = 4000k30,000k60,000k(1200 + (7200k2400) + 2000)
= 4000k30,000k60,000k5000 = 2 k⌦.
[c] Rab = [(4000 + 6000 + 2000)k8000] + 5200 = (12,000k8000) + 5200
= 4800 + 5200 = 10,000 = 10 k⌦.
[d] Rab = 1200k720k(320 + 480) = 1200k720k800 = 288 ⌦.
P 3.6
Write an expression for the resistors in series and parallel from the right side
of the circuit to the left. Then simplify the resulting expression from left to
right to find the equivalent resistance.
[a] Rab = [(26 + 10)k18 + 6]k36 = (36k18 + 6)k36 = (12 + 6)k36 = 18k36 = 12 ⌦.
[b] Rab = [(12 + 18)k10k15k20 + 16]k30 + 4 + 14 = (30k10k15k20 + 16)k30 + 4 + 14
= (4 + 16)k30 + 4 + 14 = 20k30 + 4 + 14 = 12 + 4 + 14 = 30 ⌦.
[c] 15k60 = 12 ⌦;
18 + 12 + 20 = 50 ⌦;
30k45 = 18 ⌦;
50k50 = 25 ⌦;
Rab = 25 + 25 + 10 = 60 ⌦.
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Problems
[d] 18 + 12 = 30 ⌦;
30k60 = 20 ⌦;
20 + 30 = 50 ⌦;
50k75 = 30 ⌦;
30 + 20 = 50 ⌦;
50k50 = 25 ⌦;
60k20 = 15 ⌦;
15 + 25 = 40 ⌦;
3–13
Rab = 40k40 = 20 ⌦.
P 3.7
[a] Circuit in Fig. P3.7(a):
Req = 360k(90 + 120k(160 + 200)) = 360k(90 + (120k360)) = 360k(90 + 90)
= 360k180 = 120 ⌦.
Circuit in Fig. P3.7(b):
Req = ([(750 + 250)k1000] + 100)k([(150 + 600)k500] + 300)
= [(1000k1000) + 100]k[(750k500) + 300] = (500 + 100)k(300 + 300)
= 600k600 = 300 ⌦.
Circuit in Fig. P3.7(c):
1
1
1
1
1
30
1
1
+
+
+ +
=
= ;
=
Re
20 15 20 4 12
60
2
Re = 2 ⌦;
Re + 16 = 18 ⌦;
18k18 = 9 ⌦;
Req = 10 + 8 + 9 = 27 ⌦.
Circuit in Fig. P3.7(d):
15k30 = 10 ⌦;
10 + 20 = 30 ⌦;
60k30 = 20 ⌦;
20 + 10 = 30 ⌦;
30k80k(40 + 20) = 30k80k60 = 16 ⌦;
Req = 16 + 24 + 10 = 50 ⌦.
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3–14
CHAPTER 3. Simple Resistive Circuits
[b] Note that in every case, the power delivered by the source must equal the
power absorbed by the equivalent resistance in the circuit. For the circuit
in Fig. P3.7(a):
P = Is2 Req = (0.032 )(120) = 108 mW.
For the circuit in Fig. P3.7(b):
P = Is2 (Req ) = (0.05)2 (300) = 0.75 = 750 mW.
For the circuit in Fig. P3.7(c):
P =
Vs2
1442
= 768 W.
=
Req
27
For the circuit in Fig. P3.7(d):
Vs2
0.082
P =
= 128 µ W.
=
Req
50
P 3.8
[a] From Ex. 3-1: i1 = 4 A,
i2 = 8 A,
is = 12 A;
at node b:
12 + 4 + 8 = 0,
at node d: 12 4
[b] v1
= 4is = 48 V
8 = 0.
v3 = 3i2 = 24 V;
v4 = 6i2 = 48 V.
v2 = 18i1 = 72 V
loop abda:
120 + 48 + 72 = 0;
loop bcdb:
72 + 24 + 48 = 0;
loop abcda:
120 + 48 + 24 + 48 = 0.
P 3.9
[a] p4⌦
p3⌦
= i2s 4 = (12)2 4 = 576 W
p18⌦ = (4)2 18 = 288 W;
(8)2 3 = 192 W
p6⌦ = (8)2 6 = 384 W.
=
[b] p120V (delivered) = 120is = 120(12) = 1440 W.
[c] pdiss = 576 + 288 + 192 + 384 = 1440 W.
P 3.10
[a] R + R = 2R.
[b] R + R + R + · · · + R = nR.
[c] R + R = 2R = 3000 so R = 1500 = 1.5 k⌦.
This is a resistor from Appendix H.
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Problems
3–15
[d] nR = 4000;
so if n = 4, R = 1 k⌦.
This is a resistor from Appendix H.
P 3.11
[a] Req = RkR =
R
R2
= .
2R
2
= RkRkRk · · · kR
(n R’s)
R
= Rk
n 1
R2 /(n 1)
R2
R
=
=
= .
R + R/(n 1)
nR
n
[b] Req
R
= 5000 so R = 10 k⌦.
2
This is a resistor from Appendix H.
R
= 4000 so R = 4000n.
[d]
n
If n = 3
r = 4000(3) = 12 k⌦.
This is a resistor from Appendix H. So put three 12k resistors in parallel
to get 4k⌦.
[c]
P 3.12
4=
20R2
R2 + 40
so
R2 = 10 ⌦;
3=
20Re
40 + Re
so
Re =
Thus,
P 3.13
10RL
120
=
17
10 + RL
120
⌦;
17
so
RL = 24 ⌦.
500
(75) = 15 V.
(500 + 2000)
[b] i = 75/2500 = 30 mA;
[a] vo =
PR1 = 2000(0.03)2 = 1.8 W;
PR2 = 500(0.03)2 = 0.45 W.
[c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage
requirement, first pick R1 to meet the 1 W specification
i R1 =
75
15
R1
Thus, R1
,
602
Therefore,
✓
or
3600 ⌦.
R1
60
R1
◆2
R1  1.
Now use the voltage specification:
75R2 + 15R2 + 3600(15).
Thus, R2 =
3600(15)
= 900 ⌦.
60
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3–16
P 3.14
CHAPTER 3. Simple Resistive Circuits
[a] vo =
40R2
=8
R 1 + R2
so
R 2 RL
;
R 2 + RL
Let Re = R2 kRL =
vo =
R1 = 4R2 .
40Re
= 7.5
R 1 + Re
so
Then, 4R2 = 4.33Re =
Thus, R2 = 300 ⌦
R1 = 4.33Re .
4.33(3600R2 )
.
3600 + R2
and
R1 = 4(300) = 1200 ⌦.
[b] The resistor that must dissipate the most power is R1 , as it has the largest
resistance and carries the same current as the parallel combination of R2
and the load resistor. The power dissipated in R1 will be maximum when
the voltage across R1 is maximum. This will occur when the voltage
divider has a resistive load. Thus,
vR1 = 40
pR1 =
7.5 = 32.5 V;
32.52
= 880.2 m W.
1200
Thus the minimum power rating for all resistors should be 1 W.
P 3.15
Refer to the solution to Problem 3.14. The voltage divider will reach the
maximum power it can safely dissipate when the power dissipated in R1 equals
1 W. Thus,
vR2 1
=1
1200
vo = 40
So,
so
vR1 = 34.64 V.
34.64 = 5.36 V.
40Re
= 5.36
1200 + Re
Thus,
and
(300)RL
= 185.68
300 + RL
Re = 185.68 ⌦.
and
RL = 487.26 ⌦.
The minimum value for RL from Appendix H is 560 ⌦.
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Problems
P 3.16
3–17
[a ]
120 k⌦ + 30 k⌦ = 150 k⌦;
75 k⌦k150 k⌦ = 50 k⌦;
vo1 =
240
(50,000) = 160 V;
(25,000 + 50,000)
vo =
120,000
(vo1 ) = 128 V,
(150,000)
vo = 128 V.
[b ]
i=
240
= 2.4 mA;
100,000
75,000i = 180 V;
vo =
120,000
(180) = 144 V;
150,000
vo = 144 V.
[c] It removes loading e↵ect of second voltage divider on the first voltage
divider. Observe that the open circuit voltage of the first divider is
75,000
0
(240) = 180 V.
=
vo1
(100,000)
Now note this is the input voltage to the second voltage divider when the
current controlled voltage source is used.
P 3.17
(24)2
= 80,
R 1 + R2 + R3
Therefore, R1 + R2 + R3 = 7.2 ⌦.
(R1 + R2 )24
= 12;
(R1 + R2 + R3 )
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3–18
CHAPTER 3. Simple Resistive Circuits
Therefore, 2(R1 + R2 ) = R1 + R2 + R3 .
Thus, R1 + R2 = R3 ;
2R3 = 7.2;
R3 = 3.6 ⌦.
R2 (24)
= 5;
R 1 + R2 + R3
4.8R2 = R1 + R2 + 3.6 = 7.2;
Thus, R2 = 1.5 ⌦;
P 3.18
R1 = 7.2
[a] At no load:
vo = kvs =
vo = ↵vs =
Therefore k
=
↵
=
✓
1
↵
↵
◆
[b] R1
R2
[c ]
R1 =
✓
(1
where Re =
(1
and
R1 =
and
R1 =
k)
k
(1
R2 .
↵)
↵
R o R2
.
R o + R2
Re .
(1 k)
R2 Ro
R2 .
=
R o + R2
k
k)
k
Re
vs ,
R 1 + Re
R2
R 1 + R2
Re
R 1 + Re
Solving for R2 yields
Also,
R3 = 2.1 ⌦.
R2
vs .
R 1 + R2
At full load:
Thus
R2
R2 =
R2
.·.
(k ↵)
Ro .
↵(1 k)
R1 =
(k
↵)
↵k
Ro .
◆
0.05
Ro = 2.5 k⌦;
0.68
✓
◆
0.05
=
Ro = 14.167 k⌦.
0.12
=
Maximum dissipation in R2 occurs at no load, therefore,
[(60)(0.85)]2
PR2 (max) =
= 183.6 mW.
14,167
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Problems
3–19
Maximum dissipation in R1 occurs at full load.
PR1 (max) =
[60
0.80(60)]2
= 57.60 mW.
2500
[d ]
P 3.19
PR 1
=
PR2
=
(60)2
= 1.44 W = 1440 mW;
2500
(0)2
= 0 W.
14,167
[a]
Req = (10 + 20)k[12 + (90k10)] = 30k15 = 10 ⌦;
v2.4A = 10(2.4) = 24 V;
vo = v20⌦ =
v90⌦ =
io =
20
(24) = 16 V;
10 + 20
9
90k10
(24) = (24) = 14.4 V;
6 + (90k10)
15
14.4
= 0.16 A.
90
14.4)2
= 15.36 W.
6
6
[c] p2.4A = (2.4)(24) = 57.6 W.
Thus the power developed by the current source is 57.6 W.
[b] p6⌦ =
P 3.20
(v2.4A
v90⌦ )2
=
(24
Req = 6k30k20 = 4 ⌦.
Using current division, the current in the 30 ⌦ resistor is
i30⌦ =
4
Req
(30) = (30) = 4 A.
30
30
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3–20
CHAPTER 3. Simple Resistive Circuits
Thus, the power in the 30 ⌦ resistor is
p30⌦ = 30i230⌦ = 30(4)2 = 480 W.
P 3.21
Begin by using KCL at the top node to relate the branch currents to the
current supplied by the source. Then use the relationships among the branch
currents to express every term in the KCL equation using just i2 :
0.05 = i1 + i2 + i3 + i4 = 0.6i2 + i2 + 2i2 + 4i1 = 0.6i2 + i2 + 2i2 + 4(0.6i2 ) = 6i2 .
Therefore,
i2 = 0.05/6 = 0.00833 = 8.33 mA.
Find the remaining currents using the value of i2 :
i1 = 0.6i2 = 0.6(0.00833) = 0.005 = 5 mA;
i3 = 2i2 = 2(0.00833) = 0.01667 = 16.67 mA;
i4 = 4i1 = 4(0.005) = 0.02 = 20 mA.
Since the resistors are in parallel, the same voltage, 25 V, appears across each
of them. We know the current and the voltage for every resistor so we can use
Ohm’s law to calculate the values of the resistors:
R1 = 25/i1 = 25/0.005 = 5000 = 5 k⌦;
R2 = 25/i2 = 25/0.00833 = 3000 = 3 k⌦;
R3 = 25/i3 = 25/0.01667 = 1500 = 1.5 k⌦;
R4 = 25/i4 = 25/0.02 = 1250 = 1.25 k⌦.
The resulting circuit is shown below:
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Problems
P 3.22
3–21
[a] Let vo be the voltage across the parallel branches, positive at the upper
terminal, then
ig = vo G1 + vo G2 + · · · + vo GN = vo (G1 + G2 + · · · + GN ).
It follows that
vo =
ig
.
(G1 + G2 + · · · + GN )
The current in the k th branch is
ik =
[b] i5 =
P 3.23
i k = v o Gk ;
Thus,
ig Gk
.
[G1 + G2 + · · · + GN ]
40(0.2)
= 3.2 A.
2 + 0.2 + 0.125 + 0.1 + 0.05 + 0.025
[a] The equivalent resistance of the 4 ⌦ resistor and the resistors to its right is
4k(9 + 7) = 4k16 = 3.2 ⌦.
Using voltage division,
3.2
(10) = 1.51 V.
3.2 + 6 + 12
9
(1.51) = 0.85 V.
[b] v9 =
7+9
v4 =
P 3.24
[a] The equivalent resistance of the 100 ⌦ resistor and the resistors to its right
is
100k(80 + 70) = 100k150 = 60 ⌦.
Using current division,
i50 =
[b] v70 =
P 3.25
120
(50 + 90 + 60)k300
(0.03) =
(0.03) = 0.018 = 18 mA.
50 + 90 + 60
200
60
(80 + 70)k100
(0.018) =
(0.018) = 0.0072 = 7.2 mA.
80 + 70
150
[a] Begin by finding the equivalent resistance of the 30 ⌦ resistor and all
resistors to its right:
([(12 + 18)k10k15k20] + 16)k30 = 12 ⌦.
Now use voltage division to find the voltage across the 4 ⌦ resistor:
v4 =
4
(6) = 0.8 V.
4 + 12 + 14
[b] Use Ohm’s law to find the current in the 4 ⌦ resistor:
i4 = v4 /4 = 0.8/4 = 0.2 A.
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3–22
CHAPTER 3. Simple Resistive Circuits
[c] Begin by finding the equivalent resistance of all resistors to the right of the
30 ⌦ resistor:
[(12 + 18)k10k15k20] + 16 = 20 ⌦.
Now use current division:
30k20
i16 =
(0.2) = 0.12 = 120 mA.
20
[d] Note that the current in the 16 ⌦ resistor divides among four branches –
20 ⌦, 15 ⌦, 10 ⌦, and (12 + 18) ⌦:
20k15k10k(12 + 18)
(0.12) = 0.048 = 48 mA.
10
[e] Use Ohm’s law to find the voltage across the 10 ⌦ resistor:
i10 =
v10 = 10i10 = 10(0.048) = 0.48 V.
[f ] v18 =
P 3.26
18
(0.48) = 0.288 = 288 mV.
12 + 18
[a] The equivalent resistance of the circuit to the right of the 360 ⌦ resistor is
(200 + 160)k120 + 90 = 180 ⌦.
Thus by current division,
360k180
(0.03) = 0.01 = 10 mA.
180
[b] Using Ohm’s law:
i360 =
v360 = 360i360 = 360(0.01) = 3.6 V.
[c] The voltage across the 360 ⌦ resistor divides between two resistors – the
90 ⌦ resistor and the 120k(160 + 200) = 90 ⌦ equivalent resistance. Using
voltage division,
v90 =
90
90
(v360 ) =
(3.6) = 1.8 V.
90 + 90
180
[d] The current in the 90 ⌦ resistor can be found using Ohm’s law:
v90
1.8
=
= 0.02 = 20 mA.
90
90
Note that i360 + i90 = 10 + 20 = 30 mA (checks).
i90 =
[e] The current i90 divides between the 120 ⌦ branch and the
160 + 200 = 360 ⌦ branch. Using current division,
i160 =
120k(160 + 200)
90
i90 =
(0.02) = 0.005 = 5 mA.
160 + 200
360
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Problems
P 3.27
3–23
[a] The equivalent resistance to the right of the 36 ⌦ resistor is
6 + [18k(26 + 10)] = 18 ⌦.
By current division,
36k18
(0.45) = 0.15 = 150 mA.
36
[b] Using Ohm’s law,
i36 =
v36 = 36i36 = 36(0.15) = 5.4 V.
[c] Before using voltage division, find the equivalent resistance of the 18 ⌦
resistor and the resistors to its right:
18k(26 + 10) = 12 ⌦.
Now use voltage division:
12
(5.4) = 3.6 V.
12 + 6
10
[d] v10 =
(3.6) = 1 V.
10 + 26
v18 =
P 3.28
Find the equivalent resistance of all the resistors except the 2 ⌦:
5 ⌦k20 ⌦ = 4 ⌦;
4 ⌦ + 6 ⌦ = 10 ⌦;
10k(15 + 12 + 13) = 8 ⌦ = Req .
Use Ohm’s law to find the current ig :
ig =
125
125
= 12.5 A.
=
2 + Req
2+8
Use current division to find the current in the 6 ⌦ resistor:
i6⌦ =
8
(12.5) = 10 A.
6+4
Use current division again to find io :
io =
P 3.29
5k20
5k20
i6⌦ =
(10) = 2 A.
20
20
Use current division to find the current in the 8 ⌦ resistor. Begin by finding
the equivalent resistance of the 8 ⌦ resistor and all resistors to its right:
Req = ([(20k80) + 4]k30) + 8 = 20 ⌦;
i8 =
60k20
60kReq
(0.25) = 0.1875 = 187.5 mA.
(0.25) =
Req
20
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3–24
CHAPTER 3. Simple Resistive Circuits
Use current division to find i1 from i8 :
i1 =
30k20
30k[4 + (80k20)]
(i8 ) =
(0.1875) = 0.075 = 75 mA.
30
30
Use current division to find i4⌦ from i8 :
30k[4 + (80k20)]
30k20
(i8 ) =
(0.1875) = 0.1125 = 112.5 mA.
4 + (80k20)
20
i4⌦ =
Finally, use current division to find i2 from i4⌦ :
i2 =
P 3.30
80k20
80k20
(i4⌦ ) =
(0.1125) = 0.09 = 90 mA.
20
20
The equivalent resistance of the circuit to the right of the 90 ⌦ resistor is
Req = [(150k75) + 40]k(30 + 60) = 90k90 = 45 ⌦.
Use voltage division to find the voltage drop between the top and bottom
nodes:
vReq =
45
(3) = 1 V.
45 + 90
Use voltage division again to find v1 from vReq :
v1 =
50
5
150k75
(1) = (1) = V.
150k75 + 40
90
9
Use voltage division one more time to find v2 from vReq :
v2 =
P 3.31
30
1
(1) = V.
30 + 60
3
Use current division to find the current in the branch containing the 10 k and
15 k resistors, from bottom to top
i10k+15k =
(10 k + 15 k)k(3 k + 12 k)
(18) = 6.75 mA.
10 k + 15 k
Use Ohm’s law to find the voltage drop across the 15 k resistor, positive at the
top:
v15k =
(6.75 m)(15 k) =
101.25 V.
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Problems
3–25
Find the current in the branch containing the 3 k and 12 k resistors, from
bottom to top
i10k+15k =
(10 k + 15 k)k(3 k + 12 k)
(18) = 11.25 mA.
3 k + 12 k
Use Ohm’s law to find the voltage drop across the 12 k resistor, positive at the
top:
v12k =
P 3.32
(12 k)(11.25 m) =
135 V;
vo = v15k
v12k =
[a] v20k =
20
(45) = 36 V;
20 + 5
v90k =
( 135) = 33.75 V.
90
(45) = 27 V;
90 + 60
vx = v20k
[b] v20k =
101.25
v90k = 36
27 = 9 V.
20
(Vs ) = 0.8vs ;
25
v90k =
90
(Vs ) = 0.6Vs ;
150
vx = 0.8Vs
0.6Vs = 0.2Vs .
P 3.33
Original meter:
Re =
50 ⇥ 10 3
= 0.01 ⌦.
5
Modified meter:
Re =
(0.02)(0.01)
= 0.00667 ⌦.
0.03
.·. (Ifs )(0.00667) = 50 ⇥ 10 3 ;
.·. Ifs = 7.5 A.
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3–26
P 3.34
CHAPTER 3. Simple Resistive Circuits
[a] The model of the ammeter is an ideal ammeter in parallel with a resistor
whose resistance is given by
100 µV
= 10 ⌦.
10 µA
Rs =
We can calculate the current through the real meter using current
division:
(10/99)
10
1
im =
(imeas ) =
(imeas ) =
imeas .
10 + (10/99)
990 + 10
100
[b] At full scale, imeas = 1 A or 106 µ A, and im = 10 µ A so 999,990 µ A flows
throught the resistor RA :
RA =
100
100 µ V
=
⌦;
999,990 µ A
999,990
im =
(100/999,990)
1
(imeas ) =
(imeas ).
10 + (100/999,990)
100,000
[c] Yes.
P 3.35
The current in the shunt resistor at full-scale deflection is
iA = ifullscale 3 ⇥ 10 3 A. The voltage across RA at full-scale deflection is
always 150 mV; therefore,
RA =
150 ⇥ 10 3
150
=
ifullscale 3 ⇥ 10 3
1000ifullscale
3
.
150
= 30.018 m⌦.
5000 3
[b] Let Rm be the equivalent ammeter resistance:
[a] RA =
0.15
= 0.03 = 30 m⌦.
5
150
[c] RA =
= 1.546 ⌦.
100 3
0.15
[d] Rm =
= 1.5 ⌦.
0.1
Rm =
P 3.36
At full scale the voltage across the shunt resistor will be 100 mV; therefore the
power dissipated will be
PA =
(100 ⇥ 10 3 )2
.
RA
Therefore RA
(100 ⇥ 10 3 )2
= 40 m⌦.
0.25
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Problems
3–27
Otherwise the power dissipated in RA will exceed its power rating of 0.25 W.
When RA = 40 m⌦, the shunt current will be
iA =
100 ⇥ 10 3
= 2.5 A.
40 ⇥ 10 3
The measured current will be imeas = 2.5 + 0.002 = 2.502 A.
.·. Full-scale reading is, for practical purposes, 2.5 A.
P 3.37
For all full-scale readings the total resistance is
Rv + Rmovement =
full-scale reading
.
10 3
We can calculate the resistance of the movement as follows:
Rmovement =
Therefore,
P 3.38
20 mV
= 20 ⌦.
1 mA
Rv = 1000 (full-scale reading)
[a] Rv = 1000(50)
20 = 49, 980 ⌦;
[b] Rv = 1000(5)
20 = 4980 ⌦;
[c] Rv = 1000(0.25)
20 = 230 ⌦;
[d] Rv = 1000(0.025)
20 = 5 ⌦.
20.
[a] vmeas = (50 ⇥ 10 3 )[15k45k(4980 + 20)] = 0.5612 V.
[b] vtrue = (50 ⇥ 10 3 )(15k45) = 0.5625 V;
✓
0.5612
% error =
0.5625
P 3.39
◆
1 ⇥ 100 =
0.224%.
The current in the 10 ⌦ resistor is the voltage supplied divided by the
equivalent resistance attached to the voltage source:
ig =
50
= 1.995526 A.
(60k20.1 + 10)
The current measured by the ammeter is found using current division:
imeas =
60k20.1
(1.995526) = 1.494776 A.
20.1
The true value of the current in the 10 ⌦ resistor is
ig =
50
= 2 A,
(15 + 10)
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3–28
CHAPTER 3. Simple Resistive Circuits
so the true value of the current in the 20 ⌦ resistor is
itrue =
20k60
(2) = 1.5 A.
20
%error =
P 3.40

1 ⇥ 100 =
0.348267% ⇡
0.35%.
Begin by using current division to find the actual value of the current io :
itrue =
15
(50 mA) = 12.5 mA;
15 + 45
imeas =
15
(50 mA) = 12.4792 mA;
15 + 45 + 0.1
% error =
P 3.41
1.494776
1.5

12.4792
12.5
1 100 =
0.166389% ⇡
0.17%.
[a ]
20 ⇥ 103 i1 + 80 ⇥ 103 (i1
iB ) = 7.5;
80 ⇥ 103 (i1
iB ) = 0.6 + 40iB (0.2 ⇥ 103 );
.·.
80iB = 7.5 ⇥ 10 3 ;
100i1
80i1 88iB = 0.6 ⇥ 10 3 .
Calculator solution yields iB = 225 µA.
[b] With the insertion of the ammeter the equations become
100i1
80iB = 7.5 ⇥ 10 3
80 ⇥ 103 (i1
80i1
(no change);
iB ) = 103 iB + 0.6 + 40iB (200);
89iB = 0.6 ⇥ 10 3 ;
Calculator solution yields iB = 216 µA.
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Problems
✓
216
[c] % error =
225
P 3.42
◆
1 100 =
Rmeter = Rm + Rmovement =
3–29
4%.
500 V
= 1000 k⌦;
0.5 mA
vmeas = (50 k⌦k250 k⌦k1000 k⌦)(10 mA) = (40 k⌦)(10 mA) = 400 V;
vtrue = (50 k⌦k250 k⌦)(10 mA) = (41.67 k⌦)(10 mA) = 416.67 V;
✓
400
% error =
416.67
P 3.43
◆
1 100 =
[a] R1
=
(100/0.002) = 50 k⌦;
R2
=
(10/0.002) = 5 k⌦;
R3
= (1/0.002) = 500 ⌦.
[b] Let ia
=
actual current in the movement;
id
=
design current in the movement.
✓
ia
Then % error =
id
◆
1 100.
For the 100 V scale:
100
100
ia =
=
,
50,000 + 25
50,025
50,000
ia
= 0.9995
=
id
50,025
For the 10 V scale:
5000
ia
= 0.995
=
id
5025
For the 1 V scale:
500
ia
= 0.9524
=
id
525
P 3.44
4%.
id =
100
;
50,000
% error = (0.9995
1)100 =
0.05%.
% error = (0.995
1.0)100 =
0.4975%.
% error = (0.9524
1.0)100 =
4.76%.
[a] vmeter = 180 V.
[b] Rmeter = (100)(200) = 20 k⌦;
vmeter =
[c] vmeter =
70k20
(180) = 78.75 V.
70k20 + 20
20k20
(180) = 22.5 V.
20k20 + 20
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3–30
CHAPTER 3. Simple Resistive Circuits
[d] vmeter a = 180 V;
vmeter b + vmeter c = 101.26 V.
No, because of the loading e↵ect.
P 3.45
From the problem statement we have
Vs (10)
(1) Vs in mV; Rs in M⌦;
50 =
10 + Rs
48.75 =
Vs (6)
6 + Rs
[a] From Eq (1)
(2).
10 + Rs = 0.2Vs ;
.·. Rs = 0.2Vs
10.
Substituting into Eq (2) yields
48.75 =
6Vs
0.2Vs 4
or
Vs = 52 mV.
[b] From Eq (1)
50 =
520
10 + Rs
or
50Rs = 20.
So Rs = 400 k⌦.
P 3.46
[a] Since the unknown voltage is greater than either voltmeter’s maximum
reading, the only possible way to use the voltmeters would be to connect
them in series.
[b ]
Rm1 = (300)(1000) = 300 k⌦;
Rm2 = (150)(800) = 120 k⌦;
.·. Rm1 + Rm2 = 420 k⌦;
i1 max =
300
⇥ 10 3 = 1 mA;
300
i2 max =
150
⇥ 10 3 = 1.25 mA;
120
.·. imax = 1 mA since meters are in series;
vmax = 10 3 (300 + 120)103 = 420 V.
Thus the meters can be used to measure the voltage
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Problems
[c] im =
399
= 0.95 mA;
420 ⇥ 103
vm1 = (0.95)(300) = 285 V
P 3.47
3–31
vm2 = (0.95)(120) = 114 V.
The current in the series-connected voltmeters is
im =
288
= 0.96 mA;
300
v80 k⌦ = (0.96)(80) = 76.8 V;
Vpower supply = 288 + 115.2 + 76.8 = 480 V.
P 3.48
[a] Rmovement = 50 ⌦;
R1 + Rmovement =
30
= 30 k⌦
1 ⇥ 10 3
R2 + R1 + Rmovement =
.·. R1 = 29,950 ⌦;
150
= 150 k⌦
1 ⇥ 10 3
R3 + R2 + R1 + Rmovement =
.·. R3 = 150 k⌦.
.·. R2 = 120 k⌦;
300
= 300 k⌦;
1 ⇥ 10 3
[b]
v1 = (0.96 m)(150 k) = 144 V;
imove =
i1 =
144
= 0.96 mA;
120 + 29.95 + 0.05
144
= 0.192 mA;
750 k
i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA;
vmeas = vx = 144 + 150i2 = 316.8 V.
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3–32
CHAPTER 3. Simple Resistive Circuits
[c] v1 = 150 V;
i2 = 1 m + 0.20 m = 1.20 mA;
i1 = 150/750,000 = 0.20 mA;
.·. vmeas = vx = 150 + (150 k)(1.20 m) = 330 V.
P 3.49
[a] Rmeter = 360 k⌦ + 200 k⌦k50 k⌦ = 400 k⌦;
400k600 = 240 k⌦;
Vmeter =
240
(300) = 240 V.
300
[b] What is the percent error in the measured voltage?
600
(300) = 272.73 V;
660
✓
◆
240
% error =
1 100 = 12%.
272.73
True value =
P 3.50
Since the bridge is balanced, we can remove the detector without disturbing
the voltages and currents in the circuit.
It follows that
i1 =
i2 =
ig (R2 + Rx )
ig (R2 + Rx )
X
=
;
R 1 + R2 + R3 + Rx
R
ig (R1 + R3 )
ig (R1 + R3 )
X
=
;
R 1 + R2 + R3 + Rx
R
v 3 = R 3 i1 = v x = i 2 R x ;
.·.
R3 ig (R2 + Rx )
Rx ig (R1 + R3 )
X
X
=
;
R
R
.·. R3 (R2 + Rx ) = Rx (R1 + R3 ).
From which Rx =
R 2 R3
.
R1
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Problems
P 3.51
3–33
[a]
The condition for a balanced bridge is that the product of the opposite
resistors must be equal:
(1200)(600)
= 900 ⌦.
800
[b] The source current is the sum of the two branch currents. Each branch
current can be determined using Ohm’s law, since the resistors in each
branch are in series and the voltage drop across each branch is 21 V:
(800)(Rx ) = (1200)(600)
is =
so
Rx =
21 V
21 V
+
= 25 mA.
800 ⌦ + 600 ⌦ 1200 ⌦ + 900 ⌦
[c] We can use Ohm’s law to find the current in each branch:
ileft =
21
= 15 mA;
800 + 600
iright =
21
= 10 mA.
1200 + 900
Now we can use the formula p = Ri2 to find the power dissipated by each
resistor:
p800 = (800)(0.015)2 = 180 mW
p600 = (600)(0.015)2 = 135 mW;
p1200 = (1200)(0.01)2 = 120 mW
p900 = (900)(0.01)2 = 90 mW.
Thus, the 800 ⌦ resistor absorbs the most power; it absorbs 180 mW of
power.
[d] From the analysis in part (c), the 900 ⌦ resistor absorbs the least power; it
absorbs 90 mW of power.
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3–34
P 3.52
CHAPTER 3. Simple Resistive Circuits
Redraw the circuit, replacing the detector branch with a short circuit.
6 k⌦k30 k⌦ = 5 k⌦;
12 k⌦k20 k⌦ = 7.5 k⌦;
is =
75
= 6 mA;
12,500
v1 = 0.006(5000) = 30 V;
v2 = 0.006(7500) = 45 V;
i1 =
30
= 5 mA;
6000
i2 =
45
= 3.75 mA;
12,000
id = i1
P 3.53
i2 = 1.25 mA.
Note the bridge structure is balanced, that is 10 ⇥ 18 = 30 ⇥ 6, hence there is
no current in the 50⌦ resistor. It follows that the equivalent resistance of the
circuit is
Req = 3 + (10 + 6)k(30 + 18) = 3 + 12 = 15 ⌦.
The source current is 300/15 = 20 A.
The current down through the branch containing the 30 ⌦ and 18 ⌦ resistors is
i18 =
12
(20) = 5 A;
30 + 18
.·. p18 = 18(5)2 = 450 W.
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Problems
3–35
P 3.54
In order that all four decades (1, 10, 100, 1000) that are used to set R3
contribute to the balance of the bridge, the ratio R2 /R1 should be set to 0.001.
P 3.55
Use the figure below to transform the
R1 =
(40)(25)
= 9.756 ⌦;
40 + 25 + 37.5
R2 =
(25)(37.5)
= 9.1463 ⌦;
40 + 25 + 37.5
R3 =
(40)(37.5)
= 14.634 ⌦.
40 + 25 + 37.5
Replace the
to an equivalent Y:
with its equivalent Y in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 ⌦ resistor:
(100 + 9.756)k(125 + 9.1463) + 14.634 = 75 ⌦.
The equivalent resistance seen by the source is thus 5 + 75 = 80 ⌦. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
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3–36
CHAPTER 3. Simple Resistive Circuits
Thus, the power associated with the source is
Ps =
P 3.56
(40)(0.5) =
20 W.
Use the figure below to transform the Y to an equivalent
Ra =
7500
(25)(100) + (25)(40) + (40)(100)
=
= 300 ⌦;
25
25
Rb =
(25)(100) + (25)(40) + (40)(100)
7500
=
= 187.5 ⌦;
40
40
Rc =
(25)(100) + (25)(40) + (40)(100)
7500
=
= 75 ⌦.
100
100
Replace the Y with its equivalent
:
in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 ⌦ resistor:
300k[(125k187.5) + (37.5k75)] = 75 ⌦.
The equivalent resistance seen by the source is thus 5 + 75 = 80 ⌦. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
Thus, the power associated with the source is
Ps =
(40)(0.5) =
20 W.
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Problems
P 3.57
Use the figure below to transform the Y to an equivalent
:
Ra =
8750
(25)(125) + (25)(37.5) + (37.5)(125)
=
= 233.33 ⌦;
37.5
37.5
Rb =
8750
(25)(125) + (25)(37.5) + (37.5)(125)
=
= 350 ⌦;
25
25
Rc =
8750
(25)(125) + (25)(37.5) + (37.5)(125)
=
= 70 ⌦.
125
125
Replace the Y with its equivalent
3–37
in the circuit to get the figure below:
Find the equivalent resistance to the right of the 5 ⌦ resistor:
350k[(100k233.33) + (40k70)] = 75 ⌦.
The equivalent resistance seen by the source is thus 5 + 75 = 80 ⌦. Use Ohm’s
law to find the current provided by the source:
is =
40
= 0.5 A.
80
Thus, the power associated with the source is
Ps =
(40)(0.5) =
20 W.
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3–38
P 3.58
CHAPTER 3. Simple Resistive Circuits
Begin by transforming the -connected resistors (10 ⌦, 40 ⌦, 50 ⌦) to
Y-connected resistors. Both the Y-connected and -connected resistors are
shown below to assist in using Eqs. 3.15 – 3.17:
Now use Eqs. 3.15 – 3.17 to calculate the values of the Y-connected resistors:
R1 =
(40)(10)
= 4 ⌦;
10 + 40 + 50
R2 =
(10)(50)
= 5 ⌦;
10 + 40 + 50
R3 =
(40)(50)
= 20 ⌦.
10 + 40 + 50
The transformed circuit is shown below:
The equivalent resistance seen by the 24 V source can be calculated by making
series and parallel combinations of the resistors to the right of the 24 V source:
Req = (15 + 5)k(1 + 4) + 20 = 20k5 + 20 = 4 + 20 = 24 ⌦.
Therefore, the current i in the 24 V source is given by
i=
24 V
= 1 A.
24 ⌦
Use current division to calculate the currents i1 and i2 . Note that the current
i1 flows in the branch containing the 15 ⌦ and 5 ⌦ series connected resistors,
while the current i2 flows in the parallel branch that contains the series
connection of the 1 ⌦ and 4 ⌦ resistors:
i1 =
4
4
(i) = (1 A) = 0.2 A,
15 + 5
20
and
i2 = 1 A
0.2 A = 0.8 A.
Now use KVL and Ohm’s law to calculate v1 . Note that v1 is the sum of the
voltage drop across the 4 ⌦ resistor, 4i2 , and the voltage drop across the 20 ⌦
resistor, 20i:
v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V.
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Problems
3–39
Finally, use KVL and Ohm’s law to calculate v2 . Note that v2 is the sum of
the voltage drop across the 5 ⌦ resistor, 5i1 , and the voltage drop across the
20 ⌦ resistor, 20i:
v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 V.
P 3.59
[a] After the 30 ⌦—60 ⌦—10 ⌦ delta is replaced by its equivalent wye, the
circuit reduces to
Use current division to calculate i1 :
i1 =
40
22 + 18
(5 A) = (5 A) = 4 A.
22 + 18 + 4 + 6
50
[b] Return to the original circuit and write a KVL equation around the upper
left loop:
(22 ⌦)i22⌦ + v
so
(4 ⌦)(i1 ) = 0;
v = (4 ⌦)(4 A)
(22 ⌦)(5 A
4 A) =
6 V.
[c] Write a KCL equation at the lower center node of the original circuit:
6
v
=4+
= 3.9 A.
60
60
[d] Write a KVL equation around the bottom loop of the original circuit:
i2 = i1 +
v5A + (4 ⌦)(4 A) + (10 ⌦)(3.9 A) + (1 ⌦)(5 A) = 0.
So,
v5A = (4)(4) + (10)(3.9) + (1)(5) = 60 V.
Thus,
P 3.60
p5A = (5 A)(60 V) = 300 W.
8 + 12 = 20 ⌦;
20k60 = 15 ⌦;
15 + 20 = 35 ⌦;
35k140 = 28 ⌦;
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3–40
CHAPTER 3. Simple Resistive Circuits
28 + 22 = 50 ⌦;
50k75 = 30 ⌦;
30 + 10 = 40 ⌦;
ig = 240/40 = 6 A;
io = (6)(50)/125 = 2.4 A;
i140⌦ = (6
2.4)(35)/175 = 0.72 A;
p140⌦ = (0.72)2 (140) = 72.576 W.
P 3.61
[a] The three Y-connected resistors, whose values are 30 ⌦, 60 ⌦, and 18 ⌦,
have been replaced with three -connected resistors in the figure below:
We calculate the values of the
-connected resistors using Eqs. 3.18–3.20:
Rx =
(30)(60) + (30)(18) + (18)(60)
= 190 ⌦;
18
Ry =
(30)(60) + (30)(18) + (18)(60)
= 114 ⌦;
18
(30)(60) + (30)(18) + (18)(60)
= 57 ⌦.
18
Now calculate the equivalent resistance Rab by making series and parallel
combinations of the resistors:
Rz =
Rab = 20 + [(10k190 + 30k114)k57] + 9 = 50 ⌦.
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Problems
3–41
[b] The three -connected resistors, whose values are 30 ⌦, 60 ⌦, and 18 ⌦,
have been replaced with three Y-connected resistors in the figure below:
We calculate the values of the Y-connected resistors using Eqs. 3.15–3.17:
Ra =
(60)(30)
= 16.67 ⌦;
18 + 30 + 60
Rb =
(60)(18)
= 10 ⌦;
18 + 30 + 60
Rc =
(18)(30)
= 5 ⌦.
18 + 30 + 60
Make series and parallel combinations of the resistors to find the
equivalent resistance Rab :
20 + [(10 + 16.67)k(30 + 10)] + 5 + 9 = 50 ⌦.
[c] Convert the delta connection R1 —R2 —R3 to its equivalent wye.
Convert the wye connection R1 —R3 —R4 to its equivalent delta.
P 3.62
[a] After the 20 ⌦—80 ⌦—40 ⌦ wye is replaced by its equivalent delta, the
circuit reduces to
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3–42
CHAPTER 3. Simple Resistive Circuits
Now the circuit can be reduced to
Req = 44 + 280k92.5 = 113.53 ⌦;
ig = 5/113.53 = 44.04 mA;
i = (280/372.5)(44) = 33.11 mA;
v52.5⌦ = (52.5)(33.11 m) = 1.74 V;
io = 1.74/210 = 8.28 mA.
[b] v40⌦ = (40)(33.11 m) = 1.32 V;
i1 = 1.32/56 = 23.65 mA.
[c] Now that io and i1 are known return to the original circuit
i80⌦ = 44.04 m
23.65 m = 20.39 mA;
i20⌦ = 23.65 m
8.28 m = 15.37 mA;
i2 = i80⌦ + i20⌦ = 35.76 mA.
[d] pdel = (5)(44.04 m) = 220.2 mW.
P 3.63
[a] Convert the upper delta to a wye.
R1 =
(50)(50)
= 12.5 ⌦;
200
R2 =
(50)(100)
= 25 ⌦;
200
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Problems
R3 =
3–43
(100)(50)
= 25 ⌦.
200
Convert the lower delta to a wye.
R4 =
(60)(80)
= 24 ⌦;
200
R5 =
(60)(60)
= 18 ⌦;
200
R6 =
(80)(60)
= 24 ⌦.
200
Now redraw the circuit using the wye equivalents.
Rab = 1.5 + 12.5 +
(120)(80)
+ 18 = 14 + 48 + 18 = 80 ⌦.
200
[b] When vab = 400 V,
400
ig =
= 5 A;
80
48
i31 = (5) = 3 A;
80
p31⌦ = (31)(3)2 = 279 W.
P 3.64
P 3.65
1
R1
=
Ra
R 1 R 2 + R2 R 3 + R3 R 1
1/G1
=
(1/G1 )(1/G2 ) + (1/G2 )(1/G3 ) + (1/G3 )(1/G1 )
G 2 G3
(1/G1 )(G1 G2 G3 )
=
.
=
G 1 + G2 + G3
G 1 + G2 + G3
Similar manipulations generate the expressions for Gb and Gc .
Ga
=
Subtracting Eq. 3.13 from Eq. 3.14 gives
R1
R2 = (Rc Rb
Rc Ra )/(Ra + Rb + Rc ).
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3–44
CHAPTER 3. Simple Resistive Circuits
Adding this expression to Eq. 3.12 and solving for R1 gives
R1 = Rc Rb /(Ra + Rb + Rc ).
To find R2 , subtract Eq. 3.14 from Eq. 3.12 and add this result to Eq. 3.13.
To find R3 , subtract Eq. 3.12 from Eq. 3.13 and add this result to Eq. 3.14.
Using the hint, Eq. 3.14 becomes
R 1 + R3 =
Rb (R1 + R3 )R2
Rb [(R2 /R3 )Rb + (R2 /R1 )Rb ]
.
=
(R2 /R1 )Rb + Rb + (R2 /R3 )Rb
(R1 R2 + R2 R3 + R3 R1 )
Solving for Rb gives Rb = (R1 R2 + R2 R3 + R3 R1 )/R2 . To find Ra : First use
Eqs. 3.15–3.17 to obtain the ratios (R1 /R3 ) = (Rc /Ra ) or Rc = (R1 /R3 )Ra
and (R1 /R2 ) = (Rb /Ra ) or Rb = (R1 /R2 )Ra . Now use these relationships to
eliminate Rb and Rc from Eq. 3.13. To find Rc , use Eqs. 3.15–3.17 to obtain
the ratios Rb = (R3 /R2 )Rc and Ra = (R3 /R1 )Rc . Now use the relationships to
eliminate Rb and Ra from Eq. 3.12.
P 3.66
[a] Rab = 2R1 +
Therefore
Thus
R2 (2R1 + RL )
= RL.
2R1 + R2 + RL
2R1
RL +
R2 (2R1 + RL )
= 0.
2R1 + R2 + RL
RL2 = 4R12 + 4R1 R2 = 4R1 (R1 + R2 ).
When Rab = RL , the current into terminal a of the attenuator will be
vi /RL .
Using current division, the current in the RL branch will be
R2
vi
·
.
RL 2R1 + R2 + RL
Therefore
and
vo =
vi
R2
·
RL
RL 2R1 + R2 + RL
vo
R2
=
.
vi
2R1 + R2 + RL
[b] (300)2 = 4(R1 + R2 )R1 ;
22,500 = R12 + R1 R2 ;
R2
vo
;
= 0.5 =
vi
2R1 + R2 + 300
.·. R1 + 0.5R2 + 150 = R2 ;
0.5R2 = R1 + 150;
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Problems
3–45
R2 = 2R1 + 300;
.·. 22,500 = R12 + R1 (2R1 + 300) = 3R12 + 300R1 ;
.·. R12 + 100R1
7500 = 0.
Solving,
R1 = 50 ⌦;
R2 = 2(50) + 300 = 400 ⌦.
[c] From Appendix H, choose R1 = 47 ⌦ and R2 = 390 ⌦. For these values,
Rab 6= RL , so the equations given in part (a) cannot be used. Instead
Rab = 2R1 + [R2 k(2R1 + RL )] = 2(47) + 390k(2(47) + 300)
= 94 + 390k394 = 290 ⌦;
✓
290
% error =
300
◆
1 100 =
3.33%.
Now calculate the ratio of the output voltage to the input voltage. Begin
by finding the current through the top left R1 resistor, called ia :
vi
ia =
.
Rab
Now use current division to find the current through the RL resistor,
called iL :
R2
ia .
iL =
R2 + 2R1 + RL
Therefore, the output voltage, vo , is equal to RL iL :
R2 R L v i
vo =
.
Rab (R2 + 2R1 + RL )
Thus,
vo
390(300)
R2 R L
=
= 0.5146;
=
vi
Rab (R2 + 2R1 + RL )
290(390 + 2(47) + 300)
% error =
P 3.67
✓
0.5146
0.5
◆
1 100 = 2.92%.
[a] After making the Y-to-
transformation, the circuit reduces to
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3–46
CHAPTER 3. Simple Resistive Circuits
Combining the parallel resistors reduces the circuit to
3RRL
2.25R2 + 3.75RRL
=
3R + RL
3R + RL .
Now note:
0.75R +
Therefore
2.25R2 + 3.75RRL
3R
3R + RL
3R(3R + 5RL )
! =
Rab =
.
2
15R + 9RL
2.25R + 3.75RRL
3R +
3R + RL
!
If R = RL , we have
Therefore
Rab =
3RL (8RL )
= RL .
24RL
Rab = RL .
[b] When R = RL , the circuit reduces to
io =
ii (3RL )
1
1 vi
ii =
=
,
4.5RL
1.5
1.5 RL
Therefore
P 3.68
[a] 3.5(3R
10.5R
vo
= 0.5.
vi
RL ) = 3R + RL ;
1050 = 3R + 300;
7.5R = 1350,
R2 =
1
vo = 0.75RL io = vi .
2
R = 180 ⌦;
2(180)(300)2
= 4500 ⌦.
3(180)2 (300)2
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Problems
3–47
[b ]
vo =
42
vi
=
= 12 V;
3.5
3.5
io =
12
= 40 mA;
300
i1 =
30
42 12
=
= 6.67 mA;
4500
4500
ig =
42
= 140 mA;
300
i2 = 140
6.67 = 133.33 mA;
i3 = 40
6.67 = 33.33 mA;
i4 = 133.33
33.33 = 100 mA;
p4500 top = (6.67 ⇥ 10 3 )2 (4500) = 0.2 W;
p180 left = (133.33 ⇥ 10 3 )2 (180) = 3.2 W;
p180 right = (33.33 ⇥ 10 3 )2 (180) = 0.2 W;
p180 vertical = (100 ⇥ 10 3 )2 (180) = 0.48 W;
p300 load = (40 ⇥ 10 3 )2 (300) = 0.48 W.
The 180 ⌦ resistor carrying i2 .
[c] p180 left = 3.2 W.
[d] Two resistors dissipate minimum power – the 4500 ⌦ resistor and the 180
⌦ resistor carrying i3 .
[e] They both dissipate 0.2 W.
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3–48
P 3.69
CHAPTER 3. Simple Resistive Circuits
[a ]
va =
vin R4
R o + R4 +
vb =
R3
vin ;
R 2 + R3
vo = va
vb =
R
;
R4 vin
R o + R4 +
R
R3
vin .
R 2 + R3
When the bridge is balanced,
R3
R4
vin =
vin ;
R o + R4
R 2 + R3
.·.
R3
R4
=
.
R o + R4
R 2 + R3
Thus,
vo
=
=
=
⇡
[b]
R4 vin
R o + R4 +
R4 vin
R o + R4
R
1
1
R4 vin
R o + R4 + R R o + R4
R4 vin (
R)
(Ro + R4 + R)(Ro + R4 )
( R)R4 vin
,
since R << R4 .
(Ro + R4 )2

R = 0.03Ro ;
Ro =
R 2 R4
(1000)(5000)
= 10,000 ⌦;
=
R3
500
R = (0.03)(104 ) = 300 ⌦;
.·. vo ⇡
[c] vo
=
=
=
300(5000)(6)
=
(15,000)2
40 mV.
( R)R4 vin
(Ro + R4 + R)(Ro + R4 )
300(5000)(6)
(15,300)(15,000)
39.2157 mV.
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Problems
P 3.70
( R)R4 vin
;
(Ro + R4 )2
[a] approx value =
true value =
.·.
3–49
( R)R4 vin
;
(Ro + R4 + R)(Ro + R4 )
(Ro + R4 + R)
approx value
=
;
true value
(Ro + R4 )
.·. % error =

R o + R4
R o + R4 + R
1 ⇥ 100 =
R
⇥ 100.
R o + R4
Note that in the above expression, we take the ratio of the true value to
the approximate value because both values are negative.
But Ro =
R 2 R4
;
R3
.·. % error =
[b] % error =
P 3.71
R3 R
.
R4 (R2 + R3 )
(500)(300)
⇥ 100 =
(5000)(1500)
2%.
R(R3 )(100)
= 0.5;
(R2 + R3 )R4
R(500)(100)
= 0.5;
(1500)(5000)
.·.
R = 75 ⌦;
% change =
P 3.72
75
⇥ 100 = 0.75%.
10,000
[a] Using the equation for voltage division,
Vy =
Ry
Ry + (1
)Ry
VS =
Ry
VS = VS .
Ry
[b] Since represents the touch point with respect to the bottom of the
screen, (1
) represents the location of the touch point with respect to
the top of the screen. Therefore, the y-coordinate of the pixel
corresponding to the touch point is
y = (1
)py .
Remember that the value of y is capped at (py
1).
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3–50
P 3.73
CHAPTER 3. Simple Resistive Circuits
[a] Use the equations developed in the Practical Perspective and in Problem
3.72:
Vx
1
Vx = ↵VS
so
↵=
= = 0.2;
VS
5
Vy = V S
so
=
Vy
3.75
= 0.75.
=
VS
5
[b] Use the equations developed in the Practical Perspective and in Problem
3.72:
x = (1
↵)px = (1
0.2)(480) = 384;
y = (1
)py = (1
0.75)(800) = 200.
Therefore, the touch occurred in the upper right corner of the screen.
P 3.74
Use the equations developed in the Practical Perspective and in Problem 3.72:
x = (1
↵)px
so
↵=1
x
=1
px
480
= 0.25;
640
y
=1
py
192
= 0.8125;
1024
Vx = ↵VS = (0.25)(8) = 2 V;
y = (1
)py
so
=1
Vy = VS = (0.8125)(8) = 6.5 V.
P 3.75
From the results of Problem 3.74, the voltages corresponding to the touch
point (480, 192) are
Vx1 = 2 V;
Vy1 = 6.5 V.
Now calculate the voltages corresponding to the touch point (240, 384):
x = (1
↵)px
so
↵=1
x
=1
px
240
= 0.625;
640
y
=1
py
384
= 0.625;
1024
Vx2 = ↵VS = (0.625)(8) = 5 V;
y = (1
)py
so
=1
Vy2 = VS = (0.625)(8) = 5 V.
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Problems
3–51
When the screen is touched at two points simultaneously, only the smaller of
the two voltages in the x direction is sensed. The same is true in the y
direction. Therefore, the voltages actually sensed are
Vx = 2 V;
Vy = 5 V.
These two voltages identify the touch point as (480, 384), which does not
correspond to either of the points actually touched! Therefore, the resistive
touch screen is appropriate only for single point touches.
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Techniques of Circuit Analysis
4
Assessment Problems
AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
The two node voltage equations are
v1
v 1 v2
v1
+
+
= 0;
15 +
60 15
5
v2 v2 v1
+
= 0.
5+
2
5
Place
in standard
✓ these equations
◆
✓
◆ form:
1
1
1
1
+
+
+ v2
=
v1
60 15 5
5
✓
◆
✓
◆
1
1 1
+
+ v2
=
v1
5
2 5
Solving, v1 = 60 V and v2 = 10 V;
Therefore, i1 = (v1 v2 )/5 = 10 A.
[b] p15A =
(15 A)v1 =
(15 A)(60 V) =
[c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W=
15;
5.
900 W = 900 W(delivered).
50 W(delivered).
4–1
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4–2
CHAPTER 4. Techniques of Circuit Analysis
AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:
The two node voltage equations are:
v1 v1 v2
4.5 +
+
= 0;
1
6+2
v2 v1 v2 30
v2
+
+
= 0.
12
6+2
4
Place
form:
✓ these◆equations✓in standard
◆
1
1
v1 1 +
+ v2
= 4.5;
8
8
◆
✓
◆
✓
1
1
1 1
+ +
+ v2
= 7.5.
v1
8
12 8 4
Solving, v1 = 6 V
v2 = 18 V.
To find the voltage v, first find the current i through the series-connected 6 ⌦
and 2 ⌦ resistors:
i=
6 18
v1 v2
=
=
6+2
8
1.5 A.
Using a KVL equation, calculate v:
v = 2i + v2 = 2( 1.5) + 18 = 15 V.
AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as
shown:
The node voltage equations are:
v1 50 v1 v1 v2
+
+
3i1 =
6
8
2
v2 v2 v1
+
+ 3i1 =
5+
4
2
0;
0.
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Problems
4–3
The dependent source requires the following constraint equation:
50 v1
i1 =
.
6
Place these equations in standard form:
✓
◆
✓
◆
1
50
1 1 1
v1
+ +
;
+ v2
+ i1 ( 3) =
6 8 2
2
6
✓
◆
✓
◆
1
1 1
+
v1
+ v2
+ i1 (3)
= 5;
2
4 2
✓ ◆
50
1
.
+ v2 (0)
+ i1 (1)
=
v1
6
6
v2 = 16 V;
i1 = 3 A.
Solving, v1 = 32 V;
Using these values to calculate the power associated with each source:
p50V =
50i1
=
150 W;
p5A =
5(v2 )
=
80 W;
p3i1 = 3i1 (v2
v1 )
=
144 W.
[b] All three sources are delivering power to the circuit because the power
computed in (a) for each of the sources is negative.
AP 4.4 Redraw the circuit and label the reference node and the node at which the
node voltage equation will be written:
The node voltage equation is
vo 10 vo + 20i
vo
+
+
40
10
20
= 0.
The constraint equation required by the dependent source is
i = i10 ⌦ + i30 ⌦ =
10
vo
10
+
10 + 20i
.
30
Place these equations in standard form:
✓
◆
1
1
1
vo
+
+
+ i (1)
= 1;
40 10 20
✓ ◆
✓
◆
20
10
1
+ i 1
= 1+ .
vo
10
30
30
Solving,
i =
3.2 A
and
vo = 24 V.
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4–4
CHAPTER 4. Techniques of Circuit Analysis
AP 4.5 Redraw the circuit identifying the three node voltages and the reference node:
Note that the dependent voltage source and the node voltages v and v2 form a
supernode. The v1 node voltage equation is
v1 v
v1
+
7.5
2.5
4.8 = 0.
The supernode equation is
v
v
v2
v2 12
v1
+
+
+
= 0.
2.5
10 2.5
1
The constraint equation due to the dependent source is
ix =
v1
.
7.5
The constraint equation due to the supernode is
v + ix = v2 .
Place this set of equations in standard form:
✓
◆
✓
◆
1
1
1
v1
+
+ v
+ v2 (0)
+ ix (0) = 4.8;
7.5 2.5
2.5
✓
◆
✓
◆
✓
◆
1
1
1
1
+
+1
+ v
+ v2
+ ix (0) = 12;
v1
2.5
2.5 10
2.5
✓
◆
1
+ v(0)
+ v2 (0)
+ ix (1) = 0;
v1
7.5
v1 (0)
+ v(1)
+ v2 ( 1)
+ ix (1)
=
0.
Solving this set of equations gives v1 = 15 V, v2 = 10 V, ix = 2 A, and
v = 8 V.
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Problems
4–5
AP 4.6 Redraw the circuit identifying the reference node and the two unknown node
voltages. Note that the right-most node voltage is the sum of the 60 V source
and the dependent source voltage.
The node voltage equation at v1 is
v1
60
2
+
v1
v1
+
24
(60 + 6i )
= 0.
3
The constraint equation due to the dependent source is
i =
60 + 6i
3
v1
.
Place these two equations in standard form:
✓
◆
1
1
1
+
+
+ i ( 2)
= 30 + 20;
v1
2 24 3
✓ ◆
1
+ i (1 2) = 20.
v1
3
Solving,
i =
4A
and
v1 = 48 V.
AP 4.7 [a] Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
80 + 5(i1
i2 ) + 26(i1
i3 )
= 0;
30i2 + 90(i2
i3 ) + 5(i2
i1 )
= 0;
8i3 + 26(i3
i1 ) + 90(i3
i2 )
= 0.
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4–6
CHAPTER 4. Techniques of Circuit Analysis
Place these equations in standard form:
31i1
5i2
26i3
=
80;
5i1 + 125i2
90i3
=
0;
26i1 90i2 + 124i3
Solving,
=
0.
i1 = 5 A;
i2 = 2 A;
p80V =
(80)i1 =
i3 = 2.5 A.
(80)(5) =
400 W.
Therefore the 80 V source is delivering 400 W to the circuit.
[b] p8⌦ = (8)i23 = 8(2.5)2 = 50 W, so the 8 ⌦ resistor dissipates 50 W.
AP 4.8 [a] b = 8,
n = 6,
b
n + 1 = 3.
[b] Redraw the circuit identifying the three mesh currents:
The three mesh-current equations are
25 + 2(i1
i2 ) + 5(i1
( 3v ) + 14i2 + 3(i2
1i3
10 + 5(i3
i3 ) + 10
=
0;
i3 ) + 2(i2
i1 )
= 0;
i1 ) + 3(i3
i2 )
= 0.
The dependent source constraint equation is
v = 3(i3
i2 ).
Place these four equations in standard form:
7i1
2i2
5i3 + 0v
=
15;
2i1 + 19i2
3i3 + 3v
=
0;
3i2 + 9i3 + 0v
=
10;
=
0.
5i1
0i1 + 3i2
3i3 + 1v
Solving
i1 = 4 A;
i2 =
1 A;
i3 = 3 A;
v = 12 V;
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Problems
pds =
( 3v )i2 = 3(12)( 1) =
4–7
36 W.
Thus, the dependent source is delivering 36 W, or absorbing
36 W.
AP 4.9 Redraw the circuit identifying the three mesh currents:
The mesh current equations are:
25 + 6(ia
ib ) + 8(ia
ic )
= 0;
2ib + 8(ib
ic ) + 6(ib
ia )
= 0;
5i + 8(ic
ia ) + 8(ic
ib )
= 0.
The dependent source constraint equation is i = ia . We can substitute this
simple expression for i into the third mesh equation and place the equations
in standard form:
14ia
6ib
8ic
=
25;
6ia + 16ib
8ic
=
0;
8ib + 16ic
=
0.
3ia
Solving, ia = 4 A;
Thus, vo = 8(ia
ib = 2.5 A;
ic ) = 8(4
ic = 2 A.
2) = 16 V.
AP 4.10 Redraw the circuit identifying the mesh currents:
Since there is a current source on the perimeter of the i3 mesh, we know that
i3 = 16 A. The remaining two mesh equations are
30 + 3i1 + 2(i1
i2 ) + 6i1
8i2 + 5(i2 + 16) + 4i2 + 2(i2
i1 )
=
0;
= 0.
Place these equations in standard form:
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4–8
CHAPTER 4. Techniques of Circuit Analysis
11i1
2i2
=
2i1 + 19i2
=
Solving:
30;
80.
i1 = 2 A,
i2 =
4 A,
The current in the 2 ⌦ resistor is
i3 =
16 A.
i1
i2 = 6 A .·.
p2 ⌦ = (6)2 (2) = 72 W.
Thus, the 2 ⌦ resistors dissipates 72 W.
AP 4.11 Redraw the circuit and identify the mesh currents:
There are current sources on the perimeters of both the ib mesh and the ic
mesh, so we know that
ib =
10 A;
ic =
2v
.
5
The remaining mesh current equation is
75 + 2(ia + 10) + 5(ia
0.4v ) = 0.
The dependent source requires the following constraint equation:
v = 5(ia
ic ) = 5(ia
0.4v ).
Place the mesh current equation and the dependent source equation is
standard form:
7ia
2v
=
55;
5ia
3v
=
0.
Solving:
ia = 15 A;
ib =
10 A;
ic = 10 A;
v = 25 V.
Thus, ia = 15 A.
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Problems
4–9
AP 4.12 Redraw the circuit and identify the mesh currents:
The 2 A current source is shared by the meshes ia and ib . Thus we combine
these meshes to form a supermesh and write the following equation:
10 + 2ib + 2(ib
ic ) + 2(ia
ic ) = 0.
The other mesh current equation is
6 + 1ic + 2(ic
ia ) + 2(ic
ib ) = 0.
The supermesh constraint equation is
ia
ib = 2.
Place these three equations in standard form:
2ia + 4ib
4ic
=
10;
2ia
2ib + 5ic
=
6;
ia
ib + 0ic
=
2.
Solving,
ia = 7 A;
Thus,
p1 ⌦ = i2c (1) = (6)2 (1) = 36 W.
ib = 5 A;
ic = 6 A.
AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1 :
The node voltage equation is
v1
20
15
2+
25
v1
10
= 0.
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4–10
CHAPTER 4. Techniques of Circuit Analysis
Rearranging and solving,
v1
✓
◆
20 25
1
1
+
+
=2+
15 10
15 10
p2A =
35(2) =
.·. v1 = 35 V;
70 W.
Thus the 2 A current source delivers 70 W.
AP 4.14 Redraw the circuit and identify the mesh currents:
There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. The
other two mesh current equations are
128 + 4(i1
4) + 6(i1
30ix + 5i2 + 6(i2
i2 ) + 2i1
i1 ) + 3(i2
=
0;
4) =
0.
The constraint equation due to the dependent source is
ix = i1
i3 = i 1
4.
Substitute the constraint equation into the second mesh equation and place
the resulting two mesh equations in standard form:
12i1
6i2
=
144;
24i1 + 14i2
=
132.
Solving,
i1 = 9 A;
i2 =
.·. v4A = 3(i3
p4A =
6 A;
i2 )
v4A (4) =
i3 = 4 A;
ix = 9
4 = 5 A;
4ix = 10 V;
(10)(4) =
40 W.
Thus, the 4 A current source delivers 40 W.
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Problems
4–11
AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled:
Transform the 120 V source in series with the 20 ⌦ resistor into a 6 A
source in parallel with the 20 ⌦ resistor. Also transform the 60 V source
in series with the 5 ⌦ resistor into a 12 A source in parallel with the 5 ⌦
resistor. The result is the following circuit:
Combine the three current sources into a single current source, using
KCL, and combine the 20 ⌦, 5 ⌦, and 6 ⌦ resistors in parallel. The
resulting circuit is shown on the left. To simplify the circuit further,
transform the resulting 30 A source in parallel with the 2.4 ⌦ resistor into
a 72 V source in series with the 2.4 ⌦ resistor. Combine the 2.4 ⌦ resistor
in series with the 1.6 ⌦ resisor to get a very simple circuit that still
maintains the voltage v. The resulting circuit is on the right.
Use voltage division in the circuit on the right to calculate v as follows:
8
(72) = 48 V.
12
[b] Calculate i in the circuit on the right using Ohm’s law:
v=
v
48
=
= 6 A.
8
8
Now use i to calculate va in the circuit on the left:
i=
va = 6(1.6 + 8) = 57.6 V.
Returning back to the original circuit, note that the voltage va is also the
voltage drop across the series combination of the 120 V source and 20 ⌦
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4–12
CHAPTER 4. Techniques of Circuit Analysis
resistor. Use this fact to calculate the current in the 120 V source, ia :
ia =
120 57.6
120 va
=
= 3.12 A;
20
20
p120V =
(120)ia =
(120)(3.12) =
374.40 W.
Thus, the 120 V source delivers 374.4 W.
AP 4.16 To find RTh , replace the 72 V source with a short circuit:
Note that the 5 ⌦ and 20 ⌦ resistors are in parallel, with an equivalent
resistance of 5k20 = 4 ⌦. The equivalent 4 ⌦ resistance is in series with the 8 ⌦
resistor for an equivalent resistance of 4 + 8 = 12 ⌦. Finally, the 12 ⌦
equivalent resistance is in parallel with the 12 ⌦ resistor, so
RTh = 12k12 = 6 ⌦.
Use node voltage analysis to find vTh . Begin by redrawing the circuit and
labeling the node voltages:
The node voltage equations are
v1 72 v1
v1 vTh
+
+
= 0;
5
20
8
vTh v1 vTh 72
+
= 0.
8
12
Place these equations in standard form:
✓
◆
✓
◆
1
72
1
1
1
v1
+
+
;
+ vTh
=
5 20 8
8
5
✓
◆
✓
◆
1
1
1
+
v1
+ vTh
= 6.
8
8 12
Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent
circuit is a 64.8 V source in series with a 6 ⌦ resistor.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
4–13
AP 4.17 We begin by performing a source transformation, turning the parallel
combination of the 15 A source and 8 ⌦ resistor into a series combination of a
120 V source and an 8 ⌦ resistor, as shown in the figure on the left. Next,
combine the 2 ⌦, 8 ⌦ and 10 ⌦ resistors in series to give an equivalent 20 ⌦
resistance. Then transform the series combination of the 120 V source and the
20 ⌦ equivalent resistance into a parallel combination of a 6 A source and a
20 ⌦ resistor, as shown in the figure on the right.
Finally, combine the 20 ⌦ and 12 ⌦ parallel resistors to give
RN = 20k12 = 7.5 ⌦. Thus, the Norton equivalent circuit is the parallel
combination of a 6 A source and a 7.5 ⌦ resistor.
AP 4.18 Find the Thévenin equivalent with respect to A, B using source
transformations. To begin, convert the series combination of the 36 V source
and 12 k⌦ resistor into a parallel combination of a 3 mA source and 12 k⌦
resistor. The resulting circuit is shown below:
Now combine the two parallel current sources and the two parallel resistors to
give a 3 + 18 = 15 mA source in parallel with a 12 kk60 k= 10 k⌦ resistor.
Then transform the 15 mA source in parallel with the 10 k⌦ resistor into a
150 V source in series with a 10 k⌦ resistor, and combine this 10 k⌦ resistor
in series with the 15 k⌦ resistor. The Thévenin equivalent is thus a 150 V
source in series with a 25 k⌦ resistor, as seen to the left of the terminals A,B
in the circuit below.
Now attach the voltmeter, modeled as a 100 k⌦ resistor, to the Thévenin
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4–14
CHAPTER 4. Techniques of Circuit Analysis
equivalent and use voltage division to calculate the meter reading vAB :
vAB =
100,000
(150) = 120 V.
125,000
AP 4.19 Begin by calculating the open circuit voltage, which is also vTh , from the
circuit below:
Summing the currents away from the node labeled vTh We have
vTh 24
vTh
+ 4 + 3ix +
= 0.
8
2
Also, using Ohm’s law for the 8 ⌦ resistor,
ix =
vTh
.
8
Substituting the second equation into the first and solving for vTh yields
vTh = 8 V.
Now calculate RTh . To do this, we use the test source method. Replace the
voltage source with a short circuit, the current source with an open circuit,
and apply the test voltage vT , as shown in the circuit below:
Write a KCL equation at the middle node:
iT = ix + 3ix + vT /2 = 4ix + vT /2.
Use Ohm’s law to determine ix as a function of vT :
ix = vT /8.
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Problems
4–15
Substitute the second equation into the first equation:
iT = 4(vT /8) + vT /2 = vT .
Thus,
RTh = vT /iT = 1 ⌦.
The Thévenin equivalent is an 8 V source in series with a 1 ⌦ resistor.
AP 4.20 Begin by calculating the open circuit voltage, which is also vTh , using the
node voltage method in the circuit below:
The node voltage equations are
v (vTh + 160i )
v
+
4 = 0;
60
20
v
vTh vTh vTh + 160i
+
+
= 0.
40
80
20
The dependent source constraint equation is
i =
vTh
.
40
Substitute the constraint equation into the node voltage equations and put the
two equations in standard form:
✓
◆
✓
◆
1
5
1
v
+
+ vTh
= 4;
60 20
20
✓
◆
✓
◆
1
5
1
1
+
+
+ vTh
= 0.
v
20
40 80 20
Solving, v = 172.5 V and vTh = 30 V.
Now use the test source method to calculate the test current and thus RTh .
Replace the current source with a short circuit and apply the test source to
get the following circuit:
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4–16
CHAPTER 4. Techniques of Circuit Analysis
Write a KCL equation at the rightmost node:
iT =
vT vT vT + 160i
+
+
.
80 40
80
The dependent source constraint equation is
i =
vT
.
40
Substitute the constraint equation into the KCL equation and simplify the
right-hand side:
iT =
vT
.
10
Therefore,
RTh =
vT
= 10 ⌦.
iT
Thus, the Thévenin equivalent is a 30 V source in series with a 10 ⌦ resistor.
AP 4.21 First find the Thévenin equivalent circuit. To find vTh , create an open circuit
between nodes a and b and use the node voltage method with the circuit
below:
The node voltage equations are:
vTh (100 + v ) vTh v1
+
= 0;
4
4
v1 100 v1 20 v1 vTh
+
+
= 0.
4
4
4
The dependent source constraint equation is
v = v1
20.
Place these three equations in standard form:
✓
◆
✓
◆
✓
◆
1
1
1 1
+
+ v1
+ v
= 25;
vTh
4 4
4
4
✓
◆
✓
◆
1
1 1 1
+ +
+ v1
+ v (0)
= 30;
vTh
4
4 4 4
vTh (0)
+
v1 (1)
+ v ( 1)
= 20.
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Problems
4–17
Solving, vTh = 120 V, v1 = 80 V, and v = 60 V.
Now create a short circuit between nodes a and b and use the mesh current
method with the circuit below:
The mesh current equations are
100 + 4(i1
i2 ) + v + 20
=
0;
v + 4i2 + 4(i2
isc ) + 4(i2
i1 )
= 0;
20
v + 4(isc
i2 )
= 0.
The dependent source constraint equation is
v = 4(i1
isc ).
Place these four equations in standard form:
4i1
4i2 + 0isc + v
4i1 + 12i2
0i1
=
80;
4isc
v
=
0;
4i2 + 4isc
v
=
20;
v
=
0.
4i1 + 0i2
4isc
Solving, i1 = 45 A, i2 = 30 A, isc = 40 A, and v = 20 V. Thus,
RTh =
vTh
120
= 3 ⌦.
=
isc
40
[a] For maximum power transfer, R = RTh = 3 ⌦.
[b] The Thévenin voltage, vTh = 120 V, splits equally between the Thévenin
resistance and the load resistance, so
120
vload =
= 60 V.
2
Therefore,
pmax =
2
vload
602
= 1200 W.
=
Rload
3
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4–18
CHAPTER 4. Techniques of Circuit Analysis
AP 4.22 Sustituting the value R = 3 ⌦ into the circuit and identifying three mesh
currents we have the circuit below:
The mesh current equations are:
100 + 4(i1
v + 4i2 + 4(i2
20
i2 ) + v + 20
i3 ) + 4(i2
v + 4(i3
i1 )
i2 ) + 3i3
= 0;
= 0;
=
0.
The dependent source constraint equation is
v = 4(i1
i3 .)
Place these four equations in standard form:
4i1
4i2 + 0i3 + v
4i1 + 12i2
0i1
=
80;
4i3
v
=
0;
4i2 + 7i3
v
=
20;
v
=
0.
4i1 + 0i2
4i3
Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and v = 40 V.
[a] p100V = (100)i1 = (100)(30) =
delivering 3000 W.
3000 W. Thus, the 100 V source is
[b] pdepsource = v i2 =
delivering 800 W.
800 W. Thus, the dependent source is
(40)(20) =
[c] From Assessment Problem 4.21(b), the power delivered to the load resistor
is 1200 W, so the load power is (1200/3800)100 = 31.58% of the
combined power generated by the 100 V source and the dependent source.
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Problems
4–19
Problems
P 4.1
[a] 11 branches, 8 branches with resistors, 2 branches with independent
sources, 1 branch with a dependent source.
[b] The current is unknown in every branch except the one containing the 8 A
current source, so the current is unknown in 10 branches.
[c] 9 essential branches: R4 R5 forms an essential branch as does R8 10 V.
The remaining seven branches are essential branches that contain a single
element.
[d] The current is known only in the essential branch containing the current
source, and is unknown in the remaining 8 essential branches.
[e] From the figure there are 6 nodes – three identified by rectangular boxes,
two identified with single black dots, and one identified by a triangle.
[f ] There are 4 essential nodes, three identified with rectangular boxes and
one identified with a triangle.
[g] A mesh is like a window pane, and as can be seen from the figure there are
6 window panes or meshes.
P 4.2 [a] From Problem 4.1(d) there are 8 essential branches where the current is
unknown, so we need 8 simultaneous equations to describe the circuit.
[b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at
(4 1) = 3 of these essential nodes. There would also be a dependent
source constraint equation.
[c] The remaining 4 equations needed to describe the circuit will be derived
from KVL equations.
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4–20
CHAPTER 4. Techniques of Circuit Analysis
[d] We must avoid using the topmost mesh and the leftmost mesh. Each of
these meshes contains a current source, and we have no way of
determining the voltage drop across a current source.
P 4.3 [a] There are eight circuit components, seven resistors and the voltage source.
Therefore there are eight unknown currents. However, the voltage source
and the R1 resistor are in series, so have the same current. The R4 and
R6 resistors are also in series, so have the same current. The R5 and R7
resistors are in series, so have the same current. Therefore, we only need
5 equations to find the 5 distinct currents in this circuit.
[b]
There are three essential nodes in this circuit, identified by the boxes. At
two of these nodes you can write KCL equations that will be independent
of one another. A KCL equation at the third node would be dependent
on the first two. Therefore there are two independent KCL equations.
[c] Sum the currents at any two of the three essential nodes a, b, and c. Using
nodes a and c we get
i1 + i2 + i4 = 0;
i1
i3 + i5 = 0.
[d] There are three meshes in this circuit: one on the left with the
components vs , R1 , R2 and R3 ; one on the top right with components R2 ,
R4 , and R6 ; and one on the bottom right with components R3 , R5 , and
R7 . We can write KVL equations for all three meshes, giving a total of
three independent KVL equations.
[e]
vs + R1 i1 + R2 i2 + R3 i3 = 0;
R 4 i4 + R 6 i4
R2 i2 = 0;
R3 i3 + R5 i5 + R7 i5 = 0.
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Problems
4–21
P 4.4
[a] At node a:
i1 + i2 + i4 = 0.
At node b:
i2 + i3
At node c:
i1
i4
i5 = 0.
i3 + i5 = 0.
[b] There are many possible solutions. For example, adding the equations at
nodes a and c gives the equation at node b:
( i1 + i2 + i4 ) + (i1
i3 + i5 ) = 0
so
i2
i3 + i4 + i5 = 0.
This is the equation at node b with both sides multiplied by
1.
P 4.5 [a] At node g,
i4 + i6
i5
I = 0.
[b] From Example 4.2, the equations at nodes b, c, and e are
At node b:
i1 + i2 + i6
At node c:
i1
i3
i5 = 0;
At node e:
i3 + i4
i2 = 0.
I = 0;
Add the equations at nodes b and c to give
i2 + i6
I
i3
i5 = 0.
From the equation at node e,
i2
i3 = i 4 .
Combine the two equations above and rearrange to give the equation at
node g.
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4–22
CHAPTER 4. Techniques of Circuit Analysis
P 4.6
Note that we have chosen the lower node as the reference node, and that the
voltage at the upper node with respect to the reference node is vo . Write a
KCL equation (node voltage equation) by summing the currents leaving the
upper node:
vo 25
vo
+
+ 0.04 = 0.
120 + 5 25
Solve by multiplying both sides of the KCL equation by 125 and collecting the
terms involving vo on one side of the equation and the constants on the other
side of the equation:
vo 25 + 5vo + 5 = 0
.·.
6vo = 20
so
vo = 10/2 = 3.33 V.
P 4.7 [a] From the solution to Problem 4.6 we know vo = 3.33 V; therefore
p40mA = (3.33)(0.04) = 133.33 mW.
The power developed by the 40 mA source is 133.33 mW; that is, the
40 mA source is dissipating 133.33 mW.
[b] The current into the positive terminal of the 25 V source in the figure of
Problem 4.6 is
ig = (3.33 25)/125 = 173.33 mA.
The power in the 25 V source is
p25V = (25)( 0.17333) = 4.33 W.
The power developed by the 25 V source is 4.33 W.
=
(0.17333)2 (5) = 150.22 mW;
p120⌦
=
(0.17333)2 (120) = 3.60533 W;
p25⌦
=
(3.33)2 /25 = 444.44 mW.
[c] p5⌦
X
X
pdis = 0.15022 + 3.60533 + 0.44444 + 0.13333 = 4.33 W;
pdev = 4.33 W. (checks!)
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Problems
4–23
P 4.8
[a] The node voltage equation is:
vo 25 vo
+
+ 0.04 = 0.
125
25
Solving,
vo + 25 + 5vo + 5 = 0
.·.
6vo = 20
so
vo = 3.33 V.
[b] Let vx = voltage drop across 40 mA source:
vx = vo (100)(0.04) = 3.33 4 = 0.667 V;
p40mA = ( 0.667)(0.04) = 26.67 mW.
The power developed by the 40 mA source is 26.67 mW.
[c] Let ig = current into positive terminal of 25 V source:
ig = (3.33 25)/125 = 173.33 mA;
p25V = (25)( 0.17333) = 4.33 W.
The power developed by the 25 V source is 4.33 W.
=
(0.17333)2 (5) = 150.22 mW;
p120⌦
=
(0.17333)2 (120) = 3.60533 W;
p25⌦
=
(3.33)2 /25 = 444.44 mW;
p100⌦
=
(0.04)2 (100) = 160 mW.
[d] p5⌦
X
X
pdis = 0.15022 + 3.60533 + 0.44444 + 0.160 = 4.36 W;
pdev = 0.0267 + 4.333 = 4.36 W. (checks!)
[e] vo is independent of any finite resistance connected in series with the 40
mA current source.
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4–24
CHAPTER 4. Techniques of Circuit Analysis
P 4.9
2+
vo 45
vo
+
= 0;
50
1+4
vo = 50 V;
p2A =
(50)(2) =
100 W (delivering).
The 2 A source extracts
to the circuit.
v1
v 1 v2
P 4.10 6 +
+
= 0;
40
8
v2
v1
8
+
v2
v2
+
+ 1 = 0;
80 120
Solving, v1 = 120 V;
CHECK:
v2 = 96 V.
(120)2
= 360 W;
40
p40⌦ =
p8⌦ =
100 W from the circuit, because it delivers 100 W
96)2
(120
8
= 72 W;
p80⌦ =
(96)2
= 115.2 W;
80
p120⌦ =
(96)2
= 76.8 W;
120
p6A =
(6)(120) =
720 W;
p1A = (1)(96) = 96 W;
X
X
pabs = 360 + 72 + 115.2 + 76.8 + 96 = 720 W;
pdev = 720 W (CHECKS).
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Problems
4–25
P 4.11
v1
144
v1
v 1 v2
+
+
=0
4
10
80
v 2 v 1 v2
+
=0
3+
80
5
Solving, v1 = 100 V;
so
so
29v1
v2 = 2880;
v1 + 17v2 = 240.
v2 = 20 V.
P 4.12 [a]
The two node voltage equations are:
v1 v1 44 v1 v2
+
+
= 0;
6
4
1
v2 v2 v1 v2 + 2
+
+
= 0.
3
1
2
Place these equations in standard form:
✓
◆
44
1 1
v1
+ +1
;
+ v2 ( 1)
=
6 4
4
✓
◆
1
2
1
+1+
.
v1 ( 1)
+ v2
=
3
2
2
Solving, v1 = 12 V;
v2 = 6 V.
Now calculate the branch currents from the node voltage values:
44 12
ia =
= 8A;
4
12
= 2A;
ib =
6
12 6
= 6A;
ic =
1
6
= 2A;
id =
3
6+2
= 4A.
ie =
2
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4–26
CHAPTER 4. Techniques of Circuit Analysis
[b] psources = p44V + p2V = (44)ia (2)ie = (44)(8) (2)(4) = 352 8 = 360W.
Thus, the power developed in the circuit is 360 W. Note that the resistors
cannot develop power!
P 4.13 [a]
v1 40 v1 v2
v1
+
+
=0
so 31v1 20v2 + 0v3 = 400;
40
4
2
v 2 v1 v2 v3
+
28 = 0
so
2v1 + 3v2 v3 = 112;
2
4
v3 v3 v2
+
+ 28 = 0
so 0v1 v2 + 3v3 = 112.
2
4
Solving, v1 = 60 V; v2 = 73 V; v3 = 13 V.
[b] ig =
40
60
4
=
pg = (40)( 5) =
5 A;
200 W.
Thus the 40 V source delivers 200 W of power.
P 4.14 [a]
v1
125
1
+
v1
v2
+
v1
v3
6
24
v2 v1 v2 v2 v3
+
+
6
2
12
v3 + 125 v3 v2 v3 v1
+
+
1
12
24
=
0;
=
0;
=
0.
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Problems
4–27
In standard form:
✓
◆
✓
◆
✓
◆
1
1
1
1 1
v1
+ +
+ v2
+ v3
1 6 24
6
24
✓
◆
✓
◆
✓
◆
1
1
1
1 1
+ +
+ v2
+ v3
v1
6
6 2 12
12
✓
◆
✓
◆
✓
◆
1
1
1
1
1
+
+
+ v2
+ v3
v1
24
12
1 12 24
Solving, v1 = 101.24 V;
Thus, i1 =
125
v1
1
v2 = 10.66 V;
= 23.76 A
v2
= 5.33 A
2
v3 + 125
= 18.43 A
i3 =
1
i2 =
[b]
X
i5 =
i6 =
v2
6
v2
v3
12
v1
=
0;
=
v3 =
v1
125;
v3
24
125.
106.57 V.
= 15.10 A;
= 9.77 A;
= 8.66 A.
Pdev = 125i1 + 125i3 = 5273.09 W;
X
P 4.15
i4 =
=
Pdis = i21 (1) + i22 (2) + i23 (1) + i24 (6) + i25 (12) + i26 (24) = 5273.09 W.
v 1 v2
v1 + 40 v1
+
+
+ 5 = 0;
12
25
20

v2
v1
20
5+
v2
v1
+
40
7.5 = 0;
v 3 v2
v3
+
+ 7.5 = 0.
40
40
Solving, v1 =
10 V;
p5A = 5(v1
v2 ) = 5( 10
v2 = 132 V;
132) =
v3 =
84 V;
i40V =
10 + 40
= 2.5 A.
12
710 W (del);
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4–28
CHAPTER 4. Techniques of Circuit Analysis
p7.5A = ( 84
p40V =
132)(7.5) =
(40)(2.5) =
1620 W (del);
100 W (del);
p12⌦ = (2.5)2 (12) = 75 W;
p25⌦ =
p20⌦ =
102
v12
=
= 4 W;
25
25
(v1
v2 )2
1422
=
= 1008.2 W;
20
20
p40⌦ (lower) =
p40⌦ (right) =
X
X
842
(v3 )2
=
= 176.4 W;
40
40
(v2
v3 )2
2162
=
= 1166.4 W;
40
40
pdiss = 75 + 4 + 1008.2 + 176.4 + 1166.4 = 2430 W;
pdev = 710 + 1620 + 100 = 2430 W
(CHECKS).
The total power dissipated in the circuit is 2430 W.
P 4.16 [a]
vo
v1
R
+
vo
v2
R
+
vo
v3
R
+ ··· +
vo
vn
R
.·. nvo = v1 + v2 + v3 + · · · + vn ;
1
1
.·. vo = [v1 + v2 + v3 + · · · + vn ] =
n
n
1
[b] vo = (120 + 60
3
= 0;
Xn
v .
k=1 k
30) = 50 V.
P 4.17
vo
vo
vo 150i
160
+
+
= 0;
10
100
50
i =
vo
;
100
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Problems
Solving, vo = 100 V;
io =
100
i =
4–29
1 A.
(150)( 1)
= 5 A.
50
p150i = 150i io =
750 W.
.·. The dependent voltage source delivers 750 W to the circuit.
P 4.18
[a]
v
v
v
50
vo
+
+
= 0;
500
1000
2000
v
vo
+
= 0.
750 200
vo v
2000
Solving,
v = 30 V;
vo = 10 V.
30 50
50
=
=
500
500
0.04 A;
P50V = 50i50V = 50( 0.04) =
2W
[b] i50V =
Pds =
v
vo
✓
◆
v
=
750
(10)(30/750) =
(2 W supplied);
0.4 W
(0.4 W supplied);
Ptotal = 2 + 0.4 = 2.4 W supplied.
P 4.19
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4–30
CHAPTER 4. Techniques of Circuit Analysis
v1 v2
v1
+
= 0;
1000
1250
[a] 0.02 +
v2
v2
v2 2500i
v 2 v1
+
+
+
1250
4000 2000
200
i =
= 0;
v 2 v1
.
1250
Solving,
v1 = 60 V;
v2 = 160 V;
i = 80 mA.
P20mA = (0.02)v1 = (0.02)(60) = 1.2 W (absorbed);
v2
ids =
2500i
200
=
160
(2500)(0.08)
=
200
Pds = (2500i )ids = 2500(0.08)( 0.2) =
0.2 A;
40 W (40 W developed);
Pdeveloped = 40 W.
[b] P1k =
602
v12
=
= 3.6 W;
1000
1000
P1250 = 1250i2 = 1250(0.08)2 = 8 W;
1602
v22
P4k =
=
= 6.4 W;
4000
4000
P2k =
1602
v22
=
= 12.8 W;
2000
2000
P200 = 200i2ds = 200( 0.2)2 = 8 W;
Pabsorbed = P20mA + P1k + P1250 + P4k + P2k + P200
= 1.2 + 3.6 + 8 + 6.4 + 12.8 + 8 = 40 W (check).
P 4.20 [a]
The node voltage equation is:
0.45 +
vo
vo
+
100
vo 45
6.25i
+
= 0.
5
25
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Problems
4–31
The dependent source constraint equation is:
i =
45
vo
.
25
Place these equations in standard form:
✓
◆
✓
◆
1
1
6.25
45
1
vo
+ +
+ 0.45;
+ i
=
100 5 25
5
25
✓ ◆
45
1
.
vo
+ i (1)
=
25
25
Solving,
vo = 15V;
i = 1.2 A.
vo 6.25i
15 7.5
=
= 1.5 A;
[b] ids =
5
5
pds = [6.25(1.2)](1.5) = 11.25 W.
Thus, the dependent source absorbs 11.25 W.
[c] p450mA = (0.45)(15) = 6.75 W;
pX
(1.2)(45) = 54 W;
45V =
pdev = 6.75 + 54 = 60.75 W ;
Thus the independent sources develop 60.75 W.
Also,
X
pdis = pds + p100⌦ + p5⌦ + p25⌦
= 11.25 + (15)2 /100 + (1.5)2 (5) + (1.2)2 (25)
= 11.25 + 2.25 + 11.25 + 36 = 60.75 W (checks!).
P 4.21 [a]
io =
v2
;
40
v 1 v2
v1
+
=0
20
5
v2
v 2 v3
v2 v1
+
+
5
40
10
v3 v2 v3 11.5io v3 96
+
+
=0
10
5
4
5io +
Solving, v1 = 156 V;
v2 = 120 V;
so
so
so
10v1
13v2 + 0v3 = 0.
8v1 + 13v2
0v1
4v3 = 0.
63v2 + 220v3 = 9600.
v3 = 78 V.
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4–32
CHAPTER 4. Techniques of Circuit Analysis
120
v2
=
= 3 A;
40
40
[b] io =
i3 =
ig =
v3
78
11.5io
=
5
78
96
4
p5io =
=
5io v1 =
11.5(3)
= 8.7 A;
5
4.5 A;
5(3)(156) =
2340 W(dev);
p11.5io = 11.5io i3 = 11.5(3)(8.7) = 300.15 W(abs);
p96V = 96( 4.5) =
X
pdev = 2340 + 432 = 2772 W.
CHECK
X
.·.
P 4.22
432 W(dev);
pdis
X
=
1562 (156 120)2 1202 (120 78)2
+
+
+
20
5
40
50
+(8.7)2 (5) + (4.5)2 (4) + 300.15 = 2772 W;
pdev =
X
pdis = 2772 W.
The two node voltage equations are:
v 1 v2
v1 40 v1
+
+
= 0;
5
50
10
v2
v2 40
v 2 v1
10 +
+
= 0.
10
40
8
Place these equations in standard form:
✓
◆
✓
◆
1
1
1
1
+
+
+ v2
v1
5 50 10
10
✓
◆
✓
◆
1
1
1
1
+
+
v1
+ v2
10
10 40 8
=
40
;
5
=
10 +
40
.
8
v2 = 80V.
Solving, v1 = 50V;
Thus, vo = v1 40 = 50 40 = 10V.
POWER CHECK:
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Problems
ig
=
(50
p40V
=
(40)(7) = 280W (abs);
p5⌦
=
(50
40)2 /5 = 20W (abs);
p8⌦
=
(80
40)2 /8 = 200W (abs);
p10⌦
=
(80
50)2 /10 = 90W (abs);
p50⌦
=
502 /50 = 50W (abs);
p40⌦
=
802 /40 = 160W (abs);
p10A
=
X
40)/5 + (80
(80)(10) =
40)/8 = 7A;
800W (del).
pabs = 280 + 20 + 200 + 90 + 50 + 160 = 800W =
P 4.23 [a]
There is only one node voltage equation:
va
va 80
va + 30
+
+
+ 0.01 = 0.
5000
500
1000
Solving,
va + 30 + 10va + 5va 400 + 50 = 0
so
·
..
va = 20 V.
Calculate the currents:
i1 = ( 30 20)/5000 = 10 mA;
i2
=
20/500 = 40 mA;
i4
=
80/4000 = 20 mA;
i5
=
(80
i3 + i4 + i5
4–33
X
pdel .
16va = 320;
20)/1000 = 60 mA.
10 mA = 0 so i3 = 0.01
0.02
0.06 =
0.07 =
70 mA.
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4–34
CHAPTER 4. Techniques of Circuit Analysis
[b] p30V
=
(30)( 0.01) =
p10mA
=
(20
p80V
=
(80)( 0.07) =
p5k
=
( 0.01)2 (5000) = 0.5 W;
p500⌦
=
(0.04)2 (500) = 0.8 W;
p1k
=
(80
p4k
=
(80)2 /(4000) = 1.6 W.
X
P 4.24
X
0.3 W;
80)(0.01) =
0.6 W;
5.6 W;
20)2 /(1000) = 3.6 W;
pabs = 0.5 + 0.8 + 3.6 + 1.6 = 6.5 W;
pdel = 0.3 + 0.6 + 5.6 = 6.5 W. (checks!)
The two node voltage equations are:
vb vb vc
7+
+
= 0;
3
1
vc v b v c 4
+
= 0.
2vx +
1
2
The constraint equation for the dependent source is:
vx = vc 4.
Place these equations in standard form:
✓
◆
1
vb
+1
+ vc ( 1)
+ vx (0)
=
3
✓
◆
1
+ vc 1 +
+ vx ( 2) =
vb ( 1)
2
4
;
2
vb (0)
4.
+ vc (1)
+ vx ( 1) =
7;
Solving, vc = 9 V, vx = 5 V, and vo = vb = 1.5 V.
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Problems
4–35
P 4.25 [a]
v2
230
v2
v4
+
v2
v3
= 0;
1
1
v3 v2 v3 v3 v5
+
+
= 0;
1
1
1
v4 v2 v4 230 v4 v5
+
+
= 0;
1
6
2
v5 v3 v5 v5 v4
+
+
= 0;
1
6
2
1
+
Solving, v2 = 150 V;
i2⌦ =
v5
v4
2
=
v3 = 80 V;
140
90
2
so
3v2
1v3
1v4 + 0v5 = 230.
so
1v2 + 3v3 + 0v4
so
12v2 + 0v3 + 20v4
so
0v2
v4 = 140 V;
12v3
1v5 = 0.
6v5 = 460.
6v4 + 20v5 = 0.
v5 = 90 V.
= 25 A;
p2⌦ = (25)2 (2) = 1250 W.
[b]
i230V
=
=
p230V
=
v1
v2
+
v1
v4
1
6
230 150 230 140
+
= 80 + 15 = 95 A.
1
6
(230)(95) = 21,850 W.
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4–36
CHAPTER 4. Techniques of Circuit Analysis
Check:
X
Pdis
=
(80)2 (1) + (70)2 (1) + (80)2 (1) + (15)2 (6) + (10)2 (1)
+(10)2 (1) + (25)2 (2) + (15)2 (6) = 21,850 W.
P 4.26
v1 v2 v1 20
v1
+
+
= 0;
30,000
5000
2000
v2 v1 v2 20
v2
+
+
= 0;
1000
5000
5000
Solving, v1 = 15 V;
Thus, io =
so
so
22v1
6v2 = 300.
v1 + 7v2 = 20.
v2 = 5 V.
v 1 v2
= 2 mA.
5000
P 4.27 [a]
This circuit has a supernode includes the nodes v1 , v2 and the 25 V
source. The supernode equation is
v2
v2
v1
+
+
= 0.
2+
50 150 75
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Problems
4–37
The supernode constraint equation is
v1
v2 = 25.
Place these two equations in standard form:
✓ ◆
✓
◆
1
1
1
v1
+
+ v2
=
2;
50
150 75
v1 (1)
+ v2 ( 1)
Solving, v1 =
= 25.
37.5 V and v2 =
p2A = (2)vo = (2)( 37.5) =
62.5 V, so vo = v1 =
37.5 V.
75 W.
The 2 A source delivers 75 W.
[b]
This circuit now has only one non-reference essential node where the
voltage is not known – note that it is not a supernode. The KCL
equation at v1 is
v1
v1 + 25 v1 + 25
+
+
= 0.
50
150
75
Solving, v1 = 37.5 V so vo = v1 =
2+
p2A = (2)vo = (2)( 37.5) =
37.5 V.
75 W.
The 2 A source delivers 75 W.
[c] The choice of a reference node in part (b) resulted in one simple KCL
equation, while the choice of a reference node in part (a) resulted in a
supernode KCL equation and a second supernode constraint equation.
Both methods give the same result but the choice of reference node in
part (b) yielded fewer equations to solve, so is the preferred method.
P 4.28 Place 4v inside a supernode and use the lower node as a reference. The
resulting circuit is
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4–38
CHAPTER 4. Techniques of Circuit Analysis
The supernode KCL equation is
v1
v2
v2
100 v1
+
+
+
= 0.
10
60 20 30
The dependent source constraint equation is
v1
v2 = 4v = 4v2 .
Solving,
v1 = 75 V;
v2 = 15 V.
vo = 100
v1 = 25 V.
P 4.29
Node equations:
v1 20 v3 v2
v3
v1
+
+
+
+ 3.125v = 0;
20
2
4
80
v2
v2 v3 v2 20
+
+
= 0,
40
4
1
Constraint equations:
v = 20
v1
v2 ;
35i = v3 ;
i = v2 /40.
Solving, v1 =
20.25 V;
v2 = 10 V;
v3 =
29 V.
Let ig be the current delivered by the 20 V source, then
ig =
20
(20.25) 20 10
+
= 30.125 A;
2
1
pg (delivered) = 20(30.125) = 602.5 W.
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Problems
4–39
P 4.30
[a] The left-most node voltage is 75 1250i . The right-most node voltage is
75 V. Write KCL equations at the essential nodes labeled 1 and 2.
[b] From the values given,
105 85
v 1 v2
=
= 0.04 A;
500
500
v = v2 75 = 85 75 = 10 V;
i =
vL = 75
(1250)(0.04) = 25 V;
v L v1
vL
25 105
25
+
=
+
= 0.07 A;
1000
2500
1000
2500
85 75
v2 75
ix =
+ 0.07 = 0.09 A.
iy =
500
500
Calculate the total power:
ix =
Pdstop = 1250i (ix ) = 1250(0.04)( 0.07) =
Pdsbot =
v1 (12 ⇥ 10 3 v ) =
3.5 W;
(105)(12 ⇥ 10 3 )(10) =
12.6 W;
P75V = 75iy = 75(0.09) = 6.75 W;
P1k =
(vL v1 )2
(25 105)2
=
= 6.4 W;
1000
1000
P2.5k =
252
vL2
=
= 0.25 W;
2500
2500
P500mid = 500i2 = 500(0.04)2 = 0.8 W;
P500right =
102
v2
=
= 0.2 W;
500
500
852
v22
=
= 1.7 W;
P4.25k =
4250
4250
Psupplied = 3.5 + 12.6 = 16.1 W.
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4–40
CHAPTER 4. Techniques of Circuit Analysis
[c] Pabsorbed = 6.75 + 6.4 + 0.25 + 0.8 + 0.2 + 1.7 = 16.1 W = Psupplied .
Therefore the analyst is correct.
P 4.31 From Eq. 4.13, iB = vc /(1 + )RE ,
Substitute vc from Eq. 4.14 into Eq. 4.13 to give, iB = (vb
Vo )/(1 + )RE .
Now substitute vb from
" Eq. 4.16 into the above expression
# to give,
VCC (1 + )RE R2 + Vo R1 R2
1
Vo
iB =
(1 + )RE R1 R2 + (1 + )RE (R1 + R2 )
=
[VCC R2 /(R1 + R2 )] Vo
VCC R2 Vo (R1 + R2 )
=
.
R1 R2 + (1 + )RE (R1 + R2 )
[R1 R2 /(R1 + R2 )] + (1 + )RE
P 4.32 [a]
The three mesh current equations are:
44 + 4ia + 6(ia
1ic + 3(ic
ie ) + 6(ic
2 + 3(ie
ic )
= 0;
ia )
= 0;
ic ) + 2ie
=
0.
Place these equations in standard form:
ia (4 + 6) + ic ( 6) + ie (0)
=
44;
ia ( 6) + ic (1 + 3 + 6) + ie ( 3) =
0;
ia (0) + ic ( 3) + ie (3 + 2)
2.
=
Solving, ia = 8 A;
ic = 6 A;
ie = 4 A.
Now calculate the remaining branch currents:
ib
=
ia
ic = 2 A;
id
=
iv
ie = 2 A.
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Problems
4–41
[b] p44V = (44)(8) = 352 W (dev);
p2V = (2)(4) = 8 W (dev).
Therefore, the total power developed is 360 W.
P 4.33 [a]
The three mesh current equations are:
125 + 1i1 + 6(i1
i6 ) + 2(i1
i3 )
= 0;
24i6 + 12(i6
i3 ) + 6(i6
i1 )
= 0;
125 + 2(i3
i1 ) + 12(i3
i6 ) + 1i3
=
0,
Place these equations in standard form:
i1 (1 + 6 + 2) + i3 ( 2) + i6 ( 6)
=
i1 ( 6) + i3 ( 12) + i6 (24 + 12 + 6) =
i1 ( 2) + i3 (2 + 12 + 1) + i6 ( 12)
125;
0;
= 125.
Solving, i1 = 23.76 A;
i3 = 18.43 A;
i6 = 8.66 A.
Now calculate the remaining branch currents:
i2
=
i1
i3 = 5.33 A;
i4
=
i1
i6 = 15.10 A;
i5
=
i3
i6 = 9.77 A.
[b] psources = ptop + pbottom =
=
2969.58
(125)(23.76)
2303.51 =
(125)(18.43)
5273 W.
Thus, the power developed in the circuit is 5273 W.
Now calculate the power absorbed by the resistors:
p1top = (23.76)2 (1) = 564.39 W;
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4–42
CHAPTER 4. Techniques of Circuit Analysis
p2 = (5.33)2 (2) = 56.79 W;
p1bot = (18.43)2 (1) = 339.59 W;
p6 = (15.10)2 (6) = 1367.64 W;
p12 = (9.77)2 (12) = 1145.22 W;
p24 = (8.66)2 (24) = 1799.47 W.
The power absorbed by the resistors is
564.39 + 56.79 + 339.59 + 1367.64 + 1145.22 + 1799.47 = 5273 W so the
power balances.
P 4.34 [a]
The four mesh current equations are:
230 + 1(i1
2i3 + 1(i3
i2 ) + 1(i1
i3 ) + 1(i1
i4 )
= 0;
6i2 + 1(i2
i3 ) + 1(i2
i1 )
= 0;
i4 ) + 1(i3
i1 ) + 1(i3
i2 )
= 0;
6i4 + 1(i4
i1 ) + 1(i4
i3 )
= 0.
Place these equations in standard form:
i1 (3) + i2 ( 1) + i3 ( 1) + i4 ( 1) =
i1 ( 1) + i2 (8) + i3 ( 1) + i4 (0)
= 0;
i1 ( 1) + i2 ( 1) + i3 (5) + i4 ( 1) =
i1 ( 1) + i2 (0) + i3 ( 1) + i4 (8)
230;
0;
= 0.
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Problems
Solving, i1 = 95 A;
i2 = 15 A;
i3 = 25 A;
The power absorbed by the 2 ⌦ resistor is
4–43
i4 = 15 A.
p5 = i23 (2) = (25)2 (2) = 1250 W.
[b] p230 =
(230)i1 =
(230)(95) =
21,850 W.
P 4.35
The three mesh current equations are:
20 + 2000(i1
i2 ) + 30,000(i1
i3 )
= 0;
5000i2 + 5000(i2
i3 ) + 2000(i2
i1 )
= 0;
1000i3 + 30,000(i3
i1 ) + 5000(i3
i2 )
= 0.
Place these equations in standard form:
i1 (32,000) + i2 ( 2000) + i3 ( 30,000)
= 20;
i1 ( 2000) + i2 (12,000) + i3 ( 5000)
= 0;
i1 ( 30,000) + i2 ( 5000) + i3 (36,000)
= 0.
Solving, i1 = 5.5 mA;
i2 = 3 mA;
Thus, io = i3 i2 = 2 mA.
i3 = 5 mA.
P 4.36 [a]
80 = 400i1
140 =
200i2 ;
200i1 + 600i2 ;
Solving, i1 = 0.1 A;
i2 =
0.2 A.
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4–44
CHAPTER 4. Techniques of Circuit Analysis
ia = i1 = 0.1 A;
ib = i 1
i2 = 0.3 A;
ic =
i2 = 0.2 A.
[b] If the polarity of the 140 V source is reversed, we have
80 = 400i1
140 =
200i2 ;
200i1 + 600i2 ;
i1 = 0.38 A and i2 = 0.36 A.
ia = i1 = 0.38 A;
ib = i1
i2 = 0.02 A;
ic =
i2 =
0.36 A.
P 4.37 [a]
The mesh current equations are:
230 + 1(i1
i2 ) + 2(i1
i3 ) + 115 + 4i1
6i2 + 3(i2
i3 ) + 1(i2
i1 )
460 + 5i3
115 + 2(i3
i1 ) + 3(i3
=
0;
= 0;
i2 )
=
0.
Place these equations in standard form:
i1 (1 + 2 + 4) + i2 ( 1) + i3 ( 2) =
115;
i1 ( 1) + i2 (6 + 3 + 1) + i3 ( 3) =
0;
i1 ( 2) + i2 ( 3) + i3 (5 + 2 + 3) =
345.
i2 = 10.6A;
i3 = 36.8A.
Solving,
i1 = 4.4A;
The only components that can develop power in the circuit are the
sources:
p230V =
(230)(4.4) = 1012W;
p115V
=
(115)( 36.8
4.4) = 4738W;
p460V = (460)( 36.8) = 16,928W.
X
.·.
pdev = 1012 + 16,928 = 17940W.
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Problems
4–45
[b] From part (a) we know that the 115 V source is dissipating power;
compute the power dissipated by the resistors:
p1⌦
=
(1)(4.4 + 10.6)2 = 225W;
p4⌦
=
(4)(4.4)2 = 77.44W;
p6⌦
=
(6)( 10.6)2 = 674.16W;
p2⌦
=
(2)(4.4 + 36.8)2 = 3394.88W;
p3⌦
=
(3)( 10.6 + 36.8)2 = 2059.32W;
p5⌦ = (5)( 36.8)2 = 6771.2W.
X
.·.
pdis = 4738 + 225 + 77.44 + 674.16 + 3394.88 + 2059.32 + 6771.2
= 17940W. (checks!)
P 4.38
The two KCL equations are
160 + 10i1 + 100(i1
io ) = 0;
30io + 150i + 20io + 100i = 0;
The dependent source constraint equation is
i = io
i1 .
Solving,
i1 = 6 A;
io = 5 A;
i =
pds = (150i )io = 150( 1)(5) =
1 A;
750 W.
Thus, 750 W is delivered by the dependent source.
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4–46
CHAPTER 4. Techniques of Circuit Analysis
P 4.39
65 + 4i1 + 5(i1
i2 ) + 6i1 = 0;
8i2 + 3v + 15i2 + 5(i2
i1 ) = 0;
v = 4i1 .
Solving, i1 = 4 A
i2 =
1A
v = 16 V.
p15⌦ = ( 1)2 (15) = 15 W.
P 4.40
Mesh equations:
53i + 8i1
3i2
0i
3i1 + 30i2
0i
5i1
5i3 = 0;
20i3 = 30;
20i2 + 27i3 = 30.
Constraint equations:
i = i2
i3 .
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
Solving, i1 = 110 A;
i2 = 52 A;
pdepsource = 53i i1 = (53)( 8)(110) =
i3 = 60 A;
i =
4–47
8 A.
46,640 W;
Therefore, the dependent source is developing 46,640 W.
CHECK:
p30V =
30i2 =
1560 W (left source);
p30V =
30i3 =
1800 W (right source);
X
pdev = 46,640 + 1560 + 1800 = 50 kW;
p3⌦ = (110
52)2 (3) = 10,092 W;
p5⌦ = (110
60)2 (5) = 12,500 W;
p20⌦ = ( 8)2 (20) = 1280 W;
p7⌦ = (52)2 (7) = 18,928 W;
p2⌦ = (60)2 (2) = 7200 W;
X
pdiss = 10,092 + 12,500 + 1280 + 18,928 + 7200 = 50 kW.
P 4.41
660 = 30i1
20i =
0=
10i1 + 60i2
15i1
i = i2
10i2
15i3 ;
50i3 ;
50i2 + 90i3 ;
i3 .
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4–48
CHAPTER 4. Techniques of Circuit Analysis
Solving, i1 = 42 A;
i2 = 27 A;
i3 = 22 A;
i = 5 A.
20i = 100 V;
p20i =
100i2 =
100(27) =
2700 W;
.·. p20i (developed) = 2700 W.
CHECK:
p660V =
.·.
X
660(42) =
Pdev
X
Pdis
27,720 W (dev).
=
27,720 + 2700 = 30,420 W.
=
(42)2 (5) + (22)2 (25) + (20)2 (15) + (5)2 (50)+
(15)2 (10)
=
30,420 W.
P 4.42 [a]
10 = 18i1
0=
16i2 ;
16i1 + 28i2 + 4i ;
4 = 8i ;
Solving, i1 = 1 A;
vo = 16(i1
i2 = 0.5A;
i = 0.5 A.
i2 ) = 16(0.5) = 8V.
[b] p4i = 4i i2 = (4)(0.5)(0.5) = 1W (abs);
.·. p4i (deliver) =
1W.
P 4.43
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Problems
4–49
Mesh equations:
128i1
80i2 = 240;
80i1 + 200i2 = 120.
Solving,
i1 = 3 A;
i2 = 1.8 A.
Therefore,
v1 = 40(6
3) = 120 V;
v2 = 120(1.8
1) = 96 V.
P 4.44 [a]
Mesh equations:
65i1
40i2 + 0i3
40i1 + 55i2
0i1
100io = 0;
5i3 + 11.5io = 0;
5i2 + 9i3
11.5io = 0;
1i1 + 1i2 + 0i3 + 1io = 0.
Solving,
i1 = 7.2 A;
i2 = 4.2 A;
i3 =
4.5 A;
io = 3 A.
Therefore,
v1 = 20[5(3)
7.2] = 156 V;
v2 = 40(7.2
4.2) = 120 V;
v3 = 5(4.2 + 4.5) + 11.5(3) = 78 V.
[b] p5io =
5io v1 =
p11.5io = 11.5io (i2
5(3)(156) =
2340 W;
i3 ) = 11.5(3)(4.2 + 4.5) = 300.15 W;
p96V = 96i3 = 96( 4.5) =
432 W.
Thus, the total power dissipated in the circuit, which equals the total
power developed in the circuit is 2340 + 432 = 2772 W.
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4–50
CHAPTER 4. Techniques of Circuit Analysis
P 4.45
5 + 38(i1
5) + 30(i1
67 + 40i2 + 30(i2
i1 ) + 6(i2
Solving, i1 = 2.5 A;
[a] v5A = 38(2.5
=
i2 ) + 12i1 = 0;
5) = 0.
i2 = 0.5 A.
5) + 6(0.5
5)
122 V;
p5A = 5v5A = 5( 122) =
610 W.
Therefore, the 5 A source delivers 610 W.
[b] p5V = 5(2.5) = 12.5 W;
p67V = 67(0.5) = 33.5 W.
Therefore, only the current source delivers power and the total power
delivered is 610 W.
[c]
P 4.46 [a]
X
presistors = (2.5)2 (38) + (4.5)2 (6) + (2)2 (30) + (2.5)2 (12) + (0.5)2 (40)
X
= 564 W;
pabs = 564 + 12.5 + 33.5 = 610 W =
X
pdel (CHECKS).
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Problems
4–51
The mesh current equation for the right mesh is:
5400(i1 0.005) + 3700i1 150(0.005 i1 ) = 0.
.·. i1 = 3 mA.
Solving,
9250i1 = 27.75
Then,
i = 5 3 = 2 mA.
[b] vo = (0.005)(10,000) + (5400)(0.002) = 60.8 V;
p5mA = (60.8)(0.005) = 304 mW.
Thus, the 5 mA source delivers 304 mW.
[c] pdep source = 150i i1 = ( 150)(0.002)(0.003) =
The dependent source delivers 0.9 mW.
0.9 mW.
P 4.47 [a]
Mesh equations:
50 + 6i1
4i2 + 9i = 0;
4i1 + 29i2
9i
20i3 = 0.
Constraint equations:
i = i2 ;
i3 =
Solving, i1 =
1.7v ;
5 A;
v = 2i1 .
i2 = 16 A;
i3 = 17 A;
v =
10 V.
9i = 9(16) = 144 V;
ia = i2
i1 = 21 A;
ib = i2
i3 =
vb = 20ib =
[b]
1 A;
20 V;
p50V =
50i1 = 250 W (absorbing);
p9i =
ia (9i ) =
p1.7V =
1.7v vb = i3 vb = (17)( 20) =
X
(21)(144) =
3024 W (delivering);
340 W (delivering).
Pdev = 3024 + 340 = 3364 W;
X
Pdis = 250 + ( 5)2 (2) + (21)2 (4) + (16)2 (5) + ( 1)2 (20)
= 3364 W.
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4–52
CHAPTER 4. Techniques of Circuit Analysis
P 4.48
Mesh equations:
10i
4i1 = 0;
4i + 24i1 + 6.5i = 400.
Solving, i1 = 15 A;
i = 16 A.
v20A = 1i + 6.5i = 7.5(16) = 120 V;
p20A =
20v20A =
(20)(120) =
2400 W (del);
p6.5i = 6.5i i1 = (6.5)(16)(15) = 1560 W (abs).
Therefore, the independent source is developing 2400 W, all other elements are
absorbing power, and the total power developed is thus 2400 W.
CHECK:
p1⌦ = (16)2 (1) = 256 W;
p5⌦ = (20
16)2 (5) = 80 W;
p4⌦ = (1)2 (4) = 4 W;
p20⌦ = (20
X
15)2 (20) = 500 W;
pabs = 1560 + 256 + 80 + 4 + 500 = 2400 W. (CHECKS)
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Problems
4–53
P 4.49 [a]
Supermesh equations:
1000ib + 4000(ic
ic
id ) + 500(ic
ia ) = 0;
ib = 0.01.
Two remaining mesh equations:
5500ia
500ic =
30;
4000id
4000ic =
80.
In standard form,
500ia + 1000ib + 4500ic
0ia
4000id = 0;
1ib + 1ic + 0id = 0.01;
5500ia + 0ib
0ia + 0ib
500ic + 0id =
30;
4000ic + 4000id =
80.
Solving:
ia =
10 mA;
ib =
60 mA;
ic =
50 mA;
id =
70 mA.
i3 = i d =
70 mA.
Then,
i1 = i a =
10 mA;
i2 = ia
ic = 40 mA;
[b] psources = 30( 0.01) + [1000( 0.06)](0.01) + 80( 0.07) =
6.5 W;
presistors = 1000(0.06)2 + 5000(0.01)2 + 500(0.04)2
+4000( 0.05 + 0.07)2 = 6.5 W.
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4–54
CHAPTER 4. Techniques of Circuit Analysis
P 4.50
The supermesh equation is:
20 + 4i1 + 9i2 90 + 6i2 + 1i1 = 0.
The supermesh constraint equation is :
i1 i2 = 6.
Place these equations in standard form:
i1 (4 + 1) + i2 (9 + 6)
= 20 + 90;
i1 (1) + i2 ( 1)
= 6.
Solving,
i1 = 10A;
Now find the power:
i2 = 4A.
p4⌦
=
102 (4) = 400W;
p1⌦
=
102 (1) = 100W;
p9⌦
=
42 (9) = 144W;
p6⌦
=
42 (6) = 96W;
p20V
=
(20)(10) =
v6A
=
9i2
p6A
=
( 30)(6) =
180W;
p90V
=
(90)(4) =
360W.
200W;
90 + 6i2 = (9)(4)
90 + (6)(4) =
30V;
In summary:
X
pdev = 200 + 180 + 360 = 740W;
pdiss = 400 + 100 + 144 + 96 = 740W.
Thus the power dissipated in the circuit is 740 W.
X
P 4.51 [a]
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Problems
4–55
The supermesh equation is:
60 + 4i1 + 9i2 90 + 6i2 + 1i1 = 0.
The supermesh constraint equation is :
i1 i2 = 6.
Place these equations in standard form:
i1 (4 + 1) + i2 (9 + 6)
= 60 + 90;
i1 (1) + i2 ( 1)
= 6.
Solving,
i1 = 12A;
Now find the power:
i2 = 6A.
p4⌦
=
122 (4) = 576W;
p1⌦
=
122 (1) = 144W;
p9⌦
=
62 (9) = 324W;
p6⌦
=
62 (6) = 216W;
p60V
=
(60)(15) =
vo
=
9i2
720W;
90 + 6i2 = 9(6)
90 + 6(6) = 0V
(the 6 A source acts like a short circuit carrying 6 A of current);
p6A
=
(0)(6) = 0W;
p90V
=
(90)(6) =
540W.
In summary:
X
pdev = 576 + 144 + 324 + 216 = 1260W
(note that the power of
the
X 6 A source is zero);
pdiss = 720 + 540 = 1260W.
Thus the power dissipated in the circuit is 1260 W.
[b]
Now there is no longer a supermesh. The two simple mesh current
equations are:
60 + 4i1 + 1i1
=
0;
90 + 6i2 + 9i2
=
0.
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4–56
CHAPTER 4. Techniques of Circuit Analysis
Since these equations are uncoupled, each can be solved separately:
i1 = 60/5 = 12A;
5i1 = 60 .·.
15i2 = 90 .·.
i2 = 90/15 = 6A.
Since the currents are the same as in part (a), the power will be the same
as calculated in part (a). Thus, the power dissipated in the circuit is
again 1260 W.
[c] As noted in part (a), the 6 A source has zero voltage drop, so is equivalent
to a short circuit (which has no voltage drop by definition) carrying 6 A
of current, as in the circuit of part (b).
P 4.52 [a]
The i1 mesh current equation:
100 + 5(i1 i2 ) + 10(i1 i3 ) + 2i1 = 0.
The i2 — i3 supermesh equationa:
2i2 + 20i3 + 10(i3 i1 ) + 5(i2 i1 ) = 0.
The supermesh constraint:
i3 i2 = 1.2ib = 1.2i1 .
Place these equations in standard form:
i1 (5 + 10 + 2) + i2 ( 5) + i3 ( 10)
= 100;
i1 ( 10
= 0;
5) + i2 (2 + 5) + i3 (20 + 10)
i1 (1.2) + i2 (1) + i3 ( 1)
i2 =
Solving,
i1 = 7.4A;
Solve for the requested currents:
ia
=
i2 =
ib
=
i1 = 7.4A;
ic
=
i3 = 4.68A;
id
=
i1
i2 = 11.6A;
ie
=
i1
i3 = 2.72A.
=
4.2A;
0.
i3 = 4.68A.
4.2A;
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Problems
[b] Find vcs :
2i2 + vcs + 5(i2 i1 ) = 0
Calculate the power:
vcs =
2( 4.2)
p100V
=
(100)(7.4) =
pdep source
=
(66.4)[1.2(7.4)] =
p2⌦
=
2( 4.2)2 = 35.28W;
p5⌦
=
5(7.4 + 4.2)2 = 672.8W;
p2⌦
=
2(7.4)2 = 109.52W;
p10⌦
=
10(7.4
p20⌦
=
20(4.68)2 = 438.048W.
X
P 4.53 [a]
.·.
X
5( 4.2
4–57
7.4) = 66.4V.
740W;
589.632W;
4.68)2 = 73.984W;
pdev = 740 + 589.632 = 1329.632W;
pdis = 35.28 + 672.8 + 109.52 + 73.984 + 438.048 = 1329.632W.
4id + 10(ie
id ) + 5(ie
5(ix
ie ) + 10(id
id
ix = 2ib = 2(ie
ie )
ix ) = 0;
240 + 40(ix
ix .)
Solving, id = 10 A;
ie = 18 A;
ia = 19
ib = i e
ix =
[b] va = 40ia =
7 A;
280 V;
19) = 0;
ix = 26 A.
ix =
8 A;
vb = 5ib + 40ia =
p19A =
19va = 5320 W;
p4id =
4id ie =
720 W;
p2ia =
2ib vb =
5120 W;
ic = i e
id = 8 A.
320 V;
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4–58
CHAPTER 4. Techniques of Circuit Analysis
p240V =
240id =
2400 W;
p40⌦ = (7)2 (40) = 1960 W;
p5⌦ = (8)2 (5) = 320 W;
p10⌦ = (8)2 (10) = 640 W;
X
X
Pgen = 720 + 5120 + 2400 = 8240 W;
Pdiss = 5320 + 1960 + 320 + 640 = 8240 W.
P 4.54 [a] If the mesh-current method is used, then the value of the lower left mesh
current is io = 0. This shortcut will simplify the set of KVL equations.
The node-voltage method has no equivalent simplifying shortcut, so the
mesh-current method is preferred.
[b]
Write the mesh current equations. Note that if io = 0, then i1 = 0:
23 + 5( i2 ) + 10( i3 ) + 46
30i2 + 15(i2
Vdc + 25i3
= 0;
i3 ) + 5i2
46 + 10i3 + 15(i3
=
i2 )
0;
= 0.
Place the equations in standard form:
i2 ( 5) + i3 ( 10) + Vdc (0)
=
23;
i2 (30 + 15 + 5) + i3 ( 15) + Vdc (0)
= 0;
i2 ( 15) + i3 (25 + 10 + 15) + Vdc (1)
=
46.
i3 = 2 A;
Vdc = 45 V.
Solving,
i2 = 0.6 A;
Thus, the value of Vdc required to make io = 0 is 45 V.
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Problems
4–59
[c] Calculate the power:
p23V
=
(23)(0) = 0 W;
p46V
=
(46)(2) =
92 W;
pVdc
=
( 45)(2) =
90 W;
p30⌦
=
(30)(0.6)2 = 10.8 W;
p5⌦
=
(5)(0.6)2 = 1.8 W;
p15⌦
=
(15)(2
p10⌦
=
(10)(2)2 = 40 W;
p20⌦
=
(20)(0)2 = 0 W;
p25⌦
=
(25)(2)2 = 100 W.
X
X
0.6)2 = 29.4 W;
pdev = 92 + 90 = 182 W;
pdis = 10.8 + 1.8 + 29.4 + 40 + 0 + 100 = 182 W(checks).
P 4.55 [a] There are 4 essential nodes so using the node voltage method requires 3
KCL equations. There are 4 meshes, but the currents in two of those
meshes are defined by current sources. Only 2 KVL equations are
required. Therefore the mesh current method requires fewer equations
and is thus the preferred analysis method.
[b]
The mesh current equations:
240 + 12i1 + 20(i1
20(i2
i2 )
= 0;
i1 ) + 15(i2 + 4) + 50(i2 + idc ) + 40i2
=
0.
Place these equations in standard form:
i1 (12 + 20) + i2 ( 20) + idc (0)
=
240;
i1 ( 20) + i2 (20 + 15 + 50 + 40) + idc (50)
=
60.
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4–60
CHAPTER 4. Techniques of Circuit Analysis
But if the power associated with the 4 A source is zero, the voltage drop
across the source must be zero. This means that the voltage drop across
the 15 ⌦ resistor is also zero, so the 15 ⌦ resistor is e↵ectively removed
from the circuit. Once this happens, i2 = 4 A. Substitute this value
into the first equation and solve for i1 :
32i1 20( 4) = 240
.·.
32i1 = 160 so i1 = 5A.
Now substitute this value for i1 into the second equation and solve for idc :
20(5) + 125( 4) + 50idc = 60 so 50idc = 60 + 100 + 500 = 540;
.·.
idc = 540/50 = 10.8A.
P 4.56 [a] There are three unknown node voltages and only two unknown mesh
currents. Use the mesh current method to minimize the number of
simultaneous equations.
[b]
The mesh current equations:
2500(i1
0.01) + 2000i1 + 1000(i1
5000(i2
0.01) + 1000(i2
i2 )
i1 ) + 1000i2
= 0;
=
0.
Place the equations in standard form:
i1 (2500 + 2000 + 1000) + i2 ( 1000)
=
25;
i1 ( 1000) + i2 (5000 + 1000 + 1000) =
50.
i2 = 8 mA
Solving, i1 = 6 mA;
Find the power in the 1 k⌦ resistor:
i1k = i1 i2 = 2 mA;
p1k = ( 0.002)2 (1000) = 4 mW.
[c] No, the voltage across the 10 A current source is readily available from the
mesh currents, and solving two simultaneous mesh-current equations is
less work than solving three node voltage equations.
[d] vg = 2000i1 + 1000i2 = 12 + 8 = 20 V;
p10mA = (20)(0.01) = 200 mW.
Thus the 10 mA source develops 200 mW.
P 4.57 [a] There are three unknown node voltages and three unknown mesh currents,
so the number of simultaneous equations required is the same for both
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Problems
4–61
methods. The node voltage method has the advantage of having to solve
the three simultaneous equations for one unknown voltage provided the
connection at either the top or bottom of the circuit is used as the
reference node. Therefore recommend the node voltage method.
[b]
The node voltage equations are:
v1 v2 v1 v3
v1
+
+
= 0;
5000
2500
1000
v2
v2 v1 v2 v3
+
+
= 0;
0.01 +
4000
2500
2000
v3
v3 v1 v3 v2
+
+
= 0.
1000
2000
1000
Put✓the equations in standard
form:
◆
✓
◆
✓
◆
1
1
1
1
1
+
+
+ v2
+ v3
v1
5000 2500 1000
2500
1000
◆
✓
◆
✓
◆
✓
1
1
1
1
1
+
+
+ v2
+ v3
v1
2500
4000 2500 2000
2000
✓
◆
✓
◆
✓
◆
1
1
1
1
1
+
+
+ v2
+ v3
v1
1000
2000
2000 1000 1000
Solving,
v1 = 6.67 V; v2 = 13.33 V; v3 = 5.33 V.
p10m = (13.33)(0.01) = 133.33 mW.
Therefore, the 10 mA source is developing 133.33 mW.
=
0;
=
0.01;
=
0.
P 4.58 [a] The node voltage method requires summing the currents at two
supernodes in terms of four node voltages and using two constraint
equations to reduce the system of equations to two unknowns. If the
connection at the bottom of the circuit is used as the reference node,
then the voltages controlling the dependent sources are node voltages.
This makes it easy to formulate the constraint equations. The current in
the 20 V source is obtained by summing the currents at either terminal of
the source.
The mesh current method requires summing the voltages around the two
meshes not containing current sources in terms of four mesh currents. In
addition the voltages controlling the dependent sources must be
expressed in terms of the mesh currents. Thus the constraint equations
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4–62
CHAPTER 4. Techniques of Circuit Analysis
are more complicated, and the reduction to two equations and two
unknowns involves more algebraic manipulation. The current in the 20 V
source is found by subtracting two mesh currents.
Because the constraint equations are easier to formulate in the node
voltage method, it is the preferred approach.
[b]
Node voltage equations:
v2
v1
+ 0.003v +
0.2 = 0;
100
250
v4
v3
0.2 +
+
0.003v = 0.
100 200
Constraints:
v2 = v a ;
v3 = v ;
v4
v3 = 0.4va ;
Solving, v1 = 24 V; v2 = 44 V;
v2
= 24 mA;
io = 0.2
250
v3 =
72 V;
v2
v1 = 20.
v4 =
54 V.
p20V = 20(0.024) = 480 mW.
Thus, the 20 V source absorbs 480 mW.
P 4.59 [a] Apply source transformations to both current sources to get
io =
(6.75 + 8.25)
= 20 mA.
330 + 150 + 270
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Problems
4–63
[b]
The node voltage equations:
v1 v 2
v1
+
= 0;
0.025 +
330
150
v2 v 1
v2
+
0.025 = 0.
270
150
Place
in
form:
✓ these equations
◆
✓ standard
◆
1
1
1
+
+ v2
=
0.025;
v1
330 150
150
✓
◆
✓
◆
1
1
1
+
+ v2
= 0.025.
v1
150
270 150
Solving, v1 = 1.65 V;
v2 = 1.35 V;
v 2 v1
= 20 mA.
.·. io =
150
P 4.60 [a]
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4–64
CHAPTER 4. Techniques of Circuit Analysis
io =
135/30,000 =
4.5 mA.
[b]
va
=
(7500)( 0.0045) = 33.75 V;
33.75
va
=
= 0.375 mA;
=
90,000
90,000
=
8.4 ⇥ 10 3 + 0.375 ⇥ 10 3 + 4.5 ⇥ 10 3 =
ia
ib
vb
p120V
Check:
p8.4mA
X
(6000)(3.525 ⇥ 10 3 ) 33.75 = 12.6 V;
12.6 120
= 3.315 mA;
=
40,000
= (120)( 3.315 ⇥ 10 3 ) = 397.8 mW.
=
ig
X
3.525 mA;
=
( 33.75)(8.4 ⇥ 10 3 ) =
Pdev
=
397.8 + 283.5 = 681.3mW;
Pdis
=
=
283.5mW;
( 12.6)2 ( 33.75)2
+
60,000
90,000
3 2
+(6000)( 3.525 ⇥ 10 ) + (7500)( 4.5 ⇥ 10 3 )2
(40,000)( 3.315 ⇥ 10 3 )2 +
681.3mW.
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Problems
4–65
P 4.61 [a]
vo = (12,500)(0.001) = 12.5 V.
[b]
5000i1 + 40,000i2
i2
30,000i3 = 35;
i1 = 0.008;
30,000i2 + 70,000i3 = 25.
Solving,
i1 =
5.33 mA;
vo = (25,000)(i3
i2 = 2.667 mA;
i3 = 1.5 mA;
0.001) = (25,000)(0.0005) = 12.5 V.
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4–66
CHAPTER 4. Techniques of Circuit Analysis
P 4.62 [a] Applying a source transformation to each current source yields
Now combine the 12 V and 5 V sources into a single voltage source and
the 6 ⌦, 6 ⌦ and 5 ⌦ resistors into a single resistor to get
Now use a source transformation on each voltage source, thus
which can be reduced to
.·. io =
8.5
(1) =
10
0.85 A.
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Problems
4–67
[b]
34ia
17ib = 12 + 5 + 34 = 51;
17ia + 18.5ib =
Solving, ib =
34.
0.85 A = io .
P 4.63 [a] First remove the 16 ⌦ and 260 ⌦ resistors:
Next use a source transformation to convert the 1 A current source and
40 ⌦ resistor:
which simplifies to
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4–68
CHAPTER 4. Techniques of Circuit Analysis
250
(480) = 400 V.
.·. vo =
300
[b] Return to the original circuit with vo = 400 V:
ig =
520
+ 1.6 = 3.6 A;
260
p520V =
(520)(3.6) =
1872 W.
Therefore, the 520 V source is developing 1872 W.
[c] v1 =
520 + 1.6(4 + 250 + 6) =
v g = v1
1(16) =
104
p1A = (1)( 120) =
16 =
104 V;
120 V;
120 W.
Therefore the 1 A source is developing 120 W.
[d]
X
pdev = 1872 + 120 = 1992 W;
X
.·.
P 4.64 vTh =
pdiss = (1)2 (16) +
X
pdiss =
X
(104)2 (520)2
+
+ (1.6)2 (260) = 1992 W;
40
260
pdev .
40
(60) = 48 V;
50
RTh = 8 + 10k40 = 16 ⌦.
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Problems
4–69
P 4.65 Find the open-circuit voltage:
0.075 +
v1
v1
+
= 0;
4000 5000
v1 = 166.67 V;
so
voc =
3000
v1 = 100 V.
5000
Find the short-circuit current:
isc =
4000k2000
(0.075) = 50 mA.
2000
Thus,
IN = isc = 50 mA;
RN =
voc
100
= 2 k⌦.
=
isc
0.05
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4–70
CHAPTER 4. Techniques of Circuit Analysis
P 4.66
12 + 12(i1
2(isc
8) + 6(i1
8) + 6(isc
Solving,
isc ) = 0;
i1 ) = 0.
isc = 8.67 A;
RTh = 2 + 12k6 = 6 ⌦.
P 4.67 After making a source transformation the circuit becomes
300 + 40(i1
i2 ) + 8i1 = 0;
450 + 150i2 + 10i2 + 40(i2
.·. i1 = 5.25 A and i2 =
i1 ) = 0.
1.2 A.
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Problems
4–71
vTh = 10i2 + 8i1 = 30 V;
RTh = (40k8 + 10)k150 = 15 ⌦.
P 4.68 First we make the observation that the 8 mA current source and the 20 k⌦
resistor will have no influence on the behavior of the circuit with respect to
the terminals a,b. This follows because they are in parallel with an ideal
voltage source. Hence our circuit can be simplified to
or
Therefore the Norton equivalent is
P 4.69 First, find the Thévenin equivalent with respect to Ro .
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4–72
CHAPTER 4. Techniques of Circuit Analysis
Ro (⌦) io (mA) vo (V)
100
13.64
1.364
120
12.5
1.5
150
11.11
1.667
180
10
1.8
P 4.70
12.72 = VTh
12 = VTh
2RTh .
20RTh .
Solving the above equations for VTh and RTh yields
VTh = 12.8V,
.·. IN = 320A,
RTh = 40 m⌦;
RN = 40 m⌦.
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Problems
4–73
P 4.71
i1 = 100/20 = 5 A;
100 = vTh
5RTh ,
vTh = 100 + 5RTh .
i2 = 200/50 = 4 A;
200 = vTh
4RTh ,
vTh = 200 + 4RTh ;
.·. 100 + 5RTh = 200 + 4RTh
so
RTh = 100 ⌦.
vTh = 100 + 500 = 600 V.
P 4.72 [a] First, find the Thévenin equivalent with respect to a,b using a succession
of source transformations.
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4–74
CHAPTER 4. Techniques of Circuit Analysis
.·. vTh = 54 V
vmeas =
RTh = 4.5 k⌦.
54
(85.5) = 51.3 V.
90
✓
◆
51.3 54
[b] %error =
⇥ 100 =
54
5%.
P 4.73
Use voltage division to calculate v1 and v2 :
v1 =
501
(5) = 4.168053 V;
501 + 100
v2 =
5000
(5) = 4.1666667 V.
5000 + 1000
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Problems
4–75
Now calculate VTh :
VTh = v1
v2 = 4.168053
4.1666667 = 1.3866 mV.
Calculate RTh by removing the voltage source and creating series and parallel
combinations of the resisitors:
RTh = 100k501 + 1000k5000 =
(100)(501) (1000)(5000)
+
= 916.69 ⌦.
601
6000
The resulting Thévenin equivalent circuit is shown below:
Use KVL to calculate igal :
igal =
1.3866 ⇥ 10 3
= 1.43 µA.
916.69 + 50
P 4.74
OPEN CIRCUIT
100 = 2500i1 + 625(i1 + 10 3 v2 );
v2 =
6000
(5000i1 ).
10,000
Solving,
i1 = 0.02 A;
v2 = voc = 60 V.
SHORT CIRCUIT
v2 = 0;
isc =
5000
i1 ;
4000
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4–76
CHAPTER 4. Techniques of Circuit Analysis
i1 =
100
= 0.032 A.
2500 + 625
Thus,
5
isc = i1 = 0.04 A;
4
RTh =
60
= 1.5 k⌦.
0.04
P 4.75
The node voltage equations and dependant source equation are:
v1 280
v1
v1 v2
+
+
+ 0.2i
= 0;
2000
2000
2000
v2
v2 v 1
+
0.2i
= 0;
2000
5600
280 v1
.
=
i
2000
In standard form:
✓
◆
✓
◆
1
1
1
280
1
+
+
;
v1
+ v2
+ i (0.2) =
2000 2000 2000
2000
2000
✓
◆
✓
◆
1
1
1
+
+ v2
+ i ( 0.2) = 0;
v1
2000
2000 5600
◆
✓
280
1
.
+ v2 (0) + i (1) =
v1
2000
2000
Solving,
v1 = 120 V; v2 = 112 V; i = 0.08 A
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Problems
4–77
VTh = v2 = 112 V.
The mesh current equations are:
280 + 2000i + 2000(i
2000(isc
isc )
0.2i ) + 2000(isc
i )
=
0;
= 0.
Put these equations in standard form:
i (4000) + isc ( 2000)
=
280;
i ( 2400) + isc (4000)
= 0.
Solving, i = 0.1 A;
isc = 0.06 A;
112
= 1866.67 ⌦.
RTh =
0.06
P 4.76 [a] Find the Thévenin equivalent with respect to the terminals of the
ammeter. This is most easily done by first finding the Thévenin with
respect to the terminals of the 50 ⌦ resistor.
Thévenin voltage: note i is zero.
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4–78
CHAPTER 4. Techniques of Circuit Analysis
vTh vTh vTh vTh 40
+
+
+
= 0;
80
40
240
16
Solving, vTh = 24 V.
Short-circuit current:
isc = 2.5 + 6isc ,
RTh =
24
=
0.5
.·. isc =
0.5 A;
48 ⌦.
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Problems
Rtotal =
4–79
24
= 2.4 ⌦;
10
Rmeter = 2.4
2 = 0.40 ⌦.
[b] Actual current:
iactual =
24
= 12 A;
2
% error =
10
12
12
⇥ 100 =
16.67%.
P 4.77 [a] Replace the voltage source with a short circuit and find the equivalent
resistance looking into the terminals a,b:
RTh = 10k40 + 8 = 16 ⌦.
[b] Replace the current source with an open circuit and the voltage source
with a short circuit. Find the equivalent resistance from the terminals
a,b:
RN = 12k6 + 2 = 6 ⌦.
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4–80
CHAPTER 4. Techniques of Circuit Analysis
P 4.78 [a] Open circuit:
v2 9 v2
+
20
70
1.8 = 0;
v2 = 35 V;
vTh =
60
v2 = 30 V.
70
Short circuit:
v2 9 v2
+
1.8 = 0;
20
10
.·. v2 = 15 V.
ia =
9
15
=
20
isc = 1.8
RTh =
0.3 A;
0.3 = 1.5 A;
30
= 20 ⌦.
1.5
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Problems
4–81
[b]
RTh = (20 + 10)k60 = 20 ⌦ (CHECKS).
P 4.79 VTh = 0, since circuit contains no independent sources.
v1 250i
v1 vt
v1
+
+
= 0;
100
200
150
vt v1 vt 250i
+
150
50
i =
1 = 0;
v t v1
.
150
In standard form:
v1
✓
◆
✓
1
+i
150
◆
✓
v1
✓
1
1
1
+
+ vt
+i
150
150 50
◆
✓
250
= 1;
50
◆
v1
✓
1
1
+ vt
+ i ( 1) = 0.
150
150
1
1
1
+
+
+ vt
100 200 150
◆
✓
◆
✓
◆
250
= 0;
200
◆
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4–82
CHAPTER 4. Techniques of Circuit Analysis
Solving,
v1 = 75 V;
vt = 150 V;
i = 0.5 A;
vt
= 150 ⌦.
.·. RTh =
1A
P 4.80 VTh = 0 since there are no independent sources in the circuit. To find RTh ,
apply a 1 A test source and calculate the voltage drop across the test source.
Use the mesh current method.
The mesh current equations for the two meshes on the left:
10ix + 5(ix
10ix + 20(iy
iy ) + 40ix
=
1) + 10iy + 5(iy
ix )
0;
= 0.
Place these equations in standard form:
ix ( 10 + 5 + 40) + iy ( 5)
ix (10
=
0;
5) + iy (20 + 10 + 5) =
20.
Solving,
ix = 80 mA;
iy = 560 mA.
Find the voltage drop across the 1 A source:
vT = 20(1 0.56) = 8.8 V;
.·.
RTh = vT /1A = 8.8/1 = 8.8 ⌦.
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Problems
4–83
P 4.81 VTh = 0 since there are no independent sources in the circuit. Thus we need
only find RTh .
v
250ix
500
ix =
1.5ix +
v
750
1 = 0;
v
.
750
Solving,
v = 1500 V;
RTh =
ix = 2 A;
v
= 1500 = 1.5 k⌦.
1A
P 4.82 [a]
RTh = 5000k(1600 + 2400k4800 + 1800) = 2.5 k⌦;
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4–84
CHAPTER 4. Techniques of Circuit Analysis
Ro = RTh = 2.5 k⌦.
[b]
7200i1
4800i2 = 60;
4800i1 + 4800i2 + 8400i3 = 0;
i2
i3 = 0.015.
Solving,
i1 = 19.4 mA;
i2 = 16.6 mA;
i3 = 1.6 mA;
voc = 5000i3 = 8 V.
pmax = (1.6 ⇥ 10 3 )2 (2500) = 6.4 mW.
[c] The resistor closest to 2.5 k⌦ from Appendix H has a value of 2.7 k⌦. Use
voltage division to find the voltage drop across this load resistor, and use
the voltage to find the power delivered to it:
v2.7k =
2700
(8) = 4.154 V;
2700 + 2500
p2.7k =
(4.154)2
= 6.391 mW.
2700
The percent error between the maximum power and the power delivered
to the best resistor from Appendix H is
% error =
✓
6.391
6.4
◆
1 (100) =
0.1%.
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Problems
4–85
P 4.83 Write KVL equations for the left mesh and the supermesh, place them in
standard form, and solve:
At i1 :
60 + 2400i1 + 4800(i1
Supermesh:
4800(i2
Constraint:
i2
i2 ) = 0;
i1 ) + 1600i3 + (5000k2500)i3 + 1800i3 = 0;
i3 = 0.015 = 0;
Standard form:
i1 (7200) + i2 ( 4800) + i3 (0) = 60;
i1 ( 4800) + i2 (4800) + i3 (5066.67) = 0;
i1 (0) + i2 (1) + i3 ( 1) = 0.015.
Calculator solution:
i1 = 19.933 mA;
i2 = 17.4 mA;
i3 = 2.4 mA.
Calculate voltage across the current source:
v15mA = 4800(i1
i2 ) = 12.16 V.
Calculate power delivered by the sources:
p15mA = (0.015)(12.16) = 182.4 mW (abs);
p60V =
60i1 =
60(0.019933) =
1.196 W (del);
pdelivered = 1.196 W.
Calculate power absorbed by the 2.5 k⌦ resistor and the percentage power:
i2.5k =
5000k2500
i3 = 1.6 mA
2500
p2.5k = (0.0016)2 (2500) = 6.4 mW
% delivered to Ro :
0.0064
(100) = 0.535%
1.196
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4–86
CHAPTER 4. Techniques of Circuit Analysis
P 4.84 [a] From the solution to Problem 4.69 we have
Ro (⌦) po (mW)
100
18.595
120
18.75
150
18.52
180
18
The 120 ⌦ resistor dissipates the most power, because its value is equal to
the Thévenin equivalent resistance of the circuit.
[b]
[c] Ro = 120 ⌦,
po = 18.75 mW, which is the maximum power that can be
delivered to a load resistor.
P 4.85 [a] Since 0  Ro  1 maximum power will be delivered to the 6 ⌦ resistor
when Ro = 0.
302
= 150 W.
[b] P =
6
P 4.86 [a] From the solution of Problem 4.75 we have RTh = 1866.67 ⌦ and
VTh = 112 V. Therefore
Ro = RTh = 1866.67 ⌦.
[b] p =
(56)2
= 1.68 W.
1866.67
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Problems
4–87
[c]
The node voltage equations are:
v1 280
v1
v1 v2
+
+
+ 0.2i
= 0;
2000
2000
2000
v2
v2
v2 v 1
+
+
0.2i
= 0.
2000
5600 1866.67
The dependent source constraint equation is:
280 v1
.
i =
2000
Place
standard
✓ these equations in ◆
✓ form:◆
1
1
1
280
1
+
+
;
+ v2
+ i (0.2) =
v1
2000 2000 2000
2000
2000
✓
◆
✓
◆
1
1
1
1
+
+
+ v2
+ i ( 0.2) = 0;
v1
2000
2000 5600 1866.67
v1 (1) + v2 (0) + i (2000) = 280.
v2 = 56 V;
Solving, v1 = 100 V;
Calculate the power:
p280V = (280)(0.09) = 25.2 W;
pX
v2 )(0.2i ) = 0.792 W;
dep source = (v1
pdev = 25.2 W;
i = 90 mA.
1.68
⇥ 100 = 6.67%.
25.2
[d] The 1.8 k⌦ resistor in Appendix H is closest to the Thévenin equivalent
resistance.
% delivered =
[e] Substitute the 1.8 k⌦ resistor into the original circuit and calculate the
power developed by the sources in this circuit:
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4–88
CHAPTER 4. Techniques of Circuit Analysis
The node voltage equations are:
v1
v1 v2
v1 280
+
+
+ 0.2i
= 0;
2000
2000
2000
v2 v 1
v2
v2
+
+
0.2i
= 0.
2000
5600 1800
The dependent source constraint equation is:
280 v1
i =
.
2000
Place
standard
✓ these equations in ◆
✓ form:◆
1
280
1
1
1
+
+
;
+ v2
+ i (0.2) =
v1
2000 2000 2000
2000
2000
✓
◆
✓
◆
1
1
1
1
+
+
+ v2
+ i ( 0.2) = 0;
v1
2000
2000 5600 1800
v1 (1) + v2 (0) + i (2000) = 280.
v2 = 54.98 V;
Solving, v1 = 99.64 V;
Calculate the power:
pX
(280)(0.09018) = 25.25 W;
280V =
pdev = 25.25 mW;
pL = (54.98)2 /1800 = 1.68 W; % delivered =
i = 90.18 mA.
1.68
⇥ 100 = 6.65%.
25.25
P 4.87 We begin by finding the Thévenin equivalent with respect to Ro . After making
a couple of source transformations the circuit simplifies to
i =
160
30i
;
50
i = 2 A;
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Problems
4–89
VTh = 20i + 30i = 50i = 100 V.
Using the test-source method to find the Thévenin resistance gives
iT =
v T vT
+
30
30( vT /30)
;
20
1
4
2
1
iT
+
=
= ;
=
vT
30 10
30
15
RTh =
vT
15
= 7.5 ⌦.
=
iT
2
Thus our problem is reduced to analyzing the circuit shown below:
p=
✓
100
7.5 + Ro
◆2
Ro = 250;
104
Ro = 250;
Ro2 + 15Ro + 56.25
104 Ro
= Ro2 + 15Ro + 56.25;
250
40Ro = Ro2 + 15Ro + 56.25;
Ro2
25Ro + 56.25 = 0;
Ro = 12.5 ±
p
156.25
56.25 = 12.5 ± 10;
Ro = 22.5 ⌦;
Ro = 2.5 ⌦.
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4–90
CHAPTER 4. Techniques of Circuit Analysis
P 4.88 [a] Open circuit voltage
Node voltage equations:
v1 v2
v1 60 v1 4i
+
+
2
5
4
v2 v 1
+ 2v
4
Constraint equations:
v
=
i
=
60
v1
=
0;
=
0.
v1 ;
v2
.
4
Place
the
equations
in✓standard
form:
✓
◆
◆
✓
◆
1
4
1 1 1
+ +
+ v2
+i
+ v (0)
v1
2 5 4
4
5
✓
◆
✓ ◆
1
1
+ v2
+ i (0) + v (2)
v1
4
4
=
30;
=
0;
v1 (1) + v2 (0) + i (0) + v (1)
= 60;
v1 (1) + v2 ( 1) + i ( 4) + v (0)
=
Solving,
v1 = 20 V;
Short circuit current:
v2 =
300 V;
i = 80 A;
0.
v = 40 V.
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Problems
4–91
The node voltage equation:
v1 60 v1 4i
v1
+
+
= 0.
2
5
4
The constraint equation:
i = v1 /4.
Place
in
form:
✓ these equations
◆
✓ standard
◆
4
1 1 1
v1
+ +
+i
= 30;
2 5 4
5
✓ ◆
1
+ i ( 1)
= 0.
v1
4
Solving,
v1 = 40V; i = 10 A.
Then,
v = 60 40 = 20 V
and
isc = i
2v = 10 40 = 30 A.
Thus,
Ro = RTh = 300/ 30 = 10 ⌦.
[b]
pmax =
(150)2
= 2250 W.
10
[c]
The node voltage equation:
va + 150
va 60 va 4i
+
+
= 0.
2
5
4
The constraint equation is:
va + 150
.
i =
4
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4–92
CHAPTER 4. Techniques of Circuit Analysis
Place
in ✓standard
form:
✓ the equations
◆
◆
4
150
1 1 1
va
+ +
;
+i
= 30
2 5 4
5
4
✓
◆
1
150
.
va
+ i (1)
=
4
4
Solving,
va = 30 V;
i = 45 A.
Calculate the power:
va 60
i60V =
= 15 A;
2
p60V
=
iccvs
=
pccvs
=
4(45)( 30) =
pvccs
=
( 150)[2(30)] =
X
(60)( 15) =
va
4i
5
=
900 W;
30 A;
5400 W;
9000 W.
pdev = 900 + 5400 + 9000 = 15,300 W;
% delivered =
2250
⇥ 100 = 14.7%.
15,3000
P 4.89 [a] Find the Thévenin equivalent with respect to the terminals of RL .
Open circuit voltage:
The mesh current equations are:
3600 + 45(i1
30i2 + 60(i2
i2 ) + 300(i1
i3 ) + 45(i2
150i + 15i3 + 300(i3
i3 ) + 30i1
i1 )
i1 ) + 60(i3
i2 )
=
0;
=
0;
= 0.
The dependent source constraint equation is:
i = i 1 i2 .
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Problems
4–93
Place these equations in standard form:
i1 (45 + 300 + 30) + i2 ( 45) + i3 ( 300) + i (0)
=
3600;
i1 ( 45) + i2 (30 + 60 + 45) + i3 ( 60) + i (0)
=
0;
i1 ( 300) + i2 ( 60) + i3 (15 + 300 + 60) + i ( 150)
= 0;
i1 (1) + i2 ( 1) + i3 (0) + i ( 1)
=
Solving, i1 = 99.6 A; i2 = 78 A;
VTh = 300(i1 i3 ) = 360 V.
Short-circuit current:
i3 = 100.8 A;
0.
i = 21.6 A;
The mesh current equations are:
3600 + 45(i1
30i2 + 60(i2
i2 ) + 30i1
i3 ) + 45(i2
150i + 15i3 + 60(i3
=
i1 )
i2 )
0;
= 0;
= 0.
The dependent source constraint equation is:
i = i 1 i2 .
Place these equations in standard form:
i1 (45 + 30) + i2 ( 45) + i3 (0) + i (0)
= 3600;
i1 ( 45) + i2 (30 + 60 + 45) + i3 ( 60) + i (0)
=
0;
i1 (0) + i2 ( 60) + i3 (60 + 15) + i ( 150)
=
0;
i1 (1) + i2 ( 1) + i3 (0) + i ( 1)
=
0.
Solving,
i1 = 92 A;
isc = i1
i3 =
4 A;
i2 = 73.33 A; i3 = 96 A; i = 18.67 A;
VTh
360
= 90 ⌦.
RTh =
=
isc
4
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4–94
CHAPTER 4. Techniques of Circuit Analysis
RL = RTh = 90 ⌦.
[b] pmax =
1802
= 360 W.
90
P 4.90 [a] We begin by finding the Thévenin equivalent with respect to the terminals
of Ro .
Open circuit voltage
The mesh current equations are:
100 + 4(i1
i2 ) + 80(i1
i3 ) + 16i1
=
0;
124i + 8(i2
i3 ) + 4(i2
i1 )
=
0;
=
0.
i1 (4 + 80 + 16) + i2 ( 4) + i3 ( 80) + i (0)
=
100;
i1 ( 4) + i2 (8 + 4) + i3 ( 8) + i (124)
=
0;
i1 ( 80) + i2 ( 8) + i3 (12 + 80 + 8) + i (0)
=
i1 (1) + i2 (0) + i3 ( 1) + i (1)
= 0.
50 + 12i3 + 80(i3
i1 ) + 8(i3
i2 )
The constraint equation is:
i = i 3 i1 .
Place these equations in standard form:
i2 = 10.5 A;
Solving, i1 = 4.7 A;
Also,
VTh = vab = 80i = 48 V.
Now find the short-circuit current.
i3 = 4.1 A;
50;
i =
0.6 A.
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Problems
4–95
Note with the short circuit from a to b that i is zero, hence 124i is
also zero.
The mesh current equations are:
100 + 4(i1
8(i2
i2 ) + 16i1
=
0;
i3 ) + 4(i2
i1 )
=
0;
50 + 12i3 + 8(i3
i2 )
=
0.
Place these equations in standard form:
i1 (4 + 16) + i2 ( 4) + i3 (0)
=
i1 ( 4) + i2 (8 + 4) + i3 ( 8) =
i1 (0) + i2 ( 8) + i3 (12 + 8)
100;
0;
=
50.
Solving,
i1 = 5 A; i2 = 0 A;
Then,
isc = i1 i3 = 7.5 A;
RTh = 48/7.5 = 6.4 ⌦.
i3 =
2.5 A.
For maximum power transfer Ro = RTh = 6.4 ⌦.
242
[b] pmax =
= 90 W.
6.4
[c] The problem reduces to the analysis of the following circuit. In
constructing the circuit we have used the fact that i is 0.3 A, and
hence 124i is 37.2 V.
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4–96
CHAPTER 4. Techniques of Circuit Analysis
Using the node voltage method to find v1 and v2 yields
v1
100 v1 24 v2 24 v2 50
+
+
+
= 0;
16
4
8
23
v1 = 37.2.
v2
Solving, v1 = 22.4 V;
v2 = 59.6 V.
It follows that
22.4 100
= 4.85 A;
=
ig1
16
59.6 50
= 0.8 A;
ig2
=
12
50 24
= 3.25 A;
i2
=
8
ids
=
3.25
p100V
=
100ig1 =
p50V
=
50ig2 = 40 W;
pds
=
37.2ids =
.·.
X
0.8 =
4.05 A;
485 W;
150.66 W;
pdev = 485 + 150.66 = 635.66 W;
.·. % delivered =
90
(100) = 14.16%.
635.66
.·. 14.16% of developed power is delivered to the load resistor.
[d] The resistor from Appendix H that is closest to the Thévenin resistance is
10 ⌦. To calculate the power delivered to a 10 ⌦ load resistor, calculate
the current using the Thévenin circuit and use it to find the power
delivered to the load resistor:
48
i10 =
= 2.92683 A;
6.4 + 10
p10 = 10(2.92683)2 = 85.6633 W.
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Problems
4–97
Thus, using a 10 ⌦ resistor selected from Appendix H will cause 85.6633
W of power to be delivered to the load, compared to the maximum power
of 90 W that will be delivered if a 6.4 ⌦ resistor is used.
P 4.91 [a] First find the Thévenin equivalent with respect to Ro .
Open circuit voltage: i = 0; 184 = 0.
v1 180 v1 180 v1
v1
+
+
+
16
20
10
10
v =
v1
180
(2) = 0.2v1
10
v1 = 80 V;
v =
VTh = 180 + v = 180
0.1v = 0;
36;
20 V;
20 = 160 V.
Short circuit current
v1
v1 180 v2
v2
+
+
+
16
20
8
10
0.1( 180) = 0;
v2 + 184i = v1 ;
i =
v2 =
180 v2
+
= 90 + 0.125v2 ;
2
8
640 V;
v1 = 1200 V;
i = isc = 10 A;
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4–98
CHAPTER 4. Techniques of Circuit Analysis
RTh = VTh /isc = 160/10 = 16 ⌦;
.·. Ro = 16 ⌦.
[b]
pmax = (80)2 /16 = 400 W.
[c]
v1 180 v2 80 v2
v1
+
+
+
16
20
8
10
v2 + 184i = v1 ;
180
80
2
+
180) = 0;
i = 80/16 = 5 A.
Therefore, v1 = 640 V and v2 =
ig =
0.1(80
180
640
20
280 V; thus,
= 27 A;
p180V (dev) = (180)(27) = 4860 W.
P 4.92 [a] 110 V source acting alone:
Re =
i0 =
35
10(14)
=
⌦;
24
6
110
132
=
A;
5 + 35/6
13
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Problems
✓
35
v =
6
0
◆✓
4–99
◆
770
132
V = 59.231 V;
=
13
13
4 A source acting alone:
5 ⌦k10 ⌦ = 50/15 = 10/3 ⌦;
10/3 + 2 = 16/3 ⌦;
16/3k12 = 48/13 ⌦.
Hence our circuit reduces to:
It follows that
va00 = 4(48/13) = (192/13) V
and
v 00 =
va00
(10/3) =
(16/3)
.·.
v = v 0 + v 00 =
5 00
v =
8 a
770
13
(120/13) V =
9.231 V.
120
= 50 V.
13
v2
[b] p =
= 250 W.
10
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4–100
CHAPTER 4. Techniques of Circuit Analysis
P 4.93 Voltage source acting alone:
io1 =
135
=
40 + 100k25
vo1 =
60
( 135) =
90
2.25 A;
90 V.
Current source acting alone:
v1
v1
+
+ 18 = 0
30 60
18 +
v2
v3
20
.·.
.·.
v1 =
360 V;
vo2 = 360 V;
v 2 v3
v2
+
= 0;
80
20
+
v3
v3
+
= 0;
25 40
v2 = 441.6 V;
.·. vo = vo1 + vo2 =
io = io1 + io2 =
v3 = 192 V;
io2 = 192/40 = 4.8 A.
90 + 360 = 270 V;
2.25 + 4.8 = 2.55 A.
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Problems
4–101
P 4.94 10 V source acting alone:
vo1 =
20
(10) = 5 V.
20 + 5 + 15
20 V source acting alone:
vo2 =
13.333
(20) = 5 V.
13.333 + 10 + 30
6 A current source acting alone:
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4–102
CHAPTER 4. Techniques of Circuit Analysis
Node voltage equations:
v 1 v2
v1
+
6
15
5
v 2 v1
v2
v 2 v3
+
+
5
40
10
v3
v3 v2
+
+6
10
30
=
0;
=
0;
=
0.
In standard form:
✓
◆
✓
◆
1
1
1
v1
+
+ v2
+ v3 (0)
15 5
5
✓
◆
✓
◆
✓
◆
1
1
1
1
1
+
+
+ v2
+ v3
v1
5
5 40 10
10
✓
◆
✓
◆
1
1
1
+
+ v3
v1 (0) + v2
10
10 30
Solving,
Note that
Finally,
= 6;
=
=
0;
6.
v1 = 22.5 V;
v2 = 0 V;
v3 = 45 V.
vo3 = v2 = 0 V.
vo = vo1 + vo2 + vo3 = 5 + 5 + 0 = 10 V.
P 4.95 [a] By hypothesis i0o + i00o = 1.5 mA.
i000
o = 10
(2)
= 1 mA;
(20)
.·. io = 1.5 + 1 = 2.5 mA.
[b] With all three sources in the circuit write a single node voltage equation.
vb
vb 20
+
5
18
2
.·. vb = 45 V.
io =
10 = 0;
vb
= 2.5 mA.
18
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Problems
4–103
P 4.96 4.5 A source:
14 ⌦k10 ⌦k15 ⌦ = 4.2 ⌦;
.·. i0o =
4.5(6)
=
18
1.5 A.
20 A source:
i00o =
4.2(20)
= 4.67 A.
18
50 V source:
i000
o =
4.2( 5)
=
18
1.167 A;
io = i0o + i00o + i000
o =
1.5 + 4.67
1.167 = 2 A.
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4–104
CHAPTER 4. Techniques of Circuit Analysis
P 4.97 90-V source acting alone:
2000(i1
i2 ) + 2.5vb = 90;
2000i1 + 7000i2
4000i3 = 0;
4000i2 + 6000i3
2.5vb = 0;
vb = 1000i2 .
Solving,
i1 = 37.895 mA;
i3 = 30.789 mA;
i 0 = i1
i3 = 7.105 mA.
40-V source acting alone:
2000(i1
i2 ) + 2.5vb = 0;
2000i1 + 7000i2
4000i3 = 0;
4000i2 + 6000i3
2.5vb =
40.
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Problems
4–105
vb = 1000i2
Solving,
i1 = 2.105 mA;
Hence,
i3 =
15.789 mA;
i00 = i1
i3 = 17.895 mA;
i = i0 + i00 = 7.105 + 17.895 = 25 mA.
P 4.98 Voltage source acting alone:
vo1 vo1 35
+
20
5
5
✓
35
vo1
5
◆
= 0;
.·. vo1 = 33.6 V.
Current source acting alone:
vo2
vo2
+7+
20
5
5
✓
vo2
5
◆
= 0;
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4–106
CHAPTER 4. Techniques of Circuit Analysis
.·. vo2 =
5.6 V.
vo = vo1 + vo2 = 33.6
5.6 = 28 V.
P 4.99
The mesh equations are:
i1 (36.3) + i2 ( 0.2) + i3 ( 36) + i4 (0) + i5 (0)
=
120;
i1 ( 0.2) + i2 (45.3) + i3 (0) + i4 ( 45) + i5 (0)
=
120;
i1 ( 36) + i2 (0) + i3 (72.3) + i4 ( 0.2) + i5 ( 36)
= 0;
i1 (0) + i2 ( 45) + i3 ( 0.2) + i4 (90.3) + i5 ( 45)
= 0;
i1 (0) + i2 (0) + i3 ( 36) + i4 ( 45) + i5 (108)
= 0.
Solving,
i1 = 15.226 A; i2 = 13.953 A;
11.314 A; i5 = 8.695 A.
Find the requested voltages:
v1 = 36(i3 i5 ) = 118.6 V;
v2 = 45(i4 i5 ) = 117.8 V;
v3 = 27i5 = 234.8 V.
i3 = 11.942 A;
i4 =
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Problems
4–107
P 4.100
At v1 :
At v2 :
At v3 :
At v4 :
At v5 :
v1
120 v1 v2 v1 v3
+
+
= 0;
10
20
10
v2
120 v2 120 v2 v1 v1 v3 v2 v4
+
+
+
+
= 0;
20
20
20
20
20
v3
v1
10
v4
v2
20
v5
v4
10
+
+
+
v3
v2
20
v4
v3
20
v5
v3
10
+
+
+
v3
v4
20
v4
v5
10
+
v3
v 3 v5
+
= 0;
10
10
= 0;
v5
= 0.
5
A calculator solution yields
v1 = 80 A;
v2 = 80 A;
v3 = 40 A;
v4 = 40 A;
v5 = 20 A.
v2
.·. i1 =
v1
20
= 0 A;
i1 =
v3
v4
20
= 0 A.
P 4.101 [a] In studying the circuit in Fig. P4.101 we note it contains six meshes and
six essential nodes. Further study shows that by replacing the parallel
resistors with their equivalent values the circuit reduces to four meshes
and four essential nodes as shown in the following diagram.
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4–108
CHAPTER 4. Techniques of Circuit Analysis
The node-voltage approach will require solving three node voltage
equations along with equations involving vx , vy , and ix .
The mesh-current approach will require writing one mesh equation and
one supermesh equation plus five constraint equations involving the five
sources. Thus at the outset we know the supermesh equation can be
reduced to a single unknown current. Since we are interested in the
power developed by the 50 V source, we will retain the mesh current ib
and eliminate the mesh currents ia , ic and id .
The supermesh is denoted by the dashed line in the following figure.
[b] Summing the voltages around the supermesh yields
1.25vx + (100/3)ia + (200/3)ib + 50 + 50ib + 250(ic
id ) + 50ic = 0.
The remaining mesh equation is
50ix + 350id
250ic = 0.
The constraint equations are
vy
= i b ic ;
0.9 = ic ia ;
25
vy = 50ib ;
ix = ic
id .
ib =
0.6 A;
vx =
(200/3)ib ;
Solving,
ia =
0.3 A;
ic = 0.6 A;
id = 0.4 A.
Finally,
p50V = 50ib =
30 W.
The 50 V source delivers 30 W of power.
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Problems
4–109
P 4.102 [a]
v
v1
v
v v2
+ +
= 0;
2xr
R 2r(L x)
"
#
v=
v1 RL + xR(v2 v1 )
.
RL + 2rLx 2rx2
1
1
1
v2
v1
v
+ +
+
;
=
2xr R 2r(L x)
2xr 2r(L x)
2rx2 .
[b] Let D = RL + 2rLx
dv
(RL + 2rLx 2rx2 )R(v2 v1 ) [v1 RL + xR(v2
=
dx
D2
dv
= 0 when numerator is zero.
dx
The numerator simplifies to
x2 +
v1 )]2r(L
2x)
;
RL(v2 v1 ) 2rv1 L2
2Lv1
x+
= 0.
(v2 v1 )
2r(v2 v1 )
Solving for the roots of the quadratic yields
x=
[c] x =
L
v2
L
v2
8
<
v1 :
8
<
v1 :
v1 ±
v1 ±
v2 = 1200 V,
v2
v1
=
s
v1 v2
R
(v2
2rL
v1 v2
R
(v1
2rL
v1 = 1000 V,
r = 5 ⇥ 10 5 ⌦/m;
L
s
9
=
v1 )2 ; .
9
=
v2 )2 ; ;
L = 16 km;
R = 3.9 ⌦;
16,000
= 80;
1200 1000
v1 v2 = 1.2 ⇥ 106 ;
3.9( 200)2
= 0.975 ⇥ 105 ;
(10 ⇥ 10 5 )(16 ⇥ 103 )
p
x = 80{ 1000 ± 1.2 ⇥ 106 0.0975 ⇥ 106 };
R
(v1
2rL
v2 )2 =
= 80{ 1000 ± 1050} = 80(50) = 4000 m.
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4–110
CHAPTER 4. Techniques of Circuit Analysis
[d]
vmin
=
v1 RL + R(v2 v1 )x
;
RL + 2rLx 2rx2
=
(1000)(3.9)(16 ⇥ 103 ) + 3.9(200)(4000)
(3.9)(16,000) + 10 ⇥ 10 5 (16,000)(4000) 10 ⇥ 10 5 (16 ⇥ 106 )
=
975 V .
P 4.103 [a]
voc = VTh = 75 V;
Therefore
RTh =
iL =
60
= 3 A;
20
iL =
75 60
15
=
.
RTh
RTh
15
= 5 ⌦.
3
vo
VTh vo
=
.
RL
RTh
✓
VTh vo
VTh
=
Therefore RTh =
vo /RL
vo
[b] iL =
P 4.104
◆
1 RL .
dv1
R1 [R2 (R3 + R4 ) + R3 R4 ]
=
;
dIg1
(R1 + R2 )(R3 + R4 ) + R3 R4
R1 R3 R4
dv1
=
;
dIg2
(R1 + R2 )(R3 + R4 ) + R3 R4
dv2
R1 R 3 R 4
+
;
dIg1 (R1 + R2 )(R3 + R4 ) + R3 R4
R3 R4 (R1 + R2 )
dv2
=
.
dIg2
(R1 + R2 )(R3 + R4 ) + R3 R4
P 4.105 From the solution to Problem 4.104 we have
dv1
=
dIg1
25[5(125) + 3750]
=
30(125) + 3750
175
V/A =
12
14.5833 V/A
and
(25)(50)(75)
dv2
=
=
dIg1
30(125) + 3750
12.5 V/A.
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Problems
By hypothesis,
.·.
v1 = (
Ig1 = 11
12 =
4–111
1 A;
175
175
)( 1) =
= 14.583 V.
12
12
Thus, v1 = 25 + 14.583 = 39.583 V.
Also,
v2 = ( 12.5)( 1) = 12.5 V.
Thus, v2 = 90 + 12.5 = 102.5 V.
The PSpice solution is
v1 = 39.583 V
and
v2 = 102.5 V.
These values are in agreement with our predicted values.
P 4.106 From the solution to Problem 4.104 we have
(25)(50)(75)
dv1
= 12.5 V/A
=
dIg2
30(125) + 3750
and
(50)(75)(30)
dv2
= 15 V/A.
=
dIg2
30(125) + 3750
By hypothesis,
.·.
Ig2 = 17
16 = 1 A;
v1 = (12.5)(1) = 12.5 V.
Thus, v1 = 25 + 12.5 = 37.5 V.
Also,
v2 = (15)(1) = 15 V.
Thus, v2 = 90 + 15 = 105 V .
The PSpice solution is
v1 = 37.5 V
and
v2 = 105 V.
These values are in agreement with our predicted values.
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4–112
CHAPTER 4. Techniques of Circuit Analysis
P 4.107 From the solutions to Problems 4.104 — 4.106 we have
dv1
=
dIg1
175
V/A;
12
dv1
= 12.5 V/A;
dIg2
dv2
=
dIg1
12.5 V/A;
dv2
= 15 V/A.
dIg2
By hypothesis,
Ig1 = 11
12 =
1 A;
Ig2 = 17
16 = 1 .A
Therefore,
v1 =
175
+ 12.5 = 27.0833 V;
12
v2 = 12.5 + 15 = 27.5 V.
Hence
v1 = 25 + 27.0833 = 52.0833 V;
v2 = 90 + 27.5 = 117.5 V.
The PSpice solution is
v1 = 52.0830 V
and
v2 = 117.5 V.
These values are in agreement with our predicted values.
P 4.108 By hypothesis,
R1 = 27.5
25 = 2.5 ⌦;
R2 = 4.5
5=
R3 = 55
50 = 5 ⌦;
R4 = 67.5
75 =
0.5 ⌦;
7.5 ⌦.
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Problems
4–113
So
v1 = 0.5833(2.5)
5.417( 0.5) + 0.45(5) + 0.2( 7.5) = 4.9168 V;
.·. v1 = 25 + 4.9168 = 29.9168 V.
v2 = 0.5(2.5) + 6.5( 0.5) + 0.54(5) + 0.24( 7.5) =
.·. v2 = 90
1.1 V;
1.1 = 88.9 V.
The PSpice solution is
v1 = 29.6710 V
and
v2 = 88.5260 V.
Note our predicted values are within a fraction of a volt of the actual values.
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The Operational Amplifier
5
Assessment Problems
AP 5.1 [a] Derive the expression for the output voltage using circuit analysis:
vo = ( Rf /Ri )vs = ( 80/16)vs ,
so
vo =
5vs ;
vs ( V)
0.4
2.0
3.5
0.6
1.6
2.4;
vo ( V)
2.0
10.0
15.0
3.0
8.0
10.0;
Two of the values, 3.5 V and
2.4 V, cause the op amp to saturate.
[b] Use the negative power supply value to determine the largest input
voltage:
15 =
5vs ,
vs = 3 V.
Use the positive power supply value to determine the smallest input
voltage:
10 =
5vs ,
vs =
2 V;
2 V  vs  3 V.
Therefore
AP 5.2 From Assessment Problem 5.1
vo = ( Rf /Ri )vs = ( Rx /16,000)vs = ( Rx /16,000)( 0.640)
= 0.64Rx /16,000 = 4 ⇥ 10 5 Rx .
Use the negative power supply value to determine one limit on the value of Rx :
4 ⇥ 10 5 Rx =
15
so
Rx =
15/4 ⇥ 10 5 =
375 k⌦.
5–1
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5–2
CHAPTER 5. The Operational Amplifier
Since we cannot have negative resistor values, the lower limit for Rx is 0. Now
use the positive power supply value to determine the upper limit on the value
of Rx :
4 ⇥ 10 5 Rx = 10
so
Rx = 10/4 ⇥ 10 5 = 250 k⌦.
Therefore,
0  Rx  250 k⌦.
AP 5.3 [a] This is an inverting summing amplifier so
vo = ( Rf /Ra )va + ( Rf /Rb )vb =
(250/5)va
(250/25)vb =
50va
10vb
Substituting the values for va and vb :
vo =
50(0.1)
10(0.25) =
5
2.5 =
7.5 V.
[b] Substitute the value for vb into the equation for vo from part (a) and use
the negative power supply value:
vo =
50va
10(0.25) =
50va
Therefore 50va = 7.5,
2.5 =
10 V.
so va = 0.15 V.
[c] Substitute the value for va into the equation for vo from part (a) and use
the negative power supply value:
vo =
50(0.10)
10vb =
Therefore 10vb = 5,
5
10vb =
10 V;
so vb = 0.5 V.
[d] The e↵ect of reversing polarity is to change the sign on the vb term in
each equation from negative to positive.
Repeat part (a):
vo =
50va + 10vb =
5 + 2.5 =
2.5 V.
Repeat part (b):
vo =
50va + 2.5 =
10 V;
50va = 12.5,
va = 0.25 V.
Repeat part (c), using the value of the positive power supply:
vo =
5 + 10vb = 15 V;
10vb = 20;
vb = 2.0 V.
AP 5.4 [a] Write a node voltage equation at vn ; remember that for an ideal op amp,
the current into the op amp at the inputs is zero:
vn
vn vo
+
= 0.
4500
63,000
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Problems
5–3
Solve for vo in terms of vn by multiplying both sides by 63,000 and
collecting terms:
14vn + vn
vo = 0
so
vo = 15vn .
Now use voltage division to calculate vp . We can use voltage division
because the op amp is ideal, so no current flows into the non-inverting
input terminal and the 400 mV divides between the 15 k⌦ resistor and
the Rx resistor:
vp =
Rx
(0.400.)
15,000 + Rx
Now substitute the value Rx = 60 k⌦:
vp =
60,000
(0.400) = 0.32 V.
15,000 + 60,000
Finally, remember that for an ideal op amp, vn = vp , so substitute the
value of vp into the equation for v0
vo = 15vn = 15vp = 15(0.32) = 4.8 V.
[b] Substitute the expression for vp into the equation for vo and set the
resulting equation equal to the positive power supply value:
0.4Rx
vo = 15
15,000 + Rx
!
= 5;
15(0.4Rx ) = 5(15,000 + Rx ) so Rx = 75 k⌦.
AP 5.5 [a] Since this is a di↵erence amplifier, we can use the expression for the
output voltage in terms of the input voltages and the resistor values
given in Eq. 5.22:
vo =
20(60)
vb
10(24)
50
va .
10
Simplify this expression and subsitute in the value for vb :
vo = 5(vb
va ) = 20
5va .
Set this expression for vo to the positive power supply value:
20
5va = 10 V so va = 2 V.
Now set the expression for vo to the negative power supply value:
20
5va =
10 V so va = 6 V;
Therefore 2  va  6 V.
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5–4
CHAPTER 5. The Operational Amplifier
[b] Begin as before by substituting the appropriate values into Eq. 5.8:
8(60)
vb
10(12)
vo =
5va = 4vb
5va .
Now substitute the value for vb :
vo = 4(4)
5va = 16
5va .
Set this expression for vo to the positive power supply value:
16
5va = 10 V so va = 1.2 V.
Now set the expression for vo to the negative power supply value:
16
5va =
10 V so va = 5.2 V;
Therefore 1.2  va  5.2 V.
AP 5.6 Ra = Rc = 1.5 ⌦;
Rb = 12 ⌦.
Then,
Acm =
1.5Rd 1.5(12)
;
1.5(1.5 + Rd
Adm =
Rd (13.5) + 12(1.5 + Rd )
.
2(1.5)(1.5 + Rd )
Set the CMRR equal to 100 and solve for Rd :
Rd (13.5) + 12(1.5 + Rd )
Adm
3(1.5 + Rd )
CMRR =
=
1.5Rd 18
Acm
1.5(1.5 + Rd )
=
25.5Rd + 18
= 100.
3Rd 36
Therefore,
Rd = 13.18 kV.
Set the CMRR equal to
CMRR =
25.5Rd + 18
=
3Rd 36
100 and solve for Rd :
100.
Therefore,
Rd = 11 kV.
Thus,
11 kV  Rd  13.18 kV.
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Problems
5–5
AP 5.7 [a] Replace the op amp with the more realistic model of the op amp from Fig.
5.18:
Write the node voltage equation at the left hand node:
vn vg
v n vo
vn
+
+
= 0.
500,000
5000
100,000
Multiply both sides by 500,000 and simplify:
vn + 100vn
100vg + 5vn
5v0 = 0 so 21.2vn
vo = 20vg .
Write the node voltage equation at the right hand node:
300,000( vn ) vo vn
+
= 0.
5000
100,000
Multiply through by 100,000 and simplify:
vo
20vo + 6 ⇥ 106 vn + vo
vn = 0 so 6 ⇥ 106 vn + 21vo = 0.
Use Cramer’s method to solve for vo :
=
No =
vo =
21.2
1
6 ⇥ 106 21
= 6,000,445.2;
21.2 20vg
0
120 ⇥ 106 vg ;
19.9985vg ;
so
6
6 ⇥ 10
No
=
=
vo
=
vg
19.9985.
[b] Use Cramer’s method again to solve for vn :
N1 =
vn =
20vg
1
0
21
N1
= 420vg ;
= 6.9995 ⇥ 10 5 vg ;
vg = 1 V,
vn = 69.995 µV.
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5–6
CHAPTER 5. The Operational Amplifier
[c] The resistance seen at the input to the op amp is the ratio of the input
voltage to the input current, so calculate the input current as a function
of the input voltage:
vg 6.9995 ⇥ 10 5 vg
v g vn
=
.
5000
5000
Solve for the ratio of vg to ig to get the input resistance:
ig =
Rg =
vg
=
ig
1
5000
= 5000.35 ⌦.
6.9995 ⇥ 10 5
[d] This is a simple inverting amplifier configuration, so the voltage gain is
the ratio of the feedback resistance to the input resistance:
vo
=
vg
100,000
=
5000
20.
Since this is now an ideal op amp, the voltage di↵erence between the two
input terminals is zero; since vp = 0, vn = 0
Since there is no current into the inputs of an ideal op amp, the
resistance seen by the input voltage source is the input resistance:
Rg = 5000 ⌦.
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Problems
5–7
Problems
P 5.1
[a] The five terminals of the op amp are identified as follows:
[b] The input resistance of an ideal op amp is infinite, which constrains the
value of the input currents to 0. Thus, in = 0 A.
[c] The open-loop voltage gain of an ideal op amp is infinite, which constrains
the di↵erence between the voltage at the two input terminals to 0. Thus,
(vp vn ) = 0.
[d] Write a node voltage equation at vn :
vn 2.5 vn vo
+
= 0.
10,000
40,000
But vp = 0 and vn = vp = 0. Thus,
2.5
10,000
P 5.2
vo
= 0 so vo =
40,000
10 V.
[a] Let the value of the voltage source be vs :
v n v s vn v o
+
= 0.
10,000
40,000
But vn = vp = 0. Therefore,
vo =
40,000
vs =
10,000
4vs .
When vs = 6 V, vo = 4( 6) = 24 V; saturates at vo = 15 V.
When vs = 3.5 V, vo = 4( 3.5) = 14 V.
When vs = 1.25 V, vo = 4( 1.25) = 5 V.
When vs = 1 V, vo = 4(1) = 4 V.
When vs = 2.4 V, vo = 4(2.4) = 9.6 V.
When vs = 5.4 V, vo = 4(5.4) = 21.6 V; saturates at vo = 15 V.
15
= 3.75 V;
so
vs =
[b] 4vs = 15
4
15
= 3.75 V.
so
vs =
4vs = 15
4
The range of source voltages that avoids saturation is
3.75 V  vs  3.75 V.
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5–8
P 5.3
CHAPTER 5. The Operational Amplifier
vo =
(0.5 ⇥ 10 3 )(10,000) =
5
vo
=
=
5000
5000
.·. io =
P 5.4
vb
va
20
+
vb vo
= 0,
100
therefore vo = 6vb
5va .
vb = 0 V,
vo =
16 V (sat).
[b] va = 2 V,
vb = 0 V,
vo =
10 V.
[c] va = 2 V,
vb = 1 V,
vo =
4 V.
[d] va = 1 V,
vb = 2 V,
vo = 7 V.
.·.
vo = 9.6
5va = ±16;
1.25  va  5.12 V.
240 ⇥ 10 3
= 30 µA.
8000
0 va
[b]
= ia so va = 60,000ia =
60,000
va
va v o
va
+
+
= 0;
[c]
60,000 40,000
30,000
[a] ia =
.·. vo = 2.25va =
[d] io =
P 5.6
1 mA.
[a] va = 4 V,
[e] If vb = 1.6 V,
P 5.5
5 V;
vp =
1.8 V.
4.05 V.
va v o
vo
+
= 277.5 µA.
20,000
30,000
5000
(6) = 2 V = vn ;
5000 + 10,000
vn + 5 vn vo
+
= 0;
3000
6000
2(2 + 5) + (2
vo ) = 0;
vo = 16 V.
iL =
16
vo
=
= 2000 ⇥ 10 6 ;
8000
8000
iL = 2 mA.
P 5.7
Since the current into the inverting input terminal of an ideal op-amp is zero,
the voltage across the 3.3 M⌦ resistor, positive on the left, is
(3.3 ⇥ 106 )(2.5 ⇥ 10 6 ) or 8.25 V. Therefore the voltmeter reads 8.25 V.
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Problems
P 5.8
5–9
[a] The gain of an inverting amplifier is the negative of the ratio of the
feedback resistor to the input resistor. If the gain of the inverting
amplifier is to be 2.5, the feedback resistor must be 2.5 times as large as
the input resistor. There are many possible designs that use a resistor
value chosen from Appendix H. We present one here that uses 3.3 k⌦
resistors. Use a single 3.3 k⌦ resistor as the input resistor. Then
construct a network of 3.3 k⌦ resistors whose equivalent resistance is
2.5(3.3) = 8.25 k⌦ by connecting two resistors in parallel and connecting
the parallel resistors in series with two other resistors. The resulting
circuit is shown here:
[b] To amplify signals in the range 2 V to 3 V without saturating the op
amp, the power supply voltages must be greater than or equal to the
product of the input voltage and the amplifier gain.
2.5( 2) = 5 V
and
2.5(3) =
7.5 V.
Thus, the power supplies should have values of
P 5.9
[a]
[b]
[c]
30,000
=4
Rin
so
Rin =
7.5 V and 5 V.
30,000
= 7500 = 7.5 k⌦,
4
12
= 3 V;
4
12
4vin = 12
= 3 V;
so
vin =
4
.·.
3 V  vin  3 V.
4vin = 12
so
vin =
Rf
(2) = 12
so
Rf = 45 k⌦;
7500
45,000
vo
Rf
= 6.
=
=
vin
Rin
7500
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5–10
CHAPTER 5. The Operational Amplifier
The amplifier has a gain of 6.
P 5.10
[a] Replace the combination of vg , 1.6 k⌦, and the 6.4 k⌦ resistors with its
Thévenin equivalent.
[12 + 50]
(0.20).
1.28
Then vo =
At saturation vo =
✓
5 V;
◆
12 + 50
(0.2) =
1.28
5,
therefore
or
= 0.4.
Thus for 0 
 0.40 the operational amplifier will not saturate.
(12 + 13.6)
(0.20) = 4 V.
[b] When
= 0.272,
vo =
1.28
vo
vo
+
+ io = 0;
Also
10 25.6
.·. io =
P 5.11
vo
10
vo
4
4
=
+
mA = 556.25 µA.
25.6
10 25.6
[a] Let v be the voltage from the potentiometer contact to ground. Then
0 vg 0 v
+
= 0;
2000
50,000
25vg
.·. v =
v = 0,
v
v
0
vo
v
+
+
↵R
50,000 (1 ↵)R
v
v
+ 2v +
↵
1
v
✓
1 V;
= 0;
vo
= 0;
↵
◆
1
vo
1
+2+
;
=
↵
1 ↵
1 ↵
.·. vo =
"
1 1 + 2(1
When ↵ = 0.2,
When ↵ = 1,
.·.
25(40 ⇥ 10 3 ) =
6.6 V  vo 
↵) +
vo =
vo =
(1
↵)
↵
#
.
1(1 + 1.6 + 4) =
1(1 + 0 + 0) =
6.6 V.
1 V
1 V.
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Problems
[b]
"
1 1 + 2(1
↵) +
↵ + 2↵(1
↵) + (1
↵ + 2↵
#
=
7;
↵) = 7↵;
↵ = 7↵;
1=0
↵⇠
= 0.186.
so
[a] This circuit is an example of an inverting summing amplifier.
220
220
220
[b] vo =
va
vb
vc = 8 + 15 11 = 4 V.
33
22
80
[c] vo = 19 10vb = ±6;
.·. vb =
vb =
.·.
P 5.13
↵)
↵
2↵2 + 1
.·. 2↵2 + 5↵
P 5.12
(1
5–11
1.3 V when vo =
6 V;
2.5 V when vo = 6 V;
2.5 V  vb 
1.3 V.
[a]
120,000
=8
Ra
120,000
=5
Rb
120,000
= 12
Rc
[b]
120,000
= 15 k⌦;
8
120,000
= 24 k⌦;
Rb =
5
120,000
= 10 k⌦.
Rc =
12
so
Ra =
so
so
[8(2) + 5vb + 12( 1)] =
4
4
5vb =
5vb = 11
thus
4
5vb = 15
5vb = 19
thus
15
so
so
5vb ;
11
= 2.2 V;
5
19
= 3.8 V.
vb =
5
vb =
Thus,
3.8 V  vb  2.2 V.
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5–12
P 5.14
CHAPTER 5. The Operational Amplifier
"
vo =
6=
P 5.15
#
Rf
Rf
Rf
(0.15) +
(0.1) +
(0.25) ;
3000
5000
25,000
8 ⇥ 10 5 Rf ;
Rf = 75 k⌦;
.·. 0  Rf  75 k⌦.
We want the following expression for the output voltage:
vo =
(8va + 4vb + 10vc + 6vd ).
This is an inverting summing amplifier, so each input voltage is amplified by a
gain that is the ratio of the feedback resistance to the resistance in the
forward path for the input voltage. Pick a feedback resistor with divisors of 8,
4, 10, and 6 – say 120 k⌦:
vo =
"
#
120k
120k
120k
120k
va +
vb +
vc +
vd .
Ra
Rb
Rc
Rd
Solve for each input resistance value to yield the desired gain:
.·.
Ra = 120,000/8 = 15 k⌦ Rc = 120,000/10 = 12 k⌦;
Rb = 120,000/4 = 30 k⌦
Rd = 120,000/6 = 20 k⌦.
Now create the 5 resistor values needed from the realistic resistor values in
Appendix H. Note that Rf = 120 k⌦, Ra = 15 k⌦, and Rc = 12 k⌦ are already
values from Appendix H. Create Rb = 30 k⌦ by combining two 15 k⌦ resistors
in series. Create Rd = 20 k⌦ by combining two 10 k⌦ resistors in series. Of
course there are many other acceptable possibilities. The final circuit is shown
here:
P 5.16
[a] Write a KCL equation at the inverting input to the op amp:
vd vc
vd
vd vo
v d va vd vb
+
+
+
+
= 0.
40,000
22,000
100,000 352,000 220,000
Multiply both sides by 220,000, plug in the values of the input voltages,
and rearrange to solve for vo :
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Problems
vo = 220,000
5–13
1
5
4
+
+
40,000 22,000 100,000
8
8
+
+
352,000 220,000
!
= 14 V.
[b] Write a KCL equation at the inverting input to the op amp. Use the given
values of input voltages in the equation:
8 va
8 9
8 13
8
8 vo
+
+
+
+
= 0.
40,000 22,000 100,000 352,000 220,000
Simplify and solve for vo :
44
10
5.5va
11 + 5 + 8
vo = 0 so vo = 36
5.5va .
Set vo to the positive power supply voltage and solve for va :
5.5va = 15
36
.·.
va = 3.818 V.
Set vo to the negative power supply voltage and solve for va :
5.5va =
36
15
.·.
va = 9.273 V.
Therefore,
3.818 V  va  9.273 V.
P 5.17
[a]
8 9
8 13
8
8 v0
8 4
+
+
+
+
= 0;
40,000 22,000 100,000 352,000
Rf
8
vo
Rf
=
2.7272 ⇥ 10 5
For vo = 15 V,
For vo =
[b] io =
P 5.18
15 V,
(if + i10k ) =
so Rf =
8 vo
.
2.727 ⇥ 10 5
Rf = 256.7 k⌦;
Rf < 0
"
so this solution is not possible.
#
15
15 8
+
=
3
256.7 ⇥ 10
10,000
1527 µ A.
[a] This circuit is an example of the non-inverting amplifier.
[b] Use voltage division to calculate vp :
vp =
8000
vs
vs = .
8000 + 32,000
5
Write a KCL equation at vn = vp = vs /5:
vs /5 vs /5 vo
+
= 0.
7000
56,000
Solving,
vo = 8vs /5 + vs /5 = 1.8vs .
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5–14
CHAPTER 5. The Operational Amplifier
[c] 1.8vs = 12
P 5.19
so
vs = 6.67 V;
1.8vs =
15
Thus,
8.33 V  vs  6.67 V.
so
vs =
8.33 V.
[a] The circuit shown is a non-inverting amplifier.
[b] We assume the op amp to be ideal, so vn = vp = 2 V. Write a KCL
equation at vn :
2
2 vo
+
= 0.
25,000 150,000
Solving,
vo = 14 V.
P 5.20
[a] vp = vn =
45
vg = 0.6vg ;
75
0.6vg 0.6vg vo
.·.
+
= 0;
15
48
.·. vo = 2.52vg = 2.52(3), vo = 7.56 V.
[b] vo = 2.52vg = ±10.
vg = ±3.97 V,
[c]
3.97 V  vg  3.97 V.
0.6vg 0.6vg vo
+
= 0;
15
Rf
✓
◆
0.6Rf
+ 0.6 vg = vo = ±10;
15
.·. 3Rf + 45 = ±150;
P 5.21
3Rf = 150
45;
Rf = 35 k⌦.
[a] From the equation for the non-inverting amplifier,
R s + Rf
= 2.5
Rs
so
Rs + Rf = 2.5Rs
and therefore
Rf = 1.5Rs .
Choose Rs = 22 k⌦, which is a component in Appendix H. Then
Rf = (1.5)(22) = 33 k⌦, which is also a resistor value in Appendix H.
The resulting non-inverting amplifier circuit is shown here:
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Problems
[b] vo = 2.5vg = 16
vo = 2.5vg =
so
16
5–15
vg = 6.4 V.
so
vg =
6.4 V.
Therefore,
6.4 V  vg  6.4 V.
P 5.22
[a] From Eq. 5.7,
vo =
R s + Rf
vg
Rs
so
vo
Rf
=1+
= 6.
vg
Rs
So,
Rf
= 5.
Rs
Thus,
Rs =
75,000
Rf
=
= 15 k⌦.
5
5
[b] vo = 6vg.
When vg = 2.5 V, vo = 6( 2.5) = 15 V.
When vg = 1.5 V, vo = 6(1.5) = 9 V.
The power supplies can be set at 9 V and 15 V.
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5–16
P 5.23
CHAPTER 5. The Operational Amplifier
[a] This circuit is an example of a non-inverting summing amplifier.
[b] Write a KCL equation at vp and solve for vp in terms of vs :
vp 6
vp vs
+
= 0;
15,000
30,000
2vs + vp
2vp
6=0
so
vp = 2vs /3 + 2.
Now write a KCL equation at vn and solve for vo :
vn v o
vn
+
=0
so
vo = 4vn .
20,000
60,000
Since we assume the op amp is ideal, vn = vp . Thus,
vo = 4(2vs /3 + 2) = 8vs /3 + 8.
[c] 8vs /3 + 8 = 16
so
8vs /3 + 8 =
Thus,
P 5.24
[a]
va
vp
Ra
16
vs = 3 V.
so
9 V  vs  3 V.
+
vp
vb
Rb
+
vp
Rc
vs =
vc
9 V.
= 0;
R b Rc
Ra R c
Ra R b
.·. vp =
va +
vb +
vc ;
D
D
D
where D = Rb Rc + Ra Rc + Ra Rb .
v n vo
vn
+
= 0;
20,000 100,000
!
100,000
+ 1 vn = 6vn = vo ;
20,000
6Rb Rc
6Ra Rc
6Ra Rb
.·. vo =
va +
vb +
vc .
D
D
D
By hypothesis,
6Rb Rc
= 1;
D
Then
3
6Ra Rb /D
=
6Ra Rc /D
2
6Ra Rc
= 2;
D
so
6Ra Rb
= 3.
D
Rb = 1.5Rc .
But from the circuit
Rb = 15 k⌦
so
Rc = 10 k⌦.
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Problems
5–17
Similarly,
1
6Rb Rc /D
=
6Ra Rb /D
3
so
3Rc = Ra .
Thus,
Ra = 30 k⌦.
[b] vo = 1(0.7) + 2(0.4) + 3(1.1) = 4.8 V
vn = vo /6 = 0.8 V = vp ;
ia =
v a vp
0.7 0.8
=
=
30,000
30,000
3.33 µA;
ib =
0.4 0.8
v b vp
=
=
15,000
15,000
26.67 µA;
ic =
1.1 0.8
v c vp
=
= 30 µA.
10,000
10,000
Check:
ia + ib + ic = 0?
P 5.25
3.33
26.67 + 30 = 0 (checks).
[a] This is a di↵erence amplifier circuit.
[b] Use Eq. 5.8 with Ra = 5 k⌦, Rb = 20 k⌦, Rc = 8 k⌦, Rd = 2 k⌦, and
vb = 5 V:
20
Rd (Ra + Rb )
Rb
2(5 + 20)
vb
(5)
va = 5 4va .
va =
vo =
Ra (Rc + Rd )
Ra
5(8 + 2)
5
[c]
1
1
P 5.26
Rf
5000 + Rf
(2) =
5000
5000
2000(5000 + Rf )
(5)
5000(8000 + 2000)
vp =
Rf
= 10
5000
Rf
= 10
5000
1500
( 18) =
9000
3 + 18
+
1600
But
Rf
Rf < 0 which is not a possible solution.
so
Rf = 5000(11) = 55 k⌦.
3 vo
= 0;
Rf
vo = 9 V;
9 V;
Rf
;
5000
3 V = vn ;
.·. vo = 0.009375Rf
vo =
so
2Rf
=1
5000
3.
Rf = 1280 ⌦;
Rf =
0,
640 ⌦;
.·. Rf = 1.28 k⌦.
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5–18
P 5.27
CHAPTER 5. The Operational Amplifier
[a] vo =
Rd (Ra + Rb )
vb
Ra (Rc + Rd )
vo = 5.17
[b] vn = vp =
6.70 =
Rb
47(110)
(0.80)
va =
Ra
10(80)
10(0.67);
1.53 V.
(800)(47)
= 470 mV;
80
(670 470)10 3
ia =
= 20 µA;
10 ⇥ 103
Rin a =
va
670 ⇥ 10 3
=
= 33.5 k⌦.
ia
20 ⇥ 10 6
[c] Rin b = Rc + Rd = 80 k⌦.
P 5.28
Rd (Ra + Rb )
vb
Ra (Rc + Rd )
vo =
Rb
va .
Ra
By hypothesis: Rb /Ra = 4;
Rc + Rd = 470 k⌦;
Rd (Ra + 4Ra )
=3
.·.
Ra 470,000
Rd = 282 k⌦;
so
Rd (Ra + Rb )
= 3;
Ra (Rc + Rd )
Rc = 188 k⌦.
Create Rd = 282 k⌦ by combining a 270 k⌦ resistor and a 12 k⌦ resistor in
series. Create Rc = 188 k⌦ by combining a 120 k⌦ resistor and a 68 k⌦ resistor
in series. Also, when vo = 0 we have
vn
va
Ra
+
vn
= 0;
Rb
✓
◆
va
va
0.8va
= 0.2 ;
Ra
Ra
Ra
.·. vn 1 +
Rb
ia =
.·. Ra = 4.4 k⌦;
= va ;
vn = 0.8va .
Rin =
va
= 5Ra = 22 k⌦;
ia
Rb = 17.6 k⌦.
Create Ra = 4.4 k⌦ by combining two 2.2 k⌦ resistors in series. Create
Rb = 17.6 k⌦ by combining a 12 k⌦ resistor and a 5.6 k⌦ resistor in series.
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Problems
P 5.29
5–19
[a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0,
therefore vn = 0 and we have
0
va
Ra
+
vo0
0
+ in = 0,
Rb
Therefore
va
vo0
=
,
Rb
Ra
in = 0.
Rb
va .
Ra
vo0 =
Assume vb is acting alone. Replace va with a short circuit. Now
vb Rd
;
R c + Rd
vp = v n =
vn vo00
vn
+
+ in = 0,
Ra
Rb
in = 0;
✓
vo00
= 0;
Rb
◆✓
1
1
+
Ra Rb
vo00 =
✓
◆
Rd
vb
R c + Rd
◆✓
Rb
+1
Ra
◆
Rd
Rd
vb =
R c + Rd
Ra
Rd
vo = vo0 + vo00 =
Ra
Rd
[b]
Ra
✓
R a + Rb
R c + Rd
◆
=
✓
Rb
,
Ra
Rd Ra = R b Rc ,
Rd
When
Ra
✓
◆
R a + Rb
vb
R c + Rd
◆
=
Eq. 5.8 reduces to vo =
P 5.30
◆
R a + Rb
vb ;
R c + Rd
Rb
va .
Ra
therefore Rd (Ra + Rb ) = Rb (Rc + Rd );
therefore
R a + Rb
R c + Rd
✓
Ra
Rc
=
.
Rb
Rd
Rb
.
Ra
Rb
vb
Ra
Rb
Rb
va =
(vb
Ra
Ra
va ).
v p = R b ib = v n ;
Rb ib Rb ib vo
+
2000
Rf
.·.
.·.
1
1
+
2000 Rf
R b ib
R b ib
ia = 0;
✓
Rf
1+
2000
◆
!
ia =
vo
;
Rf
R f ia = v o .
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5–20
CHAPTER 5. The Operational Amplifier
By hypopthesis, vo = 8000(ib
Rf = 8 k⌦
✓
Rb 1 +
ia ). Therefore,
(use 3.3 k⌦ and 4.7 k⌦ resistors in series);
◆
8000
= 8000
2000
so
Rb = 1.6 k⌦.
To construct the 1.6 k⌦ resistor, combine 270 ⌦, 330 ⌦, and 1 k⌦ resistors in
series.
P 5.31
[a]
vp
v p v c vp v d
+
+
= 0;
20,000
30,000
20,000
.·. 8vp = 2vc + 3vd = 8vn .
v n vo
vn va vn vb
+
+
= 0;
20,000
18,000
180,000
.·. vo
=
20vn
=
20[(1/4)vc + (3/8)vd ]
9va
=
20(0.75 + 1.5)
10(2) = 16 V.
[b] vo = 5vc + 30
9
9va
10vb
9(1)
10vb
20 = 5vc + 1;
±20 = 5vc + 1;
.·. vb =
.·.
4.2 V
and
vb = 3.8 V;
4.2 V  vb  3.8 V.
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Problems
P 5.32
P 5.33
5–21
95(100 + 5) + 100(5 + 95)
= 19.975.
2(5)(5 + 95)
(5)(95) 5(100)
[b] Acm =
= 0.05.
(5)(5 + 95)
19.975
= 399.5.
[c] CMRR =
0.05
[a] Adm =
Acm =
(20)(50) (50)Rx
;
20(50 + Rx )
Adm =
50(20 + 50) + 50(50 + Rx )
;
2(20)(50 + Rx )
Rx + 120
Adm
;
=
Acm
2(20 Rx )
.·.
Rx + 120
= ±1000 for the limits on the value of Rx .
2(20 Rx )
If we use +1000 Rx = 19.93 k⌦;
If we use
1000 Rx = 20.07 k⌦;
19.93 k⌦  Rx  20.07 k⌦.
P 5.34
↵Rg
[a] vp =
vg
↵Rg + (Rg ↵Rg )
vo
vn = vp = ↵vg
v n v g vn v o
+
=0
R1
Rf
(vn
vg )
Rf
+ vn
R1
vo = 0
↵
vo
↵
vo
↵
0.0
8V
0.4
4V
0.8 0 V
0.1
7V
0.5
3V
0.9 1 V
0.2
6V
0.6
2V
1.0 2 V
0.3
5V
0.7
1V
=
✓
◆
=
(↵vg
vg )4 + ↵vg ;
=
[(↵
1)4 + ↵]vg ;
=
(5↵
4)vg
=
(5↵
4)(2) = 10↵
Rf
1+
↵vg
R1
Rf
vg ;
R1
8.
vo
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5–22
CHAPTER 5. The Operational Amplifier
[b] Rearranging the equation for vo from (a) gives
vo =
✓
◆
Rf
+ 1 vg ↵ +
R1
✓
◆
Rf
vg .
R1
Therefore,
✓
◆
Rf
+ 1 vg ;
slope =
R1
intercept =
✓
◆
Rf
vg .
R1
[c] Using the equations from (b),
6=
✓
◆
Rf
+ 1 vg ;
R1
✓
4=
◆
Rf
vg .
R1
Solving,
vg =
P 5.35
Rf
= 2.
R1
2 V;
[a] vp = vs ,
vn =
R 1 vo
,
R 1 + R2
✓
vn = v p .
◆
✓
◆
R2
R 1 + R2
Therefore vo =
vs = 1 +
vs .
R1
R1
[b] vo = vs .
[c] Because vo = vs , thus the output voltage follows the signal voltage.
P 5.36
It follows directly from the circuit that vo = 16vg .
From the plot of vg we have vg = 0, t < 0.
vg
=
vg
=
vg
=
vg
=
vg
=
0  t  0.5;
t
t + 1 0.5  t  1.5;
t
2
1.5  t  2.5;
t + 3 2.5  t  3.5;
t
4
3.5  t  4.5,
etc.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
Therefore
vo =
vo
=
vo
=
vo
=
0  t  0.5;
16t
16
0.5  t  1.5;
16t + 32
1.5  t  2.5;
16t
16t
5–23
48
2.5  t  3.5;
vo =
16t + 64 3.5  t  4.5, etc.
These expressions for vo are valid as long as the op amp is not saturated.
Since the peak values of vo are ±5, the output is clipped at ±5. The plot is
shown below.
P 5.37
vp =
5.4
vg = 0.75vg = 3 cos(⇡/4)t V;
7.2
vn v o
vn
+
= 0;
20,000
60,000
4vn = vo ;
vn = v p ;
.·. vo = 12 cos(⇡/4)t V
0  t  1,
but saturation occurs at vo = ±10 V.
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5–24
P 5.38
CHAPTER 5. The Operational Amplifier
[a]
vn
va
R
2vn
+
vn
vo
R
= 0;
va = v o ;
va v n v a v o
va
+
= 0;
+
Ra
R
R
va

2
1
+
Ra R
vo
vn
= ;
R
R
va
✓
vn = v o
R
2+
Ra
◆
v n = v p = v a + vg ;
.·. 2vn
va = 2va + 2vg
.·. va
vo =
2va + va
✓
R
Ra
✓
2vg
(1).
va
vg = v o ;
◆
R
.·. va 1 +
Ra
va = va + 2vg ;
◆
vo = v g .
(2)
Now combining equations (1) and (2) yields
va
R
=
Ra
3vg ,
or va = 3vg
Ra
.
R
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
Hence ia =
va
3vg
=
Ra
R
5–25
Q.E.D.
[b] At saturation vo = ± Vcc ,
.·. va = ± Vcc
and
✓
R
.·. va 1 +
Ra
2vg
◆
(3)
= ± Vcc + vg .
(4)
Dividing Eq (4) by Eq (3) gives
1+
± Vcc + vg
R
=
;
Ra
± Vcc 2vg
.·.
± Vcc + vg
R
=
Ra
± Vcc 2vg
or Ra =
P 5.39
1=
(± Vcc 2vg )
R
3vg
3vg
± Vcc
2vg
,
Q.E.D.
[a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output
voltage of the amplifier on the right. Then
vo1 =
ia =
47
(1) =
10
4.7 V;
220
( 0.15) = 1.0 V;
33
vo2 vo1
= 5.7 mA.
1000
[b] ia = 0 when vo1 = vo2
Thus
47
(vL ) = 1;
10
vL =
vo2 =
10
=
47
so from (a) vo2 = 1 V.
212.77 mV.
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5–26
CHAPTER 5. The Operational Amplifier
P 5.40
i1 =
15 10
= 1 mA;
5000
i2 + i1 + 0 = 10 mA;
i2 = 9 mA;
vo2 = 10 + (400)(9) ⇥ 10 3 = 13.6 V;
i3 =
15
13.6
= 0.7 mA;
2000
i4 = i3 + i1 = 1.7 mA;
vo1 = 15 + 1.7(0.5) = 15.85 V.
P 5.41
[a] Assume the op-amp is operating within its linear range, then
iL =
8
= 2 mA.
4000
For RL = 4 k⌦
vo = (4 + 4)(2) = 16 V
Now since vo < 20 V our assumption of linear operation is correct,
therefore
iL = 2 mA.
[b] 20 = 2(4 + RL );
RL = 6 k⌦.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
5–27
[c] As long as the op-amp is operating in its linear region iL is independent of
RL . From (b) we found the op-amp is operating in its linear region as
long as RL  6 k⌦. Therefore when RL = 6 k⌦ the op-amp is saturated.
We can estimate the value of iL by assuming ip = in ⌧ iL . Then
iL = 20/(4000 + 16,000) = 1 mA. To justify neglecting the current into
the op-amp assume the drop across the 50 k⌦ resistor is negligible, since
the input resistance to the op-amp is at least 500 k⌦. Then
ip = in = (8 4)/(500 ⇥ 103 ) = 8 µA. But 8 µA ⌧ 1 mA, hence our
assumption is reasonable.
[d]
P 5.42
(320 ⇥ 10 3 )2
= 6.4 µW.
(16 ⇥ 103 )
✓ ◆
16
[b] v16 k⌦ =
(320) = 80 mV;
64
[a] p16 k⌦ =
p16 k⌦ =
(80 ⇥ 10 3 )2
= 0.4 µW.
(16 ⇥ 103 )
6.4
pa
= 16.
=
pb
0.4
[d] Yes, the operational amplifier serves several useful purposes:
[c]
• First, it enables the source to control 16 times as much power
delivered to the load resistor. When a small amount of power controls
a larger amount of power, we refer to it as power amplification.
• Second, it allows the full source voltage to appear across the load
resistor, no matter what the source resistance. This is the voltage
follower function of the operational amplifier.
• Third, it allows the load resistor voltage (and thus its current) to be
set without drawing any current from the input voltage source. This
is the current amplification function of the circuit.
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5–28
P 5.43
CHAPTER 5. The Operational Amplifier
From Eq. 5.28,
1
1
vref
1
= vn
+
+
R+ R
R+ R R
R Rf
!
vo
.
Rf
Substituting Eq. 5.30 for vp = vn :
vref
=
R+ R
(R
vref
⇣
1
+ R 1 R + R1f
R+ R
R)
Rearranging,
⇣
✓
vo
1
= vref
Rf
R
R
⌘
1
+ R 1 R + R1f
R+ R
⌘
vo
.
Rf
◆
1
.
R+ R
Thus,
vo = vref
P 5.44
!
2 R
Rf .
2
R
( R)2
[a] Replace the op amp with the model from Fig. 5.18:
Write two node voltage equations, one at the left node, the other at the
right node:
vn vg
v n vo
vn
+
+
= 0;
5000
100,000 500,000
v o vn
vo
vo + 3 ⇥ 105 vn
+
+
= 0.
5000
100,000 500
Simplify and place in standard form:
5vo = 100vg ;
106vn
(6 ⇥ 106
1)vn + 221vo = 0.
Let vg = 1 V and solve the two simultaneous equations:
vo =
19.9844 V;
vn = 736.1 µV.
Thus the voltage gain is vo /vg =
19.9844.
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Problems
5–29
[b] From the solution in part (a), vn = 736.1 µV.
[c] ig =
736.1 ⇥ 10 6 vg
;
5000
vg
v g vn
=
5000
Rg =
vg
=
ig
1
5000
= 5003.68 ⌦.
736.1 ⇥ 10 6
[d] For an ideal op amp, the voltage gain is the ratio between the feedback
resistor and the input resistor:
vo
=
vg
100,000
=
5000
20.
For an ideal op amp, the di↵erence between the voltages at the input
terminals is zero, and the input resistance of the op amp is infinite.
Therefore,
vn = vp = 0 V;
P 5.45
Rg = 5000 ⌦.
[a]
vn
0.88
vn
vn vTh
+
+
= 0;
1600
500,000
24,000
vTh + 105 vn vTh vn
+
= 0.
2000
24,000
Solving, vTh =
13.198 V.
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5–30
CHAPTER 5. The Operational Amplifier
Short-circuit current calculation:
vn 0.88 vn 0
vn
+
+
= 0;
500,000
1600
24,000
.·. vn = 0.8225 V.
isc =
vn
24,000
RTh =
105
vn =
2000
41.13 A;
vTh
= 320.9 m⌦.
isc
[b] The output resistance of the inverting amplifier is the same as the
Thévenin resistance, i.e.,
Ro = RTh = 320.9 m⌦.
[c]
✓
◆
330
vo =
( 13.2) =
330.3209
13.18 V.
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Problems
vn
5–31
vn
vn + 13.18
0.88
+
+
= 0;
1600
500,000
24,000
.·. vn = 942 µV.
ig =
P 5.46
0.88
942 ⇥ 10 6
= 549.41 µA;
1600
Rg =
0.88
= 1601.71 ⌦.
ig
[a] vTh =
24,000
(0.88) =
1600
13.2 V;
RTh = 0, since op-amp is ideal.
[b] Ro = RTh = 0 ⌦.
[c] Rg = 1.6 k⌦
since
vn = 0.
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5–32
P 5.47
CHAPTER 5. The Operational Amplifier
[a]
vn vo
vn vg
+
= 0;
15,000
135,000
.·. vo = 10vn
Also
9vg .
vo = A(vp
.·. vn =
Avn .
vo
;
A
✓
◆
10
=
.·. vo 1 +
A
vo =
vn ) =
9vg .
9A
vg .
(10 + A)
9(90)(0.4)
= 3.24 V.
(10 + 90)
[c] vo = 9(0.4) = 3.60 V.
9(0.4)A
[d] 3.42 =
;
10 + A
[b] vo =
.·. A = 190.
P 5.48
[a] Replace the op amp with the model shown in Fig. 5.18. The node voltage
equation at the inverting input:
vn
vn vg
vn vo
+
+
= 0.
40,000 500,000
80,000
Simplify:
12.5vn + vn
vg + 6.25vn
6.25vo = 0.
The node voltage equation at the op amp output:
vo
vo
+
4000
20,000(vp
5000
vn )
+
vo vn
= 0.
80,000
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Problems
5–33
Simplify:
20vo + 16vo
320,000(vp
v n ) + vo
vn = 0.
From the input,
vp
vn = 0.8(vg
vn ).
Substituting into the equation written at the output,
20vo + 16vo
256,000(vg
v n ) + vo
vn = 0.
Now let vg = 1 V; plug this value into both the input and output
equations and simplify into two simultaneous equations:
19.75vn
6.25vo = 1;
255,999vn + 37vo = 256,000.
These equations are in standard form, so solve them to yield
vo = 2.9986 V; vn = 999.571 mV.
Thus,
vo
2.9986
= 2.9986.
=
vg
1
[b] From part (a), vn = 999.571 mV. Use this value to solve for vp :
vp = 0.8(1
[c] vp
vn ) + vn = 999.914 mV.
vn = 343.6 µV.
v g vp
1 999.914 ⇥ 10 3
=
= 859 pA.
100,000
100,000
[e] For an ideal op amp, vn = vp = vg , so the KVL equation at the inverting
node is
vo
vg v o
+
= 0.
40,000
80,000
[d] ig =
Then,
vo = 3vg
so
vo
= 3.
vg
Also,
vn = vp = 1 V;
P 5.49
vp
vn = 0 V;
ig = 0 A.
[a] Use the approximation for Eq. 5.31 to solve for Rf ; note that since we are
using 1% strain gages, = 0.01:
Rf =
vo R
(5)(120)
= 2 k⌦.
=
2 vref
(2)(0.01)(15)
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5–34
CHAPTER 5. The Operational Amplifier
[b] Now solve for
=
given vo = 50 mV:
(0.05)(120)
vo R
= 100 ⇥ 10 6 .
=
2Rf vref
2(2000)(15)
The change in strain gage resistance that corresponds to a 50 mV change
in output voltage is thus
R = (100 ⇥ 10 6 )(120) = 12 m⌦.
R=
P 5.50
[a]
Let R1 = R + R
vp
vp vp vin
+
+
= 0;
Rf
R
R1
.·. vp
"
#
1
1
vin
1
+ +
=
;
Rf
R R1
R1
.·. vp =
RRf vin
= vn .
RR1 + Rf R1 + Rf R
vn vn vin vn vo
+
+
= 0;
R
R
Rf
vn
"
1
1
1
+ +
R R Rf
.·. vn
"
R + 2Rf
RRf
"
#
vo
vin
;
=
Rf
R
#
vo
vin
=
R
Rf ;
R + 2Rf
vo
=
.·.
Rf
RRf
.·.
"
#"
RRf vin
RR1 + Rf R1 + Rf R
vo
R + 2Rf
=
Rf
RR1 + Rf R1 + Rf R
#
vin
;
R
#
1
vin ;
R
[R2 + 2RRf R1 (R + Rf ) RRf ]Rf
.·. vo =
vin .
R[R1 (R + Rf ) + RRf ]
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Problems
Now substitute R1 = R +
vo =
5–35
R and get
R(R + Rf )Rf vin
.
R[(R + R)(R + Rf ) + RRf ]
R⌧R
(R + Rf )Rf (
R)vin
vo ⇡
.
2
R (R + 2Rf )
If
[b] vo ⇡
P 5.51
47 ⇥ 104 (48 ⇥ 104 )( 95)15
⇡
108 (95 ⇥ 104 )
3.384 V.
[c] vo =
95(48 ⇥ 104 )(47 ⇥ 104 )15
=
104 [(1.0095)104 (48 ⇥ 104 ) + 47 ⇥ 108 ]
[a] vo ⇡
R)vin
(R + Rf )Rf (
;
2
R (R + 2Rf )
vo =
3.368 V.
(R + Rf )(
R)Rf vin
;
R[(R + R)(R + Rf ) + RRf ]
R[(R + R)(R + Rf ) + RRf ]
approx value
=
;
.·.
true value
R2 (R + 2Rf )
R[(R +
.·. Error =
=
.·. % error =
[b] % error =
P 5.52
R)(R + Rf ) + RRf ]
R2 (R + 2Rf )
R2 (R + 2Rf )
R (R + Rf )
.
R (R + 2Rf )
R(R + Rf )
⇥ 100.
R(R + 2Rf )
95(48 ⇥ 104 ) ⇥ 100
= 0.48%.
104 (95 ⇥ 104 )
1=
R(48 ⇥ 104 )
⇥ 100;
104 (95 ⇥ 104 )
.·.
R=
9500
= 197.91667 ⌦;
48
197.19667
.·. % change in R =
⇥ 100 ⇡ 1.98%.
104
P 5.53
[a] It follows directly from the solution to Problem 5.50 that
vo =
[R2 + 2RRf R1 (R + Rf ) RRf ]Rf vin
.
R[R1 (R + Rf ) + RRf ]
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5–36
CHAPTER 5. The Operational Amplifier
Now R1 = R
vo =
(R + Rf )Rf ( R)vin
.
R[(R
R)(R + Rf ) + RRf ]
R ⌧ R and get
Now let
vo ⇡
R. Substituting into the expression gives
(R + Rf )Rf Rvin
.
R2 (R + 2Rf )
[b] It follows directly from the solution to Problem 5.50 that
.·.
R[(R
approx value
=
true value
.·. Error =
(R
=
R = 9810 ⌦
.·. vo ⇡
R)(R + Rf ) + RRf
R(R + 2Rf )
R(R + 2Rf )
R(R + Rf )
.
R(R + 2Rf )
.·. % error =
[c] R
R)(R + Rf ) + RRf ]
.
R2 (R + 2Rf )
R(R + Rf )
⇥ 100.
R(R + 2Rf )
.·.
R = 10,000
9810 = 190 ⌦;
(48 ⇥ 104 )(47 ⇥ 104 )(190)(15)
⇡ 6.768 V;
108 (95 ⇥ 104 )
[d] % error =
190(48 ⇥ 104 )(100)
=
104 (95 ⇥ 104 )
0.96%.
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Inductance, Capacitance, and
Mutual Inductance
6
Assessment Problems
AP 6.1 [a] ig = 8e 300t
v=L
8e 1200t A;
dig
=
dt
v(0+ ) =
9.6e 300t + 38.4e 1200t V,
t > 0+ ;
9.6 + 38.4 = 28.8 V.
[b] v = 0 when 38.4e 1200t = 9.6e 300t
or t = (ln 4)/900 = 1.54 ms.
[c] p = vi = 384e 1500t 76.8e 600t 307.2e 2400t W.
dp
= 0 when e1800t 12.5e900t + 16 = 0.
[d]
dt
Let x = e900t
and solve the quadratic x2
x = 1.44766,
t=
ln 1.45
= 411.05 µs;
900
x = 11.0523,
t=
ln 11.05
= 2.67 ms;
900
12.5x + 16 = 0.
p is maximum at t = 411.05 µs.
[e] pmax = 384e 1.5(0.41105)
76.8e 0.6(0.41105)
307.2e 2.4(0.41105) = 32.72 W.
[f ] W is max when i is max, i is max when di/dt is zero.
When di/dt = 0, v = 0, therefore t = 1.54 ms.
[g] imax = 8[e 0.3(1.54)
e 1.2(1.54) ] = 3.78 A;
wmax = (1/2)(4 ⇥ 10 3 )(3.78)2 = 28.6 mJ.
6–1
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6–2
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
AP 6.2 [a] i = C
dv
d
= 24 ⇥ 10 6 [e 15,000t sin 30,000t]
dt
dt
0.36 sin 30,000t]e 15,000t A,
= [0.72 cos 30,000t
✓
◆
31.66 mA,
p = vi =
649.23 mW.
⇡
[b] i
ms =
80
[c] w =
✓
i(0+ ) = 0.72 A.
◆
⇡
ms = 20.505 V,
v
80
✓ ◆
1
Cv 2 = 126.13 µJ.
2
✓
◆
1 Zt
i dx + v(0 )
AP 6.3 [a] v =
C 0
=
Z t
1
3 cos 50,000x dx = 100 sin 50,000t V.
0.6 ⇥ 10 6 0
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t
= 150 sin 100,000t W,
[c] w(max) =
p(max) = 150 W.
✓ ◆
1
2
Cvmax
= 0.30(100)2 = 3000 µJ = 3 mJ.
2
60(240)
= 48 mH.
300
[b] i(0+ ) = 3 + 5 = 2 A.
1 Zt
[c] i =
( 0.03e 5x ) dx 2 =)0.125e 5t 2.125) A.
0.048 0+
1 Zt
( 0.03e 5x ) dx + 3 = (0.1e 5t + 2.9) A;
[d] i1 =
+
0.06 0
AP 6.4 [a] Leq =
1 Zt
( 0.03e 5x ) dx
i2 =
+
0.24 0
5 = (0.025e 5t
5.025) A;
i1 + i2 = i.
AP 6.5 v1 =
Z t
1
240 ⇥ 10 6 e 10x dx
2 ⇥ 10 6 0+
10 = ( 12e 10t + 2) V;
v2 =
Z t
1
240 ⇥ 10 6 e 10x dx
8 ⇥ 10 6 0+
5 = ( 3e 10t
v1 (1) = 2 V,
W =

v2 (1) =
2) V;
2 V;
1
1
(2)(2)2 + (8)( 2)2 ⇥ 10 6 = 20 µJ.
2
2
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Problems
6–3
AP 6.6 [a] Summing the voltages around mesh 1 yields
di1
d(i2 + ig )
+8
+ 20(i1
dt
dt
or
4
di2
di1
+ 25i1 + 8
4
dt
dt
i2 ) + 5(i1 + ig ) = 0
!
dig
5ig + 8
.
dt
20i2 =
Summing the voltages around mesh 2 yields
16
d(i2 + ig )
di1
+8
+ 20(i2
dt
dt
i1 ) + 780i2 = 0
or
di2
dig
di1
20i1 + 16
+ 800i2 = 16 .
8
dt
dt
dt
[b] From the solutions given in part (b)
i1 (0) =
0.4
11.6 + 12 = 0;
i2 (0) =
0.01
0.99 + 1 = 0.
These values agree with zero initial energy in the circuit. At infinity,
i1 (1) =
0.4A;
i2 (1) =
0.01A.
When t = 1 the circuit reduces to
✓
.·. i1 (1) =
◆
7.8
7.8
+
=
20
780
0.4A;
i2 (1) =
7.8
=
780
0.01A.
From the solutions for i1 and i2 we have
di1
= 46.40e 4t
dt
60e 5t ;
di2
= 3.96e 4t
dt
5e 5t .
Also,
dig
= 7.84e 4t .
dt
Thus
di1
= 185.60e 4t
4
dt
25i1 =
10
240e 5t ;
290e 4t + 300e 5t ;
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6–4
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
8
di2
= 31.68e 4t
dt
20i2 =
19.80e 4t + 20e 5t ;
0.20
9.8e 4t ;
5ig = 9.8
8
40e 5t ;
dig
= 62.72e 4t .
dt
Test:
185.60e 4t
240e 5t
?
+0.20 + 19.80e 4t
9.8 + (300
240
20e 5t =
40
20)e 5t
?
?
9.8 + 0e 5t + (237.08
52.92e 4t =
40e 5t
9.8e 4t + 62.72e 4t ];
[9.8
290 + 31.68 + 19.80)e 4t =
+(185.60
9.8
290e 4t + 300e 5t + 31.68e 4t
10
290)e 4t =
9.8
9.8
(9.8 + 52.92e 4t ; )
52.92e 4t ;
52.92e 4t . (OK)
Also,
8
di1
= 371.20e 4t
dt
20i1 =
16
16
232e 4t + 240e 5t ;
8
di2
= 63.36e 4t
dt
800i2 =
480e 5t ;
8
80e 5t ;
792e 4t + 800e 5t ;
dig
= 125.44e 4t .
dt
Test:
371.20e 4t
8
(8
480e 5t + 8 + 232e 4t
?
792e 4t + 800e 5t =
8) + (800
480
240
+(371.20 + 232 + 63.36
(800
800)e 5t + (666.56
125.44e 4t =
240e 5t + 63.36e 4t
80e 5t
125.44e 4t ;
80)e 5t
?
792)e 4t =
?
792)e 4t =
125.44e 4t ;
125.44e 4t ;
125.44e 4t . (OK)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
6–5
AP 6.7 Since P1 = P2 ,
L1
0.025
N12
= 0.25.
=
=
2
N2
L2
0.1
Therefore,
N1 p
= 0.25 = 0.5,
N2
so
N2 =
500
N1
=
= 1000 turns.
0.5
0.5
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6–6
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Problems
P 6.1
[a] v = L
di
;
dt
di
= 18[t( 10e 10t ) + e 10t ] = 18e 10t (1
dt
v = (50 ⇥ 10 6 )(18)e 10t (1
= 0.9e 10t (1
10t);
10t)
10t) mV,
t > 0.
[b] p = vi.
v(200 ms) = 0.9e 2 (1
2) =
121.8 µV.
i(200 ms) = 18(0.2)e 2 = 487.2 mA.
p(200 ms) = ( 121.8 ⇥ 10 6 )(487.2 ⇥ 10 3 ) =
59.34 µW.
[c] delivering.
1
1
[d] w = Li2 = (50 ⇥ 10 6 )(487.2 ⇥ 10 3 )2 = 5.93 µJ.
2
2
[e] The energy is a maximum where the current is a maximum:
diL
= 18[t( 10)e 10t + e 10t ) = 18e 10t (1
dt
10t);
diL
= 0 when t = 0.1 s.
dt
imax = 18(0.1)e 1 = 662.2 mA;
1
wmax = (50 ⇥ 10 6 )(662.2 ⇥ 10 3 )2 = 10.96 µJ.
2
P 6.2
[a] 0  t  2 ms :
i=
Z t
1Zt
1
vs dx + i(0) =
5 ⇥ 10 3 dx + 0
L 0
200 ⇥ 10 6 0
t
= 25x = 25t A;
0
2 ms  t < 1 :
Z t
1
i=
(0) dx + 25(2 ⇥ 10 3 ) = 50 mA.
200 ⇥ 10 6 2⇥10 3
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Problems
6–7
[b]
P 6.3
[a] i
= 0
t < 0;
i
= 50t A
0  t  5 ms;
i
= 0.5
i
= 0
[b] v = L
5  t  10 ms;
50t A
10 ms < t.
di
= 20 ⇥ 10 3 (50) = 1 V
dt
v = 20 ⇥ 10 3 ( 50) =
1V
0  t  5 ms;
5  t  10 ms.
v
=
0
t < 0;
v
=
1V
0 < t < 5 ms;
v
=
v
=
1V
0
5 < t < 10 ms;
10 ms < t.
p = vi.
p = 0
t < 0;
p =
0 < t < 5 ms;
(50t)(1) = 50t W
p = (0.5
p = 0
50t)( 1) = 50t
0.5 W
5 < t < 10 ms;
10 ms < t.
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6–8
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
w
w
w
= 0
t < 0;
Z t
x2 t
(50x) dx = 50
= 25t2 J
2 0
0
=
=
Z t
=
25x2
0.005
(50x
0.5) dx + 0.625 ⇥ 10 3
t
0.5x
0.005
w
P 6.4
=
25t2
=
0
0 < t < 5 ms;
+0.625 ⇥ 10 3
0.5t + 2.5 ⇥ 10 3 J
5 < t < 10 ms;
10 ms < t.
i = (B1 cos 200t + B2 sin 200t)e 50t ;
i(0) = B1 = 75 mA;
di
= (B1 cos 200t + B2 sin 200t)( 50e 50t ) + e 50t ( 200B1 sin 200t + 200B2 cos 200t)
dt
= [(200B2
50B1 ) cos 200t
(200B1 + 50B2 ) sin 200t]e 50t ;
di
= [(40B2
dt
10B1 ) cos 200t
(40B1 + 10B2 ) sin 200t]e 50t ;
v(0) = 4.25 = 40B2
10B1 = 40B2
0.75
v = 0.2
.·. B2 = 125 mA.
Thus,
i = (75 cos 200t + 125 sin 200t)e 50t mA,
v = (4.25 cos 200t
i(0.025) =
4.25 sin 200t)e 50t V,
28.25 mA;
0;
t
0;
v(0.025) = 1.513 V;
p(0.025) = ( 28.25)(1.513) =
P 6.5
t
42.7 mW (delivering).
[a] i(0) = A1 + A2 = 0.04;
di
=
dt
10,000A1 e 10,000t
v=
200A1 e 10,000t
40,000A2 e 40,000t ;
800A2 e 40,000t V;
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
v(0) =
200A1
6–9
800A2 = 28.
Solving, A1 = 0.1
and A2 =
0.06.
Thus,
i = 0.1e 10,000t
v=
0.06e 40,000t A,
t
0.
20e 10,000t + 48e 40,000t V,
t
0.
[b] If p = 0 then either i = 0 or v = 0. Suppose i = 0:
i = 0.1e 10,000t
0.06e 40,000t = 0;
.·. 0.1e30,000t = 0.06 so t =
17.03 µs.
This answer is impossible! So assume that v = 0:
20e 10,000t + 48e 40,000t = 0;
v=
Then,
20e30,000t =
48
.·.
t = 29.18 µs.
This answer makes sense; therefore, the power is 0 at t = 29.18 µs.
P 6.6
[a] From Problem 6.5 we have
A1 + A2 = 0.04.
Now, we add the second equation for the coefficients:
200A1
800A2 =
68.
Solving,
Thus,
A1 =
0.06;
i=
A2 = 0.1.
0.06e 10,000t + 0.1e 40,000t A t
v = 12e 10,000t
80e 40,000t A t
0;
0.
[b] i = 0 when 0.06e 10,000t = 0.1e 40,000t ;
.·. e30,0000t = 5/3 so t = 17.03 µs.
Thus,
i > 0 for 0  t  17.03 µs
and
i < 0 for 17.03 µs  t < 1;
v = 0 when 12e 10,000t = 80e 40,000t ;
.·. e30,0000t = 20/3 so t = 63.24 µs.
Thus,
v < 0 for 0  t  63.24 µs
and
v > 0 for 63.24 µs  t < 1.
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6–10
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
Therefore,
p < 0 for 0  t  17.03 µs
and
63.24 µs  t < 1;
(inductor delivers energy).
p > 0 for 17.03 µs  t  63.24 µs (inductor stores energy).
[c] The energy stored at t = 0 is
1
1
w(0) = L[i(0)]2 = (20 ⇥ 10 3 )(40 ⇥ 10 3 )2 = 16 µJ.
2
2
The power for t > 0 is
p = vi = 6e 50,000t
8e 80,000t
0.72e 20,000t .
The energy for t > 0 is
w=
Z 1
0
=
Z 1
p dt =
0
6e 50,000x dx
8
80,000
6
50,000
Z 1
0
0.72
=
20,000
8e 80,000x dx
Z 1
0
0.72e 20,000x dx
16 µJ.
Thus, the energy stored at t = 0 equals the energy extracted for t > 0.
P 6.7
p = vi = 40t[e 10t
W=
Z 1
0
p dx =
Z 1
0
10te 20t
e 20t ].
40x[e 10x
10xe 20x
e 20x ] dx = 0.2 J.
This is energy stored in the inductor at t = 1.
P 6.8
[a] i
=
Z t
1
30 sin 500x dx
15 ⇥ 10 3 0
=
2000
=
2000
=
4(1
i =
Z t
sin 500x dx
4

cos 500x t
500
0
4
0
cos 500t)
4
4;
4 cos 500t A.
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Problems
[b]
6–11
p = vi = (30 sin 500t)( 4 cos 500t)
=
120 sin 500t cos 500t;
p =
60 sin 1000t W.
w
=
1 2
Li
2
=
1
(15 ⇥ 10 3 )16 cos2 500t
2
=
120 cos2 500t mJ;
=
[60 + 60 cos 1000t] mJ.
w
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6–12
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] Absorbing power:
P 6.9
Delivering power:
⇡  t  2⇡ ms;
0  t  ⇡ ms;
3⇡  t  4⇡ ms;
2⇡  t  3⇡ ms.
[a] v = L
v=
=
di
;
dt
25 ⇥ 10 3
d
[10 cos 400t + 5 sin 400t]e 200t
dt
25 ⇥ 10 3 ( 200e 200t [10 cos 400t + 5 sin 400t]
+e 200t [ 4000 sin 400t + 2000 cos 400t]);
v=
25 ⇥ 10 3 e 200t ( 1000 sin 400t
=
25 ⇥ 10 3 e 200t ( 5000 sin 400t)
4000 sin 400t)
= 125e 200t sin 400t V.
dv
= 125(e 200t (400) cos 400t
dt
200e 200t sin 400t)
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Problems
= 25,000e 200t (2 cos 400t
dv
=0
when
dt
.·. tan 400t = 2,
6–13
sin 400t) V/s.
2 cos 400t = sin 400t;
400t = 1.11;
t = 2.77 ms.
[b] v(2.77 ms) = 125e 0.55 sin 1.11 = 64.27 V.
P 6.10
[a] 0  t  1 s :
v=
100t;
i=
1Z t
100x dx + 0 =
5 0
i=
10t2 A.
20
x2 t
;
2 0
1s  t  3s :
v=
200 + 100t;
i(1) =
.·. i
10 .
=
1Z t
(100x
5 1
10
Z t
10
Z t
x dx
40
10(t2
1)
40(t
= 20
=
200) dx
1
= 10t2
3s  t  5s :
1
dx
1)
10
40t + 20 A.
v = 100;
i(3) = 10(9)
i
=
120 + 20 =
1Z t
100 dx
5 3
= 20t
5s  t  6s :
v=
60
10 A;
10
10 = 20t
70 A.
100t + 600;
i(5) = 100
70 = 30.
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6–14
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
i
=
[b] i(6) =
1Z t
( 100x + 600) dx + 30
5 5
Z t
Z t
=
20
=
10(t2
=
10t2 + 120t
5
x dx + 120
5
dx + 30
25) + 120(t
10(36) + 120(6)
5) + 30
320 A.
320 = 720
680 = 40 A,
[c]
P 6.11
6  t < 1.
0t<2s:
Z t
1
e 4x t
3
4x
iL =
3
⇥
10
e
dx
+
1
=
1.2
+1
2.5 ⇥ 10 3 0
4 0
= (1.3
0.3e 4t ) A,
0  t  2 s.
iL (2) = 1.3 A;
2st<1:
iL =
!
e 4(x 2) t
1.2
+ 1.3
4
2
= (0.3e 4(t 2) + 1) A,
2 s  t < 1.
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Problems
P 6.12
6–15
For 0  t  1.6 s:
iL =
1Z t
3 ⇥ 10 3 dx + 0 = 0.6 ⇥ 10 3 t;
5 0
iL (1.6 s) = (0.6 ⇥ 10 3 )(1.6) = 0.96 mA;
Rm = (20)(1000) = 20 k⌦;
vm (1.6 s) = (0.96 ⇥ 10 3 )(20 ⇥ 103 ) = 19.2 V.
P 6.13
[a] i = C
dv
= (5 ⇥ 10 6 )[500t( 2500)e 2500t + 500e 2500t ]
dt
= 2.5 ⇥ 10 3 e 2500t (1
2500t) A.
[b] v(100 µ) = 500(100 ⇥ 10 6 )e 0.25 = 38.94 mV;
i(100 µ) = (2.5 ⇥ 10 3 )e 0.25 (1
0.25) = 1.46 mA;
p(100 µ) = vi = (38.94 ⇥ 10 3 )(1.46 ⇥ 10 3 ) = 56.86 µW.
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6–16
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] p > 0, so the capacitor is absorbing power.
[d] v(100 µ) = 38.94 mV;
1
1
w = Cv 2 = (5 ⇥ 10 6 )(38.94 ⇥ 10 3 )2 = 3.79 nJ.
2
2
[e] The energy is maximum when the voltage is maximum:
dv
= 0 when (1
dt
2500t) = 0 or t = 0.4 ms;
vmax = 500(0.4 ⇥ 10 3 )2 e 1 = 73.58 mV;
1 2
pmax = Cvmax
= 13.53 nJ.
2
P 6.14
v=
10 V,
t  0;
C = 0.8 µF;
e 1000t (50 cos 500t + 20 sin 500t)V,
v = 40
t
0.
[a] i = 0, t < 0.
dv
= 1000e 1000t (50 cos 500t + 20 sin 500t)
[b]
dt
e 1000t ( 25,000 sin 500t + 10,000 cos 500t)
= e 1000t (50,000 cos 500t + 20,000 sin 500t
+25,000 sin 500t
10,000 cos 500t)
(40,000 cos 500t + 45,000 sin 500t)e 1000t ;
dv
= (32 cos 500t + 36 sin 500t)e 1000t mA.
= C
dt
=
i
[c] no.
[d] yes, from 0 to 32 mA.
[e] v(1) = 40 V;
1
1
w = Cv 2 = (0.8 ⇥ 10 6 )(40)2 = 640 µJ.
2
2
P 6.15
[a] v
=
0
t < 0;
v
=
10t A
0  t  2 s;
v
=
40
10t A
2  t  6 s;
v
=
10t
80 A
6  t  8 s;
v
=
0
8 s < t.
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Problems
[b] i = C
6–17
dv
:
dt
i
= 0
t < 0;
i
=
0 < t < 2 s;
i
=
2 mA
2 mA
i =
2 < t < 6 s;
2 mA
6 < t < 8 s;
i = 0
8 s < t.
p = vi :
p =
0
t < 0;
p =
(10t)(0.002) = 0.02t W
0 < t < 2 s;
p =
(40
10t)( 0.002) = 0.02t
p =
(10t
80)(0.002) = 0.02t
p =
0
w=
w
= 0
w
=
=
0.16 W
2 < t < 6 s;
6 < t < 8 s;
8 s < t.
Z
w
0.08 W
p dx :
Z t
0
Z t
2
t < 0;
t
(0.02x) dx = 0.01x2 = 0.01t2 J
0 < t < 2 s;
0
(0.02x
0.08) dx + 0.04
t
= (0.01x2
0.08x)
+ 0.04
2
= 0.01t2
w
=
Z t
6
0.08t + 0.16 J
(0.02x
= (0.01x2
2 < t < 6 s;
0.16) dx + 0.04
t
0.16x)
+ 0.04
6
=
w
0.01t2
= 0
0.16t + 0.64 J
6 < t < 8 s;
8 s < t.
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6–18
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c]
From the plot of power above, it is clear that power is being absorbed for
0 < t < 2 s and for 4 s < t < 6 s, because p > 0. Likewise, power is being
delivered for 2 s < t < 4 s and 6 s < t < 8 s, because p < 0.
P 6.16
12.5 ⇥ 109 (20 ⇥ 10 6 )2 = 5 V (end of first interval);
[a] v(20 µs)
=
v(20 µs)
=
106 (20 ⇥ 10 6 )
=
5 V (start of second interval);
=
106 (40 ⇥ 10 6 )
=
10 V (end of second interval).
v(40 µs)
(12.5)(400) ⇥ 10 3
(12.5)(1600) ⇥ 10 3
[b] p(10µs) = 62.5 ⇥ 1012 (10 5 )3 = 62.5 mW,
i(10µs) = 50 mA,
10
10
v(10 µs) = 1.25 V,
p(10 µs) = vi = (1.25)(50 m) = 62.5 mW. (checks)
p(30 µs) = 437.50 mW,
v(30 µs) = 8.75 V,
i(30 µs) = 0.05 A;
p(30 µs) = vi = (8.75)(0.05) = 62.5 mW. (checks)
[c] w(10 µs) = 15.625 ⇥ 1012 (10 ⇥ 10 6 )4 = 0.15625 µJ;
w = 0.5Cv 2 = 0.5(0.2 ⇥ 10 6 )(1.25)2 = 0.15625 µJ;
w(30 µs) = 7.65625 µJ;
w(30 µs) = 0.5(0.2 ⇥ 10 6 )(8.75)2 = 7.65625 µJ.
P 6.17
iC = C(dv/dt)
0 < t < 0.5 :
vc = 30t2 V;
iC = 20 ⇥ 10 6 (60)t = 1.2t mA.
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Problems
6–19
0.5 < t < 1 :
1)2 V;
vc = 30(t
iC = 20 ⇥ 10 6 (60)(t
P 6.18
1) = 1.2(t
1) mA.
1
1
[a] w(0) = C[v(0)]2 = (0.25) ⇥ 10 6 (50)2 = 312.5 µJ.
2
2
4000t
;
[b] v = (A1 + A2 t)e
v(0) = A1 = 50 V;
dv
dt
4000e 4000t (A1 + A2 t) + e 4000t (A2 )
=
=
dv
(0) = A2
dt
i=C
dv
,
dt
4000A2 t + A2 )e 4000t .
( 4000A1
4000A1 ;
i(0) = C
dv(0)
;
dt
dv(0)
i(0)
400 ⇥ 10 3
.·.
=
=
= 16 ⇥ 105 ;
dt
C
0.25 ⇥ 10 6
.·. 16 ⇥ 105 = A2
4000(50).
Thus, A2 = 16 ⇥ 105 + 2 ⇥ 105 = 18 ⇥ 105
V
.
s
[c] v = (18 ⇥ 105 t + 50)e 4000t ;
i=C
dv
d
= 0.25 ⇥ 10 6 (18 ⇥ 105 t + 50)e 4000t .
dt
dt
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6–20
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
i
=
d
[(0.45t + 12.5 ⇥ 10 6 )e 4000t ]
dt
=
(0.45t + 12.5 ⇥ 10 6 )( 4000)e 4000t + e 4000t (0.45)
0.05 + 0.45)e 4000t
= ( 1800t
P 6.19
[a]
v
w
1800t)e 4000t A,
=
(0.40
=
Z 500⇥10 6
1
50 ⇥ 10 3 e 2000t dt
0.5 ⇥ 10 6 0
=
100 ⇥ 103
=
50(1
=
1
Cv 2 = 21 (0.5)(10 6 )(11.61)2 = 33.7 µJ.
2
[b] v(1) = 50
e 2000t 500⇥10
2000 0
e 1)
t
0.
20
6
20
20 = 11.61 V;
20 = 30V;
1
w(1) = (0.5 ⇥ 10 6 )(30)2 = 225 µJ.
2
P 6.20
400 ⇥ 10 3
t = 8 ⇥ 104 t A
[a] i =
6
5 ⇥ 10
0  t  5 µs;
i = 400 ⇥ 10 3 A
5  t  20 µs;
i = 104 t
20 µs  t  50 µs.
q
=
0.5 A
Z 5⇥10 6
0
8 ⇥ 104 t dt +
2
4t
= 8 ⇥ 10
2
5⇥10
0
Z 20⇥10 6
5⇥10
6
400 ⇥ 10 3 dt +
6
3
Z 30⇥10 6
20⇥10
2
4t
6
+400 ⇥ 10 (15 ⇥ 10 ) + 10
2
6
(104 t
=
6
20⇥10
6
10 ⇥ 10 6 ]
15 ⇥ 10 6 ]
4.5 µC.
1
[b] v =
0.25 ⇥ 10 6
Z 5 µs
0
8 ⇥ 104 x dx +
Z 20 µs
5 µs
0.4x dx +
Z 50 µs
"
5 µs
20 µs
1
4 2
+(5000t2
=
4
⇥
10
t
+0.4t
0.25 ⇥ 10 6
0
5 µs
=
30⇥10
0.5t
= 8 ⇥ 104 ( 12 )(25 ⇥ 10 12 ) + 6 ⇥ 10 6 + [5000(30 ⇥ 10 6 )2
[5000(20 ⇥ 10 6 )2
!
0.5) dt
1
[1 ⇥ 10 6 + 6 ⇥ 10 6
0.25 ⇥ 10 6
20 µs
(104 x
0.5t)
50 µs
20 µs
0.5) dx
#
12.5 ⇥ 10 6 + 8 ⇥ 10 6 ] = 10V.
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Problems
6–21
1
1
[c] w = Cv 2 = (0.25 ⇥ 10 6 )(10)2 = 12.5 µJ.
2
2
P 6.21
[a] 0  t  5 µs :
1
= 2 ⇥ 105 ;
C
C = 5 µF
v = 2 ⇥ 105
Z t
0
4 dx + 12;
v = 8 ⇥ 105 t + 12 V
0  t  5 µs.
v(5 µs) = 4 + 12 = 16 V.
[b] 5 µs  t  20 µs :
5
v = 2 ⇥ 10
v=
Z t
5⇥10
2 dx + 16 =
6
4 ⇥ 105 t + 18V
4 ⇥ 105 t + 2 + 16;
5  t  20 µs.
4 ⇥ 105 (20 ⇥ 10 6 ) + 18 = 10 V.
v(20 µs) =
[c] 20 µs  t  25 µs :
v = 2 ⇥ 105
Z t
20⇥10
v = 12 ⇥ 105 t
6
6 dx + 10 = 12 ⇥ 105 t
14 V,
24 + 10;
20 µs  t  25 µs.
v(25 µs) = 12 ⇥ 105 (25 ⇥ 10 6 )
14 = 16 V.
[d] 25 µs  t  35 µs :
v = 2 ⇥ 105
Z t
25⇥10
v = 8 ⇥ 105 t
6
4 V,
4 dx + 16 = 8 ⇥ 105 t
20 + 16;
25 µs  t  35 µs.
v(35 µs) = 8 ⇥ 105 (35 ⇥ 10 6 )
4 = 24 V.
[e] 35 µs  t < 1;
5
v = 2 ⇥ 10
v = 24 V,
Z t
35⇥10
6
0 dx + 24 = 24;
35 µs  t < 1.
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6–22
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[f ]
P 6.22
[a] Combine two 10 mH inductors in parallel to get a 5 mH equivalent
inductor. Then combine this parallel pair in series with three 1 mH
inductors:
10 mk10 m + 1 m + 1 m + 1 m = 8 mH.
[b] Combine two 10 µH inductors in parallel to get a 5 µH inductor. Then
combine this parallel pair in series with four more 10 µH inductors:
10 µk10 µ + 10 µ + 10 µ + 10 µ + 10 µ = 45 µH.
[c] Combine two 100 µH inductors in parallel to get a 50 µH inductor. Then
combine this parallel pair with a 100 µH inductor and three 10 µH
inductors in series:
100 µk100 µ + 100 µ + 10 µ + 10 µ + 10 µ = 180 µH.
P 6.23
[a] 15k30 = 10 mH;
10 + 10 = 20 mH;
20k20 = 10 mH;
12k24 = 8 mH;
10 + 8 = 18 mH;
18k9 = 6 mH;
Lab = 6 + 8 = 14 mH.
[b] 12 + 18 = 30 µH;
30k20 = 12 µH;
12 + 38 = 50 µH;
30k75k50 = 15 µH;
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Problems
6–23
15 + 15 = 30 µH;
30k60 = 20 µH;
Lab = 20 + 25 = 45 µH.
P 6.24
[a] io (0) =
i1 (0)
i2 (0) = 6
1 = 5 A.
[b]
io
1Z t
2000e 100x dx + 5 =
4 0
=
=
5(e 100t
t
500
e 100x
+5
100 0
1) + 5 = 5e 100t A,
t
0.
[c]
va
= 3.2( 500e 100t ) =
vc
= va + vb =
i1
i1
[d] i2
1600e 100t + 2000e 100t
=
400e 100t V;
1Z t
400e 100x dx
1 0
=
4e 100t + 4
=
4e 100t
=
=
=
1
4
Z t
0
1600e 100t V;
6
6;
2A
t
0.
400e 100x dx + 1
e 100t + 2 A,
t
0.
1
1
1
[e] w(0) = (1)(6)2 + (4)(1)2 + (3.2)(5)2 = 60 J.
2
2
2
1
[f ] wdel = (4)(5)2 = 50 J.
2
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6–24
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[g] wtrapped = 60
1
1
wtrapped = (1)(2)2 + (4)(2)2 + 10 J. (check)
2
2
or
P 6.25
50 = 10 J
vb = 2000e 100t V;
io = 5e 100t A;
p = 10,000e 200t W;
w=
Z t
0
104 e 200x dx = 10,000
e 200x t
= 50(1
200 0
e 200t ) W;
wtotal = 50 J;
80%wtotal = 40 J.
Thus,
50
P 6.26
50e 200t = 40;
e200t = 5;
.·. t = 8.05 ms.
[a]
i(t)
=
1Z t
12e x dx + 6
2 0
= 6e x
t
+6
0
=
i(t)
[b] i1 (t)
6e t
6 + 6;
= 6e t A, t
=
1
3
Z t
= 4e x
0
0.
12e x dx + 2
t
+2
0
=
i1 (t)
4(e t
= 4e t
1) + 2;
2 A, t
0.
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Problems
1Z t
12e x dx + 4
6 0
[c] i2 (t) =
=
6–25
t
2e x
+4
0
=
i2 (t)
2(e t
1) + 4;
= 2e t + 2 A, t
0.
[d] p = vi = (12e t )(6e t ) = 72e 2t W :
w
=
Z 1
=
72
=
36 J.
0
p dt =
Z 1
0
72e 2t dt
e 2t 1
2 0
1
1
[e] w = (3)(2)2 + (6)(4)2 = 54 J.
2
2
1
1
[f ] wtrapped = (3)( 2)2 + (6)(2)2 = 18 J;
2
2
wtrapped = 54
36 = 18 J.
checks
[g] Yes, they agree.
P 6.27
From Figure 6.17(a) we have
1 Zt
1 Zt
v=
i dx + v1 (0) +
i dx + v2 (0) + · · · ;
C1 0
C2 0

Z t
1
1
v=
+
+ ···
i dx + v1 (0) + v2 (0) + · · · ;
C1 C2
0
Therefore
P 6.28

1
1
1
=
+
+ ··· ,
Ceq
C1 C2
veq (0) = v1 (0) + v2 (0) + · · · .
From Fig. 6.18(a)
i = C1
dv
dv
dv
+ C2 + · · · = [C1 + C2 + · · ·] .
dt
dt
dt
Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the
initial voltage on every capacitor must be the same. This initial voltage would
appear on Ceq .
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6–26
P 6.29
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a] Combine a 470 pF capacitor and a 10 pF capacitor in parallel to get a 480
pF capacitor:
(470 p) in parallel with (10 p) = 470 p + 10 p = 480 pF.
[b] Create a 1200 nF capacitor as follows:
(1 µ) in parallel with (0.1 µ) in parallel with (0.1 µ)
= 1000 n + 100 n + 100 n = 1200 nF.
Create a second 1200 nF capacitor using the same three resistors. Place
these two 1200 nF in series:
(1200 n)(1200 n)
(1200 n) in series with (1200 n) =
= 600 nF.
1200 n + 1200 n
[c] Combine two 220 µF capacitors in series to get a 110 µF capacitor. Then
combine the series pair in parallel with a 10 µF capacitor to get 120 µF:
[(220 µ) in series with (220 µ)] in parallel with (10 µ)
=
P 6.30
[a]
(220 µ)(220 µ)
+ 10 µ = 120 µF.
220 µ + 220 µ
1
1
1
1
+
= ;
=
C1
48 24
16
C1 = 16 nF;
C2 = 4 + 16 = 20 nF.
1
1
1
1
+
= ;
=
C3
20 30
12
C3 = 12 nF;
C4 = 12 + 8 = 20 nF.
1
1
1
1
1
+
+
= ;
=
C5
20 20 10
5
C5 = 5 nF.
Equivalent capacitance is 5 nF with an initial voltage drop of +15 V.
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Problems
[b]
1
1
1
1
+
+
=
36 18 12
6
.·. Ceq = 6 µF
6–27
24 + 6 = 30 µF.
25 + 5 = 30 µF.
1
1
3
1
+
+
=
30 30 30
30
.·. Ceq = 10 µF.
Equivalent capacitance is 10 µF with an initial voltage drop of +25 V.
P 6.31
[a]
vo
=
=
Z t
1
20 ⇥ 10 6 e x dx + 10
2 ⇥ 10 6 0
10e x
t
+10
0
=
10e t V,
t
0.
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6–28
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] v1
=
=
[c] v2
=
t
1
6
x
(20
⇥
10
)e
+4
3 ⇥ 10 6
0
6.67e t
2.67 V,
t
0.
t
1
6
x
(20
⇥
10
)e
+6
6 ⇥ 10 6
0
= 3.33e t + 2.67 V,
t
0.
[d] p = vi = (10e t )(20 ⇥ 10 6 )e t
=
w
=
Z 1
=
200 ⇥ 10 6
=
[e] w
200 ⇥ 10 6 e 2t .
0
200 ⇥ 10 6 e 2t dt
e 2t 1
2 0
100 ⇥ 10 6 (0
1) = 100 µJ.
=
1
(3 ⇥ 10 6 )(4)2 + 12 (6 ⇥ 10 9 )(6)2
2
=
132 µJ.
[f ] wtrapped
=
1
(3 ⇥ 10 6 )(8/3)2 + 21 (6 ⇥ 10 6 )(8/3)2
2
= 32 µJ.
CHECK: 100 + 32 = 132 µJ.
[g] Yes, they agree.
P 6.32
1
1
10
1 1
=
= 2;
= + +
Ce
1 5 1.25
5
.·. C2 = 0.5 µF.
vb = 20
250 + 30 =
200 V.
[a]
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Problems
vb
106 Z t
0.5 0
=
=
10,000
e 50x t
50 0
106 Z t
5 0
5 ⇥ 10 3 e 50x dx
=
20(e 50t
1)
5 ⇥ 10 3 e 50x dx
=
80(e 50t
1)
=
80e 50t
110 V.
Z t
30
=
100(e 50t
=
100e 50t + 150 V.
1) + 250
=
vc
=
200e 50t V. (checks)
= 0.2 ⇥ 10 6
=
va
0.2 ⇥ 10 6 ( 5000e 50t ) =
= 0.8 ⇥ 10 6
[a] w(0)
vd
d
[100e 50t + 150]
dt
=
5e 50t mA.
4e 50t mA.
(OK)
1
(0.2 ⇥ 10 6 )(250)2 + 12 (0.8 ⇥ 10 6 )(250)2 + 12 (5 ⇥ 10 6 )(20)2
2
+
=
e 50t mA.
d
[100e 50t + 150] =
dt
CHECK: ib = i1 + i2 =
P 6.33
30
5 ⇥ 10 3 e 50x dx + 250
0
CHECK: vb
[f ] i2
40 V.
106 Z t
1.25 0
= 106
20
20
=
[d] vd
[e] i1
200
=
= 20e 50t
[c] vc
200
200e 50t V.
=
[b] va
5 ⇥ 10 3 e 50x dx
6–29
1
(1.25 ⇥ 10 6 )(30)2
2
32,812.5 µJ.
[b] w(1) = 21 (5 ⇥ 10 6 )(40)2 + 21 (1.25 ⇥ 10 6 )(110)2 + 12 (0.2 ⇥ 10 6 )(150)2
+ 12 (0.8 ⇥ 10 6 )(150)2
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6–30
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
= 22,812.5 µJ.
1
[c] w = (0.5 ⇥ 10 6 )(200)2 = 10,000 µJ.
2
CHECK: 32,812.5 22,812.5 = 10,000 µJ.
10,000
[d] % delivered =
⇥ 100 = 30.48%.
32,812.5
[e] w
=
Z t
=
10(1
0
( 0.005e
.·. 10 2 (1
Thus, t =
P 6.34
vc
vL
)( 200e
) dx =
e 100t ) = 7.5 ⇥ 10 3 ;
Z t
0
e 100x dx
e 100t = 0.25.
ln 4
= 13.86 ms.
100
✓Z t
1.5e
1)
200(e 4000t
0
16,000x
=
150(e 16,000t
=
150e 16,000t 200e 4000t V;
dio
25 ⇥ 10 3
dt
=
50x
e 100t ) mJ.
1
0.625 ⇥ 10 6
=
50x
dx
1)
Z t
0
0.5e
4000x
dx
◆
50
50
= 25 ⇥ 10 3 ( 24,000e 16,000t + 2000e 4000t )
=
vo
P 6.35
600e 16,000t + 50e 4000t V;
= vc
vL
=
(150e 16,000t
200e 4000t )
( 600e 16,000t + 50e 4000t )
=
750e 16,000t
250e 4000t V, t > 0.
dio
dt
= 5{e 2000t [ 8000 sin 4000t + 4000 cos 4000t]
2000e 2000t [2 cos 4000t + sin 4000t]};
dio +
(0 )
dt
v2 (0+ )
v1 (0+ )
P 6.36
= 5[1(4000) + ( 2000)(2)] = 0;
dio
= 10 ⇥ 10 3 (0+ ) = 0;
dt
+
= 40io (0 ) + v2 (0+ ) = 40(10) + 0 = 400V.
[a] Rearrange by organizing the equations by di1 /dt, i1 , di2 /dt, i2 and transfer
the ig terms to the right hand side of the equations. We get
4
di1
+ 25i1
dt
8
di2
dt
20i2 = 5ig
8
dig
;
dt
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Problems
8
di1
dt
20i1 + 16
6–31
di2
dig
+ 80i2 = 16 .
dt
dt
[b] From the given solutions we have
di1
=
dt
320e 5t + 272e 4t ;
di2
= 260e 5t
dt
Thus,
4
di1
=
dt
204e 4t .
1280e 5t + 1088e 4t ;
25i1 = 100 + 1600e 5t
8
di2
= 2080e 5t
dt
1700e 4t ;
1632e 4t ;
20i2 = 20
1040e 5t + 1020e 4t ;
5ig = 80
80e 5t;
8
dig
= 640e 5t .
dt
Thus,
1280e 5t + 1088e 4t + 100 + 1600e 5t
+1632e 4t
80 + (1088
+(1600
80 + (2720
8
di1
=
dt
20 + 1040e 5t
1700 + 1632
16
2080e 5t
80e 5t
640e 5t .
1020)e 4t
?
2720)e 4t + (2640
720e 5t ;
3360)e 5t = 80
720e 5t . (OK)
2560e 5t + 2176e 4t ;
di2
= 4160e 5t
dt
80i2 = 80
?
1020e 4t = 80
2080 + 1040)e 5t = 80
1280
20i1 = 80 + 1280e 5t
16
1700e 4t
1360e 4t ;
3264e 4t ;
4160e 5t + 4080e 4t ;
dig
= 1280e 5t ;
dt
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6–32
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
2560e 5t
2176e 4t
1280e 5t + 1360e 4t + 4160e 5t
80
3264e 4t
?
4160e 5t + 4080e 4t = 1280e 5t ;
+80
( 80 + 80) + (2560
+(1360
1280 + 4160
4160)e 5t
?
3264 + 4080)e 4t = 1280e 5t ;
2176
0 + 1280e 5t + 0e 4t = 1280e 5t . (OK)
P 6.37
[a] vab = L1
di
di
di
di
di
+ L2 + M + M = (L1 + L2 + 2M ) .
dt
dt
dt
dt
dt
It follows that Lab = (L1 + L2 + 2M ).
[b] vab = L1
di
dt
M
di
di
+ L2
dt
dt
M
Therefore Lab = (L1 + L2
P 6.38
[a] vab = L1
0 = L1
di
= (L1 + L2
dt
2M )
di
.
dt
2M ).
di2
d(i1 i2 )
+M
;
dt
dt
d(i2 i1 )
dt
M
di2
d(i1 i2 )
di2
+M
+ L2 .
dt
dt
dt
Collecting coefficients of [di1 /dt] and [di2 /dt], the two mesh-current
equations become
vab = L1
di1
+ (M
dt
L1 )
di2
dt
and
di1
+ (L1 + L2
dt
Solving for [di1 /dt] gives
0 = (M
L1 )
2M )
di2
.
dt
L1 + L2 2M
di1
=
vab
dt
L1 L2 M 2
from which we have
vab =
L1 L2 M 2
L1 + L2 2M
.·. Lab =
!
!
di1
;
dt
L 1 L2 M 2
.
L1 + L2 2M
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses,
therefore
L 1 L2 M 2
.
Lab =
L1 + L2 + 2M
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Problems
P 6.39
[a] vg
= 5(ig
i1 ) + 20(i2
i1 ) + 60i2
=
5(16
16e 5t
64e 5t + 68e 4t )+
20(1
52e 5t + 51e 4t
60(1
52e 5t + 51e 4t )
60 + 5780e 4t
=
[b] vg (0) = 60 + 5780
[c] pdev
4
4
6–33
64e 5t + 68e 4t )+
5840e 5t V,
5840 = 0 V,
= vg ig
960 + 92,480e 4t
=
94,400e 5t
92,480e 9t +
93,440e 10t W,
[d] pdev (1) = 960 W,
[e] i1 (1) = 4 A;
i2 (1) = 1 A;
p5⌦ = (16
4)2 (5) = 720 W;
ig (1) = 16 A;
p20⌦ = 32 (20) = 180 W;
p60⌦ = 12 (60) = 60 W;
X
.·.
P 6.40
pabs = 720 + 180 + 60 = 960 W;
X
pdev =
X
pabs = 960 W.
[a] Yes, using KVL around the lower right loop
vo = v20⌦ + v60⌦ = 20(i2
[b] vo
=
=
vo
[c] vo
=
20(1
52e 5t + 51e 4t
60(1
52e 5t + 51e 4t )
20( 3
4
64e 5t + 68e 4t )+
116e 5t + 119e 4t ) + 60
3120e 5t + 3060e 4t ;
5440e 5t + 5440e 4t V.
= L2
d
(ig
dt
i2 ) + M
di1
dt
=
d
d
(15 + 36e 5t 51e 4t ) + 8 (4 + 64e 5t
dt
dt
2880e 5t + 3264e 4t 2560e 5t + 2176e 4t ;
=
5440e 5t + 5440e 4t V.
=
vo
i1 ) + 60i2 .
16
68e 4t )
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6–34
P 6.41
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[a] 0.5
dig
di2
+ 0.2
+ 10i2 = 0;
dt
dt
0.2
di2
+ 10i2 =
dt
[b] i2 = 625e 10t
di2
=
dt
0.2
[c] v1
dig
.
dt
250e 50t mA;
6.25e 10t + 12.5e 50t A/s;
ig = e 10t
dig
=
dt
0.5
10 A;
10e 10t A/s;
di2
+ 10i2 = 5e 10t
dt
=
5
and
0.5
dig
= 5e 10t .
dt
di2
dig
+ 0.5
dt
dt
= 5( 10e 10t ) + 0.5( 6.25e 10t + 12.5e 50t )
=
53.125e 10t + 6.25e 50t V,
t > 0.
[d] v1 (0) =
53.125 + 6.25 = 46.875 V;
Also
di2
dig
v1 (0) = 5 (0) + 0.5 (0)
dt
dt
= 5( 10) + 0.5( 6.25 + 12.5) = 46.875 V.
Yes, the initial value of v1 is consistent with known circuit behavior.
P 6.42
When the switch is opened the induced voltage is negative at the dotted
terminal. Since the voltmeter kicks downscale, the induced voltage across the
voltmeter must be negative at its positive terminal. Therefore, the voltage is
negative at the positive terminal of the voltmeter.
Thus, the upper terminal of the unmarked coil has the same instantaneous
polarity as the dotted terminal. Therefore, place a dot on the upper terminal
of the unmarked coil.
P 6.43
[a] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4.
Assign the current into terminal 4; the flux is right-to-left in coil 3-4.
Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
[b] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assign
the current into terminal 4; the flux is down in coil 3-4. Therefore, dot
terminal 4. Hence, 1 and 4 or 2 and 3.
P 6.44
✓
1
P11
[a] 2 = 1 +
k
P12
◆✓
P22
1+
P12
◆
✓
P11
= 1+
P21
◆✓
◆
P22
1+
.
P12
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Problems
6–35
Therefore
k2 =
P12 P21
.
(P21 + P11 )(P12 + P22 )
Now note that
1 =
21 = P11 N1 i1 + P21 N1 i1 = N1 i1 (P11 + P21 ),
11 +
and similarly
2 = N2 i2 (P22 + P12 ).
It follows that
(P11 + P21 ) =
1
N 1 i1
and
(P22 + P12 ) =
2
N 2 i2
!
.
Therefore
( 12 /N2 i2 )( 21 /N1 i1 )
=
k2 =
( 1 /N1 i1 )( 2 /N2 i2 )
or
v
u
u
k=t
21
1
!
12
2
12 21
1 2
!
.
[b] The fractions ( 21 / 1 ) and ( 12 / 2 ) are by definition less than 1.0,
therefore k < 1.
P 6.45
[a] w = (0.5)L1 i21 + (0.5)L2 i22 + M i1 i2 ;
q
M = 0.85 (18)(32) = 20.4 mH;
w = [9(36) + 16(81) + 20.4(54)] = 2721.6 mJ.
[b] w = [324 + 1296 + 1101.6] = 2721.6 mJ.
P 6.46
[c] w = [324 + 1296
1101.6] = 518.4 mJ.
[d] w = [324 + 1296
1101.6] = 518.4 mJ.
q
[a] M = 1.0 (18)(32) = 24 mH,
i1 = 6 A.
Therefore 16i22 + 144i2 + 324 = 0,
Therefore i2 =
✓ ◆
Therefore i2 =
4.5 A.
9
±
2
s
✓ ◆2
9
2
i22 + 9i2 + 20.25 = 0.
20.25 =
4.5 ±
p
0.
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6–36
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] No, setting W equal to a negative value will make the quantity under the
square root sign negative.
P 6.47
M2
k 2 L1
[a] L2 =
N1
=
N2
[b] P1 =
=
L1
=
L2
(0.09)2
= 50 mH;
(0.75)2 (0.288)
s
288
= 2.4.
50
L1
0.288
=
= 0.2 ⇥ 10 6 Wb/A;
2
2
N1
(1200)
P2 =
P 6.48
s
!
L2
0.05
=
= 0.2 ⇥ 10 6 Wb/A.
2
N2
(500)2
M
22.8
= 0.95.
=p
L1 L2
576
p
[b] Mmax = 576 = 24 mH.
[a] k = p
✓
N 2 P1
N1
L1
= 12
=
[c]
L2
N2 P2
N2
✓
◆2
;
◆
60
N1 2
= 6.25.
=
N2
9.6
N1 p
= 6.25 = 2.5.
N2
.·.
P 6.49
P1 =
L1
= 2 nWb/A;
N12
P12 = P21 =
P11 = P1
P 6.50
P2 =
L2
= 2 nWb/A;
N22
q
M = k L1 L2 = 180 µH;
M
= 1.2 nWb/A;
N1 N2
P21 = 0.8 nWb/A.
[a] L1 = N12 P1 ;
P1 =
P11
d 11
=
= 0.2;
d 21
P21
72 ⇥ 10 3
= 1152 nWb/A;
6.25 ⇥ 104
P21 = 2P11 ;
.·. 1152 ⇥ 10 9 = P11 + P21 = 3P11 .
P11 = 192 nWb/A;
q
q
P21 = 960 nWb/A;
M = k L1 L2 = (2/3) (0.072)(0.0405) = 36 mH;
N2 =
M
36 ⇥ 10 3
= 150 turns.
=
N1 P21
(250)(960 ⇥ 10 9 )
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Problems
6–37
L2
40.5 ⇥ 10 3
=
= 1800 nWb/A.
N22
(150)2
[c] P11 = 192. nWb/A [see part (a)]
P22
P2 P12
P2
22
[d]
=
=
=
1;
P12
P12
P12
12
[b] P2 =
P21 = P21 = 960 nWb/A;
22
12
P 6.51
1800
960
1 = 0.875.
When the touchscreen in the mutual-capacitance design is touched at the point
x, y, the touch capacitance Ct is present in series with the mutual capacitance
at the touch point, Cmxy . Remember that capacitances combine in series the
way that resistances combine in parallel. The resulting mutual capacitance is
0
Cmxy
=
P 6.52
=
P2 = 1800 nWb/A;
Cmxy Ct
.
Cmxy + Ct
[a] The self-capacitance and the touch capacitance are e↵ectively connected in
parallel. Therefore, the capacitance at the x-grid electrode closest to the
touch point with respect to ground is
Cx = Cp + Ct = 30 pF + 15 pF = 45 pF.
The same capacitance exists at the y-grid electrode closest to the touch
point with respect to ground.
[b] The mutual-capacitance and the touch capacitance are e↵ectively
connected in series. Therefore, the mutual capacitance between the
x-grid and y-grid electrodes closest to the touch point is
0
Cmxy
=
(30)(15)
Cmxy Ct
= 10 pF.
=
Cmxy + Ct
30 + 15
[c] In the self-capacitance design, touching the screen increases the
capacitance being measured at the point of touch. For example, in part
(a) the measured capacitance before the touch is 30 pF and after the
touch is 45 pF. In the mutual-capacitance design, touching the screen
decreases the capacitance being measured at the point of touch. For
example, in part (b) the measured capacitance before the touch is 30 pF
and after the touch is 10 pF.
P 6.53
[a] The four touch points identified are the two actual touch points and two
ghost touch points. Their coordinates, in inches from the upper left
corner of the screen, are
(2.1, 4.3);
(3.2, 2.5);
(2.1, 2.5);
and
(3.2, 4.3).
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6–38
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
These four coordinates identify a rectangle within the screen, shown
below.
[b] The touch points identified at time t1 are those listed in part (a). The
touch points recognized at time t2 are
(1.8, 4.9);
(3.9, 1.8);
(1.8, 1.8);
and
(3.9, 4.9).
The first two coordinates are the actual touch points and the last two
coordinates are the associated ghost points. Again, the four coordinates
identify a rectangle at time t2 , as shown here:
Note that the rectangle at time t2 is larger than the rectangle at time t1 ,
so the software would recognize the two fingers are moving toward the
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Problems
6–39
edges of the screen. This pinch gesture thus specifies a zoom-in for the
screen.
[c] The touch points identified at time t1 are those listed in part (a). The
touch points recognized at time t2 are
(2.8, 3.9);
(3.0, 2.8);
(2.8, 2.8);
and
(3.0, 3.9).
The first two coordinates are the actual touch points and the last two
coordinates are the associated ghost points. Again, the four coordinates
identify a rectangle at time t2 , as shown here:
Here, the rectangle at time t2 is smaller than the rectangle at time t1 , so
the software would recognize the two fingers are moving toward the
middle of the screen. This pinch gesture thus specifies a zoom-out for the
screen.
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Response of First-Order RL and
RC Circuits
Assessment Problems
AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a
short circuit, e↵ectively eliminating the 2 ⌦ resistor from the circuit.
First combine the 30 ⌦ and 6 ⌦ resistors in parallel:
30k6 = 5 ⌦.
Use voltage division to find the voltage drop across the parallel resistors:
5
v=
(120) = 75 V.
5+3
Now find the current using Ohm’s law:
75
v
=
= 12.5 A.
i(0 ) =
6
6
1
1
[b] w(0) = Li2 (0) = (8 ⇥ 10 3 )(12.5)2 = 625 mJ.
2
2
[c] To find the time constant, we need to find the equivalent resistance seen
by the inductor for t > 0. When the switch opens, only the 2 ⌦ resistor
remains connected to the inductor. Thus,
8 ⇥ 10 3
L
⌧=
=
= 4 ms.
R
2
t 0.
[d] i(t) = i(0 )et/⌧ = 12.5e t/0.004 = 12.5e 250t A,
[e] i(5 ms) = 12.5e 250(0.005) = 12.5e 1.25 = 3.58 A.
So w (5 ms) = 21 Li2 (5 ms) = 12 (8) ⇥ 10 3 (3.58)2 = 51.3 mJ.
7–1
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7–2
CHAPTER 7. Response of First-Order RL and RC Circuits
w (dis) = 625
51.3 = ◆573.7 mJ;
573.7
% dissipated =
100 = 91.8%.
625
✓
AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor:
Using current division,
10
(6.4) = 4 A.
i(0 ) =
10 + 6
Now use the circuit for t > 0 to find the equivalent resistance seen by the
inductor, and use this value to find the time constant:
0.32
L
= 0.1 s.
=
Req
3.2
Use the initial inductor current and the time constant to find the current
in the inductor:
i(t) = i(0 )e t/⌧ = 4e t/0.1 = 4e 10t A, t 0.
Use current division to find the current in the 10 ⌦ resistor:
4
4
io (t) =
( i) = ( 4e 10t ) = 0.8e 10t A, t 0+ .
4 + 10 + 6
20
Req = 4k(6 + 10) = 3.2 ⌦,
.·.
⌧=
Finally, use Ohm’s law to find the voltage drop across the 10 ⌦ resistor:
vo (t) = 10io = 10( 0.8e 10t ) = 8e 10t V, t 0+ .
[b] The initial energy stored in the inductor is
1
1
w(0) = Li2 (0 ) = (0.32)(4)2 = 2.56 J.
2
2
Find the energy dissipated in the 4 ⌦ resistor by integrating the power
over all time:
di
t 0+ ;
v4⌦ (t) = L = 0.32( 10)(4e 10t ) = 12.8e 10t V,
dt
p4⌦ (t) =
2
v4⌦
= 40.96e 20t W,
4
t
0+ ;
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Problems
w4⌦ (t) =
Z 1
0
7–3
40.96e 20t dt = 2.048 J.
Find the percentage of the initial energy in the inductor dissipated in the
4 ⌦ resistor:
✓
◆
2.048
% dissipated =
100 = 80%.
2.56
AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like
an open circuit.
Find the voltage drop across the open circuit by finding the voltage drop
across the 50 k⌦ resistor. First use current division to find the current
through the 50 k⌦ resistor:
80 ⇥ 103
(7.5 ⇥ 10 3 ) = 4 mA.
3
3
3
80 ⇥ 10 + 20 ⇥ 10 + 50 ⇥ 10
Use Ohm’s law to find the voltage drop:
v(0 ) = (50 ⇥ 103 )i50k = (50 ⇥ 103 )(0.004) = 200 V.
[b] To find the time constant, we need to find the equivalent resistance seen
by the capacitor for t > 0. When the switch opens, only the 50 k⌦
resistor remains connected to the capacitor. Thus,
⌧ = RC = (50 ⇥ 103 )(0.4 ⇥ 10 6 ) = 20 ms.
[c] v(t) = v(0 )e t/⌧ = 200e t/0.02 = 200e 50t V, t 0.
1
1
[d] w(0) = Cv 2 = (0.4 ⇥ 10 6 )(200)2 = 8 mJ.
2
2
1
1 2
[e] w(t) = Cv (t) = (0.4 ⇥ 10 6 )(200e 50t )2 = 8e 100t mJ.
2
2
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:
i50k =
8 ⇥ 10 3 e 100t = 2 ⇥ 10 3 ,
e100t = 4,
t = (ln 4)/100 = 13.86 ms.
AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested
voltage, vo , is the sum of the voltage drops for the two RC circuits. The
circuit for t < 0 is shown below:
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7–4
CHAPTER 7. Response of First-Order RL and RC Circuits
Find the current in the loop and use it to find the initial voltage drops
across the two RC circuits:
15
i=
= 0.2 mA,
v5 (0 ) = 4 V,
v1 (0 ) = 8 V.
75,000
There are two time constants in the circuit, one for each RC subcircuit.
⌧5 is the time constant for the 5 µF – 20 k⌦ subcircuit, and ⌧1 is the time
constant for the 1 µF – 40 k⌦ subcircuit:
⌧5 = (20 ⇥ 103 )(5 ⇥ 10 6 ) = 100 ms;
⌧1 = (40 ⇥ 103 )(1 ⇥ 10 6 ) = 40 ms.
Therefore,
v5 (t) = v5 (0 )e t/⌧5 = 4e t/0.1 = 4e 10t V, t 0;
v1 (t) = v1 (0 )e t/⌧1 = 8e t/0.04 = 8e 25t V, t 0.
Finally,
vo (t) = v1 (t) + v5 (t) = [8e 25t + 4e 10t ] V,
t 0.
[b] Find the value of the voltage at 60 ms for each subcircuit and use the
voltage to find the energy at 60 ms:
v1 (60 ms) = 8e 25(0.06) ⇠
v5 (60 ms) = 4e 10(0.06) ⇠
= 1.79 V,
= 2.20 V;
1
1
2
6
2 ⇠
w1 (60 ms) = 2 Cv1 (60 ms) = 2 (1 ⇥ 10 )(1.79) = 1.59 µJ;
w5 (60 ms) = 12 Cv52 (60 ms) = 12 (5 ⇥ 10 6 )(2.20)2 ⇠
= 12.05 µJ;
w(60 ms) = 1.59 + 12.05 = 13.64 µJ.
Find the initial energy from the initial voltage:
w(0) = w1 (0) + w2 (0) = 12 (1 ⇥ 10 6 )(8)2 + 12 (5 ⇥ 10 6 )(4)2 = 72 µJ.
Now calculate the energy dissipated at 60 ms and compare it to the
initial energy:
wdiss = w(0) w(60 ms) = 72 13.64 = 58.36 µJ;
% dissipated = (58.36 ⇥ 10 6 /72 ⇥ 10 6 )(100) = 81.05 %.
AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in
the inductor:
i(0 ) = 24/2 = 12 A = i(0+ ).
Note that i(0 ) = i(0+ ) because the current in an inductor is continuous.
[b] Use the circuit at t = 0+ , shown below, to calculate the voltage drop
across the inductor at 0+ . Note that this is the same as the voltage drop
across the 10 ⌦ resistor, which has current from two sources — 8 A from
the current source and 12 A from the initial current through the inductor.
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Problems
v(0+ ) =
10(8 + 12) =
7–5
200 V.
[c] To calculate the time constant we need the equivalent resistance seen by
the inductor for t > 0. Only the 10 ⌦ resistor is connected to the inductor
for t > 0. Thus,
⌧ = L/R = (200 ⇥ 10 3 /10) = 20 ms.
[d] To find i(t), we need to find the final value of the current in the inductor.
When the switch has been in position a for a long time, the circuit
reduces to the one below:
Note that the inductor behaves as a short circuit and all of the current
from the 8 A source flows through the short circuit. Thus,
if = 8 A.
Now,
i(t) = if + [i(0+ ) if ]e t/⌧ = 8 + [12 ( 8)]e t/0.02
= 8 + 20e 50t A, t 0.
[e] To find v(t), use the relationship between voltage and current for an
inductor:
di(t)
= (200 ⇥ 10 3 )( 50)(20e 50t ) = 200e 50t V,
t 0+ .
v(t) = L
dt
AP 7.6 [a]
From Example 7.6,
vo (t) =
60 + 90e 100t V.
Write a KCL equation at the top node and use it to find the relationship
between vo and vA :
vA vo
vA
vA + 75
+
+
= 0;
8000
160,000
40,000
20vA
20vo + vA + 4vA + 300 = 0;
25vA = 20vo
vA = 0.8vo
300;
12.
Use the above equation for vA in terms of vo to find the expression for vA :
vA (t) = 0.8( 60 + 90e 100t )
12 =
60 + 72e 100t V,
t
0+ .
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7–6
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] t
0+ , since there is no requirement that the voltage be continuous in a
resistor.
AP 7.7 For t < 0,
V0 =
For t
25,000
(60) = 15 V.
75,000 + 25,000
0,
Req = 25 k⌦,
⌧ = Req C = (25,000)(5 ⇥ 10 9 ) = 125 µs.
so
There is no source in the circuit as t ! 1 so Vf = 0. Thus,
vo (t) = Vf + (V0
Vf )e t/⌧
= 0 + (15
0)e t/125µ
= 15e 8000t V,
t
0.
AP 7.8 [a] Find the voltage across the capacitor in the direction of the capacitor
current. Call this voltage vc . The initial capacitor charge is zero, so
V0 = 0. For t 0,
Req = 20,000 + 30,000 = 50,000 ⌦;
⌧ = Req C = (50,000)(0.1 ⇥ 10 6 ) = 5 ms.
As t ! 1,
Vf = (20,000)(0.0075) = 150 V.
Therefore,
vC = Vf + (V0
Vf )e t/⌧
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Problems
150)e t/0.005
= 150 + (0
= 150
7–7
150e 200t V,
t
0.
Thus,
i=C
dvc
= (0.1 ⇥ 10 6 )[ 200( 150e 200t )]
dt
= 3e 200t mA, t
[b] i20k = (7.5
0+ .
3e 200t ) mA;
v = 20,000i20k = 150
60e 200t V,
t
0+ .
AP 7.9 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop
across the capacitor:
i=
!
40 ⇥ 103
(10 ⇥ 10 3 ) = 3.2 mA;
125 ⇥ 103
vc (0 ) = (3.2 ⇥ 10 3 )(25 ⇥ 103 ) = 80 V so vc (0+ ) = 80 V.
Now use the next circuit, valid for 0  t  10 ms, to calculate vc (t) for
that interval:
For 0  t  10 ms:
⌧ = RC = (25 ⇥ 103 )(1 ⇥ 10 6 ) = 25 ms;
vc (t) = vc (0 )et/⌧ = 80e 40t V 0  t  10 ms.
[b] Calculate the starting capacitor voltage in the interval t
the capacitor voltage from the previous interval:
vc (0.01) = 80e 40(0.01) = 53.63 V.
10 ms, using
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7–8
CHAPTER 7. Response of First-Order RL and RC Circuits
Now use the next circuit, valid for t
interval:
For t
10 ms, to calculate vc (t) for that
10 ms :
Req = 25 k⌦k100 k⌦ = 20 k⌦;
⌧ = Req C = (20 ⇥ 103 )(1 ⇥ 10 6 ) = 0.02 s.
Therefore vc (t) = vc (0.01+ )e (t 0.01)/⌧ = 53.63e 50(t 0.01) V,
t
0.01 s.
[c] To calculate the energy dissipated in the 25 k⌦ resistor, integrate the
power absorbed by the resistor over all time. Use the expression
p = v 2 /R to calculate the power absorbed by the resistor.
w25 k =
Z 0.01
0
Z 1
[80e 40t ]2
[53.63e 50(t 0.01) ]2
dt +
dt = 2.91 mJ.
25,000
25,000
0.01
[d] Repeat the process in part (c), but recognize that the voltage across this
resistor is non-zero only for the second interval:
Z 1
[53.63e 50(t 0.01) ]2
dt = 0.29 mJ.
100,000
0.01
We can check our answers by calculating the initial energy stored in the
capacitor. All of this energy must eventually be dissipated by the 25 k⌦
resistor and the 100 k⌦ resistor.
w100 k⌦ =
Check: wstored = (1/2)(1 ⇥ 10 6 )(80)2 = 3.2 mJ;
wdiss = 2.91 + 0.29 = 3.2 mJ.
AP 7.10 [a] Prior to switch a closing at t = 0, there are no sources connected to the
inductor; thus, i(0 ) = 0.
At the instant A is closed, i(0+ ) = 0.
For 0  t  1 s,
The equivalent resistance seen by the 10 V source is 2 + (3k0.8). The
current leaving the 10 V source is
10
= 3.8 A.
2 + (3k0.8)
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Problems
7–9
The final current in the inductor, which is equal to the current in the
0.8 ⌦ resistor is
3
IF =
(3.8) = 3 A.
3 + 0.8
The resistance seen by the inductor is calculated to find the time
constant:
2
L
[(2k3) + 0.8]k3k6 = 1 ⌦
⌧=
= = 2 s.
R
1
Therefore,
i = iF + [i(0+ )
iF ]e t/⌧ = 3
3e 0.5t A,
0  t  1 s.
For part (b) we need the value of i(t) at t = 1 s:
3e 0.5 = 1.18 A.
i(1) = 3
.
[b] For t > 1 s
Use current division to find the final value of the current:
9
i=
( 8) = 4.8 A
9+6
The equivalent resistance seen by the inductor is used to calculate the
time constant:
L
2
3k(9 + 6) = 2.5 ⌦
⌧=
=
= 0.8 s.
R
2.5
Therefore,
i = iF + [i(1+ )
=
iF ]e (t 1)/⌧
4.8 + 5.98e 1.25(t 1) A,
t
1 s.
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7–10
CHAPTER 7. Response of First-Order RL and RC Circuits
AP 7.11 0  t  32 ms:
1 Z 32⇥10 3
RCf 0
vo =
3
10 dt + 0 =
32⇥10
1
( 10t)
=
RCf
0
RCf = (200 ⇥ 103 )(0.2 ⇥ 10 6 ) = 40 ⇥ 10 3
1
= 25;
RCf
25( 320 ⇥ 10 3 ) = 8 V.
vo =
t
so
1
( 320 ⇥ 10 3 );
RCf
32 ms:
1 Zt
5 dy + 8 =
RCf 32⇥10 3
vo =
t
1
(5y)
+8 =
RCf
32⇥10 3
RCf = (250 ⇥ 103 )(0.2 ⇥ 10 6 ) = 50 ⇥ 10 3
vo =
20(5)(t
32 ⇥ 10 3 ) + 8 =
so
1
5(t
RCf
32 ⇥ 10 3 ) + 8;
1
= 20;
RCf
100t + 11.2.
The output will saturate at the negative power supply value:
15 =
100t + 11.2
.·.
t = 262 ms.
AP 7.12 [a] Use RC circuit analysis to determine the expression for the voltage at the
non-inverting input:
vp = Vf + [Vo
Vf ]e t/⌧ =
2 + (0 + 2)e t/⌧ ;
⌧ = (160 ⇥ 103 )(10 ⇥ 10 9 ) = 10 3 ;
1/⌧ = 625;
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Problems
vp =
2 + 2e 625t V;
7–11
vn = v p .
Write a KVL equation at the inverting input, and use it to determine vo :
vn v o
vn
+
= 0;
10,000
40,000
.·. vo = 5vn = 5vp =
10 + 10e 625t V.
The output will saturate at the negative power supply value:
10 + 10e 625t =
5;
e 625t = 1/2;
t = ln 2/625 = 1.11 ms.
[b] Use RC circuit analysis to determine the expression for the voltage at the
non-inverting input:
vp = Vf + [Vo
Vf ]e t/⌧ =
2 + (1 + 2)e 625t =
2 + 3e 625t V.
The analysis for vo is the same as in part (a):
vo = 5vp =
10 + 15e 625t V.
The output will saturate at the negative power supply value:
10 + 15e 625t =
5;
e 625t = 1/3;
t = ln 3/625 = 1.76 ms.
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7–12
CHAPTER 7. Response of First-Order RL and RC Circuits
Problems
P 7.1
[a] t < 0:
2 k⌦k6 k⌦ = 1.5k⌦.
Find the current from the voltage source by combining the resistors in
series and parallel and using Ohm’s law:
ig (0 ) =
40
= 20 mA.
(1500 + 500)
Find the branch currents using current division:
i1 (0 ) =
2000
(0.02) = 5 mA;
8000
i2 (0 ) =
6000
(0.02) = 15 mA.
8000
[b] The current in an inductor is continuous. Therefore,
i1 (0+ ) = i1 (0 ) = 5 mA;
i2 (0+ ) =
i1 (0+ ) =
5 mA.
(when switch is open)
L
0.4 ⇥ 10 3
[c] ⌧ =
=
= 5 ⇥ 10 5 s;
3
R
8 ⇥ 10
1
= 20,000;
⌧
i1 (t) = i1 (0+ )e t/⌧ = 5e 20,000t mA,
[d] i2 (t) =
i1 (t)
.·. i2 (t) =
when t
0+ ;
5e 20,000t mA,
t
t
0.
0+ .
[e] The current in a resistor can change instantaneously. The switching
operation forces i2 (0 ) to equal 15 mA and i2 (0+ ) = 5 mA.
P 7.2
[a] i(0) =
60
= 0.5 A.
120
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Problems
7–13
L
0.32
=
= 2 ms.
R
160
[c] i = 0.5e 500t A,
t 0;
[b] ⌧ =
v1 = L
d
(0.5e 500t ) =
dt
v2 =
70i =
80e 500t V
35e 500t V
t
0+ ;
0+ .
t
1
[d] w(0) = (0.32)(0.5)2 = 40 mJ;
2
w90⌦ =
Z t
0
90(0.25e
1000x
w90⌦ (1 ms) = 0.0225(1
% dissipated =
P 7.3
[a] iL (0) =
e 1000x t
) dx = 22.5
= 22.5(1
1000 0
e 1000t ) mJ;
e 1 ) = 14.22 mJ;
14.22
(100) = 35.6%.
40
12
= 2 A;
6
io (0+ ) =
12
2
io (1) =
12
= 6 A.
2
2=6
[b] iL = 2e t/⌧ ;
⌧=
2 = 4 A;
1
L
= s;
R
4
iL = 2e 4t A;
io = 6
[c] 6
iL = 6
2e 4t A,
t
0+ .
2e 4t = 5;
1 = 2e 4t ;
e4t = 2
.·. t = 173.3 ms.
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7–14
P 7.4
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] For t = 0 the circuit is:
io (0 ) = 0;
iL (0 ) =
since the switch is open.
25
= 0.1 = 100 mA;
250
vL (0 ) = 0;
since the inductor behaves like a short circuit.
[b] For t = 0+ the circuit is:
iL (0+ ) = iL (0 ) = 100 mA;
ig =
25
= 0.5 = 500 mA;
50
io (0+ ) = ig
iL (0+ ) = 500
200iL (0+ ) + vL (0+ ) = 0;
100 = 400 mA;
.·.
vL (0+ ) =
200iL (0+ ) =
20 V;
[c] As t ! 1 the circuit is:
iL (1) = 0;
io (1) =
vL (1) = 0;
25
= 500 mA.
50
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Problems
[d] ⌧ =
L
0.05
=
= 0.25 ms;
R
200
iL (t) = 0 + (0.1
[e] io (t) = ig
[f ] vL (t) = L
P 7.5
7–15
0)e 4000t = 0.1e 4000t A
iL = 0.5
0.1e 4000t A.
diL
= 0.05( 4000)(0.1)e 4000t =
dt
20e 4000t V.
t < 0:
iL (0+ ) = 8 A.
t > 0:
Re =
⌧=
(10)(40)
+ 10 = 18 ⌦;
50
L
0.072
= 4 ms;
=
Re
18
1
= 250;
⌧
.·. iL = 8e 250t A.
.·. vo =
10iL
0.072
= 64e 250t A t
diL
=
dt
80e 250t + 144e 250t
0+ .
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7–16
P 7.6
CHAPTER 7. Response of First-Order RL and RC Circuits
1
w(0) = (72 ⇥ 10 3 )(8)2 = 2304 mJ;
2
p40⌦ =
642 500t
vo2
=
e
= 102.4e 500t W;
40
40
w40⌦ =
Z 1
%diss =
P 7.7
0
102.4e 500t dt = 204.8 mJ;
204.8
(100) = 8.89%.
2304
[a] For t < 0
ig =
50
50
=
= 1 A;
20 + (75k50)
50
io (0 ) =
75k50
(1) = 0.6 A = io (0+ ).
50
For t > 0
io (t) = io (0+ )e t/⌧ A,
⌧=
0;
0.02
L
=
= 1.33 ms;
R
3 + 60k15
io (t) = 0.6e 750t A,
t
1
= 750;
⌧
0.
dio
= 0.02( 750)(0.6e 750t ) =
dt
9e 750t V;
60k15
12
vL = ( 9e 750t ) =
3 + 60k15
15
7.2e 750t V
[b] vL = L
vo =
t
t
0+ .
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Problems
P 7.8
7–17
v
400e 5t
=R=
= 40 ⌦.
i
10e 5t
1
[b] ⌧ = = 200 ms.
5
L
= 200 ⇥ 10 3
[c] ⌧ =
R
[a]
L = (200 ⇥ 10 3 )(40) = 8 H.
1
1
[d] w(0) = L[i(0)]2 = (8)(10)2 = 400 J.
2
2
[e] wdiss =
Z t
0
4000e 10x dx = 400
400e 10t ;
0.8w(0) = (0.8)(400) = 320 J;
400
.·. e10t = 5.
400e 10t = 320
Solving, t = 160.9 ms.
P 7.9
[a] Note that there are several di↵erent possible solutions to this problem,
and the answer to part (c) depends on the value of inductance chosen.
L
.
⌧
Choose a 10 mH inductor from Appendix H. Then,
R=
R=
0.01
= 10 ⌦
0.001
which is a resistor value from Appendix H.
[b] i(t) = Io e t/⌧ = 10e 1000t mA,
t 0.
1
1
[c] w(0) = LIo2 = (0.01)(0.01)2 = 0.5 µJ;
2
2
1
w(t) = (0.01)(0.01e 1000t )2 = 0.5 ⇥ 10 6 e 2000t .
2
1
So 0.5 ⇥ 10 6 e 2000t = w(0) = 0.25 ⇥ 10 6;
2
e 2000t = 0.5
then
ln 2
= 346.57 µs.
.·. t =
2000
e2000t = 2;
(for a 10 mH inductor)
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7–18
P 7.10
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] vo (t) = vo (0+ )e t/⌧ ;
3
.·. vo (0+ )e 10 /⌧ = 0.5vo (0+ );
3
.·. e10 /⌧ = 2;
3
10
L
=
;
.·. ⌧ =
R
ln 2
10 ⇥ 10
.·. L =
ln 2
3
= 14.43 mH.
[b] vo (0+ ) =
10iL (0+ ) =
vo (t) =
0.03e t/⌧ V;
10(1/10)(30 ⇥ 10 3 ) =
30 mV;
vo2
p10⌦ =
= 9 ⇥ 10 5 e 2t/⌧ ;
10
w10⌦ =
⌧=
Z 10 3
0
9 ⇥ 10 5 e 2t/⌧ dt = 4.5⌧ ⇥ 10 5 (1
1
1000 ln 2
.·.
e 2⇥10
3 /⌧
);
w10⌦ = 48.69 nJ;
1
1
wL (0) = Li2L (0) = (14.43 ⇥ 10 3 )(3 ⇥ 10 3 )2 = 64.92 nJ;
2
2
% diss in 1 ms =
P 7.11
48.69
⇥ 100 = 75%.
64.92
1
w(0) = (30 ⇥ 10 3 )(32 ) = 135 mJ;
2
1
w(0) = 27 mJ;
5
iR = 3e t/⌧ ;
pdiss = i2R R = 9Re 2t/⌧ ;
wdiss =
Z t
0
wdiss = 9R
R(9)e 2x/⌧ dx;
e 2x/⌧
2/⌧
to
=
4.5⌧ R(e 2to /⌧
1) = 4.5L(1
e 2to /⌧ );
0
4.5L = (4.5)(30) ⇥ 10 3 = 0.135;
to = 15 µs;
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Problems
1
1
e 2to /⌧ = ;
5
2to R
2to
=
= ln 1.25;
⌧
L
e2to /⌧ = 1.25;
R=
P 7.12
7–19
30 ⇥ 10 3 ln 1.25
L ln 1.25
=
= 223.14 ⌦.
2to
30 ⇥ 10 6
1
[a] w(0) = LIg2 .
2
wdiss =
Z to
0
e 2t/⌧ to
( 2/⌧ ) 0
1
e 2to /⌧ ) = Ig2 L(1 e 2to /⌧ ; )
2
Ig2 Re 2t/⌧ dt = Ig2 R
1
= Ig2 R⌧ (1
2
wdiss = w(0);
1 2
.·.
LI (1
2 g
1
e 2to /⌧ ) = ⌧
e 2to /⌧ = ;
"
#
L ln[1/(1
2to
)]
[b] R =
◆
1 2
LI .
2 g
e2to /⌧ =
1
2to
= ln
;
⌧
(1
)
R=
✓
1
(1
)
;
R(2to )
= ln[1/(1
L
); ]
.
(30 ⇥ 10 3 ) ln[1/0.8]
30 ⇥ 10 6
R = 223.14 ⌦.
P 7.13
[a] t < 0
Simplify this circuit by creating a Thévenin equivalant to the left of the
inductor and an equivalent resistance to the right of the inductor:
1 k⌦k4 k⌦ = 0.8 k⌦;
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7–20
CHAPTER 7. Response of First-Order RL and RC Circuits
20 k⌦k80 k⌦ = 16 k⌦;
(105 ⇥ 10 3 )(0.8 ⇥ 103 ) = 84 V.
84
= 5 mA.
16,800
iL (0 ) =
t > 0:
⌧=
L
6
=
⇥ 10 3 = 250 µs;
R
24
iL (t) = 5e 4000t mA,
t
1
= 4000;
⌧
0;
p4k = 25 ⇥ 10 6 e 8000t (4000) = 0.10e 8000t W;
wdiss =
Z t
0
0.10e 8000x dx = 12.5 ⇥ 10 6 [1
e 8000t ] J;
1
w(0) = (6)(25 ⇥ 10 6 ) = 75 µJ;
2
0.10w(0) = 7.5 µJ;
12.5(1
t=
e 8000t ) = 7.5;
.·. e8000t = 2.5;
ln 2.5
= 114.54 µs.
8000
[b] wdiss (total) = 75(1
e 8000t ) µJ;
wdiss (114.54 µs) = 45 µJ;
% = (45/75)(100) = 60%.
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Problems
P 7.14
7–21
t < 0;
iL (0 ) = iL (0+ ) = 4 A.
t > 0;
Find Thévenin resistance seen by inductor:
iT = 4vT ;
⌧=
vT
1
= RTh = = 0.25 ⌦;
iT
4
5 ⇥ 10 3
L
=
= 20 ms;
R
0.25
io = 4e 50t A,
vo = L
t
1/⌧ = 50.
0;
dio
= (5 ⇥ 10 3 )( 200e 50t ) =
dt
e 50t V,
t
0+ .
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–22
P 7.15
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0 :
iL (0 ) = iL (0+ ) =
i =
70
(11.84) = 11.2 A.
70 + 4
70
iT = 0.4375iT ;
160
vT = 30i + iT
(90)(70)
(90)(70)
= 30(0.4375)iT +
iT = 52.5iT ;
160
160
vT
= RTh = 52.5 ⌦.
iT
⌧=
20 ⇥ 10 3
L
=
= .·.
R
52.5
iL = 11.2e 2625t A,
[b] vL = L
t
1
= 2625;
⌧
0.
diL
= 20 ⇥ 10 3 ( 2625)(11.2e 2625t ) =
dt
588e 2625t V,
t
0+ .
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Problems
7–23
[c]
vL = 30i + 90i = 120i ;
i =
P 7.16
vL
=
120
0+ .
t
1
w(0) = (20 ⇥ 10 3 )(11.2)2 = 1254.4 mJ;
2
p30i =
30i iL =
w30i =
Z 1
0
30( 4.9e 2625t )(11.2e 2625t ) = 1646.4e 5250t W;
1646.4e 5250t dt = 1646.4
% dissipated =
P 7.17
4.9e 2625t A
e 5250t 1
= 313.6mJ;
5250 0
313.6
(100) = 25%.
1254.4
[a] t < 0 :
t = 0+ :
33 = iab + 9 + 15,
iab = 9 A,
t = 0+ .
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7–24
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] At t = 1:
t = 1.
iab = 165/5 = 33 A,
12.5 ⇥ 10 3
= 2.5 ms;
⌧1 =
5
[c] i1 (0) = 9,
i2 (0) = 15,
3.75 ⇥ 10 3
1.25 ms;
3
⌧2 =
i1 (t) = 9e 400t A,
t
0;
i2 (t) = 15e 800t A,
iab = 33
33
9e 400t
9e 400t
t
0;
15e 800t A,
t
0+ ;
15e 800t = 19;
14 = 9e 400t + 15e 800t .
Let x = e 400t
.·.
x2 = e 800t.
Substituting,
15x2 + 9x
.·.
t=
14 = 0 so x = 0.7116 = e 400t ;
[ln(1/0.7116)]
= 850.6 µs.
400
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Problems
P 7.18
7–25
[a] t < 0:
iL (0 ) =
150
(12) = 10 A.
180
t
0;
⌧=
1.6 ⇥ 10 3
= 200 ⇥ 10 6 ;
8
io =
10e 5000t A t
1/⌧ = 5000;
0.
1
[b] wdel = (1.6 ⇥ 10 3 )(10)2 = 80 mJ.
2
[c] 0.95wdel = 76 mJ;
.·. 76 ⇥ 10 3 =
.·. 76 ⇥ 10 3 =
Z to
0
8(100e 10,000t ) dt;
80 ⇥ 10 3 e 10,000t
to
0
= 80 ⇥ 10 3 (1
e 10,000to );
.·. e 10,000to = 0.05 so to = 299.57 µs;
299.57 ⇥ 10 6
to
·
=
= 1.498
..
⌧
200 ⇥ 10 6
P 7.19
so
to ⇡ 1.498⌧.
[a] t > 0:
Leq = 1.25 +
60
= 5 H.
16
iL (t) = iL (0)e t/⌧ mA;
iL (0) = 2 A;
R
7500
1
=
=
= 1500;
⌧
L
5
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7–26
CHAPTER 7. Response of First-Order RL and RC Circuits
iL (t) = 2e 1500t A,
t
0;
vR (t) = RiL (t) = (7500)(2e 1500t ) = 15,000e 1500t V,
vo =
3.75
diL
= 11,250e 1500t V,
dt
0+ ;
0+ .
t
1Z t
[b] io =
11,250e 1500x dx + 0 = 1.25e 1500t
6 0
P 7.20
t
1.25 A.
[a] From the solution to Problem 7.19,
1
1
w(0) = Leq [iL (0)]2 = (5)(2)2 = 10 J.
2
2
1
1
[b] wtrapped = (10)(1.25)2 + (6)(1.25)2 = 12.5 J.
2
2
CHECK: w(0) = 12 (1.25)(2)2 + 12 (10)(2)2 = 22.5 J;
.·. w(0) = wdiss + wtrapped .
P 7.21
[a] For t < 0:
v(0) = 150 V.
1
1
[b] w(0) = Cv(0)2 = (40 ⇥ 10 9 )(150)2 = 450 µJ.
2
2
[c] For t > 0:
Req = 5000 + 30,000k60,000 = 25 k⌦;
⌧ = Req C = (25,000)(40 ⇥ 10 9 ) = 1 ms.
[d] v(t) = v(0)e t/⌧ = 150e 1000t V,
t
0.
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Problems
P 7.22
7–27
[a] Calculate the initial voltage drop across the capacitor:
v(0) = (2.7 kk3.3 k)(40 mA) = (1485)(40 ⇥ 10 3 ) = 59.4 V.
The equivalent resistance seen by the capacitor is
Re = 3 kk(2.4 k + 3.6 k) = 3 kk6 k = 2 k⌦;
⌧ = Re C = (2000)(0.5 ⇥ 10 6 ) = 1000 µs;
1
= 1000;
⌧
v = v(0)e t/⌧ = 59.4e 1000t V
t 0.
v
= 9.9e 1000t mA, t 0+ .
[b] i =
2.4 k + 3.6 k
P 7.23
1
w(0) = (0.5 ⇥ 10 6 )(59.4)2 = 882.09 µJ;
2
i3k =
59.4e 1000t
= 19.8e 1000t mA;
3000
p3k = [(19.8 ⇥ 10 3 )e 1000t ]2 (3000) = 1.176e 2000t ;
6
e 2000x 500⇥10
1.176
(e 1
=
w3k (500 µs) = 1.176
2000 0
2000
%=
P 7.24
1) = 371.72 µJ;
371.69
⇥ 100 = 42.14%.
882.09
For t < 0:
Vo = (20,000k60,000)(20 ⇥ 10 3 ) = 300 V.
For t
0:
Req = 10,000 + (20,000k60,000) = 25 k⌦;
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7–28
CHAPTER 7. Response of First-Order RL and RC Circuits
⌧ = Req C = (25,000)(40 ⇥ 10 9 ) = 1 ms;
v(t) = Vo e t/⌧ = 300e 1000t V
P 7.25
t
0.
[a] t < 0:
i1 (0 ) = i2 (0 ) =
3
= 100 mA.
30
[b] t > 0:
i1 (0+ ) =
i2 (0+ ) =
0.2
= 100 mA;
2
0.2
=
8
25 mA.
[c] Capacitor voltage cannot change instantaneously, therefore,
i1 (0 ) = i1 (0+ ) = 100 mA.
[d] Switching can cause an instantaneous change in the current in a resistive
branch. In this circuit
i2 (0 ) = 100 mA and i2 (0+ ) = 25 mA.
[e] vc = 0.2e t/⌧ V,
t
0;
⌧ = Re C = 1.6(2 ⇥ 10 6 ) = 3.2 µs;
vc = 0.2e 312,000t V,
i1 =
t
0;
vc
= 0.1e 312,000t A,
2
t
1
= 312,500;
⌧
0.
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Problems
vc
=
8
[f ] i2 =
P 7.26
25e 312,000t mA,
7–29
0+ .
t
[a] t < 0:
Req = 12 kk8 k = 10.2 k⌦;
vo (0) =
10,200
( 120) =
10,200 + 1800
102 V.
t > 0:
⌧ = [(10/3) ⇥ 10 6 )(12,000) = 40 ms;
102e 25t V,
vo =
p=
t
1
= 25.
⌧
0;
vo2
= 867 ⇥ 10 3 e 50t W;
12,000
wdiss =
Z 12⇥10 3
0
867 ⇥ 10 3 e 50t dt
= 17.34 ⇥ 10 3 (1
✓ ◆✓
1
[b] w(0) =
2
e 50(12⇥10
3)
) = 7824 µJ.
◆
10
⇥ 10 6 (102)2 = 17.34 mJ
3
0.75w(0) = 13 mJ;
Z to
0
867 ⇥ 10 3 e 50x dx = 13 ⇥ 10 3 ;
.·. 1
P 7.27
[a] R =
e 50to = 0.75;
e50to = 4;
so to = 27.73 ms.
v
= 4 k⌦
i
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7–30
CHAPTER 7. Response of First-Order RL and RC Circuits
1
1
1
=
= 25;
C=
= 10 µF
⌧
RC
(25)(4 ⇥ 103 )
1
= 40 ms
[c] ⌧ =
25
1
[d] w(0) = (10 ⇥ 10 6 )(48)2 = 11.52 mJ
2
Z 0.06 2
Z 0.06
v
(48e 25t )2
dt =
dt
[e] wdiss (60 ms) =
R
(4 ⇥ 103 )
0
0
[b]
= 0.576
P 7.28
e 50t 0.06
=
50 0
5.74 ⇥ 10 4 + 0.01152 = 10.95 mJ
[a] Note that there are many di↵erent possible correct solutions to this
problem.
⌧
R= .
C
Choose a 100 µF capacitor from Appendix H. Then,
0.05
= 500 ⌦.
100 ⇥ 10 6
Construct a 500 ⌦ resistor by combining two 1 k⌦ resistors in parallel:
R=
[b] v(t) = Vo e t/⌧ = 50e 20t V,
[c] 50e 20t = 10
so
t
0.
e20t = 5;
ln 5
.·. t =
= 80.47 ms.
20
P 7.29
t < 0:
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Problems
7–31
t > 0:
vT =
5io
15io =
io =
P 7.30
vo
=
20
RTh =
vT
= 20 ⌦.
iT
1
= 25,000;
⌧
⌧ = RC = 40 µs;
vo = 15e 25,000t V,
.·.
20io = 20iT
t
0;
0.75e 25,000t A,
t
0+ .
[a]
vT = 20 ⇥ 103 (iT + ↵v ) + 5 ⇥ 103 iT ;
v = 5 ⇥ 103 iT ;
vT = 25 ⇥ 103 iT + 20 ⇥ 103 ↵(5 ⇥ 103 iT );
RTh = 25,000 + 100 ⇥ 106 ↵;
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7–32
CHAPTER 7. Response of First-Order RL and RC Circuits
⌧ = RTh C = 40 ⇥ 10 3 = RTh (0.8 ⇥ 10 6 ; )
RTh = 50 k⌦ = 25,000 + 100 ⇥ 106 ↵;
↵=
25,000
= 2.5 ⇥ 10 4 A/V.
100 ⇥ 106
[b] vo (0) = ( 5 ⇥ 10 3 )(3600) =
t > 0:
vo =
18e 25t V,
t
18 V,
t < 0.
0.
v
vo
v
+
+ 2.5 ⇥ 10 4 v = 0;
5000
20,000
4v + v
vo + 5v = 0;
vo
=
.·. v =
10
P 7.31
1.8e 25t V,
t
0+ .
[a]
pds = (16.2e 25t )( 450 ⇥ 10 6 e 25t ) =
7290 ⇥ 10 6 e 50t W;
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Problems
wds =
Z 1
0
pds dt =
145.8 µJ.
.·. dependent source is delivering 145.8 µJ.
[b] w5k =
Z 1
0
w20k =
3
(5000)(0.36 ⇥ 10 e
Z 1
0
7–33
25t 2
) dt = 648 ⇥ 10
6
Z 1
0
e 50t dt = 12.96 µJ;
Z 1
(16.2e 25t )2
dt = 13,122 ⇥ 10 6
e 50t dt = 262.44 µJ;
20,000
0
1
wc (0) = (0.8 ⇥ 10 6 )(18)2 = 129.6 µJ;
2
X
P 7.32
X
wdiss = 12.96 + 262.44 = 275.4 µJ;
wdev = 145.8 + 129.6 = 275.4 µJ.
[a] The equivalent circuit for t > 0:
⌧ = 2 ms;
1/⌧ = 500;
vo = 10e 500t V,
t
0;
io = e 500t mA,
t
0+ ;
i24k⌦ = e
500t
✓
◆
16
= 0.4e 500t mA,
40
t
0+ ;
p24k⌦ = (0.16 ⇥ 10 6 e 1000t )(24,000) = 3.84e 1000t mW;
w24k⌦ =
Z 1
0
3.84 ⇥ 10 3 e 1000t dt =
3.84 ⇥ 10 6 (0
1) = 3.84 µJ;
1
1
w(0) = (0.25 ⇥ 10 6 )(40)2 + (1 ⇥ 10 6 )(50)2 = 1.45 mJ;
2
2
% diss (24 k⌦) =
3.84 ⇥ 10 6
⇥ 100 = 0.26%.
1.45 ⇥ 10 3
[b] p400⌦ = 400(1 ⇥ 10 3 e 500t )2 = 0.4 ⇥ 10 3 e 1000t ;
w400⌦ =
Z 1
0
p400 dt = 0.40 µJ;
% diss (400 ⌦) =
0.4 ⇥ 10 6
⇥ 100 = 0.03%;
1.45 ⇥ 10 3
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7–34
CHAPTER 7. Response of First-Order RL and RC Circuits
500t
i16k⌦ = e
✓
◆
24
= 0.6e 500t mA,
40
0+ ;
t
p16k⌦ = (0.6 ⇥ 10 3 e 500t )2 (16,000) = 5.76 ⇥ 10 3 e 1000t W;
w16k⌦ =
Z 1
0
5.76 ⇥ 10 3 e 1000t dt = 5.76 µJ;
% diss (16 k⌦) = 0.4%.
[c]
X
wdiss = 3.84 + 5.76 + 0.4 = 10 µJ
wtrapped = w(0)
% trapped =
X
wdiss = 1.45 ⇥ 10 3
10 ⇥ 10 6 = 1.44 mJ;
1.44
⇥ 100 = 99.31%.
1.45
Check: 0.26 + 0.03 + 0.4 + 99.31 = 100%.
P 7.33
[a] v1 (0 ) = v1 (0+ ) = 40 V
v2 (0+ ) = 0;
Ceq = (1)(4)/5 = 0.8 µF.
1
= 50;
⌧
⌧ = (25 ⇥ 103 )(0.8 ⇥ 10 6 ) = 20ms;
i=
40
e 50t = 1.6e 50t mA,
25,000
v1 =
v2 =
10
1 Zt
6
0
t
0+ .
1.6 ⇥ 10 3 e 50x dx + 40 = 32e 50t + 8 V,
Z t
1
1.6 ⇥ 10 3 e 50x dx + 0 =
4 ⇥ 10 6 0
8e 50t + 8 V,
t
0;
t
0.
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Problems
7–35
1
[b] w(0) = (10 6 )(40)2 = 800 µJ.
2
1
1
[c] wtrapped = (10 6 )(8)2 + (4 ⇥ 10 6 )(8)2 = 160 µJ.
2
2
The energy dissipated by the 25 k⌦ resistor is equal to the energy
dissipated by the two capacitors; it is easier to calculate the energy
dissipated by the capacitors:
1
wdiss = (0.8 ⇥ 10 6 )(40)2 = 640 µJ.
2
Check: wtrapped + wdiss = 160 + 640 = 800 µJ;
P 7.34
w(0) = 800 µJ.
[a] At t = 0 the voltage on each capacitor will be 150 V(5 ⇥ 30), positive at
the upper terminal. Hence at t 0+ we have
150 150
+
= 1055 A.
.·. isd (0+ ) = 5 +
0.2
0.5
At t = 1, both capacitors will have completely discharged.
.·. isd (1) = 5 A.
[b] isd (t) = 5 + i1 (t) + i2 (t);
⌧1 = 0.2(10 6 ) = 0.2 µs;
⌧2 = 0.5(100 ⇥ 10 6 ) = 50 µs;
.·. i1 (t) = 750e 5⇥10 t A,
t
0+ ;
i2 (t) = 300e 20,000t A,
t
0;
6
.·. isd = 5 + 750e 5⇥10 t + 300e 20,000t A,
6
P 7.35
t
0+ .
[a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and
right of the 200 mH inductor. We get
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7–36
CHAPTER 7. Response of First-Order RL and RC Circuits
i(0 ) =
30 250
=
25 k + 30 k
i(0 ) = i(0+ ) =
4 mA;
4 mA.
[b] For t > 0, the circuit reduces to
Therefore i(1) = 30/30,000 = 1 mA.
L
200 ⇥ 10 3
[c] ⌧ =
=
= 6.67 µs.
R
30,000
[d] i(t) = i(1) + [i(0+ )
i(1)]e t/⌧
= 0.001 + [ 0.004
P 7.36
0.001]e 150,000t = 1
5e 150,000t mA,
t
0.
[a] t < 0;
KVL equation at the top node:
vo
800 vo
+
+
= 0.
120
40 10
Multiply by 120 and solve:
vo
800 = (1 + 3 + 12)vo ;
vo = 50 V;
vo
= 50/10 = 5 A.
.·. io (0 ) =
10
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Problems
7–37
t > 0:
Use voltage division to find the Thévenin voltage:
40
(400) = 333.33 V.
40 + 8
Remove the voltage source and make series and parallel combinations of
resistors to find the equivalent resistance:
VTh = vo =
RTh = 10 + 40k8 = 16.67 ⌦.
The simplified circuit is:
⌧=
40 ⇥ 10 3
L
=
= 2.4 ms;
R
16.67
io (1) =
333.33
= 20 A;
16.67
.·. io = io (1) + [io (0+ )
vo
io (1)]e t/⌧
20)e 416.67t = 20
= 20 + (5
[b] vo
1
= 416.67.
⌧
15e 416.67t A,
t
0.
dio
dt
15e 416.67t ) + 0.04( 416.67)( 15e 416.67t )
=
10io + L
=
10(20
=
200
=
200 + 100e 416.67t V,
150e 416.67t + 250e 416.67t
t
0+ .
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7–38
P 7.37
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] t < 0:
ig =
400
= 25 A;
8 + 40k10
40k10
(25) = 20 A.
.·. io (0 ) =
10
t > 0;
ig (1) =
800
= 6.25 A;
120 + 40k10
io (1) =
40k10
ig (1) = 5 A;
10
Req = 10 + 120k40 = 40 ⌦;
⌧=
40 ⇥ 10 3
L
= 1 ms;
=
Req
40
.·. io = io (1) + [io (0+ )
= 5 + (20
[b] vo
1
= 1000;
⌧
io (1)]e t/⌧
5)e 1000t = 5 + 15e 1000t A,
t
=
dio
dt
10(5 + 15e 1000t ) + 0.04( 1000)(15e 1000t )
=
50 + 150e 1000t
=
50
=
0.
10io + L
450e 1000t V,
600e 1000t
t
0+ .
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Problems
P 7.38
7–39
[a] t < 0:
32
= 1.6 A.
20
iL (0 ) =
t > 0:
32 48
=
12 + 8
iL (1) =
⌧=
0.8 A;
L
5 ⇥ 10 3
=
= 250 µs;
R
12 + 8
iL = iL (1) + [iL (0+ )
=
1
= 4000;
⌧
iL (1)]e t/⌧
0.8 + (1.6 + 0.8)e 4000t =
0.8 + 2.4e 4000t A,
t
0.
vo = 8iL + 48 = 8( 0.8 + 2.4e 4000t ) + 48 = 41.6 + 19.2e 4000t V,
[b] vL = L
diL
= 5 ⇥ 10 3 ( 4000)[2.4e 4000t ] =
dt
vL (0+ ) =
48e 4000t V,
t
t
0.
0+
48 V.
From part (a)
vo (0+ ) = 0 V.
Check: at t = 0+ the circuit is:
vo (0+ ) = 48 + (8 ⌦)(1.6 A) = 60.8 V;
.·.
vL (0+ ) =
19.2 + 32
60.8 =
vL (0+ ) + vo (0+ ) = 12( 1.6) + 32;
48 V.
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7–40
P 7.39
CHAPTER 7. Response of First-Order RL and RC Circuits
[a]
v
1Z t
Vs
+ +
v dt + Io = 0.
R R L 0
Di↵erentiating both sides,
1
1 dv
+ v = 0;
R dt L
dv R
.·.
+ v = 0.
dt
L
dv
R
[b]
=
v;
dt
L
R
dv
dt =
v dt
so
dv =
dt
L
R
dv
=
dt;
v
L
Z v(t)
dx
RZ t
=
dy;
L 0
Vo x
ln
P 7.40
R
t;
L
v(t)
=
Vo
.·.
R
v dt;
L
v(t) = Vo e (R/L)t = (Vs
RIo )e (R/L)t .
[a] From Eqs. 7.1 and 7.15
i=
✓
Vs
+ Io
R
v = (Vs
◆
Vs
e (R/L)t ;
R
Io R)e (R/L)t ;
Vs
= 4;
.·.
R
Io
Io R =
80;
Vs
Vs
= 4.
R
R
= 40;
L
Vs
= 8 A.
.·. Io = 4 +
R
Now since Vs = 4R we have
4R
8R =
Vs = 80 V;
80;
L=
R = 20 ⌦;
R
= 0.5 H.
40
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Problems
[b] i = 4 + 4e 40t ;
7–41
i2 = 16 + 32e 40t + 16e 80t ;
1
1
w = Li2 = (0.5)[16 + 32e 40t + 16e 80t ] = 4 + 8e 40t + 4e 80t ;
2
2
.·. 4 + 8e 40t + 4e 80t = 9 or e 80t + 2e 40t
1.25 = 0.
Let x = e 40t :
x2 + 2x
1.25 = 0;
But x
0 for all t. Thus,
e 40t = 0.5;
P 7.41
[a] vo (0+ ) =
Ig R2 ;
Solving, x = 0.5;
e40t = 2;
⌧=
x=
2.5.
t = 25 ln 2 = 17.33 ms.
L
;
R 1 + R2
vo (1) = 0;
vo (t) =
Ig R2 e [(R1 +R2 )/L]t V,
t
0+ .
[b] vo (0+ ) ! 1, and the duration of vo (t) ! zero.
L
⌧=
;
[c] vsw = R2 io ;
R 1 + R2
R1
.
R 1 + R2
io (0+ ) = Ig ;
io (1) = Ig
Therefore
io (t) =
Ig R1
+
R1 +R2
io (t) =
R1 Ig
2 Ig
+ (RR1 +R
e [(R1 +R2 )/L]t .
(R1 +R2 )
2)
vsw =
R1 Ig
R2 Ig
+ (1+R
e [(R1 +R2 )/L]t ,
(1+R1 /R2 )
1 /R2 )
Therefore
[d] |vsw (0+ )| ! 1;
h
Ig
Ig R1
R1 +R2
i
e [(R1 +R2 )/L]t
t
0+ .
duration ! 0.
P 7.42
Opening the inductive circuit causes a very large voltage to be induced across
the inductor L. This voltage also appears across the switch (part [d] of
Problem 7.41), causing the switch to arc over. At the same time, the large
voltage across L damages the meter movement.
P 7.43
[a] Note that there are many di↵erent possible solutions to this problem.
L
.
⌧
Choose a 1 mH inductor from Appendix H. Then,
R=
R=
0.001
= 125 ⌦.
8 ⇥ 10 6
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7–42
CHAPTER 7. Response of First-Order RL and RC Circuits
Construct the resistance needed by combining 100 ⌦, 10 ⌦, and 15 ⌦
resistors in series:
[b] i(t) = If + (Io
If )e t/⌧ ;
Io = 0 A;
If =
.·. i(t) = 200 + (0
[c] i(t) = 0.2
P 7.44
200)e 125,000t mA = 200
200e 125,000t mA,
t
0.
0.2e 125,000t = (0.75)(0.2) = 0.15;
e 125,000t = 0.25
.·. t =
25
Vf
=
= 200 mA;
R
125
e125,000t = 4;
so
ln 4
= 11.09 µs.
125,000
t > 0;
calculate vo (0+ ):
va
va
+
15
vo (0+ )
= 20 ⇥ 10 3 ;
5
.·. va = 0.75vo (0+ ) + 75 ⇥ 10 3 .
15 ⇥ 10 3 +
13vo (0+ )
i =
8va
vo (0+ )
8
.·. i =
vo (0+ )
5
va
+
vo (0+ )
8
360i =
9i + 50 ⇥ 10 3 = 0;
2600 ⇥ 10 3 ;
9i + 50 ⇥ 10 3 ;
vo (0+ )
+ 5 ⇥ 10 3 ;
80
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Problems
7–43
.·. 360i = 4.5vo (0+ ) + 1800 ⇥ 10 3 ;
8va = 6vo (0+ ) + 600 ⇥ 10 3 ;
.·. 13vo (0+ )
2.5vo (0+ ) =
6vo (0+ )
600 ⇥ 10 3
4.5vo (0+ )
1800 ⇥ 10 3 =
2600 ⇥ 10 3 ;
200 ⇥ 10 3 ;
vo (0+ ) =
80 mV;
vo (1) = 0.
Find the Thévenin resistance seen by the 4 mH inductor:
iT =
vT
vT
+
20
8
i =
vT
8
iT =
10vT
vT
+
20
80
9i ;
9i
vT
;
.·. 10i =
8
i =
vT
.
80
9vT
;
80
iT
1
5
1
1
+
=
=
S;
=
vT
20 80
80
16
.·. RTh = 16 ⌦.
⌧=
4 ⇥ 10 3
= 0.25 ms;
16
.·. vo = 0 + ( 80
1/⌧ = 4000;
0)e 4000t =
80e 4000t mV,
t
0+ .
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7–44
P 7.45
CHAPTER 7. Response of First-Order RL and RC Circuits
For t < 0:
vx
15
0.8v +
v =
vx
15
vx
0.8
vx
480
= 0;
21
480
;
21
✓
vx
◆
✓
480
vx 480
+
21
21
✓
◆
◆
vx
vs 480
+ 0.2
=
= 21vx + 3(vx
15
21
.·.
24vx = 1440 so vx = 60 V
480) = 0;
io (0 ) =
vx
= 4 A.
15
t > 0:
Find Thévenin equivalent with respect to a, b:
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Problems
VTh
320
5
0.8
✓
VTh
320
5
✓
◆
=0
vT = (iT + 0.8v )(5) = iT + 0.8
vT = 5iT + 0.8vT
vT
5
7–45
VTh = 320 V.
◆
(5);
.·. 0.2vT = 5iT ;
vT
= RTh = 25 ⌦.
iT
io (1) = 320/40 = 8 A;
⌧=
80 ⇥ 10 3
= 2 ms;
40
io = 8 + (4
8)e 500t = 8
1/⌧ = 500;
4e 500t A,
t
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–46
P 7.46
CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0:
⌧=
1
;
40
io = 5e 40t A,
t
0;
vo = 40io = 200e 40t V,
200e 40t = 100;
t > 0+ ;
e40t = 2;
1
ln 2 = 17.33 ms.
.·. t =
40
P 7.47
1
1
[a] wdiss = Le i2 (0) = (1)(5)2 = 12.5 J
2
2
1Z t
[b] i3H =
(200)e 40x dx 5
3 0
= 1.67(1
i1.5H =
=
e 40t )
5=
1.67e 40t
3.33 A
1 Zt
(200)e 40x dx + 0
1.5 0
3.33e 40t + 3.33 A
1
wtrapped = (4.5)(3.33)2 = 25 J
2
1
[c] w(0) = (3)(5)2 = 37.5 J
2
P 7.48
For t < 0,
i80mH (0) = 50 V/10 ⌦ = 5 A.
For t > 0, after making a Thévenin equivalent we have
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
✓
Vs
+ Io
i=
R
7–47
◆
Vs
e t/⌧ ;
R
R
8
1
=
=
= 80;
⌧
L
100 ⇥ 10 3
Io = 5 A;
i=
Vs
=
R
10 + (5 + 10)e 80t =
vo = 0.08
P 7.49
If =
80
=
8
10 A;
10 + 15e 80t A,
di
= 0.08( 1200e 80t ) =
dt
96e 80t V,
t
0;
t > 0+ .
[a] t < 0:
t > 0:
iL (0 ) = iL (0+ ) = 25 mA;
iL (1) =
⌧=
24 ⇥ 10 3
= 0.2 ms;
120
1
= 5000;
⌧
50 mA;
iL =
50 + (25 + 50)e 5000t =
50 + 75e 5000t mA,
vo =
120[75 ⇥ 10 3 e 5000t ] =
9e 5000t V,
t
t
0;
0+ .
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7–48
CHAPTER 7. Response of First-Order RL and RC Circuits
Z t
1
[b] i1 =
9e 5000x dx + 10 ⇥ 10 3 = (30e 5000t
3
60 ⇥ 10
0
Z t
1
[c] i2 =
9e 5000x dx + 15 ⇥ 10 3 = (45e 5000t
40 ⇥ 10 3 0
P 7.50
20) mA,
t
0.
30) mA,
t
0.
[a] Let v be the voltage drop across the parallel branches, positive at the top
node, then
v
1 Zt
1 Zt
Ig +
+
v dx +
v dx = 0;
R g L1 0
L2 0
✓
◆
1 Zt
1
v
+
+
v dx = Ig ;
Rg
L1 L2 0
v
1 Zt
+
v dx = Ig ;
R g Le 0
1 dv
v
+
= 0;
Rg dt Le
dv Rg
+
v = 0.
dt
Le
Therefore v = Ig Rg e t/⌧ ;
Thus
P 7.51
⌧ = Le /Rg .
i1 =
1 Zt
Ig Rg e x/⌧ t Ig Le
Ig Rg e x/⌧ dx =
=
(1
L1 0
L1 ( 1/⌧ ) 0
L1
i1 =
Ig L2
(1
L1 + L2
e t/⌧ ) and i2 =
[b] i1 (1) =
L2
Ig ;
L1 + L2
[a] vc (0+ ) =
50 V.
i2 (1) =
Ig L1
(1
L1 + L2
e t/⌧ );
e t/⌧ ).
L1
Ig .
L1 + L2
[b] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh = v20⌦ =
20
( 30) =
25
24 V,
RTh = 20k5 = 4 ⌦.
Therefore ⌧ = Req C = (4)(25 ⇥ 10 3 ) = 0.1 s.
[c] Use voltage division to find the final value of voltage:
20
( 30) = 24 V.
25
The simplified circuit for t > 0 is:
vc (1) =
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–49
24
( 50)
= 6.5 A.
4
[e] vc = vc (1) + [vc (0+ ) vc (1)]e t/⌧
[d] i(0+ ) =
24 + ( 50 + 24)e t/⌧ =
=
[f ] i = C
P 7.52
24
26e 10t V,
dvc
= (25 ⇥ 10 3 )( 10)( 26e 10t ) = 6.5e 10t A,
dt
t
0.
t
0+ .
[a] For t < 0:
vo (0) =
10,000
(75) = 50 V.
15,000
For t
0:
vo (1) =
40,000
( 100) =
50,000
80 V;
Req = 40 kk10 k = 8 k⌦;
⌧ = Req C = (8000)(40 ⇥ 10 9 ) = 0.32 ms;
vo (t) = vo (1) + (vo (0)
=
vo (1))e t/⌧ =
80 + (50 + 80)e 3125t
80 + 130e 3125t V.
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7–50
CHAPTER 7. Response of First-Order RL and RC Circuits
[b] For t
io =
P 7.53
0:
130e 3125t 80 + 100
= (13e 3125t + 2) mA.
10,000
[a] for t < 0:
vc (0) = 10(2) = 20 V.
For t
0:
vc (1) = 40 V;
Req = 25 ⌦
so
vc (t) = vc (1) + [vc (0)
⌧ = Req C = 25(100 ⇥ 10 6 ) = 2.5 ms;
vc (1)]e t/⌧ = 40 + (20
40)e 400t = 40
20e 400t V.
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Problems
7–51
[b] For t < 0:
vc (0) = 40 V.
For t
0:
vc (1) = 10(2) = 20 V;
Req = 40 + 10 = 50 ⌦,
vc (t) = vc (1) + [vc (0)
P 7.54
so
⌧ = Req C = 50(100 ⇥ 10 6 ) = 5 ms;
vc (1)]e t/⌧ = 20 + (40
20)e 200t = 20 + 20e 200t V.
For t < 0:
Simplify the circuit:
80/10,000 = 8 mA,
8 mA
10 k⌦k40 k⌦k24 k⌦ = 6 k⌦;
3 mA = 5 mA;
5 mA ⇥ 6 k⌦ = 30 V.
Thus, for t < 0
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7–52
CHAPTER 7. Response of First-Order RL and RC Circuits
.·. vo (0 ) = vo (0+ ) = 30 V.
t > 0:
Simplify the circuit:
8 mA + 2 mA = 10 mA;
10 kk40 kk24 k = 6 k⌦;
(10 mA)(6 k⌦) = 60 V.
Thus, for t > 0:
vo (1) =
60 V;
⌧ = RC = (10 k)(0.05 µ) = 0.5 ms;
1
= 2000;
⌧
vo = vo (1) + [vo (0+ )
60 + [30
=
P 7.55
10 ⇥ 10 3 (6 ⇥ 103 ) =
vo (1)]e t/⌧ =
60 + 90e 2000t V
t
( 60)]e 2000t
0.
[a] Use voltage division to find the initial value of the voltage:
vc (0+ ) = v9k =
9k
(120) = 90 V.
9k + 3k
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Problems
7–53
[b] Use Ohm’s law to find the final value of voltage:
vc (1) = v40k =
(1.5 ⇥ 10 3 )(40 ⇥ 103 ) =
60 V.
[c] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
VTh =
60 V,
RTh = 10 k + 40 k = 50 k⌦;
⌧ = RTh C = 1 ms = 1000 µs.
[d] vc = vc (1) + [vc (0+ )
=
P 7.56
vc (1)]e t/⌧
60 + (90 + 60)e 1000t =
60 + 150e 1000t V,
We want vc =
60 + 150e 1000t = 0:
Therefore t =
ln(150/60)
= 916.3 µs.
1000
t
0.
t < 0:
io (0 ) =
20
(10 ⇥ 10 3 ) = 2 mA;
100
vo (0 ) = (2 ⇥ 10 3 )(50,000) = 100 V.
t = 1:
io (1) =
5 ⇥ 10
3
✓
◆
20
=
100
1 mA;
RTh = 50 k⌦k50 k⌦ = 25 k⌦;
ic = C
dvo
=
dt
50 + 150e 2500t V,
6e 2500t mA,
50 V.
C = 16 nF.
1
= 2500;
⌧
⌧ = (25,000)(16 ⇥ 10 9 ) = 0.4 ms;
.·. vo (t) =
vo (1) = io (1)(50,000) =
t
t
0.
0+ ;
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7–54
CHAPTER 7. Response of First-Order RL and RC Circuits
vo
=
50,000
1 + 3e 2500t mA,
io = ic + i50k =
(1 + 3e 2500t ) mA,
t
Is R)e t/RC
i = Is
Vo
24;
i50k =
P 7.57
[a] v = Is R + (Vo
.·. Is R = 40,
0+ ;
t
0+ .
✓
Is R =
◆
Vo
e t/RC ;
R
.·. Vo = 16 V.
Is
Vo
= 3 ⇥ 10 3 ;
R
.·. Is
R=
Is
0.4Is = 3 ⇥ 10 3 ;
16
= 3 ⇥ 10 3 ;
R
R=
40
;
Is
Is = 5 mA.
40
⇥ 103 = 8 k⌦
5
1
= 2500;
RC
C=
1
10 3
=
= 50 nF;
2500R
20 ⇥ 103
⌧ = RC =
1
= 400 µs.
2500
[b] v(1) = 40 V;
1
w(1) = (50 ⇥ 10 9 )(1600) = 40 µJ;
2
0.81w(1) = 32.4 µJ;
32.4 ⇥ 10 6
= 1296;
v (to ) =
25 ⇥ 10 9
2
40
P 7.58
24e 2500to = 36;
v(to ) = 36 V;
e2500to = 6;
.·. to = 716.70 µs.
[a] Note that there are many di↵erent possible solutions to this problem.
⌧
R= .
C
Choose a 10 µH capacitor from Appendix H. Then,
0.25
= 25 k⌦.
10 ⇥ 10 6
Construct the resistance needed by combining 10 k⌦ and 15 k⌦ resistors
in series:
R=
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[b] v(t) = Vf + (Vo
Vf )e t/⌧ ;
Vf = (If )(R) = (1 ⇥ 10 3 )(25 ⇥ 103 ) = 25 V;
Vo = 100 V;
.·. v(t) = 25 + (100
25)e 4t V = 25 + 75e 4t V,
[c] v(t) = 25 + 75e 4t = 50
.·. t =
P 7.59
7–55
e 4t =
so
t
0.
t
0.
1
3
ln 3
= 274.65 ms.
4
Use voltage division to find the initial voltage:
vo (0) =
60
(50) = 30 V.
40 + 60
Use Ohm’s law to find the final value of voltage:
vo (1) = ( 5 mA)(20 k⌦) =
100 V;
⌧ = RC = (20 ⇥ 103 )(250 ⇥ 10 9 ) = 5 ms;
vo = vo (1) + [vo (0+ )
=
P 7.60
1
= 200;
⌧
vo (1)]e t/⌧
100 + (30 + 100)e 200t =
100 + 130e 200t V,
[a]
1 Zt
Is R = Ri +
i dx + Vo ;
C 0+
0=R
i
di
+ + 0;
dt C
di
i
.·.
+
= 0.
dt RC
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7–56
CHAPTER 7. Response of First-Order RL and RC Circuits
[b]
di
=
dt
i
;
RC
Z i(t)
dy
=
i(0+ ) y
ln
di
=
i
1 Zt
dx;
RC 0+
t
i(t)
=
;
+
i(0 )
RC
i(t) = i(0+ )e t/RC ;
✓
.·. i(t) = Is
P 7.61
dt
;
RC
For t < 0,
t > 0:
i(0+ ) =
✓
◆
Is R Vo
= Is
R
◆
Vo
;
R
Vo
e t/RC .
R
vo (0) = ( 3 m)(15 k) =
45 V.
10
(75) =
50
✓
VTh =
20 ⇥ 103 i +
vT =
20 ⇥ 103 i + 8 ⇥ 103 iT =
RTh =
vT
= 4 k⌦.
iT
20 ⇥ 103
◆
75
+ 15 = 45 V.
50 ⇥ 103
20 ⇥ 103 (0.2)iT + 8 ⇥ 103 iT = 4 ⇥ 103 iT ;
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Problems
7–57
t>0
45)e t/⌧ ;
vo = 45 + ( 45
✓
◆
1
⇥ 10 6 = 250 µs;
⌧ = RC = (4000)
16
90e 4000t V,
vo = 45
P 7.62
vo (0) = 45 V;
t
1
= 4000;
⌧
0.
vo (1) =
45 V;
RTh = 20 k⌦;
⌧ = (20 ⇥ 103 )
◆
1
⇥ 10 6 = 1.25 ⇥ 10 3 ;
16
45 + (45 + 45)e 800t =
v=
P 7.63
✓
45 + 90e 800t V,
1
= 800;
⌧
t
0.
For t > 0:
VTh = ( 25)(16,000)ib =
ib =
400 ⇥ 103 ib ;
33,000
(120 ⇥ 10 6 ) = 49.5 µA;
80,000
VTh =
400 ⇥ 103 (49.5 ⇥ 10 6 ) =
19.8 V;
RTh = 16 k⌦.
vo (1) =
19.8 V;
vo (0+ ) = 0;
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7–58
CHAPTER 7. Response of First-Order RL and RC Circuits
⌧ = (16, 000)(0.25 ⇥ 10 6 ) = 4 ms;
vo =
19.8 + 19.8e 250t V,
t
1/⌧ = 250;
0;
1
w(t) = (0.25 ⇥ 10 6 )vo2 = w(1)(1
2
P 7.64
0.36w(1)
= 0.36;
w(1)
(1
e 250t )2 =
1
e 250t = 0.6;
.·.
e 250t = 0.4
e 250t )2 J;
t = 3.67 ms.
[a] t < 0:
t > 0:
vo (0 ) = vo (0+ ) = 40 V;
vo (1) = 80 V;
⌧ = (0.16 ⇥ 10 6 )(6.25 ⇥ 103 ) = 1 ms;
vo = 80
[b] io =
=
C
40e 1000t V,
dvo
=
dt
t
1/⌧ = 1000;
0.
0.16 ⇥ 10 6 [40,000e 1000t ]
6.4e 1000t mA;
t
0+ .
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–59
Z t
1
[c] v1 =
6.4 ⇥ 10 3 e 1000x dx + 32
6
0.2 ⇥ 10
0
32e 1000t V,
= 64
t
0.
Z t
1
6.4 ⇥ 10 3 e 1000x dx + 8
[d] v2 =
0.8 ⇥ 10 6 0
8e 1000t V,
= 16
t
0.
1
1
[e] wtrapped = (0.2 ⇥ 10 6 )(64)2 + (0.8 ⇥ 10 6 )(16)2 = 512 µJ.
2
2
P 7.65
[a] For t > 0:
v(1) =
60
(90) = 30 V;
180
Req = 60 kk120 k = 40 k⌦;
⌧ = Req C = (40 ⇥ 103 )(20 ⇥ 10 9 ) = 0.8 ms;
vo = 30 + (120
[b] io =
1
= 1250;
⌧
30)e 1250t = 30 + 90e 1250t V,
0+ .
t
vo 90
30 + 90e 1250t 30 + 90e 1250t
vo
+
=
+
60,000 120,000
60,000
120,000
90
;
= 2.25e 1250t mA;
2.25 ⇥ 10 3 Z t 1250x
e
dx =
v1 =
60 ⇥ 10 9 0
P 7.66
30e 1250t + 30 V,
t
0.
[a] Let i be the current Zin the clockwise
direction around the circuit. Then
1 t
1 Zt
Vg = iRg +
i dx +
i dx
C1 0
C2 0
=
✓
◆
1 Zt
1Zt
1
iRg +
+
i dx = iRg +
i dx.
C1 C2
Ce 0
0
Now di↵erentiate the equation
0 = Rg
di
i
+
dt Ce
or
1
di
+
i = 0.
dt Rg Ce
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7–60
CHAPTER 7. Response of First-Order RL and RC Circuits
Therefore i =
P 7.67
Vg t/⌧
Vg t/Rg Ce
e
=
e
;
Rg
Rg
⌧ = Rg Ce .
v1 (t) =
1 Z t Vg x/⌧
Vg e x/⌧
e
dx =
C1 0 Rg
Rg C1 1/⌧
v1 (t) =
Vg C2
(1
C1 + C2
e t/⌧ );
⌧ = Rg Ce ;
v2 (t) =
Vg C1
(1
C1 + C2
e t/⌧ );
⌧ = Rg Ce .
[b] v1 (1) =
C2
Vg ;
C1 + C2
[a] Leq =
(3)(15)
= 2.5 H;
3 + 15
⌧=
Leq
2.5
1
=
= s;
R
7.5
3
v2 (1) =
io (1) =
120
= 16 A;
7.5
.·. io = 16
16e 3t A,
t
i1 =
i2 = io
i1 =
1);
0+ ;
t
40 3t
e A,
3
8 3t
e A,
3
8
3
0
Vg Ce
(e t/⌧
C1
0.
7.5io = 120e 3t V,
40
1Z t
120e 3x dx =
3 0
3
=
C1
Vg .
C1 + C2
io (0) = 0;
vo = 120
t
t
t
0;
0.
[b] io (0) = i1 (0) = i2 (0) = 0, consistent with initial conditions.
vo (0+ ) = 120 V, consistent with io (0) = 0.
vo = 3
di1
= 120e 3t V,
dt
t
0+
or
di2
= 120e 3t V,
t 0+ .
dt
The voltage solution is consistent with the current solutions.
vo = 15
1 = 3i1 = 40
40e 3t Wb-turns;
2 = 15i2 = 40
.·.
1 =
vo =
d 1
d 2
=
.
dt
dt
40e 3t Wb-turns;
2 as it must, since
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Problems
1 (1) =
7–61
2 (1) = 40 Wb-turns;
1 (1) = 3i1 (1) = 3(40/3) = 40 Wb-turns;
2 (1) = 15i2 (1) = 15(8/3) = 40 Wb-turns;
.·. i1 (1) and i2 (1) are consistent with
P 7.68
1 (1) and
2 (1).
[a] From Example 7.10,
50 25
L 1 L2 M 2
=
= 1 H;
L1 + L2 + 2M
15 + 10
Leq =
⌧=
L
1
= ;
R
20
1
= 20;
⌧
.·. io (t) = 4
4e 20t A,
t
0.
[b] vo = 80 20io = 80 80 + 80e 20t = 80e 20t V,
di2
di1
[c] vo = 5
5
= 80e 20t V;
dt
dt
t
0+ .
io = i1 + i2 ;
di1 di2
dio
=
+
= 80e 20t A/s;
dt
dt
dt
di2
= 80e 20t
.·.
dt
di1
;
dt
di1
.·. 80e 20t = 5
dt
400e 20t + 5
di1
= 480e 20t ;
.·. 10
dt
Z t1
0
dx =
i1 =
Z t
0
i1 = 4
4e 20t
1.6e 20t A,
= 1.6
di1 = 48e 20t dt.
48e 20y dy
48 20y t
e
= 2.4
20
0
[d] i2 = io
di1
;
dt
2.4e 20t A,
t
0.
2.4 + 2.4e 20t
t
0.
[e] io (0) = i1 (0) = i2 (0) = 0, consistent with zero initial stored energy.
vo = Leq
dio
= 1(80)e 20t = 80e 20t V,
dt
t
0+ . (checks)
Also,
vo = 5
di1
dt
5
di2
= 80e 20t V,
dt
t
0+ . (checks)
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7–62
CHAPTER 7. Response of First-Order RL and RC Circuits
vo = 10
di2
dt
5
di1
= 80e 20t V,
dt
0+ . (checks)
t
vo (0+ ) = 80 V, which agrees with io (0+ ) = 0 A.
io (1) = 4 A;
io (1)Leq = (4)(1) = 4 Wb-turns;
i1 (1)L1 + i2 (1)M = (2.4)(5) + (1.6)( 5) = 4 Wb-turns. (ok)
i2 (1)L2 + i1 (1)M = (1.6)(10) + (2.4)( 5) = 4 Wb-turns. (ok)
Therefore, the final values of io , i1 , and i2 are consistent with
conservation of flux linkage. Hence, the answers make sense in terms of
known circuit behavior.
P 7.69
[a] From Example 7.10,
Leq =
36 16
5
L 1 L2 M 2
=
= H;
L1 + L2 2M
20 8
3
⌧=
Leq
(5/3)
1
=
= ;
R
(50/3)
10
io =
100
(50/3)
100
e 10t = 6
(50/3)
50
io = 100
3
di2
di1
+4 ;
[c] vo = 2
dt
dt
[b] vo = 100
50
(6
3
6e 10t A t
0.
6e 10t ) = 100e 10t V t
0+ .
io = i1 + i2 ;
di1 di2
dio
=
+
;
dt
dt
dt
di2
dio
=
dt
dt
.·. 100e
di1
= 60e 10t
dt
10t
di1
;
dt
di1
+ 4 60e 10t
=2
dt
!
di1
;
dt
di1
= 70e 10t
.·.
dt
di1 = 70e 10t dt.
Z i1
0
dx = 70
e
.·. i1 = 70
Z t
0
e 10y dy;
10y
10
t
=7
7e 10t A,
t
0.
0
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Problems
[d] i2
=
io
i1
=
6
6e 10t
[e] vo
7 + 7e 10t
1 + e 10t A,
=
t
0.
di1
di2
+M
dt
dt
=
L2
=
18( 10e 10t ) + 4(70e 10t )
100e 10t V,
=
Also,
vo
7–63
0+ .
t
(checks)
di2
di1
+M
dt
dt
=
L1
=
2(70e 10t ) + 4( 10e 10t )
= 100e 10t V, t 0+ . CHECKS
i1 (0) = 7 7 = 0; agrees with initial conditions;
i2 (0) = 1 + 1 = 0; agrees with initial conditions;
The final values of io , i1 , and i2 can be checked via the conservation of
Wb-turns:
io (1)Leq = 6 ⇥ (5/3) = 10 Wb-turns;
i1 (1)L1 + i2 (1)M = 7(2)
i2 (1)L2 + i1 (1)M =
1(4) = 10 Wb-turns
1(18) + 7(4) = 10 Wb-turns.
Thus our solutions make sense in terms of known circuit behavior.
P 7.70
[a] Leq = 5 + 10
⌧=
2.5(2) = 10 H;
10
1
L
=
= ;
R
40
4
i=2
2e 4t A,
1
= 4;
⌧
t
0.
di
di
di1
2.5 = 2.5 = 2.5(8e 4t ) = 20e 4t V, t 0+ .
dt
dt
dt
di1
di
di
2.5 = 7.5 = 7.5(8e 4t ) = 60e 4t V, t 0+ .
[c] v2 (t) = 10
dt
dt
dt
[d] i(0) = 2 2 = 0, which agrees with initial conditions.
[b] v1 (t) = 5
80 = 40i1 + v1 + v2 = 40(2
2e 4t ) + 20e 4t + 60e 4t = 80 V.
Therefore, Kirchho↵’s voltage law is satisfied for all values of t 0.
Thus, the answers make sense in terms of known circuit behavior.
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7–64
P 7.71
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] Leq = 5 + 10 + 2.5(2) = 20 H;
⌧=
20
1
L
=
= ;
R
40
2
i=2
2e 2t A,
1
= 2;
⌧
t
0.
di
di
di1
+ 2.5 = 7.5 = 7.5(4e 2t ) = 30e 2t V, t 0+ .
dt
dt
dt
di
di
di1
+ 2.5 = 12.5 = 12.5(4e 2t ) = 50e 2t V, t 0+ .
[c] v2 (t) = 10
dt
dt
dt
[d] i(0) = 0, which agrees with initial conditions.
[b] v1 (t) = 5
80 = 40i1 + v1 + v2 = 40(2
2e 2t ) + 30e 2t + 50e 2t = 80 V.
Therefore, Kirchho↵’s voltage law is satisfied for all values of t 0.
Thus, the answers make sense in terms of known circuit behavior.
P 7.72
t < 0:
iL (0 ) = 10 V/5 ⌦ = 2 A = iL (0+ ).
0  t  5:
⌧ = 5/0 = 1;
iL (t) = 2e t/1 = 2e 0 = 2;
iL (t) = 2 A 0  t  5 s.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–65
5  t < 1:
⌧=
5
= 5 s;
1
1/⌧ = 0.2;
iL (t) = 2e 0.2(t 5) A,
P 7.73
t
5 s.
For t < 0:
i(0) =
10
(15) = 10 A.
15
0  t  10 ms:
i = 10e 100t A;
i(10 ms) = 10e 1 = 3.68 A.
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7–66
CHAPTER 7. Response of First-Order RL and RC Circuits
10 ms  t  20 ms:
Req =
(5)(20)
= 4 ⌦;
25
1
R
4
=
=
= 80;
⌧
L
50 ⇥ 10 3
i = 3.68e 80(t 0.01) A.
20 ms  t < 1:
i(20 ms) = 3.68e 80(0.02 0.01) = 1.65 A;
i = 1.65e 100(t 0.02) A;
vo = L
di
;
dt
L = 50 mH;
di
= 1.65( 100)e 100(t 0.02) =
dt
165e 100(t 0.02) ;
vo = (50 ⇥ 10 3 )( 165)e 100(t 0.02)
=
8.26e 100(t 0.02) V,
vo (25 ms) =
P 7.74
t > 20+ ms.
8.26e 100(0.025 0.02) =
5.013 V.
From the solution to Problem 7.73, the initial energy is
1
w(0) = (50 mH)(10 A)2 = 2.5 J;
2
0.04w(0) = 0.1 J;
1
.·.
(50 ⇥ 10 3 )i2L = 0.1;
2
so iL = 2 A.
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Problems
7–67
Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms
since
i(10 ms) = 3.68 A and i(20 ms) = 1.65 A.
For 10 ms  t  20 ms:
i = 3.68e 80(t 0.01) = 2;
e80(t 0.01) =
P 7.75
3.68
2
so t
0.01 = 0.0076
.·.
t = 17.6 ms.
[a] t < 0:
Using Ohm’s law,
800
ig =
= 12.5 A.
40 + 60k40
Using current division,
60
i(0 ) =
(12.5) = 7.5 A = i(0+ ).
60 + 40
[b] 0  t  1 ms:
i = i(0+ )e t/⌧ = 7.5e t/⌧ ;
R
40 + 120k60
1
=
=
= 1000;
⌧
L
80 ⇥ 10 3
i = 7.5e 1000t ;
3
i(200µs) = 7.5e 10 (200⇥10
6)
= 7.5e 0.2 = 6.14 A.
[c] i(1 ms) = 7.5e 1 = 2.7591 A.
1 ms  t:
R
40
1
=
=
= 500;
⌧
L
80 ⇥ 10 3
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7–68
CHAPTER 7. Response of First-Order RL and RC Circuits
i = i(1 ms)e (t 1 ms)/⌧ = 2.7591e 500(t 0.001) A;
i(6ms) = 2.7591e 500(0.005) = 2.7591e 2.5 = 226.48 mA.
[d] 0  t  1 ms:
i = 7.5e 1000t ;
v=L
di
= (80 ⇥ 10 3 )( 1000)(7.5e 1000t ) =
dt
600e 1 =
v(1 ms) =
600e 1000t V;
220.73 V.
[e] 1 ms  t  1:
i = 2.759e 500(t 0.001) ;
v=L
=
di
= (80 ⇥ 10 3 )( 500)(2.759e 500(t 0.001) )
dt
110.4e 500(t 0.001) V;
v(1+ ms) =
P 7.76
110.4 V.
0  t  10 µs:
⌧ = RC = (4 ⇥ 103 )(20 ⇥ 10 9 ) = 80 µs;
vo (0) = 0 V;
vo =
vo (1) =
20 + 20e 12,500t V
1/⌧ = 12,500;
20 V;
0  t  10 µs.
10 µs  t < 1:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–69
t ! 1:
i=
50 V
=
20 k⌦
2.5 mA;
vo (1) = ( 2.5 ⇥ 10 3 )(16,000) + 30 =
vo (10 µs) =
vo =
20 + 20 0.125 =
10 + ( 2.35 + 10)e (t
10 V;
2.35 V;
10⇥10
6 )/⌧
;
RTh = 4 k⌦k16 k⌦ = 3.2 k⌦;
⌧ = (3200)(20 ⇥ 10 9 ) = 64 µs;
vo =
P 7.77
10 + 7.65e 15,625(t
10⇥10
1/⌧ = 15,625;
6)
10 µs  t < 1.
0  t  200 µs:
Re = 150k100 = 60 k⌦;
✓
◆
10
⇥ 10 9 (60,000) = 200 µs;
⌧=
3
vc = 300e 5000t V.
vc (200 µs) = 300e 1 = 110.36 V.
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7–70
CHAPTER 7. Response of First-Order RL and RC Circuits
200 µs  t < 1:
Re = 30k60 + 120k40 = 20 + 30 = 50 k⌦;
✓
◆
10
⌧=
⇥ 10 9 (50,000) = 166.67 µs;
3
vc = 110.36e 6000(t
200 µs)
1
= 6000;
⌧
V;
vc (300 µs) = 110.36e 6000(100 µs) = 60.57 V;
io (300 µs) =
i1 =
60
2
io = io ;
90
3
isw = i1
P 7.78
60.57
= 1.21 mA;
50,000
2
i2 = io
3
i2 =
40
1
io = io ;
160
4
1
5
5
io = io = (1.21 ⇥ 10 3 ) = 0.50 mA.
4
12
12
t < 0:
vc (0 ) = (20 ⇥ 10 3 )(500) = 10 V = vc (0+ ).
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Problems
7–71
0  t  50 ms:
⌧ = 1;
t
vo = 10e 0 = 10 V.
1/⌧ = 0;
50 ms:
⌧ = (6.25 k)(0.16 µ) = 1 ms;
1/⌧ = 1000;
vo = 10e 1000(t
0.05)
V.
Summary:
P 7.79
vo = 10 V,
0  t  50 ms;
vo = 10e 1000(t
0.05)
V,
t
50 ms.
Note that for t > 0, vo = (4/6)vc , where vc is the voltage across the 0.5 µF
capacitor. Thus we will find vc first.
t < 0:
vc (0) =
3
( 75) =
15
15 V.
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7–72
CHAPTER 7. Response of First-Order RL and RC Circuits
0  t  800 µs:
⌧ = Re C,
Re =
(6000)(3000)
= 2 k⌦;
9000
⌧ = (2 ⇥ 103 )(0.5 ⇥ 10 6 ) = 1 ms,
vc =
15e 1000t V,
vc (800 µs) =
t
15e 0.8 =
1
= 1000;
⌧
0;
6.74 V.
800 µs  t  1.1 ms:
⌧ = (6 ⇥ 103 )(0.5 ⇥ 10 6 ) = 3 ms,
vc =
6.74e 333.33(t 800⇥10
6)
1
= 333.33;
⌧
V.
1.1 ms  t < 1:
⌧ = 1 ms,
1
= 1000;
⌧
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Problems
vc (1.1ms) =
6.74e 333.33(1100 800)10
6.1e 1000(t 1.1⇥10
vc =
vc (1.5ms) =
=
6.74e 0.1 =
6.1 V;
V;
6.1e 1000(1.5 1.1)10
vo = (4/6)( 4.09) =
P 7.80
3)
6
7–73
3
6.1e 0.4 =
=
4.09 V;
2.73 V.
1
w(0) = (0.5 ⇥ 10 6 )( 15)2 = 56.25 µJ.
2
0  t  800 µs:
15e 1000t ;
vc =
vc2 = 225e 2000t ;
p3k = 75e 2000t mW;
w3k =
Z 800⇥10 6
0
75 ⇥ 10 3 e 2000t dt
= 75 ⇥ 10 3
=
e 2000t 800⇥10
2000 0
37.5 ⇥ 10 6 (e 1.6
6
1) = 29.93 µJ.
1.1 ms  t  1:
6.1e 1000(t 1.1⇥10
3)
p3k = 12.4e 2000(t 1.1⇥10
3)
vc =
w3k =
Z 1
1.1⇥10
3
vc2 = 37.19e 2000(t 1.1⇥10
3e
2000(t 1.1⇥10
6.2 ⇥ 10 6 (0
3)
;
mW;
12.4 ⇥ 10 3 e 2000(t 1.1⇥10
= 12.4 ⇥ 10
=
V;
3)
2000
3)
dt
1
1.1⇥10
3
1) = 6.2 µJ;
The total energy dissipated by the 3 k⌦ resistor is
w3k = 29.93 + 6.2 = 36.13 µJ;
%=
36.13
(100) = 64.23%.
56.25
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7–74
P 7.81
CHAPTER 7. Response of First-Order RL and RC Circuits
[a] 0  t  75 µs:
io (0) = 0;
io (1) = 25 mA;
1
R
2000
=
=
⇥ 103 = 8000;
⌧
L
250
io = (25
25e 8000t ) mA;
vo = 0.25
dio
= 50e 8000t V.
dt
75 µs  t < 1:
io (75µs) = 25
25e 0.6 = 11.28 mA;
io = 11.28e 8000(t 75⇥10
vo = (0.25)
dio
=
dt
6)
io (1) = 0;
mA;
22.56e 8000(t 75µs) V.
.·. t < 0 :
vo
=
0;
0  t  75 µs :
vo
=
50e 8000t V;
vo
=
t
75+ µs :
22.56e 8000(t 75µs) V.
[b] vo (75 µs) = 50e 0.6 = 27.44 V;
vo (75+ µs) =
22.56 V.
[c] io (75 µs) = io (75+ µs) = 11.28 mA.
P 7.82
[a] 0  t  2.5 ms :
vo (0+ ) = 80 V;
vo (1) = 0;
L
= 2 ms;
R
1/⌧ = 500;
⌧=
vo (t) = 80e 500t V,
0+  t  2.5 ms;
vo (2.5 ms) = 80e 1.25 = 22.92 V;
io (2.5 ms) =
(80
22.92)
= 2.85 A;
20
vo (2.5+ ms) =
20(2.85) =
57.08 V;
vo (1) = 0;
⌧ = 2 ms;
1/⌧ = 500;
vo =
57.08e 500(t
0.0025)
V
t
2.5+ ms.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–75
[b]
[c] vo (5 ms) =
io =
P 7.83
16.35 V;
+16.35
= 817.68 mA.
20
[a] t < 0;
vo = 0.
0  t  4 ms:
⌧ = (200 ⇥ 103 )(0.025 ⇥ 10 6 ) = 5 ms;
vo = 100
100e 200t V,
vo (4 ms) = 100(1
1/⌧ = 200;
0  t  4 ms.
e 0.8 ) = 55.07 V.
4 ms  t  8 ms:
vo =
100 + 155.07e 200(t 0.004) V.
vo (8 ms) =
t
100 + 155.07e 0.8 =
30.32 V,
4 ms  t  8 ms.
8 ms:
vo =
30.32e 200(t 0.008) V,
t
8 ms.
[b]
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7–76
CHAPTER 7. Response of First-Order RL and RC Circuits
[c] t  0 :
vo = 0.
0  t  4 ms:
⌧ = (50 ⇥ 103 )(0.025 ⇥ 10 6 ) = 1.25 ms 1/⌧ = 800;
100e 800t V,
vo = 100
0  t  4 ms.
100e 3.2 = 95.92 V.
vo (4 ms) = 100
4 ms  t  8 ms:
vo =
100 + 195.92e 800(t 0.004) V,
vo (8 ms) =
t
92.01 V.
8 ms:
vo =
P 7.84
100 + 195.92e 3.2 =
4 ms  t  8 ms.
92.01e 800(t 0.008) V,
t
8 ms.
[a] 0  t  1 ms:
vc (0+ ) = 0;
vc (1) = 50 V;
RC = 400 ⇥ 103 (0.01 ⇥ 10 6 ) = 4 ms;
vc = 50
50e 250t ;
vo = 50
50 + 50e 250t = 50e 250t V,
1 ms  t < 1:
vc (1 ms) = 50
1/RC = 250;
0  t  1 ms.
50e 0.25 = 11.06 V;
vc (1) = 0 V;
⌧ = 4 ms;
1/⌧ = 250;
vc = 11.06e 250(t
vo =
vc =
0.001)
V;
11.06e 250(t
0.001)
V,
t
1 ms.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–77
[b]
P 7.85
vT = 2000iT + 4000(iT
= 6000iT
2 ⇥ 10 3 v ) = 6000iT
8v
8(2000iT );
vT
=
iT
10,000.
⌧=
10
=
10,000
1 ms;
1/⌧ =
1000;
i = 25e1000t mA;
.·. 25e1000t ⇥ 10 3 = 5;
t=
ln 200
= 5.3 ms.
1000
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7–78
P 7.86
CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0:
vT = 12 ⇥ 104 i + 16 ⇥ 103 iT ;
20
iT =
100
i =
.·. vT =
RTh =
0.2iT ;
24 ⇥ 103 iT + 16 ⇥ 103 iT .
vT
=
iT
8 k⌦;
⌧ = RC = ( 8 ⇥ 103 )(2.5 ⇥ 10 6 ) =
vc = 20e50t V;
50t = ln 1000
P 7.87
0.02 1/⌧ =
50;
20e50t = 20,000;
.·.
t = 138.16 ms.
Find the Thévenin equivalent with respect to the terminals of the capacitor.
RTh calculation:
iT =
.·.
vT
vT
+
2000 5000
4
iT
5+2 8
=
=
vT
10,000
vT
=
iT
10,000
=
1
vT
;
5000
1
.
10,000
10 k⌦.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7–79
Open circuit voltage calculation:
The node voltage equations:
voc v1
voc
+
2000
1000
4i = 0;
v1
v1 voc
+
1000
4000
5 ⇥ 10 3 = 0.
The constraint equation:
i =
v1
.
4000
Solving, voc =
vc (0) = 0;
80 V,
vc (1) =
v1 =
60 V.
80 V;
⌧ = RC = ( 10,000)(1.6 ⇥ 10 6 ) =
vc = vc (1) + [vc (0+ )
16 ms;
vc (1)]e t/⌧ =
1
=
⌧
62.5;
80 + 80e62.5t = 14,400.
Solve for the time of the maximum voltage rating:
e62.5t = 181;
62.5t = ln 181;
t = 83.09 ms.
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7–80
P 7.88
CHAPTER 7. Response of First-Order RL and RC Circuits
[a]
Using Ohm’s law,
vT = 5000i .
Using current division,
i =
20,000
(iT + i ) = 0.8iT + 0.8 i .
20,000 + 5000
Solve for i :
i (1
0.8 ) = 0.8iT ;
i =
0.8iT
;
1 0.8
vT = 5000i =
Find
such that RTh =
5 k⌦:
RTh =
4000
vT
=
=
iT
1 0.8
5000;
1
0.8 =
0.8
.·.
4000iT
.
(1 0.8 )
= 2.25.
[b] Find VTh :
Write a KCL equation at the top node:
VTh
VTh 40
+
5000
20,000
2.25i = 0.
The constraint equation is:
i =
(VTh 40)
= 0.
5000
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Problems
7–81
Solving,
VTh = 50 V.
Write a KVL equation around the loop:
50 =
5000i + 0.2
di
.
dt
Rearranging:
di
= 250 + 25,000i = 25,000(i + 0.01).
dt
Separate the variables and integrate to find i:
di
= 25,000 dt;
i + 0.01
Z i
Z t
dx
=
25,000 dx;
0 x + 0.01
0
.·. i =
10 + 10e25,000t mA.
di
= (10 ⇥ 10 3 )(25,000)e25,000t = 250e25,000t .
dt
Solve for the arc time:
di
v = 0.2 = 50e25,000t = 45,000;
dt
e25,000t = 900;
ln 900
= 272.1 µs.
.·. t =
25,000
P 7.89
[a]
⌧ = (25)(2) ⇥ 10 3 = 50 ms;
vc (0+ ) = 80 V;
1/⌧ = 20;
vc (1) = 0;
vc = 80e 20t V;
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7–82
CHAPTER 7. Response of First-Order RL and RC Circuits
.·. 80e 20t = 5;
e20t = 16;
t=
[b] 0+  t  138.63 ms:
i = (2 ⇥ 10 6 )( 1600e 20t ) =
3.2e 20t mA.
138.63+ ms:
t
⌧ = (2)(4) ⇥ 10 3 = 8 ms;
vc (138.63+ ms) = 5 V;
vc = 80
1/⌧ = 125;
vc (1) = 80 V;
75e 125(t 0.13863) V,
i = 2 ⇥ 10 6 (9375)e 125(t 0.13863)
= 18.75e 125(t 0.13863) mA,
t
80
138.63+ ms.
t=
ln 6.25 ⇠
= 14.66 ms.
125
Z t
4t
1
;
4 dx + 0 =
6
R(0.5 ⇥ 10 ) 0
R(0.5 ⇥ 10 6 )
4(15 ⇥ 10 3 )
=
R(0.5 ⇥ 10 6 )
P 7.91
138.63 ms.
68 = 75e 125 t = 12
e125 t = 6.25;
vo =
t
75e 125 t = 0.85(80) = 68;
[c] 80
P 7.90
ln 16
= 138.63 ms.
20
10;
4(15 ⇥ 10 3 )
= 12 k⌦.
10(0.5 ⇥ 10 6 )
.·.
R=
vo =
4(40 ⇥ 10 3 )
4t
+
6
=
+6=
R(0.5 ⇥ 10 6 )
R(0.5 ⇥ 10 6 )
.·.
R=
10;
4(40 ⇥ 10 3 )
= 20 k⌦.
16(0.5 ⇥ 10 6 )
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Problems
P 7.92
1
= 100;
RC
[a] RC = (25 ⇥ 103 )(0.4 ⇥ 10 6 ) = 10 ms;
vo = 0,
7–83
t < 0.
[b] 0  t  250 ms :
vo =
100
Z t
0.20 dx = 20t V.
0
[c] 250 ms  t  500 ms;
vo (0.25) = 20(0.25) = 5 V;
vo (t) =
[d] t
100
Z t
0.25
0.20 dx + 5 =
20(t
0.25) + 5 =
20t + 10 V.
500 ms :
vo (0.5) =
10 + 10 = 0 V;
vo (t) = 0 V.
P 7.93
[a] vo = 0,
t < 0;
RC = (25 ⇥ 103 )(0.4 ⇥ 10 6 ) = 10 ms
[b] Rf Cf = (5 ⇥ 106 )(0.4 ⇥ 10 6 ) = 2;
vo =
5 ⇥ 106
( 0.2)[1
25 ⇥ 103
[c] vo (0.25) = 40(1
vo =
=
=
1
= 100.
RC
1
= 0.5;
Rf Cf
e 0.5t ] = 40(1
e 0.5t ) V,
0  t  250 ms.
e 0.125 ) ⇠
= 4.70 V;
Vm Rf
Vm Rf
+
(2 e 0.125 )e 0.5(t 0.25)
Rs
Rs
40 + 40(2 e 0.125 )e 0.5(t 0.25)
40 + 44.70e 0.5(t 0.25) V,
250 ms  t  500 ms.
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7–84
CHAPTER 7. Response of First-Order RL and RC Circuits
[d] vo (0.5) =
0.55e 0.5(t 0.5) V,
vo =
P 7.94
[a]
40 + 44.70e 0.125 ⇠
=
Cdvp vp vb
+
= 0;
dt
R
vn
va
R
0.55 V;
t
500 ms.
therefore
dvp
1
vb
+
vp =
;
dt
RC
RC
d(vn vo )
= 0;
dt
+C
therefore
dvo
dvn
vn
=
+
dt
dt
RC
va
.
RC
But vn = vp .
Therefore
vn
dvp
vp
vb
dvn
+
=
+
=
.
dt
RC
dt
RC
RC
Therefore
dvo
1
=
(vb
dt
RC
va );
vo =
1 Zt
(vb
RC 0
va ) dy.
[b] The output is the integral of the di↵erence between vb and va and then
scaled by a factor of 1/RC.
1 Zt
[c] vo =
(vb va ) dx;
RC 0
RC = (50 ⇥ 103 )(10 ⇥ 10 9 ) = 0.5 ms
vb
va =
vo =
1 Zt
25 ⇥ 10 3 dx =
0.0005 0
50tsat =
P 7.95
25 mV;
6;
50t;
tsat = 120 ms.
Use voltage division to find the voltage at the non-inverting terminal:
vp =
80
( 45) =
100
36 V = vn .
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Problems
7–85
Write a KCL equation at the inverting terminal:
36 14
d
+ 2.5 ⇥ 10 6 ( 36
80,000
dt
.·.
2.5 ⇥ 10 6
vo ) = 0;
50
dvo
=
.
dt
80,000
Separate the variables and integrate:
dvo
=
dt
Z vo (t)
vo (0)
250
dx =
.·.
250
dvo =
Z t
0
250dt;
.·.
dy
vo (0) =
36 + 56 = 20 V;
vo (t) =
250t + 20.
vo (t)
vo (0) =
250t;
Find the time when the voltage reaches 0:
0=
P 7.96
.·.
250t + 20
20
= 80 ms.
250
t=
[a] RC = (1000)(800 ⇥ 10 12 ) = 800 ⇥ 10 9 ;
0  t  1 µs:
vg = 2 ⇥ 106 t;
6
vo =
1.25 ⇥ 10
=
2.5 ⇥ 1012
vo (1 µs) =
Z t
2 ⇥ 106 x dx + 0
0
x2 t
2
1
= 1,250,000.
RC
125 ⇥ 1010 t2 V,
=
0
125 ⇥ 1010 (1 ⇥ 10 6 )2 =
0  t  1 µs;
1.25 V.
1 µs  t  3 µs:
vg = 4
vo =
2 ⇥ 106 t;
125 ⇥ 104
Z t
6
1⇥10
"
(4
2 ⇥ 106 x) dx
1.25
#
x2 t
= 125 ⇥ 104 4x
2 ⇥ 106
1.25
2 1⇥10 6
1⇥10 6
= 5 ⇥ 106 t + 5 + 125 ⇥ 1010 t2 1.25 1.25
= 125 ⇥ 1010 t2 5 ⇥ 106 t + 2.5 V, 1 µs  t  3 µs.
t
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7–86
CHAPTER 7. Response of First-Order RL and RC Circuits
vo (3 µs) = 125 ⇥ 1010 (3 ⇥ 10 6 )2
=
5 ⇥ 106 (3 ⇥ 10 6 ) + 2.5
1.25 V.
3 µs  t  4 µs:
vg =
8 + 2 ⇥ 106 t;
vo =
125 ⇥ 104
=
4
125 ⇥ 10
Z t
3⇥10
"
6
( 8 + 2 ⇥ 106 x) dx
t
8x
6x
+2 ⇥ 10
2
1.25
t
#
2 3⇥10 6
7
= 10 t 30 125 ⇥ 10 t + 11.25 1.25
= 125 ⇥ 1010 t2 + 107 t 20 V, 3 µs  t  4 µs.
vo (4 µs) =
3⇥10 6
10 2
125 ⇥ 1010 (4 ⇥ 10 6 )2 + 107 (4 ⇥ 10 6 )
1.25
20 = 0.
[b]
[c] The output voltage will also repeat. This follows from the observation that
at t = 4 µs the output voltage is zero, hence there is no energy stored in
the capacitor. This means the circuit is in the same state at t = 4 µs as it
was at t = 0, thus as vg repeats itself, so will vo .
P 7.97
[a] T2 is normally ON since its base current ib2 is greater than zero, i.e.,
ib2 = VCC /R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0.
When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C
is charged to VCC , positive at the left terminal. This is a stable state;
there is nothing to disturb this condition if the circuit is left to itself.
[b] When S is closed momentarily, vbe2 is changed to VCC and T2 snaps
OFF. The instant T2 turns OFF, vce2 jumps to VCC R1 /(R1 + RL ) and ib1
jumps to VCC /(R1 + RL ), which turns T1 ON.
[c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since
vbe2 is the same as the voltage across C, it starts to increase from VCC
toward +VCC . However, T2 turns ON as soon as vbe2 = 0. The equation
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Problems
7–87
for vbe2 is vbe2 = VCC 2VCC e t/RC . vbe2 = 0 when t = RC ln 2, therefore
T2 stays OFF for RC ln 2 seconds.
P 7.98
[a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to
✓
◆
VCC
6(5)
= 1.2 V.
R1 =
vce2 =
R 1 + RL
25
T2 remains open for (23,083)(250) ⇥ 10 12 ln 2 ⇠
= 4 µs.
[b] ib2 =
VCC
= 259.93 µA,
R
ib2 = 0,
ib2
P 7.99
5  t  0 µs;
0 < t < RC ln 2;
=
VCC VCC (t RC ln 2)/RL C
+
e
R
RL
=
259.93 + 300e 0.2⇥10 (t 4⇥10
6
6)
µA,
RC ln 2 < t;
[a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal.
At the instant T1 turns ON the capacitor C2 is connected across b2 e2 ,
thus vbe2 = VCC . This negative voltage snaps T2 OFF. Now the polarity
of the voltage on C2 starts to reverse, that is, the right-hand terminal of
C2 starts to charge toward +VCC . At the same time, C1 is charging
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7–88
CHAPTER 7. Response of First-Order RL and RC Circuits
toward VCC , positive on the right. At the instant the charge on C2
reaches zero, vbe2 is zero, T2 turns ON. This makes vbe1 = VCC and T1
snaps OFF. Now the capacitors C1 and C2 start to charge with the
polarities to turn T1 ON and T2 OFF. This switching action repeats itself
over and over as long as the circuit is energized. At the instant T1 turns
ON, the voltage controlling the state of T2 is governed by the following
circuit:
It follows that vbe2 = VCC
2VCC e t/R2 C2 .
[b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the
voltage across C1 , thus
It follows that vce2 = VCC
VCC e t/RL C1 .
[c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will
become positive and turn T2 ON. vbe2 = 0 when VCC 2VCC e t/R2 C2 = 0,
or when t = R2 C2 ln 2.
[d] When t = R2 C2 ln 2,
vce2 = VCC
we have
VCC e [(R2 C2 ln 2)/(RL C1 )] = VCC
VCC e 10 ln 2 ⇠
= VCC .
[e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have
ib1 =
VCC VCC t/RL C1
+
e
.
R1
RL
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Problems
7–89
[f ] At the instant T2 turns back ON, t = R2 C2 ln 2; therefore
ib1 =
VCC VCC 10 ln 2 ⇠ VCC
+
e
.
=
R1
RL
R1
[g]
[h]
P 7.100 [a] tOFF2 = R2 C2 ln 2 = 14.43 ⇥ 103 (1 ⇥ 10 9 ) ln 2 ⇠
= 10 µs
[b] tON2 = R1 C1 ln 2 ⇠
= 10 µs
[c] tOFF1 = R1 C1 ln 2 ⇠
= 10 µs
[d] tON1 = R2 C2 ln 2 ⇠
= 10 µs
10 ⇠
10
+
= 10.69 mA
1000 14,430
10
10
[f ] ib1 =
+
e 10 ⇠
= 0.693 mA
14,430 1000
[g] vce2 = 10 10e 10 ⇠
= 10 V
[e] ib1 =
P 7.101 [a] tOFF2 = R2 C2 ln 2 = (14.43 ⇥ 103 )(0.8 ⇥ 10 9 ) ln 2 ⇠
= 8 µs
[b] tON2 = R1 C1 ln 2 ⇠
= 10 µs
[c] tOFF1 = R1 C1 ln 2 ⇠
= 10 µs
[d] tON1 = R2 C2 ln 2 = 8 µs
[e] ib1 = 10.69 mA
10
10
+
e 8⇠
[f ] ib1 =
= 0.693 mA
14,430 1000
[g] vce2 = 10 10e 8 ⇠
= 10 V
Note in this circuit T2 is OFF 8 µs and ON 10 µs of every cycle, whereas
T1 is ON 8 µs and OFF 10 µs every cycle.
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7–90
CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.102 If R1 = R2 = 50RL = 100 k⌦,
C1 =
48 ⇥ 10 6
= 692.49 pF;
100 ⇥ 103 ln 2
If R1 = R2 = 6RL = 12 k⌦,
C1 =
then
C2 =
36 ⇥ 10 6
= 519.37 pF.
100 ⇥ 103 ln 2
then
48 ⇥ 10 6
= 5.77 nF;
12 ⇥ 103 ln 2
C2 =
36 ⇥ 10 6
= 4.33 nF.
12 ⇥ 103 ln 2
Therefore 692.49 pF  C1  5.77 nF and 519.37 pF  C2  4.33 nF.
P 7.103 [a] 0  t  0.5:
i=
✓
◆
21
e t/⌧
60
21
30
+
60
60
where ⌧ = L/R.
i = 0.35 + 0.15e 60t/L ;
i(0.5) = 0.35 + 0.15e 30/L = 0.40;
.·. e30/L = 3;
L=
30
= 27.31 H.
ln 3
[b] 0  t  tr , where tr is the time the relay releases:
✓
◆
30
i=0+
60
0 e 60t/L = 0.5e 60t/L ;
.·. 0.4 = 0.5e 60tr /L ;
tr =
e60tr /L = 1.25;
27.31 ln 1.25 ⇠
= 0.10 s.
60
P 7.104 From the Practical Perspective,
vC (t) = 0.75VS = VS (1
e t/RC ).
0.25 = e t/RC
t=
so
RC ln 0.25.
In the above equation, t is the number of seconds it takes to charge the
capacitor to 0.75VS , so it is a period. We want to calculate the heart rate,
which is a frequency in beats per minute, so H = 60/t. Thus,
H=
60
.
RC ln 0.25
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Problems
7–91
P 7.105 In this problem, Vmax = 0.6VS , so the equation for heart rate in beats per
minute is
60
.
RC ln 0.4
H=
Given R = 150 k⌦ and C = 6 µF,
60
= 72.76.
(150,000)(6 ⇥ 10 6 ) ln 0.4
H=
Therefore, the heart rate is about 73 beats per minute.
P 7.106 From the Practical Perspective,
e t/RC ).
vC (t) = Vmax = VS (1
Solve this equation for the resistance R:
Vmax
=1
VS
e t/RC
Then,
t
= ln 1
RC
.·.
✓
R=
e t/RC = 1
so
✓
C ln 1
Vmax
.
VS
◆
Vmax
;
VS
t
Vmax
VS
◆.
In the above equation, t is the time it takes to charge the capacitor to a
voltage of Vmax . But t and the heart rate H are related as follows:
H=
60
.
t
Therefore,
R=
✓
60
HC ln 1
Vmax
VS
◆.
P 7.107 From Problem 7.106,
R=
✓
60
HC ln 1
Vmax
VS
◆.
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7–92
CHAPTER 7. Response of First-Order RL and RC Circuits
Note that from the problem statement,
Vmax
= 0.68.
VS
Therefore,
R=
60
(70)(2.5 ⇥ 10 6 ) ln (1
0.68)
= 301 k⌦.
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Natural and Step Responses of
RLC Circuits
8
Assessment Problems
1
1
,
therefore C = 500 nF.
=
(2RC)2
LC
1
,
therefore C = 1 µF;
[b] ↵ = 5000 =
2RC
AP 8.1 [a]
5000 ±
s1,2 =
[c] p
AP 8.2 iL
s
25 ⇥ 106
1
= 20,000,
LC
therefore C = 125 nF;
s1,2 =

s1 =
5.36 krad/s,
=
40 ±
(103 )(106 )
= ( 5000 ± j5000) rad/s.
20
q
(40)2
202 103 ,
s2 =
74.64 krad/s.
Z t
1
[ 14e 5000x + 26e 20,000x ] dx + 30 ⇥ 10 3
50 ⇥ 10 3 0
(
)
14e 5000x t 26e 20,000t t
+
+ 30 ⇥ 10 3
5000 0
20,000 0
=
20
=
56 ⇥ 10 3 (e 5000t
1)
26 ⇥ 10 3 (e 20,000t
=
[56e 5000t
56
26e 20,000t + 26 + 30] mA
=
56e 5000t
26e 20,000t mA,
t
AP 8.3 From the given values of R, L, and C, s1 =
[a] v(0 ) = v(0+ ) = 0,
1) + 30 ⇥ 10 3
0.
10 krad/s and s2 =
40 krad/s.
therefore iR (0+ ) = 0.
8–1
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8–2
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] iC (0+ ) =
(iL (0+ ) + iR (0+ )) =
( 4 + 0) = 4 A.
+
dvc (0 )
4
dvc (0+ )
= ic (0+ ) = 4,
=
= 4 ⇥ 108 V/s.
therefore
dt
dt
C
t 0+ ;
[d] v = [A1 e 10,000t + A2 e 40,000t ] V,
[c] C
dv(0+ )
=
dt
v(0+ ) = A1 + A2 ,
Therefore A1 + A2 = 0,
[e] A2 =
10,000A1
A1
40,000A2 .
4A2 = 40,000;
A1 = 40,000/3 V.
40,000/3 V.
[f ] v = [40,000/3][e 10,000t
e 40,000t ] V,
t
0.
1
= 8000,
therefore R = 62.5 ⌦.
2RC
10 V
= 160 mA;
[b] iR (0+ ) =
62.5 ⌦
AP 8.4 [a]
iC (0+ ) =
(iL (0+ ) + iR (0+ )) =
80
Therefore
dv(0+ )
=
dt
240 kV/s.
240 m
=
C
dvc (0+ )
= !d B2
dt
[c] B1 = v(0+ ) = 10 V,
Therefore 6000B2
[d] iL =
(iR + iC );
240,000,
iR = v/R;
iC = C
iC = e 8000t [ 240 cos 6000t +
B2 = ( 80/3) V.
dv
;
dt
1280
sin 6000t] mA.
3
460
sin 6000t] mA;
3
iL = 10e 8000t [8 cos 6000t +
82
sin 6000t] mA,
3
✓
therefore
◆
[c] 0.5LI02 = 12.5 ⇥ 10 3 ,
dv(0+ )
.
dt
80
sin 6000t] V.
3
Therefore iR = e 8000t [160 cos 6000t
106
1
1 2
=
,
=
2RC
LC
4
[b] 0.5CV02 = 12.5 ⇥ 10 3 ,
240 mA = C
↵B1 .
8000B1 =
v = e 8000t [10 cos 6000t
AP 8.5 [a]
160 =
t
1
= 500,
2RC
therefore V0 = 50 V.
0.
R = 100 ⌦.
I0 = 250 mA.
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Problems
dv(0+ )
= D1
dt
[d] D2 = v(0+ ) = 50,
iR (0+ ) =
↵D2 ;
50
= 500 mA.
100
Therefore iC (0+ ) =
Therefore
(500 + 250) =
dv(0+ )
=
dt
Therefore D1
[e] v = [50e 500t
iR =
8–3
750 ⇥
↵D2 =
750 mA;
10 3
=
C
75,000;
75,000 V/s;
↵=
1
= 500,
2RC
D1 =
50,000 V/s.
50,000te 500t ] V;
v
= [0.5e 500t
R
500te 500t ] A,
t
40
V0
=
= 0.08 A.
R
500
[b] iC (0+ ) = I iR (0+ ) iL (0+ ) = 1 0.08
diL (0+ )
Vo
40
[c]
=
=
= 62.5 A/s.
dt
L
0.64
1
1
= 1000;
= 1,562,500;
[d] ↵ =
2RC
LC
[e] iL = If + B10 e ↵t cos !d t + B20 e ↵t sin !d t,
0+ .
AP 8.6 [a] iR (0+ ) =
iL (0+ ) = 0.5 = if + B10 ,
0.5 =
1.58 A.
s1,2 =
1000 ± j750 rad/s.
If = I =
1 A;
therefore B10 = 1.5 A;
diL (0+ )
= 62.5 =
dt
↵B10 + !d B20 ,
Therefore iL (t) =
1 + e 1000t [1.5 cos 750t + (25/12) sin 750t] A,
[f ] v(t) =
LdiL
= 40e 1000t [cos 750t
dt
therefore B20 = (25/12) A;
(154/3) sin 750t]V
t
t
0.
0.
AP 8.7 i(0+ ) = 0, since there is no source connected to L for t < 0.
+
vC (0 ) = vC (0 ) =
R
↵=
= 8000;
2L
!d =
q
10,0002
!
15,000
(80) = 50 V;
15,000 + 9000
!0 =
s
1
= 10,000;
LC
Vf = 100 V.
the response is underdamped.
800002 = 6000;
vC = Vf + B10 e 8000t cos 6000t + B20 e 8000t sin 6000t;
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8–4
CHAPTER 8. Natural and Step Responses of RLC Circuits
Vf + B10 = V0
↵B10 + !d B20 =
B10 = V0
so
Vf = 50
100 =
50;
I0
,
C
so
B20 =
1
!d
✓
◆
1
I0
+ ↵B10 =
[0 + 8000( 50)] =
C
6000
66.67;
Therefore,
e 8000t (50 cos 6000t + 66.67 sin 6000t) V,
vC = 100
AP 8.8 i = C
t
0.
dvC
dt
= 2 ⇥ 10 6 [8000e 8000t (50 cos 6000t + 66.67 sin 6000t)
e 8000t (6000)( 50 sin 6000t + 66.67 cos 6000t)]
= 1.67e 8000t sin 6000t A,
t > 0.
AP 8.9 [a] From AP 8.7, I0 = 0, V0 = 50 V, and Vf = 100 V. Then,
R
= 10,000;
↵=
2L
!0 =
s
1
= 10,000;
LC
the response is critically damped.
vC = Vf + D10 te 10,000t + D20 e 10,000t ;
Vf + D20 = V0
D10
↵D20 =
so
D20 = V0
Vf = 50
100 =
50;
I0
,
C
so
I0
+ ↵D20 = 0 + 10,000( 50)] =
C
Therefore,
D10 =
vC = 100
[b] i = C
500,000te 10,000t
5 ⇥ 105 ;
50e 10,000t V,
t
0.
dvC
dt
= 2 ⇥ 10 6 [ 500,000e 10,000t + 5 ⇥ 109 te 10,000t + 500,000e 10,000t ]
= 10,000te 10,000t A,
t > 0.
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Problems
8–5
Problems
P 8.1
[a] ↵ =
1
1
=
= 250;
2RC
2(1000)(2 ⇥ 10 6 )
1
1
=
= 40,000;
LC
(12.5)(2 ⇥ 10 6 )
!o2 =
q
2502
s1,2 =
250 ±
s1 =
100 rad/s;
40,000 =
s2 =
250 ± 150;
400 rad/s.
[b] Overdamped.
[c] Note — we want !d = 120 rad/s:
q
!o2
!d =
↵2 ;
.·. ↵2 = !o2
!d2 = 40,000
(120)2 = 25,600.
↵ = 160;
1
= 160;
2RC
P 8.2
q
40,000 =
[a] iR (0) =
q
40,000 =
1
;
2RC
.·. R =
160 ± j120 rad/s.
1
= 1250 ⌦.
2(200)(2 ⇥ 10 6 )
15
= 75mA;
200
iL (0) =
45mA;
iC (0) =
iL (0)
[b] ↵ =
1
= 1562.5 ⌦.
2(160)(2 ⇥ 10 6 )
1602
160 ±
[d] s1 , s2 =
[e] ↵ =
.·. R =
iR (0) = 45
75 =
30 mA.
1
1
=
= 12,500;
2RC
2(200)(0.2 ⇥ 10 6 )
1
1
=
= 108 ;
3
LC
(50 ⇥ 10 )(0.2 ⇥ 10 6 )
p
s1,2 = 12,500 ± 1.5625 ⇥ 108 108 = 12,500 ± 7500;
!o2 =
s1 =
5000 rad/s;
s2 =
20,000 rad/s;
v = A1 e 5000t + A2 e 20,000t .
v(0) = A1 + A2 = 15;
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8–6
CHAPTER 8. Natural and Step Responses of RLC Circuits
dv
(0) =
dt
5000A1
20,000A2 =
Solving,
A1 = 10;
A2 = 5;
v = 10e 5000t + 5e 20,000t V,
[c] iC
= C
30 ⇥ 10 3
=
0.2 ⇥ 10 6
t
0.
dv
dt
= 0.2 ⇥ 10 6 [ 50,000e 5000t
10e 5000t
=
15 ⇥ 104 V/s;
100,000e 20,000t ]
20e 20,000t mA;
iR = 50e 5000t + 25e 20,000t mA;
iL =
P 8.3
↵=
iC
40e 5000t
iR =
5e 20,000t mA,
t
0.
1
1
=
= 104 ;
2RC
2(250)(0.2 ⇥ 10 6 )
.·. ↵2 = !o2 .
↵2 = 108 ;
Critical damping:
v = D1 te ↵t + D2 e ↵t ;
iR (0+ ) =
15
= 60 mA;
250
iC (0+ ) =
[iL (0+ ) + iR (0+ )] =
[ 45 + 60] =
15 mA;
v(0) = D2 = 15;
dv
= D1 [t( ↵e ↵t ) + e ↵t ]
dt
↵D2 e ↵t ;
15 ⇥ 10 3
iC (0)
=
=
C
0.2 ⇥ 10 6
dv
(0) = D1
dt
↵D2 =
D1 = ↵D2
75,000 = (104 )(15)
v = (75,000t + 15)e 10,000t V,
75,000;
75,000 = 75,000;
t
0.
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Problems
P 8.4
8–7
1
1
=
= 8000;
2RC
2(312.5)(0.2 ⇥ 10 6 )
1
1
=
= 108 ;
3
LC
(50 ⇥ 10 )(0.2 ⇥ 10 6 )
s1,2 =
8000 ±
p
80002
108 =
8000 ± j6000 rad/s;
.·. response is underdamped.
v(t) = B1 e 8000t cos 6000t + B2 e 8000t sin 6000t;
v(0+ ) = 15 V = B1 ;
iR (0+ ) =
iC (0+ ) = [ iL (0+ ) + iR (0+ )] =
3 ⇥ 10 3
dv(0+ )
=
=
dt
0.2 ⇥ 10 6
dv(0)
=
dt
[ 45 + 48] =
3 mA;
15,000 V/s;
8000B1 + 6000B2 =
6000B2 = 8000(15)
15
= 48 mA;
312.5
15,000;
.·. B2 = 17.5 V;
15,000;
v(t) = 15e 8000t cos 6000t + 17.5e 8000t sin 6000t V,
P 8.5
[a] ↵ =
1
= 1250,
2RC
s1 =
500,
s2 =
!o = 103 ,
t
0.
therefore overdamped.
2000,
therefore v = A1 e 500t + A2 e 2000t .
"
+
v(0 ) = 0 = A1 + A2 ;
Therefore
A1 =
+980
,
15
v(t) =

500A1
A2 =
980
[e 500t
15
#
iC (0+ )
dv(0+ )
= 98,000 V/s,
=
dt
C
2000A2 = 98,000.
980
15;
e 2000t ] V,
t
0.
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8–8
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b]
Example 8.4: vmax ⇠
= 74.1 V at 1.4 ms;
Example 8.5: vmax ⇠
= 36.1 V at 1.0 ms.
P 8.6
[a]
q
↵+
↵2
!o2 =
1000;
q
!o2 =
4000;
↵2
↵
Adding the above equations,
2↵ =
5000
so
q
↵ = 2500 rad/s;
2500 ±
.·.
[b] iR =
P 8.7
25002
!02 =
!02 = 15002
1000
25002
1
1
=
= 20002
LC
(0.01)C
25002
thus
!02 = 1500;
!0 = 2000.
C=
1
= 25 µF;
(0.01)20002
1
1
=
= 2500
2RC
2R(25 ⇥ 10 6 )
so
R=
v(t)
= 5e 1000t
R
t
0+ ;
!02 =
↵=
q
so
11.25e 4000t A,
iC = C
dv(t)
= 9e 4000t
dt
iL =
(iR + iC ) = 2.25e 4000t
e 1000t A,
t
4e 1000t , A,
1
= 8 ⌦.
2(25 ⇥ 10 6 )(2500)
0+ ;
t
0.
↵ = 1000/2 = 500;
R=
1
1
=
= 400 ⌦;
2↵C
2(500)(2.5 ⇥ 10 6 )
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Problems
8–9
v(0+ ) = 3(1 + 1) = 6 V;
iR (0+ ) =
dv
=
dt
6
= 15 mA;
400
300e 100t
dv(0+ )
=
dt
P 8.8
300
2700e 900t ;
2700 =
3000 V/s;
iC (0+ ) = 2.5 ⇥ 10 6 ( 3000) =
7.5 mA;
iL (0+ ) =
[15
[iR (0+ ) + iC (0+ )] =
[a] ↵ = 400;
!d =
7.5] =
7.5 mA.
!d = 300;
q
!o2
↵2 ;
.·. !o2 = !d2 + ↵2 = 9 ⇥ 104 + 16 ⇥ 104 = 25 ⇥ 104 .
1
= 25 ⇥ 104 ;
LC
L=
[b] ↵ =
1
(25 ⇥ 104 )(250 ⇥ 10 6 )
= 16 mH.
1
;
2RC
.·. R =
1
1
=
= 5 ⌦.
2↵C
(800)(250 ⇥ 10 6 )
[c] V0 = v(0) = 120 V.
[d] I0 = iL (0) =
iR (0) =
iR (0)
iC (0);
120
= 24 A;
5
iC (0) = C
dv
(0) = 250 ⇥ 10 6 [ 400(120) + 300(80)] =
dt
.·. I0 =
24 + 6 =
6 A;
18 A.
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8–10
CHAPTER 8. Natural and Step Responses of RLC Circuits
[e] iC (t) = 250 ⇥ 10 6
dv(t)
= e 400t ( 17 sin 300t
dt
6 cos 300t) A;
iR (t) =
v(t)
= e 400t (24 cos 300t + 16 sin 300t) A;
5
iL (t) =
iR (t)
iC (t)
= e 400t ( 18 cos 300t + sin 300t) A,
t
0.
Check:
diL
L
= 16 ⇥ 10 3 e 400t [7500 cos 300t + 5000 sin 300t];
dt
v(t) = e 400t [120 cos 300t + 80 sin 300t] V.
P 8.9
[a]
✓
1
2RC
◆2
.·. C =
=
1
= (500)2 ;
LC
1
= 1 µF.
(500)2 (4)
1
= 500;
2RC
.·. R =
1
= 1 k⌦.
2(500)(10 6 )
v(0) = D2 = 8 V;
iR (0) =
8
= 8mA;
1000
iC (0) =
8 + 10 = 2 mA;
2 ⇥ 10 3
dv
(0) = D1 500D2 =
= 2000 V/s;
dt
10 6
.·. D1 = 2000 + 500(8) = 6000 V/s.
[b] v = 6000te 500t + 8e 500t V,
t
0;
dv
= [ 3 ⇥ 106 t + 2000]e 500t ;
dt
iC = C
P 8.10
[a] !o =
↵=
s
dv
= ( 3000t + 2)e 500t mA,
dt
1
=
LC
s
t
0+ .
1
= 10,000;
(20 ⇥ 10 3 )(500 ⇥ 10 9 )
1
= 10,000;
2RC
R=
1
= 100 ⌦.
2(10,000)(500 ⇥ 10 9 )
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Problems
8–11
[b] v(t) = D1 te 10,000t + D2 e 10,000t
v(0) = 40 V = D2 ;
✓
◆
dv
(0) = D1
dt
1
↵D2 =
C
.·.
1
10,000D2 =
500 ⇥ 10 9
D1
.·. v(t) = (40
[c] iC (t) = 0 when
V0
;
R
I0
✓
640,000t)e 10,000t V,
40
100
0.12
t
◆
so
D1 =
640,000;
0.
dv
(t) = 0;
dt
dv
= (64 ⇥ 108 t
dt
1040 ⇥ 103 )e 10,000t ;
dv
= 0 when 640 ⇥ 108 t1 = 1040 ⇥ 103 ;
dt
v(162.5µs) = ( 64)e 1.625 =
.·. t1 = 162.5 µs.
12.6 V.
1
1
[d] w(0) = (500 ⇥ 10 9 )(40)2 + (0.02)(0.012)2 = 544 µJ;
2
2
✓
1
1
12.6
w(162.5 µs) = (500 ⇥ 10 9 )(12.6)2 + (0.02)
2
2
100
% remaining =
P 8.11
[a]
◆2
= 198.5 µJ;
198.5
(100) = 36.5%.
544
1
= 50002 .
LC
There are many possible solutions. This one begins by choosing
L = 10 mH. Then,
C=
1
(10 ⇥ 10 3 )(5000)2
= 4 µF.
We can achieve this capacitor value using components from Appendix H
by combining four 1 µF capacitors in parallel.
Critically damped:
.·. R =
↵ = !0 = 5000
so
1
= 5000;
2RC
1
= 25 ⌦.
2(4 ⇥ 10 6 )(5000)
We can create this resistor value using components from Appendix H by
combining a 10 ⌦ resistor and a 15 ⌦ resistor in series. The final circuit:
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8–12
CHAPTER 8. Natural and Step Responses of RLC Circuits
q
[b] s1,2 = ↵ ± ↵2 !02 = 5000 ± 0;
Therefore there are two repeated real roots at
P 8.12
5000 rad/s.
[a] Underdamped response:
so
↵ < !0
↵ < 5000.
Therefore we choose a larger resistor value than the one used in Problem
8.11. Choose R = 100 ⌦:
1
↵=
= 1250;
2(100)(4 ⇥ 10 6 )
p
s1,2 = 1250 ± 12502 50002 = 1250 ± j4841.23 rad/s.
[b] Overdamped response:
↵ > !0
so
↵ > 5000.
Therefore we choose a smaller resistor value than the one used in
Problem 8.11. Choose R = 20 ⌦:
1
= 6250;
↵=
2(20)(4 ⇥ 10 6 )
p
s1,2 = 6250 ± 62502 50002 = 1250 ± 3750;
=
P 8.13
2500 rad/s;
[a] 2↵ = 1000;
q
2 ↵2
and
10,000 rad/s.
↵ = 500 rad/s;
!o2 = 600;
!o = 400 rad/s;
C=
1
1
=
= 4 µF ;
2↵R
2(500)(250)
L=
1
1
=
= 1.5625 H;
!o2 C
(400)2 (4 ⇥ 10 6 )
iC (0+ ) = A1 + A2 = 45 mA;
diC diL diR
+
+
= 0;
dt
dt
dt
diC (0)
=
dt
diL (0)
dt
diR (0)
;
dt
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Problems
8–13
0
diL (0)
=
= 0 A/s;
dt
1.5625
1 dv(0)
1 iC (0)
45 ⇥ 10 3
diR (0)
=
=
=
= 45 A/s;
dt
R dt
R C
(250)(4 ⇥ 10 6 )
diC (0)
= 0 45 = 45 A/s;
.·.
dt
.·. 200A1 + 800A2 = 45;
A1 + A2 = 0.045.
Solving, A1 =
.·. iC =
15 mA;
A2 = 60 mA;
15e 200t + 60e 800t mA,
0+ .
t
[b] By hypothesis
v = A3 e 200t + A4 e 800t ,
t
0;
v(0) = A3 + A4 = 0;
45 ⇥ 10 3
dv(0)
=
= 11,250 V/s;
dt
4 ⇥ 10 6
200A3 800A4 = 11,250;
.·. A3 = 18.75 V;
v = 18.75e 200t 18.75e 800t V,
t
v
= 75e 200t 75e 800t mA,
[c] iR (t) =
250
[d] iL = iR iC ;
iL =
P 8.14
t<0:
60e 200t + 15e 800t mA,
Vo = 15 V,
Io =
t
A4 =
18.75 V;
0.
t
0+ .
0.
60 mA.
t > 0:
iR (0) =
15
= 150 mA;
100
iL (0) =
60 mA;
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8–14
CHAPTER 8. Natural and Step Responses of RLC Circuits
iC (0) =
↵=
150
( 60) =
90 mA;
1
1
=
= 5000 rad/s;
2RC
2(100)(10 6 )
1
1
=
= 16 ⇥ 106 ;
LC
(62.5 ⇥ 10 3 )(10 6 )
p
s1,2 = 5000 ± 25 ⇥ 106 16 ⇥ 106 = 5000 ± 3000;
!o2 =
s1 =
2000 rad/s;
s2 =
8000 rad/s;
.·. vo = A1 e 2000t + A2 e 8000t .
A1 + A2 = vo (0) = 15;
dvo
(0) =
dt
Solving,
2000A1
8000A2 =
A1 = 5 V,
90 ⇥ 10 3
=
10 6
A2 = 10 V;
.·. vo = 5e 2000t + 10e 8000t V,
P 8.15
!o2 =
↵=
90,000.
t
0.
1
1
=
= 16 ⇥ 106 ;
LC
(62.5 ⇥ 10 3 )(10 6 )
1
1
=
= 4000;
2RC
2(125)(10 6 )
.·. ↵2 = !o2 (critical damping).
vo (t) = D1 te 4000t + D2 e 4000t ;
vo (0) = D2 = 15 V;
iR (0) =
15
= 120 mA;
125
iL (0) =
60 mA;
iC (0) =
60 mA;
dvo
(0) =
dt
4000D2 + D1 ;
iC (0)
=
C
60 ⇥ 10 3
=
10 6
D1
4000D2 =
60,000;
60,000;
vo (t) = 15e 4000t V,
t
D1 = 0;
0.
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Problems
P 8.16
1
1
=
= 16 ⇥ 106 ;
3
6
LC
(62.5 ⇥ 10 )(10 )
!o2 =
↵=
8–15
1
1
=
= 2500;
2RC
2(200)(10 6 )
2500 ±
s1,2 =
p
25002
16 ⇥ 106 =
2500 ± j3122.5rad/s;
vo (t) = B1 e 2500t cos 3122.5t + B2 e 2500t sin 3122.5t.
vo (0) = B1 = 15 V;
iR (0) =
15
= 75 mA;
200
iL (0) =
60 mA;
iC (0) =
iR (0)
dvo
(0) =
dt
2500B1 + 3122.5B2 =
.·.
iL (0) =
.·.
15 mA
iC (0)
=
C
15,000;
15,000;
B2 = 7.21;
vo (t) = 15e 2500t cos 3122.5t + 7.21e 2500t sin 3122.5t V,
t
0.
P 8.17
vT =
10i + iT
(150)(60)
210
!
=
10
iT (150)
9000
+ iT
;
210
210
1500 + 9000
vT
= 50 ⌦;
=
iT
210
Vo =
4000
(50) = 20 V;
10,000
Io = 0;
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8–16
CHAPTER 8. Natural and Step Responses of RLC Circuits
iC (0) =
iR (0)
0.4
iC (0)
=
=
C
8 ⇥ 10 6
s
!o =
↵=
1
=
LC
s
20
=
50
iL (0) =
0.4 A;
50,000;
1
= 1562.5 rad/s;
(51.2 ⇥ 10 3 )(8 ⇥ 10 6 )
1
1
=
= 1250 rad/s;
2RC
(2)(50)(8 ⇥ 10 6 )
↵2 < !02
!d =
p
so the response is underdamped.
1562.52
12502 = 937.5;
vo = B1 e 1250t cos 937.5t + B2 e 1250t sin 937.5t.
vo (0) = B1 = 20 V;
iC (0)
;
C
dvo
(0) =
dt
↵B1 + !d B2 =
.·.
1250(20) + 937.5B2 =
vo = 20e 1250t cos 937.5t
50,000
so
26.67e 1250t sin 937.5t V,
B2 =
t
26.67;
0.
P 8.18
vT =
10i + iT
(150)(60)
210
!
=
10
iT (150)
9000
+ it
;
210
210
1500 + 9000
vT
= 50 ⌦;
=
iT
210
Vo =
4000
(50) = 20 V;
10,000
Io = 0;
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Problems
s
!o =
↵=
1
=
LC
s
1
3 )(8 ⇥ 10 6 )
(80 ⇥ 10
8–17
= 1250 rad/s;
1
109
=
= 1250 rad/s;
2RC
(2)(50)(8 ⇥ 10 6 )
↵2 = !02
so the response is critically damped.
vo = D1 te 1250t + D2 e 1250t ;
vo (0) = D2 = 20 V;
✓
◆
dvo
(0) = D1
dt
1
↵D2 =
C
.·.
D1
1
0
1250(20) =
8 ⇥ 10 6
vo =
25,000te 1250t + 20e 1250t V,
I0
V0
;
R
✓
20
50
t
◆
so
D1 =
25,000;
0.
P 8.19
vT =
10i + iT
(150)(60)
210
!
=
10
iT (150)
9000
+ it
;
210
210
1500 + 9000
vT
= 50 ⌦;
=
iT
210
Vo =
!o =
↵=
4000
(50) = 20 V;
10,000
s
1
=
LC
s
Io = 0;
1
(125 ⇥ 10
3 )(8 ⇥ 10 6 )
= 1000 rad/s;
1
109
=
= 1250 rad/s;
2RC
(2)(50)(8 ⇥ 10 6 )
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8–18
CHAPTER 8. Natural and Step Responses of RLC Circuits
↵2 > !02
so the response is overdamped.
1250 ±
s1,2 =
p
12502
10002 =
500, 2000;
✓
V0
;
R
vo = A1 e 500t + A2 e 1000t .
vo (0) = A1 + A2 = 20 V;
dvo
(0) =
dt
.·.
500A1
500A1
1
2000A2 =
0
8 ⇥ 10 6
A1 =
Solving,
vo =
P 8.20
I0
✓
6.67,
◆
20
=
50
50,000;
A2 = 26.67
6.67e 500t + 26.67e 2000t V,
[a] ↵ =
◆
1
2000A2 =
C
t
0.
1
= 0.5 rad/s;
2RC
!o2 =
1
= 25.25;
LC
!d =
q
25.25
(0.5)2 = 5 rad/s;
.·. v = B1 e t/2 cos 5t + B2 e t/2 sin 5t.
v(0) = B1 = 0;
v = B2 e t/2 sin 5t;
iR (0+ ) = 0 A;
iC (0+ ) = 4 A;
50 =
↵B1 + !d B2 =
4
dv +
(0 ) =
= 50 V/s;
dt
0.08
0.5(0) + 5B2 ;
.·. B2 = 10;
.·. v = 10e t/2 sin 5t V,
[b]
dv
=
dt
t
0.
5e t/2 sin 5t + 10e t/2 (5 cos 5t);
dv
= 0 when 10 cos 5t = sin 5t or
dt
t1 = 294.23 ms.
.·. 5t1 = 1.47,
5t2 = 1.47 + ⇡,
t2 = 922.54 ms;
5t3 = 1.47 + 2⇡,
t3 = 1550.86 ms.
tan 5t = 10;
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Problems
8–19
2⇡
2⇡
= 1256.6 ms.
=
!d
5
1256.6
Td
=
= 628.3 ms.
[d] t2 t1 = 628.3 ms;
2
2
[e] v(t1 ) = 10e (0.147115) sin 5(0.29423) = 8.59 V;
[c] t3
t1 = 1256.6 ms;
Td =
v(t2 ) = 10e (0.46127) sin 5(0.92254) =
6.27 V;
v(t3 ) = 10e (0.77543) sin 5(1.55086) = 4.58 V.
[f ]
P 8.21
[a] ↵ = 0;
!d = ! o =
p
25.25 = 5.02 rad/s;
v = B1 cos !o t + B2 sin !o t;
C
dv
(0) =
dt
v(0) = B1 = 0;
v = B2 sin !o t;
iL (0) = 4;
p
↵B1 + !d B2 = 0 + 25.25B2 ;
p
.·. B2 = 50/ 25.25 = 9.95 V;
50 =
v = 9.95 sin 5.02t V,
[b] 2⇡f = 5.02;
f=
t
0.
5.02 ⇠
= 0.80 Hz.
2⇡
[c] 9.95 V.
P 8.22
From the form of the solution we have
v(0) = A1 + A2 ;
dv(0+ )
=
dt
↵(A1 + A2 ) + j!d (A1
A2 .
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8–20
CHAPTER 8. Natural and Step Responses of RLC Circuits
We know both v(0) and dv(0+ )/dt will be real numbers. To facilitate the
algebra we let these numbers be K1 and K2 , respectively. Then our two
simultaneous equations are
K1 = A1 + A2 ;
K2 = ( ↵ + j!d )A1 + ( ↵
j!d )A2 .
The characteristic determinant is
1
=
1
( ↵ + j!d ) ( ↵
=
j!d )
j2!d .
The numerator determinants are
N1 =
K1
1
K2 ( ↵
and N2 =
=
1
K1
( ↵ + j!d ) K2
It follows that A1 =
and A2 =
(↵ + j!d )K1
j!d )
N2
=
N1
=
= K2 + (↵
!d K1
j!d )K1 .
j(↵K1 + K2 )
;
2!d
!d K1 + j(↵K1 + K2 )
.
2!d
We see from these expressions that
P 8.23
K2
A1 = A⇤2 .
By definition, B1 = A1 + A2 . From the solution to Problem 8.22 we have
A1 + A2 =
2!d K1
= K1 .
2!d
But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. 8.16.
By definition, B2 = j(A1 A2 ). From Problem 8.22 we have
B2 = j(A1
A2 ) =
j[ 2j(↵K1 + K2 )]
↵K1 + K2
=
.
2!d
!d
It follows that
K2 =
↵K1 + !d B2 ,
but K2 =
dv(0+ )
dt
and K1 = B1 .
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Problems
8–21
Thus we have
dv +
(0 ) =
dt
↵B1 + !d B2 ,
which is identical to Eq. 8.17.
P 8.24
P 8.25
!
diL
[a] v = L
= 16[e 20,000t e 80,000t ] V,
t 0.
dt
v
[b] iR =
= 40[e 20,000t e 80,000t ] mA,
t 0+ .
R
t
[c] iC = I iL iR = [ 8e 20,000t + 32e 80,000t ] mA,
diL
[a] v = L
dt
!
[b] iC (t) = I
iR
= 40e 32,000t sin 24,000t V,
iL = 24 ⇥ 10 3
= [24e 32,000t cos 24,000t
P 8.26
diL
v=L
dt
P 8.27
↵=
!
v
625
t
0+ .
0.
iL
32e 32,000t sin 24,000t] mA,
= 960,000te 40,000t kV,
t
t
0+ .
0.
1
1
=
= 1000;
2RC
2(400)(1.25 ⇥ 10 6 )
1
1
=
= 64 ⇥ 104 ;
LC
(1.25 ⇥ 10 6 )(1.25)
!o2 =
p
s1,2 =
1000 ±
s1 =
400 rad/s;
10002
64 ⇥ 104 =
s2 =
1000 ± 600 rad/s;
1600 rad/s;
io = If + A01 e 400t + A02 e 1600t ;
If =
12
= 30mA;
400
io (0) = 0;
.·. A01 + A02 =
0 = 30 ⇥ 10 3 + A01 + A02 ,
12
dio
(0) =
=
dt
1.25
Solving,
io = 30
A01 =
400A01
30 ⇥ 10 3 ;
1600A02 ;
32 mA;
32e 400t + 2e 1600t mA,
A02 = 2 mA;
t
0.
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8–22
P 8.28
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] vo (1) = 0 = Vf ;
.·. vo = A01 e 400t + A02 e 1600t .
vo (0) = 12 = A01 + A02 ;
iC (0+ ) = 0;
Note:
.·.
dvo
(0) = 0 =
dt
Solving,
[b]
400A01
1600A02 .
A01 = 16 V,
A02 =
vo (t) = 16e 400t
4e 1600t V,
dio
= [12.8e 400t
dt
3.2e 1600t ];
vo = L
dio
= 16e 400t
dt
4V
t > 0.
4e 1600t V,
t
0.
This agrees with the solution to Problem 8.27.
P 8.29
iL (0 ) = iL (0+ ) = 37.5 mA.
For t > 0:
iL (0 ) = iL (0+ ) = 37.5 mA;
1
= 100 rad/s;
2RC
↵=
s1 =
40 rad/s
s2 =
!o2 =
1
= 6400;
LC
160 rad/s;
vo (1) = 0 = Vf ;
vo = A01 e 40t + A02 e 160t ;
iC (0+ ) =
.·.
37.5 + 37.5 + 0 = 0;
dvo
= 0.
dt
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Problems
dvo
(0) =
dt
40A01
160A02 ;
.·. A01 + 4A02 = 0;
.·. A01 = 0;
A01 + A02 = 0;
A02 = 0;
.·. vo = 0 for t
0.
vo (0) = 0;
Note:
8–23
vo (1) = 0;
dvo (0)
= 0;
dt
Hence the 37.5 mA current circulates between the current source and the ideal
inductor in the equivalent circuit. In the original circuit the 7.5 V source
sustains a current of 37.5 mA in the inductor. This is an example of a circuit
going directly into steady state when the switch is closed. There is no
transient period, or interval.
P 8.30
For
t > 0:
↵=
1
= 640;
2RC
!d =
p
8002
1
= 64 ⇥ 104 ;
LC
6402 = 480;
io = If + B10 e 640t cos 480t + B20 e 640t sin 480t;
If =
25
= 0.2 A;
125
io (0) = 0.2 + B10 = 0
so
B10 =
dio
(0) =
dt
↵B10 + !d B20 =
V0
L
so
Solving,
B20 =
io (t) = 0.2
0.2e 640t cos 480t
0.2;
640( 0.2) + 480B20 = 0;
0.267;
0.267e 640t sin 480t A,
t
0.
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8–24
P 8.31
CHAPTER 8. Natural and Step Responses of RLC Circuits
1
1
= 640;
= 64 ⇥ 104 ;
2RC
LC
p
!d = 8002 6402 = 480;
[a] ↵ =
vo = Vf + B10 e 640t cos 480t + B20 e 640t sin 480t;
Vo (0+ ) = 0;
Vf = 0;
vo (0) = 0 + B10 = 0
dvo
(0) =
dt
.·.
iC (0+ ) = 0.2 A;
B10 = 0;
so
640(0) + 480B20 =
iC (0+ )
= 32,000
6.25 ⇥ 10 6
vo (t) = 66.67e 640t sin 480t V,
t
so
B20 = 66.67;
0.
[b] From the solution to Problem 8.30,
io (t) = 0.2
vo = L
0.2e 640t cos 480t
0.267e 640t sin 480t A;
dio
= (0.25)(266.67e 640t sin 480t) = 66.67e 640t sin 480t V,
dt
t
0.
Thus the solutions to Problems 8.30 and 8.31 are consistent.
P 8.32
s
!o =
↵=
1
=
LC
s
1
(25 ⇥ 10
3 )(62.5 ⇥ 10 6 )
= 800 rad/s;
1
1
=
= 640 rad/s
2RC
2(12.5)(62.5 ⇥ 10 6 )
!d =
p
8002
.·. underdamped.
6402 = 480;
If = 2 A;
iL = 2 + B10 e 640t cos 480t + B20 e 640t sin 480t;
iL (0) = 2 + B10 = 1
B10 =
so
1;
V0
;
L
diL
(0) =
dt
↵B10 + !d B20 =
.·.
640( 1) + 480B20 =
iL (t) = 2
e 640t cos 480t + 2.83e 640t sin 480t A,
50
,
25 ⇥ 10 3
B20 = 2.83;
so
t
0.
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Problems
P 8.33
↵=
!o =
8–25
1
1
=
= 1000 rad/s;
2RC
2(8)(62.5 ⇥ 10 6 )
s
1
=
LC
s
1
(25 ⇥ 10
3 )(62.5 ⇥ 10 6 )
1000 ±
s1,2 =
Overdamped:
p
= 800 rad/s;
10002
8002 =
400, 1600 rad/s;
If = 2 A;
iL = 2 + A01 e 400t + A02 e 1600t;
iL (0) = 2 + A01 + A02 = 1
diL
(0) =
dt
400A01
Solving,
1
A01 = ,
3
1
iL (t) = 2 + e 400t
3
P 8.34
↵=
!o =
A01 + A02 =
so
1600A02 =
1;
50
V0
=
= 2000;
L
25 ⇥ 10 3
A02 =
4
;
3
4 1600t
e
A,
3
t
0.
1
1
=
= 800;
2RC
2(10)(62.5 ⇥ 10 6 )
s
1
=
LC
↵2 = !02
s
1
= 800 rad/s;
(25 ⇥ 10 3 )(62.5 ⇥ 10 6 )
Critically damped.
If = 2 A;
iL = 2 + D10 te 800t + D20 e 800t ;
iL (0) = 2 + D20 = 1;
.·. D20 =
V0
;
L
diL
(0) = D10
dt
↵D20 =
.·.
800( 1) =
D10
iL = 2 + 1200te 800t
1 A;
50
25 ⇥ 10 3
e 800t A,
t
so
D10 = 1200;
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–26
P 8.35
CHAPTER 8. Natural and Step Responses of RLC Circuits
vC (0+ ) =
3.75 ⇥ 103
(150) = 50 V;
11.25 ⇥ 103
iL (0+ ) = 100 mA;
↵=
150
= 20 mA;
7500
1
1
=
= 800;
2RC
2(2500)(0.25 ⇥ 10 6 )
1
1
=
= 106 ;
LC
(4)(0.25 ⇥ 10 6 )
!o2 =
↵2 = 64 ⇥ 104 ;
s1,2 =
iL
iL (1) =
.·.
↵2 < !o2 ;
p
800 ± j 8002
106 =
underdamped
800 ± j600 rad/s;
= If + B10 e ↵t cos !d t + B20 e ↵t sin !d t
= 20 + B10 e 800t cos 600t + B20 e 800t sin 600t;
iL (0) = 20 ⇥ 10 3 + B10 ;
B10 = 100 m
diL
(0) = 600B20
dt
50
= 12.5;
4
800B10 =
.·. 600B2 = 800(80 ⇥ 10 3 ) + 12.5;
20 m = 80 mA;
B20 = 127.5 mA;
.·. iL = 20 + 80e 800t cos 600t + 127.5e 800t sin 600t mA,
P 8.36
t
0.
t < 0:
V0 = vo (0 ) = vo (0+ ) =
3000
(100) = 75 V;
4000
I0 = iL (0 ) = iL (0+ ) = 100 mA.
t > 0:
↵=
1
1
=
= 500 rad/s;
2RC
2(40)(25 ⇥ 10 6 )
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Problems
!o =
s
s
1
=
LC
.·. ↵2 > !o2
s1,2 =
1
(250 ⇥ 10
8–27
= 400;
3 )(25 ⇥ 10 6 )
overdamped.
500 ±
p
5002
4002 =
200, 800.
[a] iL = If + A1 e 200t + A2 e 800t
If = 100 mA;
iL (0) = 0.1 + A1 + A2 = 0.1
diL
(0) =
dt
200A1
Solving,
A1 = 0.5,
.·.
A2 =
25e 200t + 100e 800t V,
Z 1
pdt =
vL =
0
Z 1
0.5e 800t A.
t
0.
vL iL dt;
0
25e 200t + 100e 800t V;
iL = 0.1 + 0.5e 200t
p=
2.5e 200t
wL =
2.5
Z 1
0
+62.5
2.5
0.5e 800t A;
12.5e 400t + 10e 800t + 62.5e 1000t
Z 1
e 200t dt
Z 1
=
0.5;
diL
= (0.25)( 100e 200t + 400e 800t )
dt
=
[a] wL =
A1 + A2 = 0;
75
V0
=
= 300;
L
0.25
800A2 =
iL (t) = 0.1 + 0.5e 200t
[b] vC (t) = vL (t) = L
P 8.37
so
0
12.5
e 1000t dt
1
e 200t
+
200 0
1
0
Z 1
50
12.5
0
Z 1
e 400t dt + 10
0
50e 1600t W;
e 800t dt
e 1600t dt
1
e 400t
400 0
1
e 800t
e 1000t
+ 10
+ 62.5
800 0
1000 0
1
e 1600t
50
1600 0
All the upper limits evaluate to zero hence
wL =
2.5
200
12.5
10
62.5
+
+
400
800 1000
50
= 0 J.
1600
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8–28
CHAPTER 8. Natural and Step Responses of RLC Circuits
Since the initial and final values of the current in the inductor are the
same, the initial and final values of the energy in the inductor are the
same. Thus, there is no net energy delivered to the inductor.
25e 200t + 100e 800t V;
[b] vR =
vR
=
40
iR =
0.625e 200t + 2.5e 800t A;
pR = vR iR = 15.625e 400t
wR =
Z 1
0
= 15.625
125e 1000t + 250e 1600t W;
pR dt;
Z 1
0
e
400t
Z 1
dt
e 400t 1
= 15.625
400 0
125
0
e
1000t
Z 1
dt + 250
0
e 1600t dt
e 1000t 1
e 1600t 1
125
+ 250
.
1000 0
1600 0
Since all the upper limits evaluate to zero we have
250
15.625 125
+
= 70.3125 mJ.
400
400 1600
[c] 0.1 = iR + iC + iL (mA);
wR =
( 0.625e 200t + 2.5e 800t )
iC = 0.1
= 0.125e 200t
(0.1 + 0.5e 200t
2e 800t A
pC = vC iC = [ 25e 200t + 100e 800t ][0.125e 200t
=
3.125e 400t + 62.5e 1000t
Z 1
wC =
3.125
=
3.125
0
0.5e 800t )
400t
e
dt + 62.5
2e 800t ]
200e 1600t W.
Z 1
0
e
1000t
e 400t 1
e 1000t 1
+ 62.5
400 0
1000 0
dt
200
200
Z 1
0
e 1600t dt
e 1600t 1
.
1600 0
Since all upper limits evaluate to zero we have
62.5
3.125
+
400
1000
[d] is = 100 mA;
wC =
ps (del) = 0.1v =
Z 1
ws =
2.5
=
2.5
0
e
200
=
1600
70.3125 mJ.
2.5e 200t + 10e 800t W;
200t
Z 1
dt + 10
0
e 800t dt
e 200t 1
10
e 800t 1
2.5
+
= 0.
+10
=
200 0
800 0
200
800
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Problems
8–29
[e] wL = 0 J;
wR = 70.3125 mJ (absorbed);
wC = 70.3125 mJ (delivered);
wS = 0 mJ;
X
P 8.38
wdel = wabs = 70.3125 mJ.
36
t<0:
iL (0 ) =
= 0.12 A;
300
The circuit reduces to:
vC (0 ) = 0 V.
iL (1) = 0.1 A;
!o =
↵=
s
1
=
LC
s
1
(20 ⇥ 10
3 )(500 ⇥ 10 9 )
= 10,000 rad/s;
1
1
=
= 10,000 rad/s.
2RC
(100)(500 ⇥ 10 9 )
Critically damped:
iL = 0.1 + D10 te 10,000t + D20 e 10,000 ;
iL (0) = 0.1 + D20 = 0.12
diL
(0) = D10
dt
Solving,
↵D20 =
V0
L
so
so
D20 = 0.02;
D10
(10,000)(0.02) = 0;
D10 = 200.
iL (t) = 0.1 + 200te 10,000t + 0.02e 10,000t A,
P 8.39
[a] !o2 =
↵=
t
0.
1
1
=
= 25 ⇥ 106 ;
3
6
LC
(80 ⇥ 10 )(0.5 ⇥ 10 )
R
= !o = 5000 rad/s;
2L
.·. R = (5000)(2)L = 800 ⌦.
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8–30
CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] i(0) = iL (0) = 30 mA;
vc (0) = 800i(0) = 80 ⇥ 10 3
di(0)
;
dt
di(0)
800(30 ⇥ 10 3 )
;
=
80 ⇥ 10 3
dt
20
di(0)
=
dt
.·.
50 A/s.
[c] vC = D1 te 5000t + D2 e 5000t ;
vC (0) = D2 = 20 V;
dvC
(0) = D1
dt
D1
5000D2 =
iC (0)
=
C
30 ⇥ 10 3
=
0.5 ⇥ 10 6
100,000 =
60,000
.·.
t
0+ .
vC = 40,000te 5000t + 20e 5000t V,
P 8.40
[a] 2↵ = 5000;
q
↵2
↵=
D1 = 40,000 V/s;
↵ = 2500 rad/s;
!o2 = 4 ⇥ 106 ;
!o2 = 1500;
R
= 2500;
2L
!o2 =
iL (0)
;
C
!o = 2000 rad/s;
R = 5000L;
1
= 4 ⇥ 106 ;
LC
L=
R = 25,000 ⌦.
109
= 5 H;
4 ⇥ 106 (50)
[b] i(0) = 0;
L
di(0)
= vc (0);
dt
1
(50) ⇥ 10 9 vc2 (0) = 90 ⇥ 10 6 ;
2
.·. vc2 (0) = 3600;
vc (0) = 60 V;
60
di(0)
=
= 12 A/s.
dt
5
[c] i(t) = A1 e 1000t + A2 e 4000t ;
i(0) = A1 + A2 = 0;
di(0)
= 1000A1
dt
Solving,
4000A2 = 12.
A1 = 4 mA;
A2 =
i(t) = 4e 1000t
4e 4000t mA
4 mA;
t
0.
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Problems
[d]
di(t)
=
dt
8–31
4e 1000t + 16e 4000t
di
= 0 when 16e 4000t = 4e 1000t
dt
or e3000t = 4
.·. t =
ln 4
µs = 462.10 µs
3000
[e] imax = 4e 0.4621 4e 1.8484 = 1.89 mA
di
[f ] vL (t) = 5 = [ 20e 1000t + 80e 4000t ] V,
dt
P 8.41
↵ = 2000 rad/s;
t
0+
!d = 1500 rad/s;
!o2
↵2 = 225 ⇥ 104 ;
↵=
R
= 2000;
2L
!o2 = 625 ⇥ 104 ;
wo = 25,000 rad/s;
R = 4000L;
1
= 625 ⇥ 104 ;
LC
L=
1
(625 ⇥ 104 )(80 ⇥ 10 9 )
= 2 H;
.·. R = 8 k⌦;
i(0+ ) = B1 = 7.5 mA;
at t = 0+ .
60 + vL (0+ )
.·.
di(0+ )
=
dt
.·.
30 = 0;
30
=
2
vL (0+ ) =
30 V;
15 A/s;
di(0+ )
= 1500B2
dt
2000B1 =
15;
.·. 1500B2 = 2000(7.5 ⇥ 10 3 )
15;
.·. i = 7.5e 2000t cos 1500t mA,
t
.·. B2 = 0 A;
0.
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8–32
P 8.42
CHAPTER 8. Natural and Step Responses of RLC Circuits
From Prob. 8.41 we know vc will be of the form
vC = B3 e 2000t cos 1500t + B4 e 2000t sin 1500t.
From Prob. 8.41 we have
vC (0) =
30 V = B3
and
iC (0)
7.5 ⇥ 10 3
dvC (0)
=
=
= 93.75 ⇥ 103 .
9
dt
C
80 ⇥ 10
dvC (0)
= 1500B4
dt
2000B3 = 93,750;
.·. 1500B4 = 2000( 30) + 93,750;
vC (t) =
P 8.43
[a]
B4 = 22.5 V;
30e 2000t cos 1500t + 22.5e 2000t sin 1500t V
t
0.
1
= 20,0002 .
LC
There are many possible solutions. This one begins by choosing
L = 1 mH. Then,
C=
1
(1 ⇥ 10
3 )(20,000)2
= 2.5 µF.
We can achieve this capacitor value using components from Appendix H
by combining four 10 µF capacitors in series.
Critically damped:
↵ = !0 = 20,000
so
R
= 20,000;
2L
.·. R = 2(10 3 )(20,000) = 40 ⌦.
We can create this resistor value using components from Appendix H by
combining a 10 ⌦ resistor and two 15 ⌦ resistors in series. The final
circuit:
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Problems
q
[b] s1,2 = ↵ ± ↵2 !02 = 20,000 ± 0.
Therefore there are two repeated real roots at
P 8.44
8–33
20,000 rad/s.
[a] Underdamped response:
so
↵ < !0
↵ < 20,000.
Therefore we choose a larger resistor value than the one used in Problem
8.43 to give a smaller value of ↵. For convenience, pick ↵ = 16,000 rad/s:
R
= 16,000
so
R = 2(16,000)(10 3 ) = 32 ⌦.
2L
We can create a 32 ⌦ resistance by combining a 10 ⌦ resistor and a 22 ⌦
resistor in series.
↵=
16,000 ±
s1,2 =
q
16,0002
20,0002 =
16,000 ± j12,000 rad/s.
[b] Overdamped response:
↵ > !0
so
↵ > 20,000.
Therefore we choose a smaller resistor value than the one used in
Problem 8.43. Choose R = 50 ⌦, which can be created by combining two
100 ⌦ resistors in parallel:
↵=
R
= 25,000;
2L
s1,2 =
25,000 ±
=
P 8.45
t<0:
q
25,0002
10,000 rad/s;
I0 =
20,0002 =
and
75 mA;
25,000 ± 15,000
40,000 rad/s.
V0 = 0.
t>0:
!0 =
↵=
s
1
=
LC
s
1
(80 ⇥ 10
3 )(200 ⇥ 10 6 )
= 250;
40
R
=
= 250.
2L
2(80 ⇥ 10 3 )
Critically damped:
i = D1 te 250t + D2 e 250t ;
i(0) = D2 = I0 =
0.075;
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8–34
CHAPTER 8. Natural and Step Responses of RLC Circuits
↵D2 =
So
(250)( 0.075) =
D1
Solving,
.·.
P 8.46
1
( V0
L
di
(0) = D1
dt
RI0 ).
1
(0
80 ⇥ 10 3
(40)( 0.075)).
75e 250t mA,
t
D1 = 18.75;
i(t) = 18,750te 250t
0.
[a] For t > 0:
Since i(0 ) = i(0+ ) = 0,
va (0+ ) = 72 V.
[b] va = 5000i +
Z t
1
i dx + 72;
0.1 ⇥ 10 6 0
di
dva
= 5000 + 10 ⇥ 106 i;
dt
dt
di(0+ )
di(0+ )
dva (0+ )
= 5000
+ 10 ⇥ 106 i(0+ ) = 5000
;
dt
dt
dt
L
di(0+ )
= 72;
dt
di(0+ )
=
dt
.·.
[c] ↵ =
72
=
2.5
dva (0+ )
=
dt
28.8 A/s;
144,000 V/s.
R
12,500
=
= 2500 rad/s;
2L
2(2.5)
1
1
=
= 4 ⇥ 106 ;
LC
(2.5)(0.1 ⇥ 10 6 )
p
s1,2 = 2500 ± 25002 4 ⇥ 106 = 2500 ± 1500 rad/s.
!o2 =
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Problems
8–35
Overdamped:
va = A1 e 1000t + A2 e 4000t ;
va (0) = 72 = A1 + A2 ;
dva (0)
=
dt
144,000 =
Solving, A1 = 48;
1000A1
A2 = 24;
va = 48e 1000t + 24e 4000t V,
P 8.47
4000A2 .
0+ .
t
[a] t < 0:
100
= 2 A;
V0 = 4(100) =
50
t > 0:
500
R
=
= 625 rad/s;
↵=
2L
2(0.4)
I0 =
!o =
s
1
=
LC
.·.
↵2 < !o2
625 ±
s1,2 =
s
400 V.
1
= 500 rad/s;
(0.4)(10 ⇥ 10 6 )
overdamped.
p
6252
5002 =
250, 1000 rad/s;
io = A1 e 250t + A2 e 1000t ;
io (0) = A1 + A2 = 2;
dio
(0) =
dt
250A1
Solving,
2
A1 = ;
3
1000A2 =
1
( V0
L
[b] vo (t) =
Z t
1
io (x) dx
10 ⇥ 10 6 0
5
= 10
5
= 10
=
1500.
4
A2 = ;
3
2
4
io (t) = e 250t + e 1000t A,
3
3
.·.
RI0 ) =
t
0.
400
◆
400
(2/3)e 250x t (4/3)e 1000x t
+
250
1000
0
0
400
✓Z t
Z t
4 1000x
2 250x
e
e
dx +
dx
03
03
266.67e 250t
133.33e 1000t V,
t
!
0.
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8–36
P 8.48
CHAPTER 8. Natural and Step Responses of RLC Circuits
t < 0:
i(0) =
240
240
=
= 6 A;
8 + 30k70 + 11
40
vo (0) = 240
70
(6)(20) = 108 V.
100
8(6)
t > 0:
↵=
20
R
=
= 10,
2L
2(1)
!o2 =
↵2 = 100;
1
1
=
= 200;
LC
(1)(5 ⇥ 10 3 )
!o2 > ↵2
s1,2 =
underdamped.
p
100 ± 100 200 =
10 ± j10 rad/s;
vo = B1 e 10t cos 10t + B2 e 10t sin 10t;
vo (0) = B1 = 108 V;
C
dvo
(0) =
dt
6,
dvo
6
=
=
dt
5 ⇥ 10 3
dvo
(0) =
dt
10B1 + 10B2 =
10B2 =
1200 + 10B1 =
.·. vo = 108e 10t cos 10t
1200 V/s;
1200;
1200 + 1080;
12e 10t sin 10t V,
B2 =
t
120/10 =
12 V;
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 8.49
iC (0) = 0;
↵=
8–37
vo (0) = 50 V;
8000
R
=
= 25,000 rad/s;
2L
2(160 ⇥ 10 3 )
!o2 =
1
1
=
= 625 ⇥ 106 ;
3
LC
(160 ⇥ 10 )(10 ⇥ 10 9 )
.·. ↵2 = !o2 ;
critical damping.
vo (t) = Vf + D10 te 25,000t + D20 e 25,000t ;
Vf = 250 V;
vo (0) = 250 + D20 = 50;
dvo
(0) =
dt
D10 =
D20 =
25,000D20 + D10 = 0;
25,000D20 = 5 ⇥ 106 V/s;
vo = 250 + 5 ⇥ 106 te 25,000t
P 8.50
↵=
200 V ;
200e 25,000t V,
t
0.
200
R
=
= 400 rad/s;
2L
2(0.025)
!o =
s
1
=
LC
↵2 < !02 :
!d =
p
5002
s
1
= 500 rad/s;
(250 ⇥ 10 3 )(16 ⇥ 10 6 )
underdamped.
4002 = 300 rad/s;
vo = Vf + B10 e 400t cos 300t + B20 e 400t sin 300t;
vo (1) = 200(0.08) = 16 V;
vo (0) = 0 = Vf + B10 = 0
so
B10 =
16 V;
B20 =
dvo
(0) = 0 =
dt
400B10 + 300B20
so
.·. vo (t) = 16
16e 400t cos 300t
21.33e 400t sin 300t V,
21.33 V;
t
0.
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8–38
P 8.51
CHAPTER 8. Natural and Step Responses of RLC Circuits
↵=
!o =
R
250
=
= 500 rad/s;
2L
2(0.025)
s
1
=
LC
↵2 = !02 :
s
1
= 500 rad/s;
(250 ⇥ 10 3 )(16 ⇥ 10 6 )
critically damped.
vo = Vf + D10 te 500t + D20 e 500t ;
vo (0) = 0 = Vf + D20 ;
.·. D20 =
vo (1) = (250)(0.08) = 20 V;
dvo
(0) = 0 = D10
dt
.·.
P 8.52
↵=
!o =
↵D20
D10 = (500)( 20) =
so
10,000te 500t
vo (t) = 20
20 V;
20e 500t V,
t
10,000 V/s;
0.
R
312.5
=
= 625 rad/s;
2L
2(0.025)
s
1
=
LC
↵2 > !02 :
s1,2 =
s
1
(250 ⇥ 10
3 )(16 ⇥ 10 6 )
= 500 rad/s;
overdamped.
625 ±
p
6252
5002 =
250, 1000 rad/s;
vo = Vf + A01 e 250t + A02 e 100;t
vo (0) = 0 = Vf + A01 + A02 ;
vo (1) = (312.5)(08) = 25 V;
dvo
(0) = 0 =
dt
250A01
.·. A01 + A02 =
25 V;
1000A02 ;
Solving,
A01 =
vo (t) = 25
33.33e 250t + 8.33e 1000t V,
33.33 V;
A02 = 8.33 V;
t
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 8.53
8–39
t < 0:
150
=
30
iL (0) =
5 A;
vc (0) = 18iL (0) =
90 V.
t > 0:
↵=
10
R
=
= 50 rad/s;
2L
2(0.1)
!o2 =
1
1
=
= 5000;
LC
(0.1)(2 ⇥ 10 3 )
!o > ↵2
.·.
s1,2 =
50 ±
underdamped.
p
502
5000 =
50 ± j50;
vc = 60 + B10 e 50t cos 50t + B20 e 50t sin 50t;
vc (0) =
C
90 = 60 + B10
dvc
(0) =
dt
5;
.·.
B10 =
150;
dvc
5
(0) =
=
dt
2 ⇥ 10 3
2500;
dvc
(0) =
dt
50B10 + 50B2 =
2500
vc = 60
150e 50t cos 50t
200e 50t sin 50t V,
.·.
B20 =
200;
t
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–40
P 8.54
CHAPTER 8. Natural and Step Responses of RLC Circuits
t < 0:
20 + 28
= 75 mA;
160 + 480
io (0 ) =
vo (0 ) = 20
t
480(0.075) =
16 V.
0:
As t ! 1, Vf = 20 V.
Req = 960k480 = 320 ⌦;
↵=
320
Req
=
= 320,000 rad/s;
2L
2(0.5 ⇥ 10 3 )
!o =
s
1
=
LC
↵2 < !02 :
!d =
s
1
(0.5 ⇥ 10
3 )(12.5 ⇥ 10 9 )
= 400,000 rad/s;
underdamped.
q
400,0002
320,0002 = 240,000 rad/s;
vo = 20 + B10 e 320,000t cos 240,000t + B20 e 320,000t sin 240,000t;
vo (0) = 20 + B10 =
16
dvo
(0) =
dt
↵B10 + !d B20 =
solving,
B20 =
.·.
vo (t) = 20
so
B10 =
36 V;
I0
C
so
320,000( 36) + 240,000B20 =
75 ⇥ 10 3
.
12.5 ⇥ 10 9
23;
36e 320,000t cos 240,000t
23e 320,000t sin 240,000t V t
0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 8.55
8–41
[a] Let i be the current in the direction of the voltage drop vo (t). Then by
hypothesis
i = if + B10 e ↵t cos !d t + B20 e ↵t sin !d t
.
if = i(1) = 0,
i(0) =
Vg
= B10 .
R
Therefore i = B10 e ↵t cos !d t + B20 e ↵t sin !d t.
di(0)
= 0,
dt
therefore
di
= [(!d B20
dt
↵B10 ) cos !d t
L
Therefore !d B20
di(0)
= 0;
dt
(↵B20 + !d B10 ) sin !d t] e ↵t .
↵B10 = 0;
B20 =
↵ Vg
↵ 0
.
B1 =
!d
!d R
Therefore
(
[b]
!
)
↵2 Vg !d Vg
+
L
sin !d t e ↵t
!d R
R
di
vo = L =
dt
!
)
=
(
=
Vg L
R
↵2 + !d2
e ↵t sin !d t
!d
=
Vg L
R
!o2
e ↵t sin !d t
!d
=
Vg L
R!d
↵2
+ !d sin !d t e ↵t
!d
LVg
R
!
!
✓
◆
1
e ↵t sin !d t;
LC
vo =
Vg
e ↵t sin !d t V,
RC!d
dvo
=
dt
Vg
{!d cos !d t
!d RC
dvo
= 0 when
dt
tan !d t =
t
0.
↵ sin !d t}e ↵t ;
!d
.
↵
Therefore !d t = tan 1 (!d /↵) (smallest t).
✓
◆
1
!d
t=
tan 1
.
!d
↵
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8–42
P 8.56
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] From Problem 8.55 we have
vo =
Vg
e ↵t sin !d t;
RC!d
↵=
4800
R
=
= 37,500 rad/s;
2L
2(64 ⇥ 10 3 )
!o2 =
!d =
1
1
=
= 3906.25 ⇥ 106 ;
3
LC
(64 ⇥ 10 )(4 ⇥ 10 9 )
q
!o2
↵2 = 50 krad/s;
( 72)
Vg
= 75;
=
RC!d
(4800)(4 ⇥ 10 9 )(50 ⇥ 103 )
.·. vo = 75e 37,500t sin 50,000t V.
[b] From Problem 8.55
✓
1
!d
td =
tan 1
!d
↵
◆
!
1
50,000
tan 1
.
=
50,000
37,500
td = 18.55 µs.
[c] vmax = 75e 0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V.
[d] R = 480 ⌦;
↵ = 3750 rad/s;
!d = 62,387.4 rad/s;
vo = 601.08e 3750t sin 62,387.4t V,
t
0;
td = 24.22 µs;
vmax = 547.92 V.
P 8.57
[a] vC = V + [B10 cos !d t + B20 sin !d t] e ↵t ;
dvC
= [(!d B20
dt
↵B10 ) cos !d t
(↵B20 + !d B10 ) sin !d t]e ↵t .
Since the initial stored energy is zero,
vC (0+ ) = 0 and
It follows that
dvC (0+ )
= 0;
dt
B10 =
V
and
↵B10
0
.
B2 =
!d
When these values are substituted into the expression for [dvC /dt], we get
dvC
=
dt
!
↵2
+ !d V e ↵t sin !d t.
!d
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Problems
But
↵2
↵2 + !d2
!2
+ !d =
= o.
!d
!d
!d
!
!o2
V e ↵t sin !d t.
!d
dvC
=
dt
Therefore
[b]
8–43
dvC
= 0 when
dt
sin !d t = 0,
or !d t = n⇡,
where n = 0, 1, 2, 3, . . . .
Therefore t =
[c] When tn =
and
n⇡
.
!d
n⇡
,
!d
cos !d tn = cos n⇡ = ( 1)n ,
sin !d tn = sin n⇡ = 0.
Therefore vc (tn ) = V [1
( 1)n e ↵n⇡/!d ].
[d] It follows from [c] that
v(t1 ) = V + V e (↵⇡/!d )
vC (t1 )
vC (t3 )
Therefore
But
P 8.58
2⇡
= t3
!d
(
1
vc (t1 )
ln
Td
vc (t3 )
V
V
)
and vc (t3 ) = V + V e (3↵⇡/!d ) .
e (↵⇡/!d )
V
= (3↵⇡/! ) = e(2↵⇡/!d ) .
d
V
e
t1 = Td ,
;
thus ↵ =
Td = t3

7000
63.84
↵=
ln
= 1000;
2⇡
26.02
t1 =
!d =
3⇡
7
1
[vC (t1 )
ln
Td
[vC (t3 )
V]
.
V]
⇡
2⇡
=
ms;
7
7
2⇡
= 7000 rad/s;
Td
!o2 = !d2 + ↵2 = 49 ⇥ 106 + 106 = 50 ⇥ 106 ;
L=
P 8.59
1
(50 ⇥ 106 )(0.1 ⇥ 10 6 )
= 200 mH;
R = 2↵L = 400 ⌦.
At t = 0 the voltage across each capacitor is zero. It follows that since the
operational amplifiers are ideal, the current in the 500 k⌦ is zero. Therefore
there cannot be an instantaneous change in the current in the 1 µF capacitor.
Since the capacitor current equals C(dvo /dt), the derivative must be zero.
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8–44
P 8.60
CHAPTER 8. Natural and Step Responses of RLC Circuits
[a] From Example 8.13
d2 vo
= 2;
dt2
dg(t)
= 2,
dt
therefore
g(t) =
dvo
;
dt
g(t)
g(0) = 2t;
iR =
5
⇥ 10 3 = 10 µA = iC =
500
10 ⇥ 10 6
=
1 ⇥ 10 6
dvo (0)
=
dt
dvo
= 2t
dt
[b] t2
t2
dvo (0)
.
dt
dvo (0)
;
dt
10 = g(0; )
10 dt;
vo (0) = t2
vo = t2
C
g(0) =
10;
dvo = 2t dt
vo
g(t) = 2t + g(0);
10t;
0  t  tsat .
10t + 8,
10t + 8 =
vo (0) = 8 V;
9;
10t + 17 = 0;
t⇠
= 2.17 s.
P 8.61
Part (1) — Example 8.14, with R1 and R2 removed:
[a] Ra = 100 k⌦;
✓
d2 vo
1
=
dt2
Ra C1
C1 = 0.1 µF;
◆✓
◆
1
vg ;
Rb C2
vg = 250 ⇥ 10 3 ;
therefore
Rb = 25 k⌦;
1
= 100
Ra C1
C2 = 1 µF;
1
= 40;
Rb C2
d2 vo
= 1000.
dt2
dvo (0)
, our solution is vo = 500t2 .
dt
The second op-amp will saturate when
[b] Since vo (0) = 0 =
vo = 6 V,
or tsat =
q
6/500 ⇠
= 0.1095 s.
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Problems
dvo1
1
=
vg = 25.
dt
Ra C1
[d] Since vo1 (0) = 0, vo1 =
25t V;
vo1 ⇠
=
2.74 V.
8–45
[c]
At t = 0.1095 s,
Therefore the second amplifier saturates before the first amplifier
saturates. Our expressions are valid for 0  t  0.1095 s. Once the second
op-amp saturates, our linear model is no longer valid.
Part (2) — Example 8.14 with vo1 (0) =
2 V and vo (0) = 4 V:
[a] Initial conditions will not change the di↵erential equation; hence the
equation is the same as Example 8.14.
[b] vo = 5 + A01 e 10t + A02 e 20t
(from Example 8.14);
vo (0) = 4 = 5 + A01 + A02 .
4
+ iC (0+ )
100
iC (0+ ) =
2
= 0;
25
dvo (0+ )
4
mA = C
;
100
dt
0.04 ⇥ 10 3
dvo (0+ )
=
= 40 V/s;
dt
10 6
dvo
=
dt
10A01 e 10t
dvo +
(0 ) =
dt
10A01
20A02 e 20t ;
20A02 = 40.
Therefore A01 2A02 = 4 and A01 + A02 =
Thus, A01 = 2 and A02 = 3.
vo = 5 + 2e 10t
1.
3e 20t V.
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8–46
CHAPTER 8. Natural and Step Responses of RLC Circuits
[c] Same as Example 8.14:
dvo1
+ 20vo1 = 25.
dt
[d] From Example 8.14:
vo1 (1) =
1.25 V;
v1 (0) =
2 V (given).
Therefore
vo1 =
P 8.62
1.25 + ( 2 + 1.25)e 20t =
1.25
0.75e 20t V.
[a]
2C
dva va vg va
+
+
= 0.
dt
R
R
dva
va
vg
+
=
.
dt
RC
2RC
(1) Therefore
0
va
R
+C
d(0
vb )
dt
(2) Therefore
=0
dvb
va
+
= 0,
dt
RC
va =
RC
dvb
.
dt
dvb
d(vb vo )
2vb
+C
+C
= 0.
R
dt
dt
(3) Therefore
From (2) we have
dvb
vb
1 dvo
+
=
.
dt
RC
2 dt
dva
=
dt
RC
d2 vb
dt2
and va =
RC
dvb
.
dt
When these are substituted into (1) we get
(4)
RC
d2 vb
dt2
vg
dvb
=
.
dt
2RC
Now di↵erentiate (3) to get
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Problems
(5)
8–47
d2 vb
1 d2 vo
1 dvb
=
+
.
dt2
RC dt
2 dt2
But from (4) we have
(6)
d2 vb
1 dvb
=
+
2
dt
RC dt
vg
.
2R2 C 2
Now substitute (6) into (5)
d2 vo
=
dt2
vg
.
R2 C 2
d2 vo
vg
= 2 2.
2
dt
R C
The two equations are the same except for a reversal in algebraic sign.
[b] When R1 C1 = R2 C2 = RC :
[c] Two integrations of the input signal with one operational amplifier.
P 8.63
[a]
1
d2 vo
=
vg ;
dt2
R1 C1 R2 C2
10 6
1
=
= 250;
R1 C1 R2 C2
(100)(400)(0.5)(0.2) ⇥ 10 6 ⇥ 10 6
d2 vo
.·.
= 250vg .
dt2
0  t  0.5 :
vg = 80 mV;
d2 vo
= 20.
dt2
Let g(t) =
dvo
,
dt
Z g(t)
Z t
g(0)
dx = 20
0
then
dg
= 20 or dg = 20 dt;
dt
g(0) =
dvo
(0) = 0;
dt
dy;
g(t)
g(0) = 20t,
g(t) =
dvo
= 20t;
dt
dvo = 20t dt;
Z vo (t)
vo (0)
dx = 20
Z t
vo (t) = 10t2 V,
0
x dx;
vo (t)
vo (0) = 10t2 ,
vo (0) = 0;
0  t  0.5 .
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8–48
CHAPTER 8. Natural and Step Responses of RLC Circuits
dvo1
=
dt
1
vg =
R1 C1
dvo1 =
1.6 dt.
Z vo1 (t)
vo1 (0)
dx =
1.6
vo1 (t)
vo1 (0) =
vo1 (t) =
1.6t V,
20vg =
Z t
dy;
0
1.6t,
vo1 (0) = 0;
0  t  0.5 .
0.5+  t  tsat :
d2 vo
=
dt2
10,
let g(t) =
dg(t)
=
dt
10;
dg(t) =
Z g(t)
dx =
g(t)
g(0.5+ ) =
g(0.5+ )
g(0.5+ ) =
C
10
1.6.
Z t
dvo
;
dt
10 dt;
dy;
0.5
10(t
0.5) =
10t + 5;
dvo (0.5+ )
;
dt
0 vo1 (0.5+ )
dvo
(0.5+ ) =
;
dt
400 ⇥ 103
vo1 (0.5+ ) = vo (0.5 ) =
1.6(0.5) =
0.80 V;
0.80
dvo1 (0.5+ )
·
=
= 2 µA.
.. C
dt
0.4 ⇥ 103
2 ⇥ 10 6
dvo1
(0.5+ ) =
= 10 V/s;
dt
0.2 ⇥ 10 6
.·. g(t) =
10t + 5 + 10 =
.·. dvo =
10t dt + 15 dt.
Z vo (t)
dx =
vo (t)
vo (0.5+ ) =
vo (0.5+ )
Z t
0.5+
vo (t) = vo (0.5+ )
10t + 15 =
10y dy +
5y 2
Z t
0.5+
dvo
;
dt
15 dy;
t
t
+ 15y
0.5
5t2 + 1.25 + 15t
;
0.5
7.5;
vo (0.5+ ) = vo (0.5 ) = 2.5 V;
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Problems
.·. vo (t) =
dvo1
=
dt
5t2 + 15t
0.5+  t  tsat ;
20( 0.04) = 0.8,
dvo1 = 0.8 dt;
vo1 (0.5+ )
vo1 (0.5+ ) = 0.8t
vo1 (t)
.·. vo1 (t) = 0.8t
0.5+  t  tsat .
3.75 V,
Z vo1 (t)
8–49
dx = 0.8
Z t
0.4;
vo1 (0.5+ ) = vo1 (0.5 ) =
0.5+
dy;
0.8 V;
0.5+  t  tsat .
1.2 V,
Summary:
0  t  0.5 s :
vo1 =
0.5+ s  t  tsat :
[b]
vo1 = 0.8t
vo = 10t2 V;
1.2 V,
vo =
⌧2 = (5 ⇥ 106 )(0.2 ⇥ 10 6 ) = 1 s;
.·.
12.5 =
5t2sat + 15tsat
.·. 5t2sat
15tsat
Solving,
tsat = 3.5 sec.
5t2 + 15t
3.75 V.
3.75;
8.75 = 0.
vo1 (tsat ) = 0.8(3.5)
P 8.64
1.6t V,
1.2 = 1.6 V.
⌧1 = (106 )(0.5 ⇥ 10 6 ) = 0.50 s;
1
= 2;
⌧1
.·.
1
= 1;
⌧2
dvo
d2 v o
+ 2vo = 20
+
3
dt2
dt
s2 + 3s + 2 = 0.
(s + 1)(s + 2) = 0;
s1 =
vo = Vf + A01 e t + A02 e 2t ;
1,
s2 =
Vf =
2;
20
= 10 V;
2
vo = 10 + A01 e t + A02 e 2t ;
vo (0) = 0 = 10 + A01 + A02 ;
.·. A01 =
20,
dvo
(0) = 0 =
dt
A01
2A02 ;
A02 = 10 V.
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8–50
CHAPTER 8. Natural and Step Responses of RLC Circuits
20e t + 10e 2t V,
vo (t) = 10
0  t  0.5 s.
dvo1
+ 2vo1 =
dt
1.6;
vo (0.5) = 10
20e 0.5 + 10e 1 = 1.55 V;
.·. vo1 =
0.8 + 0.8e 1 =
vo1 (0.5) =
0.8 + 0.8e 2t V,
0  t  0.5 s.
0.51 V.
At t = 0.5 s:
iC =
C
0 + 0.51
= 1.26 µA;
400 ⇥ 103
dvo
1.26
=
= 6.32 V/s.
dt
0.2
dvo
= 1.26 µA;
dt
0.5 s  t  1:
dvo
d2 vo
+2=
+3
2
dt
dt
10;
vo (1) =
5;
.·. vo =
5 + A01 e (t 0.5) + A02 e 2(t 0.5) .
1.55 =
5 + A01 + A02 ;
dvo
(0.5) = 6.32 =
dt
A01
.·. A01 + A02 = 6.55;
2A02 ;
A01
2A02 = 6.32.
Solving,
A01 = 19.42;
.·. vo =
A02 =
12.87;
5 + 19.42e (t 0.5)
12.87e 2(t 0.5) V,
0.5  t  1.
dvo1
+ 2vo1 = 0.8;
dt
.·. vo1 = 0.4 + ( 0.51
0.4)e 2(t 0.5) = 0.4
0.91e 2(t 0.5) V,
0.5  t  1.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 8.65
[a] f (t)
8–51
= inertial force + frictional force + spring force
= M [d2 x/dt2 ] + D[dx/dt] + Kx.
✓
f
d2 x
=
[b]
2
dt
M
Given vA =
D
M
◆
dx
dt
d2 x
,
dt2
!
✓
◆
K
x;
M
then
!
vB =
1 Z t d2 x
R1 C1 0 dy 2
vC =
1 Zt
1
x;
vB dy =
R2 C2 0
R1 R2 C1 C2
vD =
R3
R3 dx
;
· vB =
R4
R4 R1 C1 dt

1 dx
.
R1 C1 dt
dy =

1
R 5 + R6
R 5 + R6
vE =
· x;
vC =
·
R6
R6
R1 R2 C1 C2
vF =

R8
f (t),
R7
Therefore
vA =
(vD + vE + vF ).

d2 x
R8
=
f (t)
2
dt
R7
Therefore M =
R7
,
R8
D=

R3
dx
R4 R1 C1 dt
R3 R7
R8 R4 R1 C1

R 5 + R6
x;
R6 R1 R2 C1 C2
and K =
R7 (R5 + R6 )
.
R8 R6 R1 R2 C1 C2
Box Number Function
P 8.66
[a] !0 =
.·.
s
1
inverting and scaling
2
summing and inverting
3
integrating and scaling
4
integrating and scaling
5
inverting and scaling
6
noninverting and scaling
1
=
LC
f0 =
s
1
(5 ⇥ 10
9 )(2 ⇥ 10 12 )
= 1010 rad/sec;
1010
!0
=
= 1.59 ⇥ 109 Hz = 1.59 GHz.
2⇡
2⇡
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8–52
CHAPTER 8. Natural and Step Responses of RLC Circuits
10
= 0.4 A;
25
1
1
w(0) = LI02 = (5 ⇥ 10 9 )(0.4)2 = 4 ⇥ 10 10 J = 0.4 nJ.
2
2
[c] Because the inductor and capacitor are assumed to be ideal, none of the
initial energy will ever be dissipated, so for all t 0 the 0.4 nJ will be
continually exchanged between the inductor and capacitor.
[b] I0 =
P 8.67
[a] !0 = 2⇡f0 = 2⇡(2 ⇥ 109 ) = 4⇡ ⇥ 109 rad/s;
!02 =
1
;
LC
.·.
C=
1
1
=
= 6.33 ⇥ 10 12 = 6.33 pF.
2
L!0
(10 9 )(4⇡ ⇥ 109 )2
V
4
sin !0 t =
sin 4⇡ ⇥ 109 t
!0 RC
(4⇡ ⇥ 109 )(10)(6.33 ⇥ 10 12 )
[b] vo (t) =
= 5.03 sin 4⇡ ⇥ 109 t V,
P 8.68
[a] ↵ =
0.
R
0.01
=
= 106 rad/s;
2L
2(5 ⇥ 10 9 )
!0 =
s
1
=
LC
[b] !02 > ↵2
[c] !d =
t
q
s
1
(5 ⇥ 10
9 )(2 ⇥ 10 12 )
= 1010 rad/s.
so the response is underdamped.
!02
↵2 =
q
(1010 )2
(106 )2 ⇡ 1010 rad/s;
1010
!0
=
= 1.59 ⇥ 109 Hz = 1.59 GHz.
2⇡
2⇡
Therefore the addition of the 10 m⌦ resistance does not change the
frequency of oscillation.
f0 =
[d] Because of the added resistance, the oscillation now occurs within a
decaying exponential envelope (see Fig. 8.9). The form of the
6
exponential envelope is e ↵t = e 10 t . Let’s assume that the oscillation is
6
described by the function Ke 10 t cos 1010 t, where the maximum
magnitude, K, exists at t = 0. How long does it take before the
magnitude of the oscillation has decayed to 0.01K?
6
Ke 10 t = 0.01K
.·.
t=
so
6
e 10 t = 0.01;
ln 0.01
= 4.6 ⇥ 10 6 s = 4.6 µs.
6
10
Therefore, the oscillations will persist for only 4.6 µs, due to the presence
of a small amount of resistance in the circuit. This is why an LC
oscillator is not used in the clock generator circuit.
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Sinusoidal Steady State Analysis
Assessment Problems
AP 9.1 [a] V = 170/ 40 V.
[b] 10 sin(1000t + 20 ) = 10 cos(1000t
.·.
70 ; )
I = 10/ 70 A.
[c] I = 5/36.87 + 10/ 53.13
= 4 + j3 + 6
j8 = 10
j5 = 11.18/ 26.57 A.
[d] sin(20,000⇡t + 30 ) = cos(20,000⇡t
Thus,
60 ).
100/ 60 = 212.13 + j212.13
V = 300/45
(50
j86.60)
= 162.13 + j298.73 = 339.90/61.51 mV.
AP 9.2 [a] v = 18.6 cos(!t
[b] I = 20/45
=
54 ) V.
50/
30 = 14.14 + j14.14
43.3 + j25
29.16 + j39.14 = 48.81/126.68 .
Therefore i = 48.81 cos(!t + 126.68 ) mA.
[c] V = 20 + j80
=
30/15 = 20 + j80
28.98
j7.76
8.98 + j72.24 = 72.79/97.08 .
v = 72.79 cos(!t + 97.08 ) V.
AP 9.3 [a] !L = (104 )(20 ⇥ 10 3 ) = 200 ⌦.
[b] ZL = j!L = j200 ⌦.
9–1
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9–2
CHAPTER 9. Sinusoidal Steady State Analysis
[c] VL = IZL = (10/30 )(200/90 ) ⇥ 10 3 = 2/120 V.
[d] vL = 2 cos(10,000t + 120 ) V.
1
1
=
= 50 ⌦.
!C
4000(5 ⇥ 10 6 )
[b] ZC = jXC = j50 ⌦.
30/25
V
=
= 0.6/115 A.
[c] I =
ZC
50/ 90
[d] i = 0.6 cos(4000t + 115 ) A.
AP 9.4 [a] XC =
AP 9.5 I1 = 100/25 = 90.63 + j42.26;
I2 = 100/145 =
81.92 + j57.36;
I3 = 100/ 95 =
8.72
I4 =
j99.62;
(I1 + I2 + I3 ) = (0 + j0) A,
AP 9.6 [a] I =
125
125/ 60
/( 60
=
|Z|/✓z
|Z|
But
60
✓Z =
therefore i4 = 0 A.
✓Z ) ;
.·. ✓Z = 45 ;
105
Z = 90 + j160 + jXC ;
.·. XC =
.·. C =
[b] I =
70 ⌦;
XC =
1
=
!C
70;
1
= 2.86 µF.
(70)(5000)
Vs
125/ 60
=
= 0.982/ 105 A;
Z
(90 + j90)
.·. |I| = 0.982 A.
AP 9.7 [a]
! = 2000 rad/s;
!L = 10 ⌦,
1
=
!C
20 ⌦;
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Problems
20(j10)
+5
(20 + j10)
Zxy = 20kj10 + 5 + j20 =
= 4 + j8 + 5
1
=
!C
[b] !L = 40 ⌦,
Zxy = 5
=5
j20 = (9
9–3
j20;
j12) ⌦.
5 ⌦;
j5 + 20kj40 = 5
"
(20)(j40)
j5 +
20 + j40
#
j5 + 16 + j8 = (21 + j3) ⌦.
"
#
20(j!L)
+ 5
[c] Zxy =
20 + j!L
j106
25!
!
20! 2 L2
j400!L
j106
.
+
+
5
400 + ! 2 L2 400 + ! 2 L2
25!
The impedance will be purely resistive when the j terms cancel, i.e.,
=
106
400!L
.
=
400 + ! 2 L2
25!
Solving for ! yields ! = 4000 rad/s.
20! 2 L2
[d] Zxy =
+ 5 = 10 + 5 = 15 ⌦.
400 + ! 2 L2
AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 ⌦ in Assessment
Problem 9.7. Thus,
V = 150/0 ,
Is =
V
150/0
= 10/0 A.
=
Zxy
15
Using current division,
IL =
20
(10) = 5
20 + j20
iL = 7.07 cos(4000t
j5 = 7.07/ 45 A;
45 ) A,
Im = 7.07 A.
AP 9.9 After replacing the wye made up of the j40 ⌦, 50 ⌦, and 40 ⌦ impedances with
its equivalent delta, the circuit becomes
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9–4
CHAPTER 9. Sinusoidal Steady State Analysis
where
Za =
j40(50) + 50(40) + 40(j40)
= 90
j40
Zb =
j40(50) + 50(40) + 40(j40)
= 40 + j72 ⌦;
50
Zc =
j40(50) + 50(40) + 40(j40)
= 50 + j90 ⌦.
40
j50 ⌦;
The circuit is further simplified by combining the impedances to the right of
the 14 ⌦ resistor into a single equivalent impedance:
Zeq = Zb k[(Zc k
j15) + (Za k10)] = (16
Therefore I =
136/0
= 4/28.07 A.
14 + 16 j16
j16) ⌦.
AP 9.10
V1 = 240/53.13 = 144 + j192 V;
V2 = 96/ 90 =
j96 V;
j!L = j(4000)(15 ⇥ 10 3 ) = j60 ⌦;
1
=
j!C
j
6 ⇥ 106
=
(4000)(25)
j60 ⌦.
Perform a source transformation:
144 + j192
V1
=
= 3.2
j60
j60
V2
=
20
j
96
=
20
j2.4 A;
j4.8 A.
Combine the parallel impedances:
Y =
1
1
+
+
j60 30
1
1
j5
1
+
=
= ;
j60 20
j60
12
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Problems
Z=
9–5
1
= 12 ⌦.
Y
Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87 V;
vo = 48 cos(4000t + 36.87 ) V.
AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the
20 ⌦ resistor and VTh = node voltage across the capacitor. Writing the node
voltage equations gives us
V1
20
2/45 +
V1
j10
10Ix
= 0 and VTh =
(10Ix ).
j10
10 j10
We also have
Ix =
V1
.
20
Solving these equations for VTh gives VTh = 10/45 V. To find the Thévenin
impedance, we remove the independent current source and apply a test
voltage source at the terminals a, b. Thus
It follows from the circuit that
10Ix = (20 + j10)Ix .
Therefore
Ix = 0 and IT =
ZTh =
VT
,
IT
VT
VT
+
j10 10.
therefore ZTh = (5
j5) ⌦.
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9–6
CHAPTER 9. Sinusoidal Steady State Analysis
AP 9.12 The phasor domain circuit is as shown in the following diagram:
The node voltage equation is
10 +
V
+
5
100/ 90
= 0;
20
V
V V
+
+
j(20/9) j5
j30 = 31.62/ 71.57 .
Solving, V = 10
Therefore v = 31.62 cos(50,000t
71.57 ) V.
AP 9.13 Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to
right. Summing the voltages around meshes a and b gives
33.8 = (1 + j2)Ia + (3
j5)(Ia
Ib )
and
0 = (3
j5)(Ib
Ia ) + 2(Ib
Ic ).
But
Vx =
j5(Ia
Ib ),
therefore
Ic =
0.75[ j5(Ia
Ib )].
Solving for I = Ia = 29 + j2 = 29.07/3.95 A.
p
!M = 80 ⌦;
AP 9.14 [a] M = 0.4 0.0625 = 0.1 H,
Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) ⌦.
Therefore |Z22 | = 500 ⌦,
✓
80
Zr =
500
◆2
(400
⇤
Z22
= (400
j300) = (10.24
j300) ⌦.
j7.68) ⌦.
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Problems
[b] I1 =
245.20
= 0.50/
184 + 100 + j400 + Z⌧
i1 = 0.5 cos(800t
✓
9–7
53.13 A;
53.13 ) A.
◆
j80
j!M
[c] I2 =
I1 =
(0.5/
Z22
500/36.87
53.13 ) = 0.08/0 A;
i2 = 80 cos 800t mA.
AP 9.15
I1 =
25 ⇥ 103 /0
Vs
=
Z1 + 2s2 Z2
1500 + j6000 + (25)2 (4
j14.4)
= 4 + j3 = 5/36.87 A;
V1 = Vs
Z1 I1 = 25,000/0
= 37,000
V2 =
I2 =
1
V1 =
25
(4 + j3)(1500 + j6000)
j28,500;
1480 + j1140 = 1868.15/142.39 V;
V2
1868.15/142.39
= 125/216.87 A.
=
Z2
4 j14.4
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9–8
CHAPTER 9. Sinusoidal Steady State Analysis
Problems
P 9.1
[a] 40 V.
[b] 2⇡f = 100⇡;
f = 50Hz.
[c] ! = 100⇡ = 314.159 rad/s.
⇡
2⇡
(60 ) = = 1.05 rad.
[d] ✓(rad) =
360
3
[e] ✓ = 60 .
1
1
= 20 ms.
[f ] T = =
f
50
[g] v = 40 when
⇡
100⇡t + = ⇡;
.·. t = 6.67 ms.
3

✓
◆
0.01
⇡
[h] v = 40 cos 100⇡ t
+
3
3
= 40 cos[100⇡t
(⇡/3) + (⇡/3)]
= 40 cos 100⇡t V.
[i] 100⇡(t + to ) + (⇡/3) = 100⇡t + (3⇡/2);
.·. 100⇡to =
7⇡
;
6
to = 11.67 ms.
11.67 ms to the left.
P 9.2
[a] ! = 2⇡f = 3769.91 rad/s,
f=
!
= 600 Hz.
2⇡
[b] T = 1/f = 1.67 ms.
[c] Vm = 10 V.
[d] v(0) = 10 cos( 53.13 ) = 6 V.
53.13 (2⇡)
[e] = 53.13 ;
= 0.9273 rad.
=
360
[f ] v(t) = 0 when 3769.19t 53.13 = 90 . Now resolve the units:
(3769.19 rad/s)t =
143.13
= 2.498 rad,
57.3 /rad
[g] (dv/dt) = ( 10)3769.91 sin(3769.91t
(dv/dt) = 0 when 3769.91t
or
3769.91t =
t = 662.62 µs.
53.13 ; )
53.13 = 0 ,
53.13
= 0.9273 rad;
57.3 /rad
Therefore t = 245.97 µs.
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Problems
P 9.3
[a]
9–9
T
= 8 + 2 = 10 ms;
T = 20 ms;
2
1
1
= 50 Hz.
f= =
T
20 ⇥ 10 3
[b] v = Vm sin(!t + ✓);
! = 2⇡f = 100⇡ rad/s;
⇡
.·. ✓ = rad = 36 ;
5
100⇡( 2 ⇥ 10 3 ) + ✓ = 0;
v = Vm sin[100⇡t + 36 ];
80.9 = Vm sin 36 ;
Vm = 137.64 V;
v = 137.64 sin(100⇡t + 36 ) = 137.64 cos(100⇡t
54 ) V.
P 9.4
[a] Left as
becomes more positive.
[b] Right.
P 9.5
[a] By hypothesis
v = 80 cos(!t + ✓);
dv
= 80! sin(!t + ✓);
dt
.·. 80! = 80,000;
! = 1000 rad/s.
[b] f =
!
= 159.155 Hz;
2⇡
2⇡/3
=
6.28
0.3335,
T =
1
= 6.28 ms;
f
.·. ✓ =
90
( 0.3335)(360) = 30 ;
.·. v = 80 cos(1000t + 30 ) V.
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9–10
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.6
Vm =
P 9.7
Z to +T
to
p
2Vrms =
p
2(120) = 169.71 V.
Vm2 cos2 (!t + ) dt = Vm2
=
P 9.8
Vrms =
Z T /2
0
s
Vm2
t
1 1
+ cos(2!t + 2 ) dt
2 2
(o
R to +T
)
Z to +T
dt +
cos(2!t + 2 ) dt
to
2
to
⇢
i
Vm2
1 h
=
T+
sin(2!t + 2 ) |ttoo +T
2
2!
2 ⇢
V
1
[sin(2!to + 4⇡ + 2 ) sin(2!to + 2 )]
= m T+
2
2!
✓ ◆
✓ ◆
1
T
T
= Vm2
(0) = Vm2
+
.
2
2!
2
1 Z T /2 2 2 2⇡
t dt;
Vm sin
T 0
T
Vm2 sin2
✓
◆
✓
Vm2 Z T /2
2⇡
t dt =
1
T
2 0
Therefore Vrms =
P 9.9
Z to +T
s
◆
V 2T
4⇡
cos t dt = m ;
T
4
Vm
1 Vm2 T
=
.
T 4
2
[a] From Eq. 9.7 we have
L
Vm R cos(
di
✓) (R/L)t
= p 2
e
2
2
dt
R +! L
!LVm sin(!t +
p
R 2 + ! 2 L2
✓)
;
Vm R cos(
✓)e (R/L)t Vm R cos(!t +
✓)
p
p
+
;
2
2
2
2
2
2
R +! L
R +! L
Ri =
"
di
R cos(!t +
L + Ri = Vm
dt
✓) !L sin(!t +
p
R 2 + ! 2 L2
✓)
#
.
But
p
R
R 2 + ! 2 L2
= cos ✓
and
p
!L
R 2 + ! 2 L2
= sin ✓.
Therefore the right-hand side reduces to
Vm cos(!t + ).
At t = 0, Eq. 9.7 reduces to
i(0) =
V cos(
✓)
✓) Vm cos(
pm
+ p 2
= 0.
2
2
2
2
R +! L
R + ! L2
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Problems
[b] iss = p
Vm
R 2 + ! 2 L2
cos(!t +
9–11
✓).
Therefore
diss
!LVm
L
=p 2
sin(!t +
dt
R + ! 2 L2
✓)
and
Riss = p
Vm R
cos(!t +
✓).
"
✓) !L sin(!t +
p
R 2 + ! 2 L2
R 2 + ! 2 L2
diss
R cos(!t +
+ Riss = Vm
L
dt
✓)
#
t
0.
= Vm cos(!t + ).
P 9.10
[a] The numerical values of the terms in Eq. 9.6 are
Vm = 20,
R/L = 1066.67,
p
R2 + ! 2 L2 = 100;
!L = 60;
✓ = tan 1 60/80,
= 25 ,
✓ = 36.87 .
Substitute these values into Equation 9.7:
i=
h
195.72e 1066.67t + 200 cos(800t
i
11.87 ) mA,
[b] Transient component = 195.72e 1066.67t mA;
Steady-state component = 200 cos(800t 11.87 ) mA.
[c] By direct substitution into the equation in part (a),
i(1.875 ms) = 28.39 mA.
[d] 200 mA,
800 rad/s,
11.87 .
[e] The current lags the voltage by 36.87 .
P 9.11
[a] Y = 50/60 + 100/
y = 111.8 cos(500t
[b] Y = 200/50
30 = 111.8/
3.43 ;
3.43 ).
100/60 = 102.99/40.29 ;
y = 102.99 cos(377t + 40.29 ).
[c] Y = 80/30
100/
225 + 50/
y = 161.59 cos(100t
29.96 ).
[d] Y = 250/0 + 250/120 + 250/
90 = 161.59/
29.96 ;
120 = 0;
y = 0.
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9–12
P 9.12
CHAPTER 9. Sinusoidal Steady State Analysis
[a] Vg = 300/78 ;
.·. Z =
Ig = 6/33 ;
300/78
Vg
=
= 50/45 ⌦.
Ig
6/33
[b] ig lags vg by 45 :
P 9.13
2⇡f = 5000⇡;
f = 2500 Hz;
T = 1/f = 400 µs;
.·. ig lags vg by
45
(400 µs) = 50 µs.
360
[a] 400 Hz.
[b] ✓v = 0 ;
I=
100
100/0
/
=
j!L
!L
90 ;
✓i =
90 .
100
= 20;
!L = 5 ⌦.
!L
5
= 1.99 mH.
[d] L =
800⇡
[e] ZL = j!L = j5 ⌦.
[c]
P 9.14
[a] ! = 2⇡f = 314,159.27 rad/s.
10 ⇥ 10 3 /0
V
= j!C(10 ⇥ 10 3 )/0 = 10 ⇥ 10 3 !C /90 ;
=
[b] I =
ZC
1/j!C
.·. ✓i = 90 .
[c] 628.32 ⇥ 10 6 = 10 ⇥ 10 3 !C;
10 ⇥ 10 3
1
=
= 15.92 ⌦,
!C
628.32 ⇥ 10 6
[d] C =
.·. XC =
15.92 ⌦.
1
1
=
;
15.92(!)
(15.92)(100⇡ ⇥ 103 )
C = 0.2 µF.
✓
1
[e] Zc = j
!C
P 9.15
◆
=
j15.92 ⌦.
[a] ZL = j(8000)(5 ⇥ 10 3 ) = j40 ⌦;
ZC =
j
=
(8000)(1.25 ⇥ 10 6 )
j100 ⌦.
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Problems
9–13
600/20
= 8.32/76.31 A.
40 + j40 j100
[c] i = 8.32 cos(8000t + 76.31 ) A.
[b] I =
P 9.16
[a] j!L = j(2 ⇥ 104 )(300 ⇥ 10 6 ) = j6 ⌦;
1
=
j!C
j
1
(2 ⇥ 104 )(5 ⇥ 10 6 )
=
j10 ⌦;
Ig = 922/30 A.
[b] Vo = 922/30 Ze ;
Ze =
1
;
Ye
Ye =
1
1
1
+j +
;
10
10 8 + j6
Ye = 0.18 + j0.04 S;
Ze =
1
= 5.42/
0.18 + j0.04
Vo = (922/30 )(5.42/
12.53 ⌦;
12.53 ) = 5000.25/17.47 V.
[c] vo = 5000.25 cos(2 ⇥ 104 t + 17.47 ) V.
P 9.17
[a] Y =
1
1
1
+
+
3 + j4 16 j12
j4
= 0.12
j0.16 + 0.04 + j0.03 + j0.25
= 160 + j120 = 200/36.87 mS.
[b] G = 160 mS.
[c] B = 120 mS.
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9–14
CHAPTER 9. Sinusoidal Steady State Analysis
[d] I = 8/0 A,
V=
8
I
=
= 40/ 36.87 V;
Y
0.2/36.87
40/ 36.87
V
=
= 10/53.13 A;
ZC
4/ 90
IC =
iC = 10 cos(!t + 53.13 ) A,
P 9.18
Im = 10 A.
[a] Z1 = R1 + j!L1 ;
R2 (j!L2 )
! 2 L22 R2 + j!L2 R22
=
;
R2 + j!L2
R22 + ! 2 L22
Z2 =
Z1 = Z2
! 2 L22 R2
when R1 = 2
R2 + ! 2 L22
R22 L2
and L1 = 2
.
R2 + ! 2 L22
(4000)2 (1.25)2 (5000)
= 2500 ⌦;
[b] R1 =
50002 + 40002 (1.25)2
(5000)2 (1.25)
L1 =
= 625 mH.
50002 + 40002 (1.25)2
P 9.19
[a] Y2 =
1
R2
j
;
!L2
1
R1 j!L1
= 2
.
R1 + j!L1
R1 + ! 2 L21
Y1 =
Therefore
R2 =
P 9.20
Y 2 = Y1
R12 + ! 2 L21
R1
when
and L2 =
R12 + ! 2 L21
.
! 2 L1
[b] R2 =
80002 + 10002 (4)2
= 10 k⌦;
8000
L2 =
80002 + 10002 (4)2
= 20 H.
10002 (4)
[a] Z1 = R1
Z2 =
j
1
;
!C1
R2
R2 /j!C2
R2 j!R22 C2
=
=
;
R2 + (1/j!C2 )
1 + j!R2 C2
1 + ! 2 R22 C22
Z1 = Z 2
when R1 =
!R22 C2
1
=
!C1
1 + ! 2 R22 C22
R2
1 + ! 2 R22 C22
or C1 =
and
1 + ! 2 R22 C22
.
! 2 R22 C2
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Problems
1000
[b] R1 =
1 + (40 ⇥ 103 )2 (1000)2 (50 ⇥ 10 4 )2
C1 =
P 9.21
[a] Y2 =
9–15
= 200 ⌦;
1 + (40 ⇥ 103 )2 (1000)2 (50 ⇥ 10 9 )2
= 62.5 nF.
(40 ⇥ 103 )2 (1000)2 (50 ⇥ 10 9 )
1
+ j!C2 ;
R2
Y1 =
j!C1
! 2 R1 C12 + j!C1
1
=
=
;
R1 + (1/j!C1 )
1 + j!R1 C1
1 + ! 2 R12 C12
Therefore
R2 =
Y1 = Y2
1 + ! 2 R12 C12
! 2 R1 C12
when
and C2 =
C1
.
1 + ! 2 R12 C12
1 + (50 ⇥ 103 )2 (1000)2 (40 ⇥ 10 9 )2
= 1250 ⌦;
[b] R2 =
(50 ⇥ 103 )2 (1000)(40 ⇥ 10 9 )2
C2 =
P 9.22
40 ⇥ 10 9
= 8 nF.
1 + (50 ⇥ 103 )2 (1000)2 (40 ⇥ 10 9 )2
[a] R = 300 ⌦ = 120 ⌦ + 180 ⌦;
!L
1
=
!C
400 so 10,000L
1
=
10,000C
400.
Choose L = 10 mH. Then,
1
1
= 100 + 400 so C =
= 0.2 µF.
10,000C
10,000(500)
We can achieve the desired capacitance by combining two 0.1 µF
capacitors in parallel. The final circuit is shown here:
[b] 0.01! =
1
!(0.2 ⇥ 10 6 )
so ! 2 =
1
= 5 ⇥ 108 ;
0.01(0.2 ⇥ 10 6 )
.·. ! = 22,360.7 rad/s.
P 9.23
[a] Using the notation and results from Problem 9.19:
RkL = 40 + j20 so R1 = 40,
L1 =
20
= 4 mH;
5000
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9–16
CHAPTER 9. Sinusoidal Steady State Analysis
R2 =
402 + 50002 (0.004)2
= 50 ⌦;
40
L2 =
402 + 50002 (0.004)2
= 20 mH;
50002 (0.004)
R2 kj!L2 = 50kj100 = 40 + j20 ⌦. (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
[b] Using the notation and results from Problem 9.21:
RkC = 40
j20 so R1 = 40,
C1 = 10 µF;
1 + 50002 (40)2 (10 µ)2
R2 =
= 50 ⌦;
50002 (40)(10 µ)2
C2 =
10 µ
1 + 50002 (40)2 (10 µ)2
= 2 µF;
R2 k( j/!C2 ) = 50k( j100) = 40
j20 ⌦. (checks)
The circuit, using combinations of components from Appendix H, is
shown here:
P 9.24
[a] (40 + j20)k( j/!C) = 50kj100k( j/!C).
To cancel out the j100 ⌦ impedance, the capacitive impedance must be
j100 ⌦:
j
=
5000C
j100 so C =
1
= 2 µF.
(100)(5000)
Check:
Rkj!Lk( j/!C) = 50kj100k( j100) = 50 ⌦.
Create the equivalent of a 2 µF capacitor from components in Appendix
H by combining two 1 µF capacitors in parallel.
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Problems
9–17
[b] (40 j20)k(j!L) = 50k( j100)k(j!L).
To cancel out the j100 ⌦ impedance, the inductive impedance must be
j100 ⌦:
j5000L = j100 so L =
100
= 20 mH.
5000
Check:
Rkj!Lk( j/!C) = 50kj100k( j100) = 50 ⌦.
Create the equivalent of a 20 mH inductor from components in Appendix
H by combining two 10 mH inductors in series.
P 9.25
First find the admittance of the parallel branches
Yp =
Zp =
1
6
j2
+
1
1
1
+ +
= 0.375
4 + j12 5 j10
j0.125 S;
1
1
= 2.4 + j0.8 ⌦;
=
Yp
0.375 j0.125
Zab =
j12.8 + 2.4 + j0.8 + 13.6 = 16
Yab =
1
1
= 0.04 + j0.03 S
=
Zab
16 j12
j12 ⌦;
= 40 + j30 mS = 50/36.87 mS.
P 9.26
Zab = 1
=1
P 9.27
j8 + (2 + j4)k(10
j20) + (40kj20)
j8 + 3 + j4 + 8 + j16 = 12 + j12 ⌦.
[a] j!L = Rk( j/!C) = j!L +
j!L +
j!L +
jR
!CR j
jR(!CR + j)
;
! 2 C 2 R2 + 1
Im(Zab ) = !L
.·.
L=
jR/!C
R j/!C
!CR2
= 0;
! 2 C 2 R2 + 1
CR2
;
! 2 C 2 R2 + 1
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9–18
CHAPTER 9. Sinusoidal Steady State Analysis
.·.
! 2 C 2 R2 + 1 =
.·.
(CR2 /L)
! =
C 2 R2
CR2
;
L
1
2
(25⇥10 9 )(100)2
1
160⇥10 6
=
= 900 ⇥ 108 .
(25 ⇥ 10 9 )2 (100)2
! = 300 krad/s.
[b] Zab (300 ⇥ 103 ) = j48 +
P 9.28
(100)( j133.33)
= 64 ⌦.
100 j133.33
1
1
=
=
6
j!C
(1 ⇥ 10 )(50 ⇥ 103 )
j20 ⌦;
j!L = j50 ⇥ 103 (1.2 ⇥ 10 3 ) = j60 ⌦;
Vg = 40/
Ze =
Ig =
90 =
j40 V.
j20 + 30kj60 = 24
j40
= 0.5
24 j8
Vo = (30kj60)Ig =
j1.5 A;
30(j60)
(0.5
30 + j60
vo = 42.43 cos(50,000t
P 9.29
Z = 4 + j(50)(0.24)
Io =
j8 ⌦;
j30 = 42.43/
45 V;
45 ) V.
j
1
= 5.66/45 ⌦;
(50)(0.0025)
0.1/ 90
V
=
= 17.67/
Z
5.66/45
io (t) = 17.67 cos(50t
j1.5) = 30
135 mA;
135 ) mA.
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Problems
P 9.30
Vs = 25/
1
=
j!C
9–19
90 V;
j20 ⌦;
j!L = j10 ⌦.
Zeq = 5 + j10k(10 + 20k
Io =
Vo
25/ 90
=
=
Zeq
10 + j10
io = 1.77 cos(4000t
P 9.31
j20) = 10 + j10 ⌦;
1.25
j1.25 = 1.77/
135 A;
135 ) A.
ZL = j(2000)(60 ⇥ 10 3 ) = j120 ⌦;
ZC =
j
=
(2000)(12.5 ⇥ 10 6 )
j40 ⌦.
Construct the phasor domain equivalent circuit:
Using current division:
I=
(120 j40)
(0.5) = 0.25
120 j40 + 40 + j120
j0.25 A;
Vo = j120I = 30 + j30 = 42.43/45 ;
vo = 42.43 cos(2000t + 45 ) V.
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9–20
P 9.32
CHAPTER 9. Sinusoidal Steady State Analysis
[a]
Vb = (2000
Ia =
Ic =
j1000)(0.025) = 50
50 j25
= 60
500 + j250
50
j25 V;
j80 mA = 100/
53.13 mA;
j25 + j50
= 50 + j25 mA = 55.9/26.57 mA;
1000
Ig = Ia + Ib + Ic = 135
[b] ia = 100 cos(1500t
j55 mA = 145.77/
22.17 mA.
53.13 ) mA;
ic = 55.9 cos(1500t + 26.57 ) mA;
ig = 145.77 cos(1500t
P 9.33
[a]
1
=
j!C
22.17 ) mA.
j250 ⌦;
j!L = j200 ⌦;
Ze = j200k(100 + 500k
j250) = 200 + j200 ⌦;
Ig = 0.025/0 ;
Vg = Ig Ze = 0.025(200 + j200) = 5 + j5 V.
Vo =
100
200
j200
(5 + j5) = 5 + j2.5 = 5.59/26.57 V;
j200
vo = 5.59 cos(50,000t + 26.57 ) V.
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Problems
[b] ! = 2⇡f = 50,000;
f=
9–21
25,000
;
⇡
T =
1
⇡
=
= 40⇡ µs;
f
25,000
.·.
26.57
(40⇡ µs) = 9.27 µs;
360
.·. vo leads ig by 9.27 µs.
P 9.34
Va
(100
20
j50)
+
Va Va (140 + j30)
+
= 0.
j5
12 + j16
Solving,
Va = 40 + j30 V;
IZ + (30 + j20)
140 + j30 (40 + j30) (140 + j30)
+
= 0.
j10
12 + j16
Solving,
IZ =
Z=
30
j10 A;
(100
j50)
30
(140 + j30)
= 2 + j2 ⌦.
j10
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9–22
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.35
V1 = j5( j2) = 10 V;
25 + 10 + (4
I1 =
15
= 2.4 + j1.8 A;
4 j3
I b = I1
j5 = (2.4 + j1.8)
j5 = 2.4
j3.2 A;
VZ =
j5I2 + (4
j5(2.4
j3.2) + (4
j3)I1 =
25 + (1 + j3)I3 + ( 1
IZ = I3
Z=
P 9.36
.·.
j3)I1 = 0;
I2 = (6.2
1
VZ
=
IZ
3.8
I3 = 6.2
j3.2) = 3.8
j12 V;
j6.6 A;
j3.4 A;
j1.88 ⌦.
53.13 ⌦ = 3000
j4000 ⌦;
!
32 ⇥ 106
;
!
Z = 3000 + j !
32 ⇥ 106
=
!
.·. ! 2 + 4000!
(2.4
.·.
1
Ig = 0.1/83.13 mA;
Vg
= 5000/
Ig
.·. !
j6.6)
j12
= 1.42
j3.4
Vg = 500/30 V;
Z=
j12) = 0
j3)(2.4 + j1.8) =
4000;
32 ⇥ 106 = 0;
.·. ! = 4000 rad/s.
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Problems
P 9.37
9–23
[a] The voltage and current are in phase when the impedance to the left of
the 1 k⌦ resistor is purely real:
R + j!L
1
k(R + j!L) =
j!C
1 + j!RC ! 2 LC
Zeq =
=
(R + j!L)(1 ! 2 LC j!RC)
.
(1 + ! 2 LC)2 + ! 2 R2 C 2
The denominator in the above expression is purely real, so set the
imaginary part of the expression’s numerator to zero and solve for !:
!R2 C + !L
!2 =
L
.·.
! 3 L2 C = 0;
(0.01) 2402 (62.5 ⇥ 10 9 )
R2 C
=
= 1024 ⇥ 106 ;
2
2
9
LC
(0.01) (62.5 ⇥ 10 )
! = 32,000 rad/s.
[b] Zt = 1000 + ( j500)k(240 + j320) = 1666.67 ⌦;
15/0
Vo
= 9/0 mA;
=
ZT
1666.67
Io =
io = 9 cos 32,000t mA.
P 9.38
[a] In order for vg and ig to be in phase, the impedance to the right of the
500 ⌦ resistor must be purely real:
Zeq = j!Lk(R + 1/!C) =
=
j!L(R + 1/j!C)
j!L + R + 1/j!C
j!L(j!RC + 1)
j!RC ! 2 LC + 1
( ! 2 RLC + j!L)(1 ! 2 LC j!RC)
.
=
(1 ! 2 LC + j!RC)(1 ! 2 LC j!RC)
The denominator of the above expression is purely real. Now set the
imaginary part of the numerator in that expression to zero and solve for
!:
!L(1
! 2 LC) + ! 3 R2 LC 2 = 0.
So
!2 =
.·.
! = 2500 rad/s
1
1
=
= 6,250,000;
2
2
6
LC R C
(0.2)(10 ) 2002 (10 6 )2
and
f = 397.9 Hz.
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9–24
CHAPTER 9. Sinusoidal Steady State Analysis
[b] Zeq = 500 + j500k(200
j400) = 1500 ⌦;
90/0
= 60/0 mA;
1500
Ig =
ig (t) = 60 cos 2500t mA.
P 9.39
[a] For ig and vg to be in phase, the impedance to the right of the 480 ⌦
resistor must be purely real.
j!L +
(1/j!C)(200)
200
= ⌘!L +
(1/j!C) + 200
1 + 200j!C
= j!L +
=
200(1 200j!C)
1 + 2002 ! 2 C 2
j!L(1 + 2002 ! 2 C 2 ) + 200(1
1 + 2002 ! 2 C 2
200j!C)
.
In the above expression the denominator is purely real. So set the
imaginary part of the numerator to zero and solve for !:
!L(1 + 2002 ! 2 C 2 )
2002 !C = 0;
!2 =
2002 (3.125 ⇥ 10 6 ) 0.1
2002 C L
=
= 640,000;
2002 C 2 L
2002 (3.125 ⇥ 10 6 )2 (0.1)
.·.
! = 800 rad/s.
[b] When ! = 800 rad/s
Zg = 480k[j80 + (200k
j400)] = 120 ⌦;
.·. Vg = Zg Ig = 120(0.06) = 7.2 V;
vo = 7.2 cos 800t V.
P 9.40
R
R
j!C
=
[a] Zp =
R + (1/j!C)
1 + j!RC
=
10,000
10,000
=
1 + j(5000)(10,000)C
1 + j50 ⇥ 106 C
=
10,000(1 j50 ⇥ 106 C)
1 + 25 ⇥ 1014 C 2
=
10,000
1 + 25 ⇥ 1014 C 2
j
5 ⇥ 1011 C
.
1 + 25 ⇥ 1014 C 2
j!L = j5000(0.8) = j4000;
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Problems
.·. 4000 =
.·. 1014 C 2
.·. C 2
9–25
5 ⇥ 1011 C
;
1 + 25 ⇥ 1014 C 2
125 ⇥ 106 C + 1 = 0;
5 ⇥ 10 8 C + 4 ⇥ 10 16 = 0.
Solving,
C1 = 40 nF
C2 = 10 nF.
10,000
[b] Re =
.
1 + 25 ⇥ 1014 C 2
When C = 40 nF
Re = 2000 ⌦;
80/0
= 40/0 mA;
ig = 40 cos 5000t mA.
2000
When C = 10 nF
Re = 8000 ⌦;
Ig =
Ig =
P 9.41
80/0
= 10/0 mA;
8000
[a] Z1 = 400
j
106
= 400
500(2.5)
ig = 10 cos 5000t mA.
j800 ⌦;
Z2 = 2000kj500L =
j106 L
;
2000 + j500L
ZT = Z1 + Z2 = 400
j800 +
= 400
500 ⇥ 106 L2
20002 + 5002 L2
j106 L
2000 + j500L
j800 + j
2 ⇥ 109 L
.
20002 + 5002 L2
ZT is resistive when
2 ⇥ 109 L
= 800 or 5002 L2
20002 + 5002 L2
Solving, L1 = 8 H and L2 = 2 H.
25 ⇥ 105 L + 20002 = 0.
[b] When L = 8 H:
ZT = 400 +
Ig =
500 ⇥ 106 (8)2
= 2000 ⌦;
20002 + 5002 (8)2
200/0
= 100/0 mA;
2000
ig = 100 cos 500t mA.
When L = 2 H:
500 ⇥ 106 (2)2
ZT = 400 +
= 800 ⌦;
20002 + 500(2)2
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9–26
CHAPTER 9. Sinusoidal Steady State Analysis
Ig =
200/0
= 250/0 mA;
800
ig = 250 cos 500t mA.
P 9.42
Simplify the top triangle using series and parallel combinations:
(1 + j1)k(1
j1) = 1 ⌦.
Convert the lower left delta to a wye:
Z1 =
(j1)(1)
= j1 ⌦;
1 + j1 j1
Z2 =
( j1)(1)
=
1 + j1 j1
Z3 =
(j1)( j1)
= 1 ⌦.
1 + j1 j1
j1 ⌦;
Convert the lower right delta to a wye:
Z4 =
( j1)(1)
=
1 + j1 j1
Z5 =
( j1)(j1)
= 1 ⌦;
1 + j1 j1
Z6 =
(j1)(1)
= j1 ⌦.
1 + j1 j1
j1 ⌦;
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Problems
9–27
The resulting circuit is shown below:
Simplify the middle portion of the circuit by making series and parallel
combinations:
P 9.43
(1 + j1
j1)k(1 + 1) = 1k2 = 2/3 ⌦;
Zab =
j1 + 2/3 + j1 = 2/3 ⌦.
Step 1 to Step 2:
j60
j48 = j12 ⌦;
240
=
j12
j20 A.
Step 2 to Step 3:
j12k36 = (3.6 + j10.8) ⌦;
P 9.44
( j20)(3.6 + j10.8) = (216
j72) V.
Step 1 to Step 2:
(4)(50) = 200 V.
Step 2 to Step 3:
50 + 30 + j60 = 80 + j60 ⌦;
200
= 1.6
80 + j60
j1.2 A.
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9–28
CHAPTER 9. Sinusoidal Steady State Analysis
Step 3 to Step 4:
ZN = (80 + j60)k
P 9.45
j100 = 100
j50 ⌦;
IN = 1.6
j1.2 A.
[a] j!L = j(400)(400) ⇥ 10 3 = j160 ⌦;
j
1
=
=
j!C
(400)(31.25 ⇥ 10 6 )
j80 ⌦.
Using voltage division,
Vab =
j80kj160
( j50) =
320 + ( j80kj160)
VTh = Vab =
20
20
j10 V;
j10 V.
[b] Remove the voltage source and combine impedances in parallel to find
ZTh = Zab :
ZTh = Zab = 320k
j80kj160 = 64
j128 ⌦.
[c]
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Problems
P 9.46
9–29
Open circuit voltage:
j20Ia + 40Ia + 20(Ia
0.4
j0.2) = 0.
Solving,
Ia =
20(0.4 + j0.2)
= 0.1 + j0.1 A;
60 j20
Voc = 40Ia + j16(0.4 + j0.2) = 0.8 + j10.4 V.
Short circuit current:
j20Ia + 40(Ia
Isc ) + 20(Ia
Ia ) + j16(Isc
40(Isc
0.4
0.4
j0.2) = 0;
j0.2) = 0.
Solving,
IN = Isc = 0.3 + j0.5 A;
ZTh =
VTh
0.8 + j10.4
= 16 + j8 ⌦.
=
IN
0.3 + j0.5
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9–30
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.47
IrmN =
IrmN =
5
j15
ZN
j3) mA,
18 j13.5
+ +4.5
ZrmN
5 j15
+1
ZrmN
j3 =
23 j15
= 3.5
ZrmN
IrmN =
+ (1
j3
5 j15
+1
4 + j3
ZrmN in k⌦.
j6 mA,
18 j13.5
+ (4.5
ZrmN
ZrmN in k⌦;
j6);
.·.
ZrmN = 4 + j3 k⌦;
j3 =
j6 mA.
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Problems
P 9.48
9–31
Short circuit current
Vx
j10(I1
5Vx
Isc ) =
j10(Isc
40 + j40;
I1 ) = 0;
Vx = 10I1 .
Solving,
IrmN = Isc = 6 + j4 A.
Open circuit voltage
I=
40 + j40
=
10 j10
4 A;
Vx = 10I =
40 V;
Voc = 5Vx
j10I =
ZrmN =
200 + j40
=
6 + j4
200 + j40 V;
20 + j20 ⌦.
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9–32
P 9.49
CHAPTER 9. Sinusoidal Steady State Analysis
Open circuit voltage:
V2 15 V2
V2
+ 88I +
= 0;
10
j50
I =
5
(V2 /5)
.
200
Solving,
V2 =
66 + j88 = 110/126.87 V = VTh .
Find the Thévenin (Norton) equivalent impedance using a test source:
IT =
VT
0.8VT
+ 88I +
;
10
j50
I =
VT /5
;
200
!
1/5
0.8
88
+
;
200
j50
IT = VT
1
10
.·.
ZN =
VT
= 30
IT
IN =
VTh
66 + j88
=
=
ZN
30 j40
j40.
2.2 + j0 A.
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Problems
9–33
The Norton equivalent circuit:
P 9.50
Open circuit voltage:
V1
= 0;
50 j100
V1 250
20 + j10
0.03Vo +
.·. Vo =
j100
V1 .
50 j100
j3V1
V1
250
V1
+
+
=
;
20 + j10 50 j100 50 j100
20 + j10
V1 = 500
j250 V;
Vo = 300
j400 V = VTh .
Short circuit current:
Isc =
250/0
= 3.5
70 + j10
j0.5 A;
VTh
300
=
Isc
3.5
j400
= 100
j0.5
ZTh =
j100 ⌦.
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9–34
CHAPTER 9. Sinusoidal Steady State Analysis
The Thévenin equivalent circuit:
P 9.51
! = 2⇡(200/⇡) = 400 rad/s;
Zc =
j
=
400(10 6 )
VT = (10,000
ZTh =
P 9.52
j2500 ⌦.
j2500)IT + 100(200)IT ;
VT
= 30
IT
j2.5 k⌦.
j!L = j100 ⇥ 103 (0.6 ⇥ 10 3 ) = j60 ⌦;
j
1
=
=
j!C
(100 ⇥ 103 )(0.4 ⇥ 10 6 )
VT =
j25IT + 5I
I =
j60
IT ;
30 + j60
VT =
j25IT + 25
VT
= Zab = 20
IT
j25 ⌦.
30I ;
j60
IT ;
30 + j60
j15 = 25/
36.87 ⌦.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 9.53
9–35
[a]
IT =
VT ↵VT
VT
+
;
1000
j1000
(1 ↵)
j 1+↵
=
j1000
j1000;
1
IT
=
VT
1000
.·. ZTh =
VT
j1000
.
=
IT
↵ 1+j
ZTh is real when ↵ = 1.
[b] ZTh = 1000 ⌦.
[c] ZTh = 500
=
j500 =
j1000
↵ 1+j
1000(↵ 1)
1000
+j
.
2
(↵ 1) + 1
(↵ 1)2 + 1
Equate the real parts:
1000
= 500
(↵ 1)2 + 1
.·.
(↵
.·.
(↵
1)2 + 1 = 2;
1)2 = 1 so ↵ = 0.
Check the imaginary parts:
(↵
(↵
1)1000
=
1)2 + 1 ↵=1
500.
Thus, ↵ = 0.
1000(↵ 1)
1000
[d] ZTh =
+j
.
2
(↵ 1) + 1
(↵ 1)2 + 1
For Im(ZTh ) > 0, ↵ must be greater than 1. So ZTh is inductive for
1 < ↵  10.
P 9.54
j!L = j(400)(50 ⇥ 10 3 ) = j20 ⌦;
j
1
=
=
j!C
(400)(50 ⇥ 10 6 )
j50 ⌦;
Vg1 = 25/53.13 = 15 + j20 V;
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9–36
CHAPTER 9. Sinusoidal Steady State Analysis
Vg2 = 18.03/33.69 = 15 + j10 V.
Vo
Vo
(15 + j20)
+
+
j50
150
Vo
(15 + j10)
= 0.
j20
Solving,
Vo = 15/0 ;
vo (t) = 15 cos 400t V.
P 9.55
V1
V1
240 V1
+
+
= 0.
j10
50
30 + j10
Solving for V1 yields
V1 = 198.63/
Vo =
P 9.56
24.44 V;
30
(V1 ) = 188.43/
30 + j10
42.88 V.
j!L = j(2500)(1.6 ⇥ 10 3 ) = j4 ⌦;
j
1
=
=
j!Ctop
(2500)(100 ⇥ 10 6 )
j4 ⌦;
j
1
=
=
j!Cmid
(2500)(50 ⇥ 10 6 )
j8 ⌦;
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Problems
Ig = 5/0 A;
5+
V2
V1
9–37
Vg = 20/90 V.
V2 V1 j20
+
= 0;
j8
j4
V1 V2 V2 j20
+
+
= 0.
j8
j4
12
Solving,
V2 =
8 + j4 V;
Io =
V2
= 1 + j2 = 2.24/63.43 A;
j4
io = 2.24 cos(2500t + 63.43 ) A.
P 9.57
Make a source transformation on the left side:
Write a KCL equation on the right hand side:
16Io +
Vo
Vo
+
= 0.
50
j25
Write a KVL equation on the left hand side:
1 + (40 + j20)Io +
Vo
= 0.
8
Solving,
Vo = 4 + j4 = 5.66/45 V;
Io =
1
Vo /8
=
j20
25
j25 mA = 35.36/
135 mA.
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9–38
P 9.58
CHAPTER 9. Sinusoidal Steady State Analysis
Write a KCL equation at the top node:
Vo 2.4I
Vo
Vo
+
+
j8
j4
5
(10 + j10) = 0.
The constraint equation is:
I =
Vo
.
j8
Solving,
Vo = j80 = 80/90 V.
P 9.59
Transform the circuit to the phasor domain; the sources are
Va = 50/
90 =
j50 V;
Vb = 25/90 = j25 V.
The impedances are
j!L = j106 (10 ⇥ 10 6 ) = j10 ⌦;
j
1
=
=
j!C
(106 )(0.1 ⇥ 10 6 )
j10 ⌦.
The phasor domain circuit is
The KVL equations are
(10
j10)I1 + j10I2
j10I1 + 10I2
10I3 =
j50;
10I3 = 0;
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Problems
10I1
9–39
10I2 + 20I3 = j25.
Solving,
I1 = 0.5
j1.5 A;
Ia =
I1 =
Ib =
I3 = 1
I3 =
1 + j0.5 A;
0.5 + j1.5 = 1.58/108.43 A;
j0.5 = 1.12/
26.57 A.
Inverse phasor-transform back to the time domain to get
ia = 1.58 cos(106 t + 108.43 ) A;
ib = 1.12 cos(106 t
26.57 ) A.
P 9.60
(10 + j5)Ia
j5Ia
j5Ib = 100/0
j5Ib = j100
Solving,
Ia =
I o = Ia
j10 A;
Ib = 20
Ib =
20 + j10 A
j20 = 28.28/
io (t) = 28.28 cos(50,000t
45 A
45 ) A
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9–40
P 9.61
CHAPTER 9. Sinusoidal Steady State Analysis
The circuit with the mesh currents identified is shown below:
The mesh current equations are:
(15 + j20)
j50I1 + 150(I1
15 + j10 + 150(I2
I2 ) = 0;
I1 ) + j20I2 = 0.
In standard form:
I1 (150
j50) + I2 ( 150) = 15 + j20;
I1 ( 150) + I2 (150 + j20) =
(15 + j10).
Solving on a calculator yields:
I1 =
0.4 A;
I2 =
Vo = 150(I1
I2 ) = 15 V
0.5 A.
Thus,
and
vo (t) = 15 cos 400t V.
P 9.62
(12
j12)Ia
12Ig
5( j8) = 0;
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Problems
12Ia + (12 + j4)Ig + j20
9–41
5(j4) = 0.
Solving,
Ig = 4
j2 = 4.47/
26.57 A
so
Io = 5
Ig = 1 + j2 = 2.24/63.43 A;
io = 2.24 cos(2500t + 63.43 ) A.
P 9.63
10/0 = (1
5/0 =
1 = j1I1
j1)I1
1I2 + j1I3 ;
1I1 + (1 + j1)I2
j1I3 ;
j1I2 + I3 .
Solving,
I1 = 11 + j10 A;
I2 = 11 + j5 A;
Ia = I 3
1 = 5 A;
Ib = I1
I3 = 5 + j10 A;
Ic = I 2
I3 = 5 + j5 A;
I d = I1
I2 = j5 A.
I3 = 6 A;
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9–42
P 9.64
CHAPTER 9. Sinusoidal Steady State Analysis
j!L = j5000(60 ⇥ 10 3 ) = j300 ⌦;
j
1
=
=
j!C
(5000)(2 ⇥ 10 6 )
400/0 + (50 + j300)Ia
(150
j100)Ib
j100 ⌦.
50Ib
150(Ia
50Ia + 150(Ia
Ib ) = 0.
Ib ) = 0;
Solving,
Ia =
0.8
1.6 A;
Ib =
1.6 + j0.8 A;
160 + j80 = 178.89/153.43 ;
Vo = 100Ib =
vo = 178.89 cos(5000t + 153.43 ) V.
P 9.65
Vo = Vg
500 j1000
Zo
(100/0 ) = 111.8/
=
ZT
300 + j1600 + 500 j1000
vo = 111.8 cos(8000t
P 9.66
100.3 V;
100.3 ) V.
j
1
=
=
j!C
(20,000)(125 ⇥ 10 9 )
j400 ⌦;
j!L = j(20,000)(0.06) = j1200 ⌦;
Let Z1 = 200
j400 ⌦;
Z2 = 60 + j1200 ⌦;
Ig = 400/0 mA;
Io =
Z1 kZ2
(200
(Ig ) =
Z1
j400)k(60 + j1200)
(0.4);
(200 j400)
= 551.5 + j149.24 mA = 571.33/15.14 mA;
io = 571.33 cos(20,000t + 15.14 ) mA.
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Problems
P 9.67
9–43
[a] Superposition must be used because the frequencies of the two sources are
di↵erent.
[b] For ! = 2000 rad/s:
10k
j5 = 2
so
Vo1 =
j4 ⌦;
2
2
j4
(20/
j4 + j2
36.87 ) = 31.62/
55.3 V.
For ! = 5000 rad/s:
j5k10 = 2 + j4 ⌦;
Vo2 =
2 + j4
(10/16.26 ) = 15.81/34.69 V.
2 + j4 j2
Thus,
vo (t) = [31.62 cos(2000t
P 9.68
55.3 ) + 15.81 cos(5000t + 34.69 )] V.
[a] Superposition must be used because the frequencies of the two sources are
di↵erent.
[b] For ! = 16,000 rad/s:
V0o
V0o
V0o 10
+
+
= 0;
j100
j400 400
V0o
1
1
1
+
+
j100 j400 400
!
=
10
;
j100
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9–44
CHAPTER 9. Sinusoidal Steady State Analysis
.·. V0o = 12 + j4 = 12.65/18.435 V.
For ! = 4000 rad/s:
V00
V00 20
V00o
+ o + o
= 0;
j400 j100
400
V00o (j
j4 + 1) = 20
so
V00o =
20
= 2 + j6 = 6.32/71.565 V.
1 j3
Thus,
vo (t) = [12.65 cos(16,000t + 18.435 ) + 6.32 cos(4000t + 71.565 )] V.
P 9.69
[a] Vg = 25/0 V;
Vp =
20
Vg = 5/0 ;
100
Vn = Vp = 5/0 V;
5 Vo
5
+
= 0;
80,000
Zp
Zp =
j80,000k40,000 = 32,000
j16,000 ⌦;
Vo =
5Zp
+5=7
80,000
j = 7.07/
8.13 ;
vo = 7.07 cos(50,000t
8.13 ) V.
[b] Vp = 0.2Vm /0 ;
Vn = Vp = 0.2Vm /0 ;
0.2Vm Vo
0.2Vm
+
= 0;
80,000 32,000 j16,000
32,000 j16,000
Vm (0.2) = Vm (0.28
.·. Vo = 0.2Vm +
80,000
.·. |Vm (0.28
j0.04);
j0.04)|  10;
.·. Vm  35.36 V.
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Problems
P 9.70
1
=
j!C1
j10 k⌦;
1
=
j!C2
j100 k⌦.
Va 2
+
5000
.·.
Va
Va
Va Vo
+
+
= 0;
j10,000 20,000
100,000
40 + j10Va + 5Va + Va
20Va
(26 + j10)Va
0 Va
+
20,000
j5Va
9–45
Vo = 0;
Vo = 40.
0 Vo
= 0;
j100,000
Vo = 0.
Solving,
Vo = 1.43 + j7.42 = 7.56/79.09 V;
vo (t) = 7.56 cos(106 t + 79.09 ) V.
P 9.71
1
= j20 k⌦.
j!C
Let Va = voltage across the capacitor, positive at upper terminal.
Then:
Vg = 4/0 V;
Va 4/0
+
20,000
Va
Va
+
= 0;
j20,000 20,000
0 Va 0 Vo
+
= 0;
20,000
10,000
.·. Vo =
Vo =
.·. Va = (1.6
j0.8) V;
Va
;
2
0.8 + j0.4 = 0.89/153.43 V.
vo = 0.89 cos(200t + 153.43 ) V.
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9–46
P 9.72
CHAPTER 9. Sinusoidal Steady State Analysis
[a]
Va
Va 4/0
+ j!Co Va +
= 0;
20,000
20,000
Va =
Vo =
Vo =
4
;
2 + j20,000!Co
Va
2
(see solution to Prob. 9.71)
2
2/180
=
;
6
2 + j4 ⇥ 10 Co
2 + j4 ⇥ 106 Co
.·. denominator angle = 45
so 4 ⇥ 106 Co = 2
[b] Vo =
.·.
C = 0.5 µF.
2/180
= 0.707/135 V;
2 + j2
vo = 0.707 cos(200t + 135 ) V.
P 9.73
[a]
1
=
j!C
j20 ⌦;
Vn Vn Vo
+
= 0;
20
j20
Vn
Vn
Vo
=
+
;
j20
20
j20
Vo =
jVn + Vn = (1
j)Vn ;
Vp =
Vg
Vg (1/j!Co )
=
;
5 + (1/j!Co )
1 + j(5)(105 )Co
Vg = 6/0 V;
Vp =
6/0
= Vn ;
1 + j5 ⇥ 105 Co
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Problems
.·. Vo =
|Vo | = q
(1 j)6/0
.
1 + j5 ⇥ 105 Co
p
2(6)
1 + 25 ⇥ 1010 Co2
9–47
= 6.
Solving,
Co = 2 µF.
6(1 j)
=
1+j
j6 V;
vo = 6 cos(105 t
90 ) V.
[b] Vo =
P 9.74
[a] j!L1 = j(5000)(2 ⇥ 10 3 ) = j10 ⌦;
j!L2 = j(5000)(8 ⇥ 10 3 ) = j40 ⌦;
j!M = j10 ⌦.
70 = (10 + j10)Ig + j10IL ;
0 = j10Ig + (30 + j40)IL .
Solving,
Ig = 4
j3 A;
IL =
1 A;
ig = 5 cos(5000t
36.87 ) A;
iL = 1 cos(5000t
180 ) A.
2
M
= p = 0.5.
L1 L2
16
[c] When t = 100⇡ µs,
[b] k = p
5000t = (5000)(100⇡) ⇥ 10 6 = 0.5⇡ = ⇡/2 rad = 90 ;
ig (100⇡µs) = 5 cos(53.13 ) = 3 A;
iL (100⇡µs) = 1 cos( 90 ) = 0 A;
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9–48
CHAPTER 9. Sinusoidal Steady State Analysis
1
1
1
w = L1 i21 + L2 i22 + M i1 i2 = (2 ⇥ 10 3 )(9) + 0 + 0 = 9 mJ.
2
2
2
When t = 200⇡ µs,
5000t = ⇡ rad = 180 ;
ig (200⇡µs) = 5 cos(180
53.13) =
iL (200⇡µs) = 1 cos(180
180) = 1 A;
4 A;
1
1
w = (2 ⇥ 10 3 )(16) + (8 ⇥ 10 3 )(1) + 2 ⇥ 10 3 ( 4)(1) = 12 mJ.
2
2
P 9.75
Remove the voltage source to find the equivalent impedance:
20
ZTh = 45 + j125 +
|5 + j5|
!2
(5
j5) = 85 + j85 ⌦.
Using voltage division:
425
VTh = Vcd = j20I1 = j20
5 + j5
P 9.76
!
= 850 + j850 V.
j!L1 = j50 ⌦;
j!L2 = j32 ⌦;
1
=
j!C
j20 ⌦;
q
j!M = j(4 ⇥ 103 )k (12.5)(8) ⇥ 10 3 = j40k ⌦;
Z22 = 5 + j32
⇤
Z22
=5
"
j20 = 5 + j12 ⌦;
j12 ⌦;
40k
Zr =
|5 + j12|
#2
(5
j12) = 47.337k 2
j113.609k 2 ;
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Problems
Zab = 20 + j50 + 47.337k 2
j113.609k 2 = (20 + 47.337k 2 ) + j(50
9–49
113.609k 2 ).
Zab is resistive when
50
113.609k 2 = 0
or
k 2 = 0.44 so k = 0.66;
.·. Zab = 20 + (47.337)(0.44) = 40.83 ⌦.
P 9.77
[a] j!L1 = j(200 ⇥ 103 )(10 3 ) = j200 ⌦;
j!L2 = j(200 ⇥ 103 )(4 ⇥ 10 3 ) = j800 ⌦;
j
1
=
=
3
j!C
(200 ⇥ 10 )(12.5 ⇥ 10 9 )
.·. Z22 = 100 + 200 + j800
⇤
.·. Z22
= 300
j400 ⌦;
j400 = 300 + j400 ⌦;
j400 ⌦.
q
M = k L1 L2 = 2k ⇥ 10 3 ;
!M = (200 ⇥ 103 )(2k ⇥ 10 3 ) = 400k;
"
400k
Zr =
500
#2
(300
j400) = k 2 (192
Zin = 200 + j200 + 192k 2
j256) ⌦;
j256k ; 2
|Zin | = [(200 + 192k)2 + (200
1
256k)2 ] 2 ;
1
d|Zin |
= [(200 + 192k)2 + (200
dk
2
1
256k)2 ] 2 ⇥
[2(200 + 192k 2 )384k + 2(200
256k 2 )( 512k)];
d|Zin |
= 0 when
dk
768k(200 + 192k 2 )
.·. k 2 = 0.125;
1024k(200 256k 2 ) = 0.
p
.·. k = 0.125 = 0.3536.
[b] Zin (min) = 200 + 192(0.125) + j[200
0.125(256)]
= 224 + j168 = 280/36.87 ⌦.
I1 (max) =
560/0
= 2/
224 + j168
36.87 A;
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9–50
CHAPTER 9. Sinusoidal Steady State Analysis
.·. i1 (peak) = 2 A.
Note — You can test that the k value obtained from setting d|Zin |/dt = 0
leads to a minimum by noting 0  k  1. If k = 1,
Zin = 392
j56 = 395.98/
8.13 ⌦.
Thus,
|Zin |k=1 > |Zin |k=p0.125 .
If k = 0,
Zin = 200 + j200 = 282.84/45 ⌦.
Thus,
|Zin |k=0 > |Zin |k=p0.125 .
P 9.78
[a] j!LL = j100 ⌦
j!L2 = j500 ⌦
Z22 = 300 + 500 + j100 + j500 = 800 + j600 ⌦
⇤
Z22
= 800
j600 ⌦
!M = 270 ⌦
✓
270
Zr =
1000
◆2
[800
j600] = 58.32
j43.74 ⌦
[b] Zab = R1 + j!L1 + Zr = 41.68 + j180 + 58.32
j43.74 = 100 + j136.26 ⌦
P 9.79
ZL =
V3
;
I3
V3
V2
=
;
1
20
1I2 =
V2
V1
=
;
50
1
50I1 = 1I2 ;
Zab =
20I3 ;
V1
.
I1
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Problems
9–51
45 ) = 1250/
45 ⌦.
Substituting,
Zab =
502 V2
V1
50V2
=
=
I1
I2 /50
I2
502 ( V3 /20)
(50)2 V3
=
=
= 6.25ZL = 6.25(200/
20I3
(20)2 I3
P 9.80
In Eq. 9.34 replace ! 2 M 2 with k 2 ! 2 L1 L2 and then write Xab as
k 2 ! 2 L1 L2 (!L2 + !LL )
2
R22
+ (!L2 + !LL )2
Xab = !L1
(
k 2 !L2 (!L2 + !LL )
}
2
R22
+ (!L2 + !LL )2
= !L1 1
For Xab to be negative requires
2
R22
+ (!L2 + !LL )2 < k 2 !L2 (!L2 + !LL ),
or
2
R22
+ (!L2 + !LL )2
k 2 !L2 (!L2 + !LL ) < 0,
which reduces to
2
R22
+ ! 2 L22 (1
k 2 ) + !L2 !LL (2
k 2 ) + ! 2 L2L < 0.
But k  1, so it is impossible to satisfy the inequality. Therefore Xab can
never be negative if XL is an inductive reactance.
P 9.81
[a]
Zab =
Vab
V1 + V2
=
;
I1
I1
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9–52
CHAPTER 9. Sinusoidal Steady State Analysis
V1
V2
=
,
N1
N2
V2 =
N1 I1 = N2 I2 ,
N2
V1 ;
N1
N1
I1 ;
N2
I2 =
✓
V2 = (I1 + I2 )ZL = I1 1 +
◆
✓
◆
N1
ZL ;
N2
✓
N1
N1
V1 + V2 =
+ 1 V2 = 1 +
N2
N2
◆2
ZL I1 ;
2
(1 + N1 /N2 ) ZL I1
.
.·. Zab =
I1
✓
Zab = 1 +
N1
N2
◆2
ZL .
[b] Assume dot on N2 is moved to the lower terminal, then
V2
V1
=
,
N1
N2
N1 I1 =
N2 I2 ,
V1 =
N1
V2 ;
N2
I2 =
N1
I1 .
N2
As in part [a]
V2 = (I2 + I1 )ZL
Zab =
(1
Zab = [1
P 9.82
and Zab =
N1 /N2 )V2
(1
=
I1
V1 + V2
;
I1
N1 /N2 )(1 N1 /N2 )ZL I1
;
I1
(N1 /N2 )]2 ZL .
[a]
N1 I1 = N2 I2 ,
Zab =
I2 =
N1
I1 ;
N2
Vab
V2
V2
=
=
;
I1 + I2
I1 + I2
(1 + N1 /N2 )I1
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Problems
V1
N1
=
,
V2
N2
Zab =
N1
V2 ;
N2
V1 =
V1 + V2 = ZL I1 =
✓
9–53
◆
N1
+ 1 V2 ;
N2
I1 ZL
;
(N1 /N2 + 1)(1 + N1 /N2 )I1
.·. Zab =
ZL
.
[1 + (N1 /N2 )]2
[b] Assume dot on the N2 coil is moved to the lower terminal. Then
N1
V2
N2
V1 =
As before
V2
Zab =
I1 + I2
.·. Zab =
Zab =
P 9.83
[a] I =
and I2 =
N1
I1 .
N2
and V1 + V2 = ZL I1 ;
ZL I1
.
(N1 /N2 )]2 I1
V2
=
N1 /N2 )I1
[1
(1
ZL
.
(N1 /N2 )]2
[1
240 240
+
= (10
24
j32
j7.5) A;
Vs = 240/0 + (0.1 + j0.8)(10
j7.5) = 247 + j7.25 = 247.11/1.68 V.
[b] Use the capacitor to eliminate the j component of I, therefore
Ic = j7.5 A,
Zc =
240
=
j7.5
The capacitive reactance is
j32 ⌦.
32 ⌦.
Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90 V.
[c] Let Ic denote the magnitude of the current in the capacitor branch. Then
I = (10
j7.5 + jIc ) = 10 + j(Ic
7.5) A.
Vs = 240/↵ = 240 + (0.1 + j0.8)[10 + j(Ic
= (247
7.5)]
0.8Ic ) + j(7.25 + 0.1Ic ).
It follows that
240 cos ↵ = (247
0.8Ic ) and 240 sin ↵ = (7.25 + 0.1Ic ).
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9–54
CHAPTER 9. Sinusoidal Steady State Analysis
Now square each term and then add to generate the quadratic equation
Ic2
605.77Ic + 5325.48 = 0;
Ic = 302.88 ± 293.96.
Therefore
Ic = 8.92 A (smallest value) and Zc = 240/j8.92 =
Therefore, the capacitive reactance is
P 9.84
j26.90 ⌦.
26.90 ⌦.
[a]
I` =
120 120
+
= 16
7.5
j12
V` = (0.15 + j6)(16
j10 A;
j10) = 62.4 + j94.5 = 113.24/56.52 V(rms);
Vs = 120/0 + V` = 205.43/27.39 V.
[b]
[c] I` =
120 120
+
= 48
2.5
j4
V` = (0.15 + j6)(48
j30 A;
j30) = 339.73/56.56 ;
Vs = 120 + V` = 418.02/42.7 .
The amplitude of Vs must be increased from 205.43 to 418.02 (more than
doubled) to maintain the load voltage at 120 V.
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Problems
[d] I` =
9–55
120
120 120
+
+
= 48 + j30 A;
2.5
j4
j2
V` = (0.15 + j6)(48 + j30) = 339.73/120.57 ;
Vs = 120 + V` = 297.23/100.23 .
The amplitude of Vs must be increased from 205.43 to 297.23 to
maintain the load voltage at 120 V.
P 9.85
The phasor domain equivalent circuit is
Vo =
Vm
2
IRx ;
I=
Vm
.
Rx jXC
As Rx varies from 0 to 1, the amplitude of vo remains constant and its phase
angle increases from 0 to 180 , as shown in the following phasor diagram:
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9–56
P 9.86
CHAPTER 9. Sinusoidal Steady State Analysis
[a] The circuit is redrawn, with mesh currents identified:
The mesh current equations are:
120/0 = 23Ia
120/0 =
0=
2Ib
20Ic ;
2Ia + 43Ib
20Ia
40Ic ;
40Ib + 70Ic .
Solving,
Ib = 21.96/0 A;
Ia = 24/0 A;
Ic = 19.40/0 A.
The branch currents are:
I1 = Ia = 24/0 A;
I2 = I b
Ia = 2.04/180 A;
I3 = Ib = 21.96/0 A;
I4 = Ic = 19.40/0 A;
I5 = Ia
Ic = 4.6/0 A;
I6 = Ib
Ic = 2.55/0 A.
[b] Let N1 be the number of turns on the primary winding; because the
secondary winding is center-tapped, let 2N2 be the total turns on the
secondary. Therefore,
240
13,200
=
N1
2N2
or
N2
1
.
=
N1
110
The ampere turn balance requires
N1 IP = N2 I1 + N2 I3 .
Therefore,
IP =
N2
1
(24 + 21.96) = 0.42/0 A.
(I1 + I3 ) =
N1
110
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Problems
P 9.87
9–57
[a]
The three mesh current equations are
120/0 = 23Ia
120/0 =
0=
2Ib
20Ic ;
2Ia + 23Ib
20Ia
20Ic ;
20Ib + 50Ic .
Solving,
Ia = 24/0 A;
.·. I2 = Ib
[b] IP =
Ib = 24/0 A;
Ic = 19.2/0 A;
Ia = 0 A.
N2
N2
(I1 + I3 ) =
(Ia + Ib )
N1
N1
1
(24 + 24) = 0.436/0 A.
110
[c] Yes; when the two 120 V loads are equal, there is no current in the
“neutral” line, so no power is lost to this line. Since you pay for power,
the cost is lower when the loads are equal.
=
P 9.88
[a]
125 = (R + 0.05 + j0.05)I1
125 =
(0.03 + j0.03)I2
(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2
RI3 ;
RI3 .
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9–58
CHAPTER 9. Sinusoidal Steady State Analysis
Subtracting the above two equations gives
0 = (R + 0.08 + j0.08)I1
(R + 0.08 + j0.08)I2 ;
.·. I1 = I2
so
I1 = 0 A.
[b] V1 = R(I1
I3 );
In = I2
V2 = R(I2
I3 ).
Since I1 = I2 (from part [a]) V1 = V2 .
[c]
250 = (440.04 + j0.04)Ia
0=
440Ib ;
440Ia + 448Ib .
Solving,
Ia = 31.656207
j0.160343 A;
Ib = 31.090917
j0.157479 A;
I = Ia
Ib = 0.56529
V1 = 40I = 22.612
j0.002864 A;
j0.11456 = 22.612/
0.290282 V;
V2 = 400I = 226.116
j1.1456 = 226.1189/
125 = (40.05 + j0.05)Ix
(0.03 + j0.03)Iy
0.290282 V.
[d]
40Iz ;
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Problems
125 =
0=
(0.03 + j0.03)Ix + (400.05 + j0.05)Iy
40Ix
9–59
400Iz ;
400Iy + 448Iz .
Solving,
Ix = 34.19
j0.182 A;
Iy = 31.396
j0.164 A;
Iz = 31.085
j0.163 A;
V1 = 40(Ix
Iz ) = 124.2/
0.35 V;
V2 = 400(Iy
Iz ) = 124.4/
0.18 V.
[e] Because an open neutral can result in severely unbalanced voltages across
the 125 V loads.
P 9.89
[a] Let N1 = primary winding turns and 2N2 = secondary winding turns.
Then
250
N2
1
14,000
= a.
=
;
.·.
=
N1
2N2
N1
112
In part c),
IP = 2aIa ;
.·. IP =
=
2N2 Ia
1
= Ia
N1
56
1
(31.656
56
IP = 565.3
j0.16);
j2.9 mA.
In part d),
IP N1 = I1 N2 + I2 N2 ;
N2
(I1 + I2 )
.·. IP =
N1
=
1
(34.19
112
j0.182 + 31.396
=
1
(65.586
112
j0.346);
IP = 585.6
j0.164)
j3.1 mA.
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9–60
CHAPTER 9. Sinusoidal Steady State Analysis
[b] Yes, because the neutral conductor carries non-zero current whenever the
load is not balanced.
P 9.90
No, the motor current drops to 5 A, well below its normal running value of
22.86 A.
P 9.91
After fuse A opens, the current in fuse B is only 15 A.
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Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/ 45 V,
I = 20/15 A.
Therefore
1
P = (100)(20) cos[ 45 (15)] = 500 W,
2
Q = 1000 sin 60 =
[b] V = 100/
45 ,
B ! A.
866.03 VAR,
I = 20/165 ;
P = 1000 cos( 210 ) =
B ! A;
866.03 W,
A ! B.
Q = 1000 sin( 210 ) = 500 VAR,
[c] V = 100/
45 ,
I = 20/
105 ;
A ! B;
P = 1000 cos(60 ) = 500 W,
A ! B.
Q = 1000 sin(60 ) = 866.03 VAR,
[d] V = 100/0 ,
A ! B;
I = 20/120;
B ! A;
P = 1000 cos( 120 ) =
500 W,
Q = 1000 sin( 120 ) =
866.03 VAR,
B ! A.
AP 10.2
pf = cos(✓v
✓i ) = cos[15
(75)] = cos( 60 ) = 0.5 leading;
rf = sin(✓v
✓i ) = sin( 60 ) =
0.866.
10–1
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10–2
AP 10.3
CHAPTER 10. Sinusoidal Steady State Power Calculations
0.18
I⇢
From Ex. 9.4 Irms = p = p A;
3
3
2
P = Irms
R=
✓
◆
0.0324
(5000) = 54 W.
3
AP 10.4 [a] ZL = j(2500)(0.405) = j1012.5 ⌦.
The phasor domain circuit is
P =
1 902
1 |V|2
=
= 3 W.
2 R
2 1350
1 |V|2
1 902
[b] Q =
=
= 4 VAR.
2 X
2 1012.5
[c] S = P + jQ = 3 + j4 = 5/53.13 VA;
so
[d] pf = cos 53.13 = 0.6 lagging.
AP 10.5 [a] Z = (39 + j26)k( j52) = 48
Therefore I` =
IL =
22.62 ⌦;
250/0
= 4.85/18.08 A (rms).
j20 + 1 + j4
48
VL = ZI` = (52/
j20 = 52/
|S| = 5 VA.
22.62 )(4.85/18.08 ) = 252.20/
VL
= 5.38/
39 + j26
[b] SL = VL I⇤L = (252.20/
4.54 V (rms);
38.23 A (rms).
4.54 )(5.38/ + 38.23 ) = 1357/33.69
= (1129.09 + j752.73) VA;
PL = 1129.09 W;
QL = 752.73 VAR;
[c] P` = |I` |2 1 = (4.85)2 · 1 = 23.52 W;
Q` = |I` |2 4 = 94.09 VAR.
[d] Sg (delivering) = 250I⇤` = (1152.62 j376.36) VA.
Therefore the source is delivering 1152.62 W and absorbing 376.36
magnetizing VAR.
(252.20)2
|VL |2
[e] Qcap =
=
= 1223.18 VAR.
52
52
Therefore the capacitor is delivering 1223.18 magnetizing VAR.
Check:
94.09 + 752.73 + 376.36 = 1223.18 VAR and
1129.09 + 23.52 = 1152.62 W.
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Problems
10–3
AP 10.6 Series circuit derivation:
S = 250I⇤ = (40,000
j30,000);
Therefore I⇤ = 160
j120 = 200/
36.87 A (rms).
I = 200/36.87 A (rms);
Z=
250
V
=
= 1.25/
I
200/36.87
Therefore R = 1 ⌦,
36.87 = (1
XC =
j0.75) ⌦;
0.75 ⌦.
Parallel circuit derivation
P =
(250)2
;
R
therefore R =
(250)2
= 1.5625 ⌦;
40,000
Q=
(250)2
;
XC
therefore XC =
(250)2
=
30,000
2.083 ⌦.
AP 10.7
S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA;
S2 = 6000(0.8)
j6000(0.6) = 4800
j3600 VA;
ST = S1 + S2 = 13,800 + j8400 VA;
ST = 200I⇤ ;
therefore I⇤ = 69 + j42
I = 69
j42 A;
Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91 V (rms).
AP 10.8 [a] The phasor domain equivalent circuit and the Thévenin equivalent are
shown below:
Phasor domain equivalent circuit:
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10–4
CHAPTER 10. Sinusoidal Steady State Power Calculations
Thévenin equivalent:
VTh = 3
j800
= 48
20 j40
ZTh = 4 + j18 +
j24 = 53.67/
26.57 V;
j800
= 20 + j10 = 22.36/26.57 ⌦.
20 j40
For maximum power transfer, ZL = (20
[b] I =
53.67/
26.57
= 1.34/
40
Therefore P =
1.34
p
2
!2
Therefore P =
!2
26.57 A;
20 = 18 W.
[c] RL = |ZTh | = 22.36 ⌦.
53.67/ 26.57
[d] I =
= 1.23/
42.36 + j10
1.23
p
2
j10) ⌦.
39.85 A;
(22.36) = 17 W.
AP 10.9
Mesh current equations:
660 = (34 + j50)I1 + j100(I1
0 = j100(I2
I1 )
I2 ) + j40I1 + j40(I1
I2 );
j40I1 + 100I2 .
Solving,
I2 = 3.5/0 A;
1
.·. P = (3.5)2 (100) = 612.50 W.
2
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Problems
10–5
AP 10.10 [a]
248 = j400I1
0 = 375(I2
j500I2 + 375(I1
I1 ) + j1000I2
I2 );
j500I1 + 400I2 .
Solving,
I1 = 0.80
j0.62 A;
j0.3 = 0.5/
I2 = 0.4
36.87 ;
1
.·. P = (0.25)(400) = 50 W.
2
[b] I1
I2 = 0.4
j0.32 A;
1
P375 = |I1 I2 |2 (375) = 49.20 W.
2
1
[c] Pg = (248)(0.8) = 99.20 W;
2
X
Pabs = 50 + 49.2 = 99.20 W. (checks)
AP 10.11 [a] VTh = 210 V;
V2 = 14 V1 ;
Short circuit equations:
840 = 80I1
20I2 + V1 ;
0 = 20(I2
I1 )
V2 ;
210
= 15 ⌦.
14
.·. I2 = 14 A;
RTh =
✓
15 = 735 W.
210
[b] Pmax =
30
AP 10.12 [a] VTh =
◆2
4(146/0 ) =
V2 = 4V1 ;
I1 = 14 I2 .
I1 =
584/0 V (rms);
4I2 .
Short circuit equations:
146/0 = 80I1
20I2 + V1 ;
0 = 20(I2
I1 )
V2 ;
.·. I2 =
146/365 =
0.40 A;
RTh =
584
= 1460 ⌦.
0.4
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10–6
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] P =
✓
584
2920
◆2
1460 = 58.40 W.
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Problems
10–7
Problems
P 10.1
1
[a] P = (100)(10) cos(50
2
15) = 500 cos 35 = 409.58 W (abs);
Q = 500 sin 35 = 286.79 VAR (abs).
1
[b] P = (40)(5) cos( 15
2
Q = 100 sin( 75 ) =
1
[c] P = (400)(10) cos(30
2
60) = 100 cos( 75 ) = 25.88 W (abs);
96.59 VAR (del).
150) = 2000 cos( 120 ) =
Q = 2000 sin( 120 ) =
1732.05 VAR (del).
1
[d] P = (200)(5) cos(160
2
40) = 500 cos(120 ) =
1000 W (del);
250 W (del);
Q = 500 sin(120 ) = 433.01 VAR (abs).
P 10.2
[a] To find the power used by an appliance, divide the yearly kW-hours used
by the product of the number of hours per month used and the number
of months:
1080 k
central AC =
= 3000 W;
(120)(3)
dishwasher =
293 k
= 813.89 W;
(30)(12)
refrigerator =
533 k
= 61.7 W;
(720)(12)
microwave =
101 k
= 935.2 W.
(9)(12)
The total power consumed by these four appliances is
P4 = 3000 + 813.89 + 61.7 + 035.2 = 4810.76 W.
When 120 V(rms) is used, the amount of current required is
4810.8
= 40.1 A(rms).
120
Therefore, the 50 A current breaker will not interrupt the current.
I4 =
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10–8
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] the power used by the dryer is
dryer =
901 k
= 3128.47 W.
(24)(12)
With this fifth appliance added, the total power consumed is
P5 = 4810.76 + 3128.47 = 7939.2 W.
Now the current required is
7939.2
= 66.2 A(rms).
120
Now the 50 A current breaker will interrupt the current.
I5 =
P 10.3
dp
=
dt
2!P sin 2!t
p = P + P cos 2!t
Q sin 2!t;
dp
= 0 when
dt
2!P sin 2!t = 2!Q cos 2!t or
cos 2!t = p
P
P 2 + Q2
;
p
sin 2!t =
Q
P 2 + Q2
2!Q cos 2!t;
tan 2!t =
Q
.
P
.
Let ✓ = tan 1 ( Q/P ), then p is maximum when 2!t = ✓ and p is minimum
when 2!t = (✓ + ⇡).
Therefore pmax = P + P · p
and pmin = P
P 10.4
[a] P =
P·p
P
2
P + Q2
P
2
P + Q2
Q· p
q
Q( Q)
p 2
= P + P 2 + Q2 ,
P + Q2
Q
=P
2
P + Q2
q
P 2 + Q2 .
1 (240)2
= 60 W;
2 480
9 ⇥ 106
1
=
=
!C
(5000)(5)
Q=
1 (240)2
=
2 ( 360)
pmax = P +
q
360 ⌦;
80 VAR;
P 2 + Q2 = 60 +
q
(60)2 + (80)2 = 160 W (del).
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Problems
[b] pmin = 60
p
602 + 802 =
10–9
40 W (abs).
[c] P = 60 W from (a).
[d] Q =
80 VAR from (a).
[e] Generates, because Q < 0.
[f ] pf = cos(✓v
I=
✓i );
240
240
+
= 0.5 + j0.67 = 0.83/53.13 A;
480
j360
.·. pf = cos(0
53.13 ) = 0.6 leading.
[g] rf = sin( 53.13 ) =
P 10.5
0.8.
[a] From the solution to Problem 9.60 we have:
Ia =
j10 A
S100V =
and
Ib =
1
(100)(j10) =
2
P100V = 0 W;
j500 VA;
and
Sj100V =
1
(j100)( 20
2
Pj100V =
500 W;
P j10⌦ = 0 W;
Q100V =
500 VAR.
j10) =
500 + j1000 VA;
and
[b]
[c]
X
X
and
Pgen = 500 W =
X
Qj100V = 1000 VAR.
1
Q j10⌦ = ( 10)|Ib |2 =
2
and
1
P10⌦ = (10)|Ia |2 = 500 W;
2
Pj5⌦ = 0 W;
20 + j10 A.
and
1
Qj5⌦ = (5)|Ia
2
2500 VAR.
Q10⌦ = 0 VAR.
Ib |2 = 2000 VAR.
Pabs .
Qgen = 500 + 2500 = 3000 VAR.
X
Qabs = 1000 + 2000 = 3000 VAR (check).
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10–10
P 10.6
CHAPTER 10. Sinusoidal Steady State Power Calculations
[a] From the solution to Problem 9.64 we have
Ia =
0.8
S400V =
j1.6 A;
Ib =
1
(400)( 0.8 + j1.6) = 160
2
P400V = 160 W;
[b]
[c]
P 10.7
1.6 + j0.8 A;
and
and
320 VAR.
Sdep.source =
1
(150I )(I )⇤ =
2
480 + j0 VA;
Pdep.source =
480 W;
Qdep.source = 0 VAR.
P j100⌦ = 0 W;
and
1
Q j100⌦ = ( 100)|Ib |2 =
2
Pj300⌦ = 0 W;
and
1
Qj300⌦ = (300)|Ia |2 = 480 VAR.
2
160 VAR.
1
P50⌦ = (50)|I |2 = 160 W;
2
and
Q50⌦ = 0 VAR.
1
P100⌦ = (100)|Ib |2 = 160 W;
2
and
Q100⌦ = 0 VAR.
X
X
Pabs = 160 + 160 + 160 = 480 W =
Qdev = 320 + 160 = 480 VAR =
Ig = 40/0 mA;
X
j2.4 A.
j320 VA;
Q400V =
and
I = 0.8
X
Pdev .
Qabs .
j!L = j10,000 ⌦;
1
=
j!C
Zeq = j10,000k(5000
j10,000) = 20,000 + j10,000 ⌦;
j10,000 ⌦.
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Problems
10–11
Vo = (20,000 + j10,000)(0.04) = 800 + j400 V;
S=
(0.04)Vo =
(0.04)(800 + j400) =
32
j16 VA;
So the average power delivered by the source is 32 W.
P 10.8
Zf =
j10,000k20,000 = 4000
Zi = 2000
.·.
P 10.9
j2000 ⌦;
4000
Zf
=
Zi
2000
Vo =
Zf
Vg ;
Zi
Vo =
(3
P =
j8000 ⌦;
j8000
=3
j2000
j1.
Vg = 1/0 V;
j1)(1) =
3 + j1 = 3.16/161.57 V;
1 Vm2
1 (10)
=
= 5 ⇥ 10 3 = 5 mW.
2 R
2 1000
j!L = j105 (0.5 ⇥ 10 3 ) = j50 ⌦;
4+
I =
Vo
Vo
+
j50
1
1
=
=
5
j!C
j10 [(1/3) ⇥ 10 6 ]
j30 ⌦.
50(Vo 50I )
= 0;
40 j30
Vo
.
j50
Place the equations in standard form:
!
Vo
1
1
+
+I
j50 40 j30
Vo
1
+ I ( 1) = 0.
j50
50
40 j30
!
= 4;
!
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10–12
CHAPTER 10. Sinusoidal Steady State Power Calculations
Solving,
Vo = 200
Io = 4
j400 V;
( 8
I =
8
j4 A;
j4) = 12 + j4 A;
1
1
P40⌦ = |Io |2 (40) = (160)(40) = 3200 W.
2
2
P 10.10 [a] line loss = 7500
2500 = 5 k⌦;
.·. |Ig |2 = 250.
line loss = |Ig |2 20
|Ig | =
p
250 A;
.·. RL = 10 ⌦;
|Ig |2 RL = 2500
|Ig |2 XL =
5000
.·. XL =
20 ⌦.
Thus,
|Z| =
q
(30)2 + (X`
.·. 900 + (X`
Solving,
20)2
20)2 =
(X`
Thus, X` = 10 ⌦
500
|Ig | = q
900 + (X`
20)2
;
25 ⇥ 104
= 1000.
250
20) = ±10.
or
X` = 30 ⌦.
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Problems
10–13
[b] If X` = 30 ⌦:
Ig =
500
= 15
30 + j10
Sg =
500I⇤g =
j5 A;
7500
j2500 VA.
Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars.
Qj30 = |Ig |2 X` = 250(30) = 7500 VAR.
Therefore the line reactance is absorbing 7500 magnetizing vars.
Q j20 = |Ig |2 XL = 250( 20) =
5000 VAR.
Therefore the load reactance is generating 5000 magnetizing vars.
X
Qgen = 7500 VAR =
X
Qabs .
If X` = 10 ⌦:
500
Ig =
= 15 + j5 A;
30 j10
500I⇤g =
Sg =
7500 + j2500 VA.
Thus, the voltage source is delivering 7500 W and absorbing 2500
magnetizing vars.
Qj10 = |Ig |2 (10) = 250(10) = 2500 VAR.
Therefore the line reactance is absorbing 2500 magnetizing vars. The
load continues to generate 5000 magnetizing vars.
X
Qgen = 5000 VAR =
Irms =
s
600t A,
1
0.1
=
Qabs .
0  t  75 ms;
P 10.11 i(t) = 200t A,
i(t) = 60
X
⇢Z 0.075
0
75 ms  t  100 ms;
(200)2 t2 dt +
Z 0.1
0.075
q
10(5.625) + 10(1.875) =
(60
p
600t)2 dt
75 = 8.66 A(rms).
3
3 ⇥ 10
= 40 ⌦.
.·. R =
75
2
P 10.12 P = Irms
R
P 10.13 [a] A = 402 (5) + ( 40)2 )(5) = 16,000;
mean value =
Vrms =
p
A
= 800;
20
800 = 28.28 V(rms).
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10–14
CHAPTER 10. Sinusoidal Steady State Power Calculations
2
Vrms
800
=
= 20 W.
R
40
V2
800
[c] R = rms =
= 80 k⌦.
P
0.01
[b] P =
P 10.14 [a] Area under one cycle of vg2 :
A = 2(5)2 (30 ⇥ 10 6 ) + 2(2)2 (37.5 ⇥ 10 6 )
= 1800 ⇥ 10 6 .
Mean value of vg2 :
A
1800 ⇥ 10 6
=
= 9;
200 ⇥ 10 6
200 ⇥ 10 6
p
.·. Vrms = 9 = 3 V(rms).
M.V. =
2
Vrms
32
=
= 4 W.
R
2.25
P 10.15 [a] Irms = 40/115 ⇠
= 0.35 A.
[b] P =
[b] Irms = 130/115 ⇠
= 1.13 A.
P 10.16 Wdc =
Vdc2
T;
R
Ws =
Z to +T
to
vs2
dt;
R
Z to +T 2
Vdc2
vs
·
..
T =
dt.
R
R
to
Vdc2 =
Vdc =
1 Z to +T 2
vs dt;
T to
s
1 Z to +T 2
vs dt = Vrms = Vrms .
T to
P 10.17 [a] Let VL = Vm /0 :
SL = 2500(0.8 + j0.6) = 2000 + j1500 VA;
I⇤` =
2000
1500
+j
;
Vm
Vm
I` =
2000
Vm
j
1500
;
Vm
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Problems
250/✓ = Vm +
✓
2000
Vm
10–15
◆
1500
j
(1 + j2);
Vm
250Vm /✓ = Vm2 + (2000
j1500)(1 + j2) = Vm2 + 5000 + j2500;
250Vm cos ✓ = Vm2 + 5000;
250Vm sin ✓ = 2500;
(250)2 Vm2 = (Vm2 + 5000)2 + 25002 ;
62,500Vm2 = Vm4 + 10,000Vm2 + 31.25 ⇥ 106
or
Vm4
52,500Vm2 + 31.25 ⇥ 106 = 0.
Solving,
Vm2 = 26,250 ± 25,647.86;
Vm = 227.81 V and Vm = 24.54 V.
If Vm = 227.81 V:
sin ✓ =
2500
= 0.044;
(227.81)(250)
.·. ✓ = 2.52 .
If Vm = 24.54 V:
sin ✓ =
2500
= 0.4075;
(24.54)(250)
.·. ✓ = 24.05 .
[b]
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10–16
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.18 [a]
1
=
j!C
j40 ⌦;
Zeq = 40k
j!L = j80 ⌦.
j40 + j80 + 60 = 80 + j60 ⌦;
Ig =
40/0
= 0.32
80 + j60
Sg =
1
Vg I⇤g =
2
1
40(0.32 + j0.24) =
2
P = 6.4 W (del);
Q = 4.8 VAR (del);
j0.24 A;
6.4
j4.8 VA;
|S| = |Sg | = 8 VA.
[b] I1 =
j40
Ig = 0.04
40 j40
j0.28 A;
1
P40⌦ = |I1 |2 (40) = 1.6 W;
2
1
P60⌦ = |Ig |2 (60) = 4.8 W;
2
X
Pdiss = 1.6 + 4.8 = 6.4 W =
[c] I j40⌦ = Ig
X
Pdev .
I1 = 0.28 + j0.04 A;
1
Q j40⌦ = |I j40⌦ |2 ( 40) =
2
1.6 VAR (del);
1
Qj80⌦ = |Ig |2 (80) = 6.4 VAR (abs);
2
X
Qabs = 6.4
1.6 = 4.8 VAR =
X
Qdev .
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Problems
10–17
P 10.19 [a]
The mesh equations are:
(10
j20)I1 + (j20)I2 = 170;
(j20)I1 + (12
j4)I2 = 0.
Solving,
I1 = 4 + j1 A;
S=
Vg I⇤1 =
I2 = 3.5
(170)(4
j5.5 A;
j1) =
680 + j170 VA.
[b] Source is delivering 680 W.
[c] Source is absorbing 170 magnetizing VAR.
p
[d] P10⌦ = ( 17)2 (10) = 170 W;
p
P12⌦ = ( 42.5)2 (12) = 510 W;
p
Q j20⌦ = ( 42.5)2 (20) = 850 VAR;
p
Qj16⌦ = ( 42.5)2 (16) = 680 VAR.
[e]
X
Pdel = 680 W;
X
.·.
[f ]
X
Pdiss = 170 + 510 = 680 W;
X
Pdel =
X
Pdiss = 680 W.
Qabs = 170 + 680 = 850 VAR;
X
.·.
Qdev = 850 VAR;
X
mag VAR dev =
X
mag VAR abs = 850.
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10–18
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.20 j!L = j25 ⌦;
Io =
1
=
j!C
j50 ⌦.
150
= 2.4 + j1.2 A;
50 j25
1
1
P = |Io |2 (50) = (7.2)(50) = 180 W;
2
2
1
Q = |Io |2 (25) = 90 VAR;
2
S = P + jQ = 180 + j90 VA;
|S| = 201.25 VA.
P 10.21 [a] S1 = 60,000
S2 =
j70,000 VA;
|VL |2
(2500)2
= 240,000 + j70,000 VA;
=
Z2⇤
24 j7
S1 + S2 = 300,000 VA;
2500I⇤L = 300,000;
.·. IL = 120 A(rms);
Vg = VL + IL (0.1 + j1) = 2500 + (120)(0.1 + j1)
= 2512 + j120 = 2514.86/2.735 Vrms.
[b] T =
1
1
=
= 16.67 ms;
f
60
t
2.735
;
=
360
16.67 ms
.·. t = 126.62 µs.
[c] VL lags Vg by 2.735 or 126.62 µs.
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Problems
P 10.22 ST = 4500
j
S1 = 2700 + j
S2 = ST
4500
(0.28) = 4500
0.96
10–19
j1312.5 VA;
2700
(0.6) = 2700 + j2025 VA;
0.8
S1 = 1800
j3337.5 = 3791.95/
61.66 VA;
pf = cos( 61.66 ) = 0.4747 leading.
P 10.23
250I⇤1 = 6000
I⇤1 = 24
j32;
j8000;
.·. I1 = 24 + j32 A(rms);
250I⇤2 = 9000 + j3000;
I⇤2 = 36 + j12;
I3 =
.·. I2 = 36
250/0
= 10 + j0 A;
25
j12 A(rms);
I4 =
250/0
= 0 + j50 A;
j5
Ig = I1 + I2 + I3 + I4 = 70 + j70 A;
Vg = 250 + (70 + j70)(j0.1) = 243 + j7 = 243.1/1.65 V (rms).
P 10.24 [a] S1 = 16 + j18 kVA;
S2 = 6
j8 kVA;
S3 = 8 + j0 kVA;
ST = S1 + S2 + S3 = 30 + j10 kVA;
250I⇤ = (30 + j10) ⇥ 103 ;
Z=
.·. I = 120
j40 A;
250
= 1.875 + j0.625 ⌦ = 1.98/18.43 ⌦.
120 j40
[b] pf = cos(18.43 ) = 0.9487 lagging.
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10–20
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.25 [a] From the solution to Problem 10.24 we have
IL = 120
j40 A(rms);
.·. Vs = 250/0 + (120
j40)(0.01 + j0.08) = 254.4 + j9.2
= 254.57/2.07 V(rms).
[b] |IL | =
q
16,000;
P` = (16,000)(0.01) = 160 W
[c] Ps = 30,000 + 160 = 30.16 kW
30
[d] ⌘ =
(100) = 99.47%.
30.16
Q` = (16,000)(0.08) = 1280 VAR.
Qs = 10,000 + 1280 = 11.28 kVAR.
P 10.26 [a] From Problem 9.78,
Zab = 100 + j136.26
so
I1 =
50
50
=
= 0.16
100 + j13.74 + 100 + j136.26
200 + j150
I2 =
j!M
j270
(0.16
I1 =
Z22
800 + j600
j0.12 A.
j0.12) = 51.84 + j15.12 mA.
VL = (300 + j100)(0.05184 + j0.01512) = 14.04 + j9.72 V;
|VL | = 17.0763 V.
[b] Pg (ideal) = 50(0.16) = 8 W;
Pg (practical) = 8
|I1 |2 (100) = 4 W.
PL = |I2 |2 (300) = 874.8 mW;
% delivered =
0.8748
(100) = 21.87%.
4
P 10.27 [a] Z1 = 240 + j70 = 250/16.26 ⌦;
pf = cos(16.26 ) = 0.96 lagging;
rf = sin(16.26 ) = 0.28;
Z2 = 160
j120 = 200/
36.87 ⌦;
pf = cos( 36.87 ) = 0.8 leading;
rf = sin( 36.87 ) =
Z3 = 30
j40 = 50/
0.6;
53.13 ⌦;
pf = cos( 53.13 ) = 0.6 leading;
rf = sin( 53.13 ) =
0.8.
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Problems
10–21
[b] Y = Y1 + Y2 + Y3 ;
Y1 =
1
;
250/16.26
1
;
200/ 36.87
Y2 =
Y3 =
1
;
50/ 53.13
Y = 19.84 + j17.88 mS;
Z=
1
= 37.44/
Y
42.03 ⌦;
pf = cos( 42.03 ) = 0.74 leading;
rf = sin( 42.03 ) =
0.67.
P 10.28 [a]
250I⇤1 = 7500 + j2500;
.·. I1 = 30
250I⇤2 = 2800
.·. I2 = 11.2 + j38.4 A(rms);
I3 =
j9600;
500
500
+
= 40
12.5 j50
Ig1 = I1 + I3 = 70
Sg1 =
j10 A(rms);
j10 A(rms);
j20 A;
250(70 + j20) =
17,500
j5000 VA.
Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars.
Ig2 = I2 + I3 = 51.2 + j28.4 A(rms);
Sg2 =
250(51.2
j28.4) =
12,800 + j7100 VA.
Thus the Vg2 source is delivering 12.8 kW and absorbing 7100
magnetizing vars.
[b]
X
Pgen = 17.5 + 12.8 = 30.3 kW;
X
X
X
Pabs = 7500 + 2800 +
X
(500)2
= 30.3kW =
Pgen .
12.5
Qdel = 9600 + 5000 = 14.6 kVAR;
Qabs = 2500 + 7100 +
X
(500)2
= 14.6 kVAR =
Qdel .
50
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10–22
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.29 S1 =
60 k
101 k
336 k
+
+
= 1140.74 + j0 VA;
(720)(12) (30)(12) (9)(12)
.·. I1 =
S2 = 2
360 k
72 k
54 k
+
+
= 355 + j0 VA;
(60)(12) (240)(12) (150)(6)
.·. I2 =
S3 =
1140.74
= 9.506 A.
120
355
= 2.96 A.
120
108 k
901 k
+
= 3449.9 + j0 VA;
(28)(12) (24)(12)
3449.9
= 14.37 A.
.·. I3 =
240
Ig1 = I1 + I3 = 23.88 A;
Ig2 = I2 + I3 = 17.33 A.
The breaker protecting the upper service conductor will trip because its feeder
current exceeds 20 A. The breaker protecting the lower service conductor will
not trip because its feeder current is less than 20 A. Thus, service to Loads 1
and 3 will be interrupted.
P 10.30 [a]
I1 =
I2 =
I3 =
4000
j1000
= 32
125
j8 A (rms);
5000
j2000
= 40
125
j16 A (rms);
10,000 + j0
= 40 + j0 A (rms);
250
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Problems
.·. Ig1 = I1 + I3 = 72
I n = I1
I2 =
10–23
j8 A (rms).
8 + j8 A (rms);
Ig2 = I2 + I3 = 80
j16 A(rms);
Vg1 = 0.05Ig1 + 125 + 0.15In = 127.4 + j0.8 V(rms);
Vg2 =
0.15In + 125 + 0.05Ig2 = 130.2
Sg1 =
[(127.4 + j0.8)(72 + j8)] =
[9166.4 + j1076.8] VA;
Sg2 =
[(130.2
[10,448 + j1923.2] VA.
j2)(80 + j16)] =
j2 V(rms);
Note: Both sources are delivering average power and magnetizing VAR to
the circuit.
[b] P0.05 = |Ig1 |2 (0.05) = 262.4 W;
P0.15 = |In |2 (0.14) = 19.2 W;
P0.05 = |Ig2 |2 (0.05) = 332.8 W;
X
X
X
X
Pdis = 262.4 + 19.2 + 332.8 + 4000 + 5000 + 10,000 = 19,614.4 W;
Pdev = 9166.4 + 10,448 = 19,614.4 W =
Qabs = 1000 + 2000 = 3000 VAR;
Qdel = 1076.8 + 1923.2 = 3000 VAR =
X
X
Pdis .
Qabs .
P 10.31 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA;
125I⇤L = (17,000 + j10,535.65);
.·. IL = 136
I⇤L = 136 + j84.29 A(rms);
j84.29 A(rms).
Vs = 125 + (136
j84.29)(0.01 + j0.08) = 133.10 + j10.04
= 133.48/4.31 V(rms);
|Vs | = 133.48 V(rms).
[b] P` = |I` |2 (0.01) = (160)2 (0.01) = 256 W.
[c]
(125)2
=
XC
10,535.65;
XC =
1
=
!C
1.48306;
C=
1.48306 ⌦;
1
= 1788.59 µF.
(1.48306)(120⇡)
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10–24
CHAPTER 10. Sinusoidal Steady State Power Calculations
[d] I` = 136 + j0 A(rms);
Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88
= 126.83/4.92 V(rms);
|Vs | = 126.83 V(rms).
[e] P` = (136)2 (0.01) = 184.96 W.
P 10.32 [a] I =
465/0
= 2.4
124 + j93
j1.8 = 3/
36.87 A(rms);
P = (3)2 (4) = 36 W.
[b] YL =
1
= 5.33
120 + j90
j4 mS;
1
=
4 ⇥ 10 3
250 ⌦.
.·. XC =
1
= 187.5 ⌦.
5.33 ⇥ 10 3
465/0
= 2.43/ 0.9 A;
[d] I =
191.5 + j3
[c] ZL =
P = (2.43)2 (4) = 23.62 W.
23.62
(100) = 65.6%.
36
Thus the power loss after the capacitor is added is 65.6% of the power
loss before the capacitor is added.
[e] % =
P 10.33
IL =
153,600 j115,200
= 32
4800
IC =
4800
4800
=j
= jIC ;
jXC
XC
j24 A(rms);
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Problems
I` = 32
j24 + jIC = 32 + j(IC
24);
Vs = 4800 + (2 + j10)[32 + j(IC
24)]
10–25
10IC ) + j(272 + 2IC );
= (5104
|Vs |2 = (5104
10IC )2 + (272 + 2IC )2 = (4800)2 ;
.·. 104IC2
100,992IC + 3,084,800 = 0.
Solving,
IC = 31.57 A(rms);
IC = 939.51 A(rms).
*Select the smaller value of IC to minimize the magnitude of I` .
.·. XC =
.·. C =
4800
=
31.57
152.04;
1
= 17.45 µF.
(152.04)(120⇡)
P 10.34 [a] So = original load = 1600 + j
1600
(0.6) = 1600 + j1200 kVA;
0.8
Sf = final load = 1920 + j
1920
(0.28) = 1920 + j560 kVA;
0.96
.·. Qadded = 560
640 kVAR.
1200 =
[b] Deliver.
[c] Sa = added load = 320
j640 = 715.54/
63.43 kVA;
pf = cos( 63.43) = 0.447 leading.
[d] I⇤L =
(1600 + j1200) ⇥ 103
= 666.67 + j500 A;
2400
IL = 666.67
j500 = 833.33/
36.87 A(rms);
|IL | = 833.33 A(rms).
[e] I⇤L =
(1920 + j560) ⇥ 103
= 800 + j233.33;
2400
IL = 800
j233.33 = 833.33/
16.26 A(rms);
|IL | = 833.33 A(rms).
P 10.35 [a] Pbefore = Pafter = (833.33)2 (0.05) = 34,727.22 W.
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10–26
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] Vs (before) = 2400 + (666.67
j500)(0.05 + j0.4)
= 2633.33 + j241.67 = 2644.4/5.24 V(rms).
|Vs (before)| = 2644.4 V(rms);
Vs (after) = 2400 + (800 + j233.33)(0.05 + j0.4)
= 2346.67 + j331.67 = 2369.99/8.04 V(rms);
|Vs (after)| = 2369.99 V(rms).
P 10.36 [a]
10 = j1(I1
I2 ) + j1(I3
I2 )
j1(I1
I3 );
0 = 1I2 + j2(I2
I3 ) + j1(I2
I1 ) + j1(I2
I1 ) + j1(I2
0 = I3
I1 ) + j2(I3
I2 ) + j1(I1
I2 ).
j1(I3
I3 ; )
Solving,
I1 = 6.25 + j7.5 A(rms);
I2 = 5 + j2.5 A(rms);
Ia = I1 = 6.25 + j7.5 A
I b = I1
I2 = 1.25 + j5 A;
Ic = I2 = 5 + j2.5 A
Id = I3
I2 =
I e = I1
I f = I3 = 5
I3 = 1.25 + j10 A
I3 = 5
j2.5 A(rms).
j5 A;
j2.5 A.
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Problems
10–27
[b]
Va = 10 V
Vb = j1Ib + j1Id = j1.25 V;
Vc = 1Ic = 5 + j2.5 V
Vd = j2Id
Ve =
j1Ie = 10
Vf = 1If = 5
Sa =
10I⇤a =
j1.25 V
j1Ib = 5 + j1.25 V;
j2.5 V.
62.5 + j75 VA;
Sb = Vb I⇤b = 6.25 + j1.5625 VA;
Sc = Vc I⇤c = 31.25 + j0 VA;
Sd = Vd I⇤d =
6.25 + j25 VA;
Se = Ve I⇤e = 0
j101.5625 VA;
Sf = Vf I⇤f = 31.25 VA.
[c]
X
Pdev = 62.5 W;
X
Pabs = 31.25 + 31.25 = 62.5 W.
Note that the total power absorbed by the coupled coils is zero:
6.25 6.25 = 0 = Pb + Pd .
[d]
X
Qdev = 101.5625 VAR;
Both the source and the capacitor are developing magnetizing vars.
X
X
Qabs = 75 + 1.5625 + 25 = 101.5625 VAR;
Q absorbed by the coupled coils is Qb + Qd = 26.5625.
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10–28
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.37 [a]
272/0 = 2Ig + j10Ig + j14(Ig
+j14Ig
I2 )
j8I2 + j20(Ig
j6I2
I2 )
0 = j20(I2
Ig )
j14Ig + j8I2 + j4I2
+j8(I2
Ig )
j6Ig + 8I2 .
Solving,
Ig = 20
I2 = 24/0 A(rms);
j4 A(rms);
P8⌦ = (24)2 (8) = 4608 W.
[b] Pg (developed) = (272)(20) = 5440 W.
272
Vg
[c] Zab =
2 = 11.08 + j2.62 = 11.38/13.28 ⌦.
2=
Ig
20 j4
[d] P2⌦ = |Ig |2 (2) = 832 W;
P 10.38 [a]
X
Pdiss = 832 + 4608 = 5440 W =
300 = 60I1 + V1 + 20(I1
0 = 20(I2
1
V2 = V1 ;
4
X
Pdev .
I2 );
I1 ) + V2 + 40I2 ;
I2 =
4I1 .
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Problems
10–29
Solving,
V1 = 260 V (rms);
V2 = 65 V (rms);
I1 = 0.25 A (rms);
I2 =
V5A = V1 + 20(I1
.·. P =
1.0 A (rms);
I2 ) = 285 V (rms);
(285)(5) =
1425 W.
Thus 1425 W is delivered by the current source to the circuit.
[b] I20⌦ = I1
I2 = 1.25 A(rms);
.·. P20⌦ = (1.25)2 (20) = 31.25 W.
✓
N1
P 10.39 [a] Zab = 1 +
N2
.·. I1 =
I2 =
◆2
(1
j2) = 25
100/0
15 + j50 + 25
j50
j50 ⌦;
= 2.5/0 A.
N1
I1 = 10/0 A;
N2
.·. IL = I1 + I2 = 12.5/0 A(rms).
P1⌦ = (12.5)2 (1) = 156.25 W;
P15⌦ = (2.5)2 (15) = 93.75 W.
[b] Pg =
X
100(2.5/0 ) =
250 W;
Pabs = 156.25 + 93.75 = 250 W =
P 10.40 [a] 25a21 + 4a22 = 500;
I25 = a1 I;
P25 = a21 I2 (25);
I4 = a2 I;
P4 = a22 I2 (4; )
P4 = 4P25 ;
.·.
X
Pdev .
a22 I2 4 = 100a21 I2 ;
100a21 = 4a22 .
25a21 + 100a21 = 500;
a1 = 2;
25(4) + 4a22 = 500;
a2 = 10.
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10–30
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =
2000/0
= 2/0 A(rms);
500 + 500
I25 = a1 I = 4 A;
P25⌦ = (16)(25) = 400 W.
[c] I4 = a2 I = 10(2) = 20 A(rms);
V4 = (20)(4) = 80/0 V(rms).
P 10.41 [a] From Problem 9.75, ZTh = 85 + j85 ⌦ and VTh = 850 + j850 V. Thus, for
⇤
maximum power transfer, ZL = ZTh
= 85 j85 ⌦:
I2 =
850 + j850
= 5 + j5 A.
170
425/0 = (5 + j5)I1
j20(5 + j5; )
325 + j100
= 42.5
.·. I1 =
5 + j5
j22.5 A.
Sg (del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA;
Pg = 18,062.5 W.
[b] Ploss = |I1 |2 (5) + |I2 |2 (45) = 11,562.5 + 2250 = 13,812.5 W;
% loss in transformer =
P 10.42 [a]
115.2 + j33.6
ZTh
240
.·. ZTh = 40
j100 ⌦;
+
13,812.5
(100) = 76.47%.
18,062.5
115.2 + j33.6
= 0;
80 j60
.·. ZL = 40 + j100 ⌦.
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Problems
[b] I =
10–31
240
= 30 A(rms);
80
P = (30)2 (40) = 36 kW.
100
= 15.9 mH;
2⇡(1000)
Construct the 40 ⌦ resistor by connecting four 10 ⌦ resistors in series.
Construct a 16 mH inductor by connecting two 10 mH inductors in
parallel, then connecting that parallel pair in series with a 10 mH and a 1
mH inductor:
[c] L =
0.01k0.01 + 0.01 + 0.001 = 0.016 mH.
Instead of using the series-connected 1 mH inductor, you can combine
additional inductors in series and in parallel to get an equivalent
inductance of 0.9 mH.
P 10.43 [a] ZTh = j40k40
j40 = 20
j20;
⇤
.·. ZL = ZTh
= 20 + j20 ⌦.
[b] VTh =
I=
40
(120) = 60
40 + j40
60
j60
= 1.5
40
j60 V.
j1.5 A;
1
Pload = |I|2 (20) = 45 W.
2
[c] The closest resistor value from Appendix H is 22 ⌦ . Find the inductor
value:
10,000L = 20
so
L = 2 mH.
The closest inductor value is 1 mH. The current in this load is
60 j60
I=
= 1.965/ 31.61 A
20 j20 + 22 + j10
Pload = 0.5(1.965)2 (22) = 42.5 W
(instead of 45 W)
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10–32
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.44 [a] Open circuit voltage:
V1 = 5I = 5
100 5I
;
25 + j10
(25 + j10)I = 100
5I ;
100
=3
30 + j10
j A;
I =
VTh =
j3
(5I ) = 15 V.
1 + j3
Short circuit current:
V2 = 5I =
I =3
100 5I
;
25 + j10
j1 A;
5I
= 15 j5 A;
1
15
= 0.9 + j0.3 ⌦;
ZTh =
15 j5
Isc =
⇤
ZL = ZTh
= 0.9
IL =
j0.3 ⌦.
15
= 8.33 A(rms);
1.8
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Problems
10–33
P = |IL |2 (0.9) = 62.5 W.
[b] VL = (0.9
VL
=
j3
j0.3)(8.33) = 7.5
j2.5 V(rms).
0.833
j2.5 A(rms);
I2 = I1 + IL = 7.5
j2.5 A(rms);
I1 =
5I = I2 + VL
.·.
Id.s. = I
4.5 + j1.5 A;
Sg =
I2 =
100(3 + j1) =
Sd.s. = 5(3
I =3
300
j1)( 4.5
j1 A;
j100 VA;
j1.5) =
75 + j0 VA;
Pdev = 300 + 75 = 375 W;
% developed =
62.5
(100) = 16.67%.
375
Checks:
P25⌦ = (10)(25) = 250 W;
P1⌦ = (67.5)(1) = 67.5 W;
P0.9⌦ = 62.5 W;
X
Pabs = 230 + 62.5 + 67.5 = 375 =
Qj10 = (10)(10) = 100 VAR;
X
Pdev .
Qj3 = (6.94)(3) = 20.82 VAR;
Q j0.3 = (69.4)( 0.3) =
Qsource =
X
20.82 VAR;
100 VAR;
Q = 100 + 20.82
20.82
100 = 0.
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10–34
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.45 ZL = |ZL |/✓ = |ZL | cos ✓ + j|ZL | sin ✓ .
|VTh |
.
Thus |I| = q
(RTh + |ZL | cos ✓)2 + (XTh + |ZL | sin ✓)2
0.5|VTh |2 |ZL | cos ✓
Therefore P =
.
(RTh + |ZL | cos ✓)2 + (XTh + |ZL | sin ✓)2
Let D = demoninator in the expression for P, then
(0.5|VTh |2 cos ✓)(D · 1
dP
=
d|ZL |
D2
|ZL |dD/d|ZL |)
;
dD
= 2(RTh + |ZL | cos ✓) cos ✓ + 2(XTh + |ZL | sin ✓) sin ✓;
d|ZL |
!
dD
dP
= 0 when D = |ZL |
.
d|ZL |
d|ZL |
Substituting the expressions for D and (dD/d|ZL |) into this equation gives us
2
2
the relationship RTh
+ XTh
= |ZL |2 or |ZTh | = |ZL |.
P 10.46 [a] ZTh = [(3 + j4)k
j8] + 7.32
j17.24 = 15
j15 ⌦;
.·. R = |ZTh | = 21.21 ⌦.
[b] VTh =
I=
j8
(112.5) = 144
3 j4
144 j108
= 4.45
36.21 j15
j108 V(rms).
j1.14 A(rms);
P = |I|2 (21.21) = 447.35 W.
[c] Pick the 22 ⌦ resistor from Appendix H for the closest match:
I=
144
37
j108
= 4.36
j15
j1.15 A(rms);
P = |I|2 (22) = 447.177 W.
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Problems
10–35
P 10.47 [a] Open circuit voltage:
V
100
5
+
V
j5
0.1V = 0;
.·. V = 40 + j80 V(rms).
VTh = V + 0.1V ( j5) = V (1
j0.5) = 80 + j60 V(rms).
Short circuit current:
Isc = 0.1V +
100
V
5
+
V
= (0.1 + j0.2)V ;
j5
V
V
+
= 0;
j5
j5
.·. V = 100 V(rms).
Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms);
ZTh =
80 + j60
VTh
=4
=
Isc
10 + j20
j2 ⌦;
.·. Ro = |ZTh | = 4.47 ⌦.
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10–36
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b]
80 + j60
p
= 7.36 + j8.82 A (rms);
4 + 20 j2
p
P = (11.49)2 ( 20) = 590.2 W.
I=
[c]
I=
80 + j60
= 10 + j7.5 A (rms);
8
P = (102 + 7.52 )(4) = 625 W.
[d]
V
100
5
+
V
Vo
+
j5
(25 + j50)
= 0;
j5
V = 50 + j25 V (rms);
0.1V = 5 + j2.5 V (rms);
5 + j2.5 + IC = 10 + j7.5;
IC = 5 + j5 A (rms);
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Problems
IL =
V
=5
j5
j10 A (rms);
IR = IC + IL = 10
j5 A (rms);
Ig = IR + 0.1V = 15
j2.5 A (rms);
100I⇤g =
j250 VA;
Sg =
10–37
1500
100 = 5(5 + j2.5) + Vcs + 25 + j50
.·.
Scs = (50
j437.5 VA.
j62.5)(5
j2.5) = 93.75
Vcs = 50
j62.5 V (rms);
Thus,
X
Pdev = 1500;
% delivered to Ro =
625
(100) = 41.67%.
1500
P 10.48 [a] The real power lost in the line is minimum if the line current is minimum.
The line current is minimum if the power factor of the load is 1, making
the load purely resistive. This will happen if the impedance of the
capacitor is 100 ⌦:
1
= 100 ⌦;
!C
[b] Iwo =
C=
1
= 26.53 µF.
(100)(120⇡)
13,800 13,800
+
= 46
300
j100
Vswo = 13,800 + (46
j138 A(rms);
j138)(1.5 + j12) = 15,525 + j345
= 15,528.83/1.27 V(rms);
Iw =
13,800
= 46 A(rms);
300
Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28 V(rms);
% increase =
[c] P`wo = |46
!
15,528.82
13,879.98
1 (100) = 11.88%.
j138|2 1.5 = 31.74 kW;
P`w = 462 (1.5) = 3174 W;
✓
31,740
% increase =
3174
◆
1 (100) = 900%.
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10–38
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.49 [a] First find the Thévenin equivalent:
j!L = j3000 ⌦;
ZTh = 6000k12,000 + j3000 = 4000 + j3000 ⌦;
VTh =
12,000
(180) = 120 V;
6000 + 12,000
j
=
!C
j1000 ⌦.
I=
120
= 18
6000 + j2000
j6 mA;
1
P = |I|2 (2000) = 360 mW.
2
[b] Set Co = 0.1 µF so
j/!C =
Set Ro as close as possible to
Ro =
q
40002 + (3000
j2000 ⌦.
2000)2 = 4123.1 ⌦;
.·. Ro = 4000 ⌦.
[c] I =
120
= 14.77
8000 + j1000
j1.85 mA;
1
P = |I|2 (4000) = 443.1 mW;
2
Yes;
443.1 mW > 360 mW.
120
[d] I =
= 15 mA;
8000
1
P = (0.015)2 (4000) = 450 mW.
2
[e] Ro = 4000 ⌦;
[f ] Yes;
Co = 66.67 nF.
450 mW > 443.1 mW.
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Problems
P 10.50 [a] Set Co = 0.1 µF, so
I=
j/!C =
10–39
j2000 ⌦; also set Ro = 4123.1 ⌦.
120
= 14.55
8123.1 + j1000
j1.79 mA;
1
P = |I|2 (4123.1) = 443.18 mW.
2
[b] Yes;
443.18 mW > 360 mW.
[c] Yes;
443.18 mW < 450 mW.
P 10.51 [a] j!L1 = j!L2 = j(400)(625 ⇥ 10 3 ) = j250 ⌦;
j!M = j(400)(312.5 ⇥ 10 3 ) = j125 ⌦.
400 = (125 + j250)Ig
0=
j125IL ;
j125Ig + (375 + j250)IL .
Solving,
Ig = 0.8
j1.2 A;
IL = 0.4 A.
Thus,
ig = 1.44 cos(400t
56.31 ) A;
iL = 0.4 cos 400t A.
M
0.3125
= 0.5.
=
0.625
L1 L2
[c] When t = 1.25⇡ ms:
[b] k = p
400t = (400)(1.25⇡) ⇥ 10 3 = 0.5⇡ rad = 90 ;
ig (1.25⇡ ms) = 1.44 cos(90
56.31 ) = 1.2 A;
iL (1.25⇡ ms) = 0.4 cos(90 ) = 0 A;
1
1
w = L1 i21 + L2 i22
2
2
1
M i1 i2 = (0.625)(1.2)2 + 0
2
0 = 450 mJ.
When t = 2.5⇡ ms:
400t = (400)(2.5⇡) ⇥ 10 3 = ⇡ = 180 ;
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10–40
CHAPTER 10. Sinusoidal Steady State Power Calculations
ig (2.5⇡ ms) = 1.44 cos(180
56.31 ) =
iL (2.5⇡ ms) = 0.4 cos(180) =
0.4 A;
1
1
w = (0.625)(0.8)2 + (0.625)( 0.4)2
2
2
0.8 A;
(0.3125)( 0.8)( 0.4) = 150 mJ.
[d] From (a), IL = 0.4 A,
1
.·. P = (0.4)2 (375) = 30 W.
2
[e] Open circuit:
VTh =
400
(j125) = 160 + j80 V.
125 + j250
Short circuit:
400 = (125 + j250)I1
0=
j125Isc ;
j125I1 + j250Isc .
Solving,
Isc = 0.4923
ZTh =
j0.7385;
160 + j80
VTh
= 25 + j200 ⌦;
=
Isc
0.4923 j0.7385
.·. RL = 201.56 ⌦.
[f ]
I=
160 + j80
= 0.592/
226.56 + j200
14.87 A;
1
P = (0.592)2 (201.56) = 35.3 W.
2
⇤
[g] ZL = ZTh
= 25 j200 ⌦.
160 + j80
[h] I =
= 3.58/26.57 ;
50
1
P = (3.58)2 (25) = 160 W.
2
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Problems
10–41
P 10.52 [a]
54 = I1 + j2(I1
I2 ) + j3I2 ;
0 = 7I2 + j2(I2
I1 )
j3I2 + j8I2 + j3(I1
I2 ).
Solving,
I1 = 12
j21 A (rms);
I2 =
21/0 V(rms)
Vo = 7I2 =
3 A (rms);
so
|Vo | = 21 V(rms).
[b] P = |I2 |2 (7) = 63 W.
[c] Pg =
(54)(12) =
648 W;
% delivered =
63
(100) = 9.72%.
648
54 = I1 + j2(I1
I2 ) + j4kI2 ;
0 = 7I2 + j2(I2
I1 )
P 10.53 [a]
j4kI2 + j8I2 + j4k(I1
I2 ).
Place the equations in standard form:
54 = (1 + j2)I1 + j(4k
0 = j(4k
I1 =
54
2)I2 ;
2)I1 + [7 + j(10
8k)]I2 ;
I2 j(4k 2)
.
(1 + j2)
Substituting,
I2 =
[7 + j(10
j54(4k 2)
8k)](1 + j2)
For Vo = 0, I2 = 0, so if 4k
(4k
2)
.
2 = 0, then k = 0.5.
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10–42
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] When I2 = 0,
54
= 10.8
I1 =
1 + j2
j21.6 A(rms);
Pg = (54)(10.8) = 583.2 W.
Check:
Ploss = |I1 |2 (1) = 583.2 W.
P 10.54 [a]
Open circuit:
VTh =
I1 =
j3I1 + j2I1 =
54
= 10.8
1 + j2
VTh =
21.6
jI1 ;
j21.6;
j10.8 V.
Short circuit:
54 = I1 + j2(I1
Isc ) + j3Isc;
0 = j2(Isc
j3Isc + j8Isc + j3(I1
I1 )
Isc .)
Solving,
Isc =
3.32 + j5.82;
ZTh =
VTh
=
Isc
21.6 j10.8
= 0.2 + j3.6 = 3.6/86.86 ⌦;
3.32 + j5.82
.·. RL = |ZTh | = 3.6 ⌦.
[b]
I=
21.6 j10.8
= 4.614/163.1 ;
3.8 + j3.6
P = |I|2 (3.6) = 76.6 W,
which is greater than when RL = 7 ⌦.
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Problems
10–43
P 10.55 Open circuit voltage:
I1 =
30/0
= 0.4
15 + j30
VTh = j30I1
j0.8 A;
j18I1 = j12I1 = 9.6 + j4.8 = 10.73/26.57 .
Short circuit current:
30/0 = 15I1 + j30(I1
0 = j15Isc
j18(Isc
Isc ) + j18Isc ;
I1 ) + j30(Isc
I1 )
j18Isc .
Solving,
Isc = 1.95/
ZTh =
43.025 A;
9.6 + j4.8
= 1.92 + j5.16 ⌦;
1.95/ 43.025
.·. I2 =
9.6 + j4.8
= 2.795/26.57 A.
3.84
30/0 = 15I1 + j30(I1
I2 ) + j18I2 ;
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10–44
CHAPTER 10. Sinusoidal Steady State Power Calculations
30 + j12(2.795/26.57 )
30 + j12I2
=
= 1 A;
15 + j30
15 + j30
.·.
I1 =
Zg =
30/0
= 30 + j0 = 30/0 ⌦.
1
P 10.56 [a]
180 = 3I1 + j4I1 + j3(I2
0 = 9I2 + j9(I2
I1 ) + j9(I1
I2 )
j3I1 ;
I1 ) + j3I1 .
Solving,
I1 = 18
j18 A(rms);
I2 = 12/0 A(rms);
.·. Vo = (12)(9) = 108 V(rms).
[b] P = (12)2 (9) = 1296 W.
[c] Sg =
(180)(18 + j18) =
% delivered =
3240
j3240 VA
.·. Pg =
3240 W;
1296
(100) = 40%.
3240
P 10.57 [a] Open circuit voltage:
180 = 3I1 + j4I1
j3I1 + j9I1
180
= 9.31
.·. I1 =
3 + j7
VTh = j9I1
j3I1 ;
j21.72 A(rms).
j3I1 = j6I1 = 130.34 + j55.86 V.
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Problems
10–45
Short circuit current:
180 = 3I1 + j4I1 + j3(Isc
0 = j9(Isc
I1 ) + j9(I1
Isc )
j3I1 ;
I1 ) + j3I1 .
Solving,
Isc = 20
ZTh =
IL =
j20 A;
130.34 + j55.86
VTh
= 1.86 + j4.66 ⌦.
=
Isc
20 j20
130.34 + j55.86
= 35.04 + j15.02 = 38.12/23.2 A;
3.72
PL = (38.12)2 (1.86) = 2702.83 W.
[b] From Problem 10.56(a),
I1 =
Zo + j9
1.86
I2 =
j6
j4.66 + j9
(35.04 + j15.02) = 30/0 A(rms);
j6
Pdev = (180)(30) = 5400 W.
[c] Begin by choosing the capacitor value from Appendix H that is closest to
the required reactive impedance, assuming the frequency of the source is
60 Hz:
1
1
4.66 =
so
C=
= 569.22 µF.
2⇡(60)C
2⇡(60)(4.66)
Choose the capacitor value closest to this capacitance from Appendix H,
which is 470 µF. Then,
XL =
1
=
2⇡(60)(470 ⇥ 10 6 )
5.6438 ⌦.
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10–46
CHAPTER 10. Sinusoidal Steady State Power Calculations
Now set RL as close as possible to
RL =
q
q
2
RTh
+ (XL + XTh )2 :
1.862 + ( 5.6438 + 4.66)2 = 2.1 ⌦.
The closest single resistor value from Appendix H is 10 ⌦. The resulting
real power developed by the source is calculated below, using the
Thévenin equivalent circuit:
IL =
130.34 + j55.86
= 11.916/27.97 ;
1.86 + j4.66 + 10 j5.6438
PL = (11.916)2 (10) = 1419.9 W
(instead of 2702.83 W).
P 10.58 [a]
Open circuit:
VTh =
120
(j10) = 36 + j48 V.
16 + j12
Short circuit:
(16 + j12)I1
j10Isc = 120;
j10I1 + (11 + j23)Isc = 0.
Solving,
Isc = 2.4 A;
ZTh =
36 + j48
= 15 + j20 ⌦;
2.4
⇤
.·. ZL = ZTh
= 15
IL =
j20 ⌦.
36 + j48
VTh
= 1.2 + j1.6 A(rms);
=
ZTh + ZL
30
PL = |IL |2 (15) = 60 W.
[b] With the load impedance attached, the mesh current equations are
(16 + j12)I1
j10I2 = 120;
j10I1 + (26 + j3)I2 = 0.
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Problems
Solving,
I1 = 4.52
j2.64 A (rms);
10–47
I2 = 1.2 + j1.6 A (rms).
Ptransformer = 12|I1 |2 + 11|I2 |2 = 12(27.4) + 11(4) = 372.8 W;
% delivered =
60
(100) = 16.1%.
372.8
P 10.59
Vo
Va
=
;
1
50
.·.
1Ia =
Vo /50
Vo /Io
5000
=
=
= 2 ⌦.
50Io
502
502
Va
=
Ia
Va
Vb
=
;
25
1
50Io ;
25Ib = 1Ia ;
.·.
Vb /Ib
Va
Vb /25
=
=
;
2
25Ib
25
Ia
.·.
Vb
Va
= 252
= 252 (2) = 1250 ⌦.
Ib
Ia
Thus Ib = [145/(200 + 1250)] = 100 mA (rms); since the ideal transformers are
lossless, P5k⌦ = P1250⌦ , and the power delivered to the 1250 ⌦ resistor is
(0.1)2 (1250) or 12.5 W.
252 (5000)
Vb
=
= 200 ⌦;
therefore a2 = 15,625,
a = 125.
2
Ib
a
145
= 362.5 mA;
P = (0.3625)2 (200) = 26.28125 W.
[b] Ib =
400
P 10.60 [a]
P 10.61 [a] Zab = 50
.·. ZL =
✓
N1
N2
(50
j400) = 2
j400 = 1
1
(1
6)2
◆2
ZL ;
j16 ⌦.
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10–48
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b]
I1 =
24
= 240/0 mA.
100
N1 I1 =
I2 =
N2 I2 ;
1.44/0 A;
6I1 =
IL = I1 + I2 =
VL = (2
1.68/0 A;
3.36 + j26.88 = 27.1/97.13 V(rms).
j16)IL =
P 10.62 [a]
For maximum power transfer, Zab = 90 k⌦
✓
Zab = 1
N1
N2
◆2
ZL ;
◆2
=
90,000
= 225.
400
.·.
✓
1
N1
= ±15;
N2
N1
N2
1
2
[b] P = |Ii | (90,000) =
N1
= 15 + 1 = 16.
N2
180
180,000
!2
(90,000) = 90 mW.
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Problems
180
[c] V1 = Ri Ii = (90,000)
180,000
!
10–49
= 90 V.
[d]
Vg = (2.25 ⇥ 10 3 )(100,000k80,000) = 100 V;
Pg (del) = (2.25 ⇥ 10 3 )(100) = 225 mW;
% delivered =
90
(100) = 40%.
225
P 10.63 [a] Replace the circuit to the left of the primary winding with a Thévenin
equivalent:
VTh = (15)(20kj10) = 60 + j120 V;
ZTh = 2 + 20kj10 = 6 + j8 ⌦.
Transfer the secondary impedance to the primary side:
Zp =
1
(100
25
jXC ) = 4
j
XC
⌦.
25
Now maximize I by setting (XC /25) = 8 ⌦:
.·. C =
[b] I =
1
= 0.25 µF.
200(20 ⇥ 103 )
60 + j120
= 6 + j12 A;
10
P = |I|2 (4) = 720 W.
Ro
= 6 ⌦;
.·. Ro = 150 ⌦.
25
60 + j120
= 5 + j10 A;
[d] I =
12
[c]
P = |I|2 (6) = 750 W.
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10–50
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.64 [a] Open circuit voltage:
40/0 = 4(I1 + I3 ) + 12I3 + VTh ;
I1
=
4
I3 ;
I1 =
4I3 .
Solving,
VTh = 40/0 V.
Short circuit current:
40/0 = 4I1 + 4I3 + I1 + V1 ;
4V1 = 16(I1 /4) = 4I1 ;
.·. V1 = I1 ;
.·. 40/0 = 6I1 + 4I3 .
Also,
40/0 = 4(I1 + I3 ) + 12I3 .
Solving,
I1 = 6 A;
RTh =
I3 = 1 A;
Isc = I1 /4 + I3 = 2.5 A;
VTh
40
= 16 ⌦.
=
Isc
2.5
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Problems
I=
10–51
40/0
= 1.25/0 A(rms);
32
P = (1.25)2 (16) = 25 W.
[b]
40 = 4(I1 + I3 ) + 12I3 + 20;
4V1 = 4I1 + 16(I1 /4 + I3 );
.·. V1 = 2I1 + 4I3 ;
40 = 4I1 + 4I3 + I1 + V1 ;
.·. I1 = 6 A;
I3 =
0.25 A;
I1 + I3 = 5.75/0 A.
P40V (developed) = 40(5.75) = 230 W;
.·. % delivered =
[c] PRL = 25 W;
25
(100) = 10.87%.
230
P16⌦ = (1.5)2 (16) = 36 W;
P4⌦ = (5.75)2 (4) = 132.25 W;
P1⌦ = (6)2 (1) = 36 W;
P12⌦ = ( 0.25)2 (12) = 0.75 W;
X
Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W =
P 10.65 [a] Open circuit voltage:
X
Pdev .
500 = 100I1 + V1 ;
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10–52
CHAPTER 10. Sinusoidal Steady State Power Calculations
V2 = 400I2 ;
V2
V1
=
1
2
.·.
V2 = 2V1 ;
I1 = 2I2 .
Substitute and solve:
2V1 = 400I1 /2 = 200I1
500 = 100I1 + 100I1
.·.
.·.
.·.
V1 = 100I1 ;
I1 = 500/200 = 2.5 A;
1
I2 = I1 = 1.25 A;
2
V1 = 100(2.5) = 250 V;
VTh = 20I1 + V1
V2 = 2V1 = 500 V;
V2 + 40I2 =
150 V(rms).
Short circuit current:
500 = 80(Isc + I1 ) + 360(Isc + 0.5I1 ; )
2V1 = 40
I1
+ 360(Isc + 0.5I1 );
2
500 = 80(I1 + Isc ) + 20I1 + V1 .
Solving,
Isc =
1.47 A;
RTh =
VTh
=
Isc
P =
150
= 102 ⌦.
1.47
752
= 55.15 W.
102
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Problems
10–53
[b]
500 = 80[I1
(75/102)]
75 + 360[I2
575 +
6000 27,000
+
= 80I1 + 180I2 ;
102
102
.·.
I1 = 3.456 A.
Psource = (500)[3.456
% delivered =
(75/102)];
(75/102)] = 1360.35 W;
55.15
(100) = 4.05%.
1360.35
[c] P80⌦ = 80(I1 + IL )2 = 592.13 W;
P20⌦ = 20I21 = 238.86 W;
P40⌦ = 40I22 = 119.43 W;
P102⌦ = 102I2L = 55.15 W;
P360⌦ = 360(I2 + IL )2 = 354.73 W;
X
Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W =
✓
200
P 10.66 [a] ZTh = 720 + j1500 +
50
◆2
(40
X
Pdev .
j30) = 1360 + j1020 = 1700/36.87 ⌦;
.·. Zab = 1700 ⌦.
Zab =
ZL
;
(1 + N1 /N2 )2
(1 + N1 /N2 )2 = 6800/1700 = 4;
.·. N1 /N2 = 1
or
N2 = N1 = 1000 turns.
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10–54
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] VTh =
IL =
255/0
(j200) = 1020/53.13 V.
40 + j30
1020/53.13
= 0.316/34.7 A(rms).
3060 + j1020
Since the transformer is ideal, P6800 = P1700 .
P = |I|2 (1700) = 170 W.
[c]
255/0 = (40 + j30)I1
.·. I1 = 4.13
j200(0.26 + j0.18);
j1.80 A(rms).
Pgen = (255)(4.13) = 1053 W;
Pdiss = 1053
170 = 883 W;
% dissipated =
883
(100) = 83.85%.
1053
30[5(44.28) + 19(15.77)]
= 15.63 kWh.
1000
30[5(44.28) + 19(8.9)]
[b]
= 11.72 kWh.
1000
30[5(44.28) + 19(4.42)]
[c]
= 9.16 kWh.
1000
30[5(44.28) + 19(0)]
[d]
= 6.64 kWh.
1000
Note that this is about 40 % of the amount of total power consumed in
part (a).
P 10.67 [a]
P 10.68 [a]
30[0.2(1433) + 23.8(3.08)]
= 10.8 kWh.
1000
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Problems
10–55
[b] The standby power consumed in one month by the microwave oven when
in the ready state is
30[23.8(3.08)]
= 2.2 kWh.
1000
This is (2.2/10.8) ⇤ 100 = 20.4% of the total power consumed by the
microwave in one month. Since it is not practical to unplug the
microwave when you are not using it, this is the cost associated with
having a microwave oven.
P 10.69 j!L1 = j(2⇡)(60)(0.25) = j94.25 ⌦;
I=
120
= 1.27/
5 + j94.25
86.96 A(rms);
P = R1 |I|2 = 5(1.27)2 = 8.06 W.
P 10.70 j!L1 = j(2⇡)(60)(0.25) = j94.25 ⌦;
I=
120
= 1.27/
0.05 + j94.25
86.97 A(rms);
P = R1 |I|2 = 0.05(1.27)2 = 81.1 mW.
Note that while the current supplied by the voltage source is virtually
identical to that calculated in Problem 10.69, the much smaller value of
transformer resistance results in a much smaller value of real power consumed
by the transformer.
P 10.71 An ideal transformer has no resistance, so consumes no real power. This is one
of the important characteristics of ideal transformers.
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Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to
find VAB , and use the known phase angle relationship between VAB and VBC
to find VBC .
VAB = VAN + VNB = VAN
VBN .
Since VAN , VBN , and VCN form a balanced set, and VAN = 240/
the phase sequence is positive,
VBN = |VAN |//VAN
120 = 240/
30
120 = 240/
30 V, and
150 V.
Then,
VAB = VAN
VBN = (240/
30 )
(240/
150 ) = 415.46/0 V.
Since VAB , VBC , and VCA form a balanced set with a positive phase sequence,
we can find VBC from VAB :
VBC = |VAB |/(/VAB
120 ) = 415.69/0
120 = 415.69/
120 V.
11–1
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11–2
CHAPTER 11. Balanced Three-Phase Circuits
Thus,
VBC = 415.69/
120 V.
AP 11.2 Make a sketch:
We know VCN and wish to find VAB . To do this, write a KVL equation to
find VBC , and use the known phase angle relationship between VAB and VBC
to find VAB .
VBC = VBN + VNC = VBN
VCN .
Since VAN , VBN , and VCN form a balanced set, and VCN = 450/
the phase sequence is negative,
VBN = |VCN |//VCN
120 = 450/
23
120 = 450/
25 V, and
145 V.
Then,
VBC = VBN
VCN = (450/
145 )
(450/
25 ) = 779.42/
175 V.
Since VAB , VBC , and VCA form a balanced set with a negative phase
sequence, we can find VAB from VBC :
VAB = |VBC |//VBC
120 = 779.42/
295 V.
But we normally want phase angle values between +180 and
360 to the phase angle computed above. Thus,
180 . We add
VAB = 779.42/65 V
AP 11.3 Sketch the a-phase circuit:
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Problems
11–3
[a] We can find the line current using Ohm’s law, since the a-phase line
current is the current in the a-phase load. Then we can use the fact that
IaA , IbB , and IcC form a balanced set to find the remaining line currents.
Note that since we were not given any phase angles in the problem
statement, we can assume that the phase voltage given, VAN , has a phase
angle of 0 .
2400/0 = IaA (16 + j12)
so
IaA =
2400/0
= 96
16 + j12
j72 = 120/
36.87 A.
With an acb phase sequence,
/IbB = /IaA + 120
and /IcC = /IaA
120
so
IaA = 120/
36.87 A;
IbB = 120/83.13 A;
IcC = 120/
156.87 A.
[b] The line voltages at the source are Vab Vbc , and Vca . They form a
balanced set. To find Vab , use the a-phase circuit to find VAN , and use
the relationship between phase voltages and line voltages for a
y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL:
Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0
j72) + 2400/0 = 2467.2 + j69.6
= (0.1 + j0.8)(96
= 2468.18/1.62 V.
From Fig. 11.9(b),
p
Vab = Van ( 3/ 30 ) = 4275.02/
28.38 V.
With an acb phase sequence,
/Vbc = /Vab + 120
and /Vca = /Vab
120
so
Vab = 4275.02/
28.38 V;
Vbc = 4275.02/91.62 V;
Vca = 4275.02/
148.38 V.
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11–4
CHAPTER 11. Balanced Three-Phase Circuits
[c] Using KVL on the a-phase circuit
Va0 n = Va0 a + Van = (0.2 + j0.16)IaA + Van
= (0.02 + j0.16)(96
j72) + (2467.2 + j69.9)
= 2480.64 + j83.52 = 2482.05/1.93 V.
With an acb phase sequence,
/Vb0 n = /Va0 n + 120
and /Vc0 n = /Va0 n
120
so
Va0 n = 2482.05/1.93 V;
Vb0 n = 2482.05/121.93 V;
Vc0 n = 2482.05/
AP 11.4
p
IcC = ( 3/
118.07 V.
p
30 )ICA = ( 3/
30 ) · 8/
15 = 13.86/
45 A.
AP 11.5
120 ) = 12/ 55 ;
IaA = 12/(65
!
#
!
"
/ 30
1
p / 30 IaA =
p
· 12/
IAB =
3
3
= 6.93/
AP 11.6 [a] IAB =
"
85 A.
!
1
p /30
3
Therefore Z =
[b] IAB =
"
55
!
1
p /
3
#
[69.28/
10 ] = 40/20 A.
4160/0
= 104/
40/20
30
#
[69.28/
20 ⌦.
10 ] = 40/
40 A.
Therefore Z = 104/40 ⌦.
AP 11.7
I =
110
110
+
= 30
3.667 j2.75
Therefore |IaA | =
AP 11.8 [a] |S| =
Q=
p
q
p
j40 = 50/
3I =
p
53.13 A.
3(50) = 86.60 A.
3(208)(73.8) = 26,587.67 VA;
(26,587.67)2
(22,659)2 = 13,909.50 VAR.
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Problems
[b] pf =
11–5
22,659
= 0.8522 lagging.
26,587.67
!
2450
/0 V;
AP 11.9 [a] VAN = p
VAN I⇤aA = S = 144 + j192 kVA.
3
Therefore
(144 + j192)1000
p
= (101.8 + j135.7) A;
I⇤aA =
2450/ 3
IaA = 101.8
j135.7 = 169.67/
53.13 A;
|IaA | = 169.67 A.
[b] P =
(2450)2
;
R
therefore R =
(2450)2
;
X
therefore X =
Q=
(2450)2
= 41.68 ⌦.
144,000
(2450)2
= 31.26 ⌦.
192,000
p
VAN
2450/ 3
[c] Z =
=
= 8.34/53.13 = (5 + j6.67) ⌦;
IaA
169.67/ 53.13
.·. R = 5 ⌦,
X = 6.67 ⌦.
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11–6
CHAPTER 11. Balanced Three-Phase Circuits
Problems
P 11.1
[a] First, convert the cosine waveforms to phasors:
Va = 180/27 ;
Vb = 180/147 ;
Vc = 180/
93 .
Subtract the phase angle of the a-phase from all phase angles:
/V0a = 27
27 = 0 ;
/V0b = 147
27 = 120 ;
/V0c =
27 =
93
120 .
Compare the result to Eqs. 11.1 and 11.2:
Therefore acb.
[b] First, convert the cosine waveforms to phasors:
Va = 4160/
18 ;
Vb = 4160/
138 ;
Vc = 4160/ + 102 .
Subtract the phase angle of the a-phase from all phase angles:
/V0a =
18 + 18 = 0 ;
/V0b =
138 + 18 =
120 ;
/V0c = 102 + 18 = 120 .
Compare the result to Eqs. 11.1 and 11.2:
Therefore abc.
P 11.2
[a] Va = 180/0 V;
Vb = 180/
120 V;
Vc = 180/ 240 = 180/120 V.
Balanced, positive phase sequence.
[b] Va = 180/
90 V;
Vb = 180/30 V;
Vc = 180/ 210 V = 180/150 V.
Balanced, negative phase sequence.
[c] Va = 400/
270 V = 400/90 V;
Vb = 400/120 V;
Vc = 400/ 30 V.
Unbalanced, phase angle in b-phase.
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Problems
11–7
[d] Va = 200/30 V;
Vb = 201/150 V;
Vc = 200/270 V = 200/
90 V.
Unbalanced, unequal amplitude in the b-phase.
[e] Va = 208/42 V;
Vb = 208/
78 V;
Vc = 208/
201 V = 208/159 V.
Unbalanced, phase angle in the c-phase.
[f ] Unbalanced; the frequencies of the waveforms are not the same.
P 11.3
Va = Vm /0 = Vm + j0;
Vb = Vm /
120 =
Vm (0.5 + j0.866);
Vc = Vm /120 = Vm ( 0.5 + j0.866);
Va + Vb + Vc = (Vm )(1 + j0
0.5
j0.866
0.5 + j0.866)
= Vm (0) = 0.
180/
90 + 180/30 + 180/150
= 0.
3(RW + jXW )
P 11.4
I=
P 11.5
I=
P 11.6
[a] Unbalanced, because the load impedance in every phase is di↵erent.
200
b] IaA =
= 8 A;
25
400/90 + 400/120 + 400/
3(RW + jXW )
30
IbB =
200/ 120
= 4/
30 j40
IcC =
200/120
= 2/83.13 A;
80 + j60
[a] IaA =
277/0
= 2.77/
80 + j60
IbB =
188.56/75
.
RW + jXW
66.87 A;
Io = IaA + IbB + IcC = 9.96/
P 11.7
=
9.79 A.
36.87 A (rms);
277/ 120
= 2.77/
80 + j60
156.87 A (rms);
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11–8
CHAPTER 11. Balanced Three-Phase Circuits
IcC =
277/120
= 2.77/83.13 A (rms);
80 + j60
Io = IaA + IbB + IcC = 0.
[b] VAN = (78 + j54)IaA = 262.79/
[c] VAB = VAN
2.17 V (rms).
VBN ;
VBN = (77 + j56)IbB = 263.73/
VAB = 262.79/
2.17
263.73/
120.84 V (rms);
120.84 = 452.89/28.55 V (rms).
[d] Unbalanced — see conditions for a balanced circuit in the text.
P 11.8
Zga + Zla + ZLa = 60 + j80 ⌦;
Zgb + Zlb + ZLb = 40 + j30⌦;
Zgc + Zlc + ZLc = 20 + j15⌦;
Let n be the reference node. Then,
VN 240/ 120
VN
VN 240 VN 240/120
+
+
+
= 0.
60 + j80
40 + j30
20 + j15
10
Solving for VN yields
VN = 42.94/
Io =
P 11.9
156.32 V;
VN
= 4.29/
10
156.32 A.
Make a sketch of the load in the frequency domain. Note that we convert the
time domain line-to-neutral voltages to phasors:
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Problems
11–9
Note that these three voltages form a balanced set with an abc phase
sequence. First, use KVL to find VAB :
VAB = VAN + VNB = VAN
= (169.71/26 )
VBN
(169.71/
94 ) = 293.95/56 V.
With an abc phase sequence,
/VBC = /VAB
120
and /VCA = /VAB + 120
so
VAB = 293.95/56 V;
VBC = 293.95/
64 V;
VCA = 293.95/176 V.
To get back to the time domain, perform an inverse phasor transform of the
three line voltages, using a frequency of !:
vAB (t) = 293.95 cos(!t + 56 ) V;
vBC (t) = 293.95 cos(!t
64 ) V;
vCA (t) = 293.95 cos(!t + 176 ) V.
p
P 11.10 [a] Van = 1/ 3/ 30 Vab = 110/
The a-phase circuit is
[b] IaA =
110/ 90
= 2.2/
40 + j30
90 V (rms).
126.87 A (rms).
[c] VAN = (37 + j28)IaA = 102.08/ 89.75 V (rms);
p
VAB = 3/30 VAN = 176.81/ 59.75 A (rms).
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11–10
CHAPTER 11. Balanced Three-Phase Circuits
P 11.11 Make a sketch of the three-phase line and load:
Z` = 0.25 + j2 ⌦/ ;
ZL = 30.48 + j22.86 ⌦/ .
[a] The line currents are IaA , IbB , and IcC . To find IaA , first find VAN and use
Ohm’s law for the a-phase load impedance. Since we are only concerned
with finding voltage and current magnitudes, the phase sequence doesn’t
matter and we arbitrarily assume a positive phase sequence. Since we are
not given any phase angles in the problem statement, we can assume the
angle of VAB is 0 . Use Fig. 11.9(a) to find VAN from VAB .
660
VAN = p /(0 30 ) = 381.05/
3
Now find IaA using Ohm’s law:
30 V.
VAN
381.05/ 30
= 8 j6 = 10/
=
ZL
30.48 + j22.86
Thus, the magnitude of the line current is
IaA =
36.87 V.
|IaA | = 10 A.
[b] The line voltage at the source is Vab . From KVL on the top loop of the
three-phase circuit,
Vab = VaA + VAB + VBb
= Z` IaA + VAB + Z` IBb
= Z` IaA + VAB
= (0.25 + j2)(10/
Z` IbB
36.87 ) + 660/0
(0.25 + j2)(10/
156.87 )
= 669.3/2.9 V.
Thus, the magnitude of the line voltage at the source is
|Vab | = 669.3 V.
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Problems
11–11
P 11.12 Make a sketch of the a-phase:
[a] Find the a-phase line current from the a-phase circuit:
IaA =
125/0
125/0
=
0.1 + j0.8 + 19.9 + j14.2
20 + j15
=4
j3 = 5/
36.87 A (rms).
Find the other line currents using the acb phase sequence:
IbB = 5/
36.87 + 120 = 5/83.13 A (rms);
IcC = 5/
36.87
120 = 5/
156.87 A (rms).
[b] The phase voltage at the source is Van = 125/0 V. Use Fig. 11.9(b) to
find the line voltage, Van , from the phase voltage:
p
Vab = Van ( 3/ 30 ) = 216.51/ 30 V (rms).
Find the other line voltages using the acb phase sequence:
Vbc = 216.51/
30 + 120 = 216.51/90 V (rms);
Vca = 216.51/
30
120 = 216.51/
150 V (rms).
[c] The phase voltage at the load in the a-phase is VAN . Calculate its value
using IaA and the load impedance:
VAN = IaA ZL = (4
j3)(19.9 + j14.2) = 122.2
j2.9 = 122.23/
1.36 V (rms).
Find the phase voltage at the load for the b- and c-phases using the acb
sequence:
VBN = 122.23/
1.36 + 120 = 122.23/118.64 V (rms);
VCN = 122.23/
1.36
120 = 122.23/
121.36 V (rms).
[d] The line voltage at the load in the a-phase is VAB . Find this line voltage
from the phase voltage at the load in the a-phase, VAN , using Fig,
11.9(b):
p
VAB = VAN ( 3/ 30 ) = 211.72/ 31.36 V (rms).
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11–12
CHAPTER 11. Balanced Three-Phase Circuits
Find the line voltage at the load for the b- and c-phases using the acb
sequence:
VBC = 211.72/
31.36 + 120 = 211.72/88.64 V (rms);
VCA = 211.72/
31.36
480
= 6.4/
60 + j45
P 11.13 [a] IAB =
IBC = 6.4/
120 = 211.72/
151.36 V (rms).
36.87 A;
156.87 A;
ICA = 6.4/83.13 A.
p
[b] IaA = 3/ 30 IAB = 11.09/
66.87 A;
IbB = 11.09/173.13 A;
IcC = 11.09/53.13 A.
[c] Transform the
-connected load to a Y-connected load:
60 + j45
Z
=
= 20 + j15 ⌦.
3
3
The single-phase equivalent circuit is:
ZY =
Van = 277.25/
= 288.34/
Vab =
p
30 + (0.8 + j0.6)(11.09/
66.87 )
30 V;
3/30 Van = 499.42/0 V;
Vbc = 499.42/
120 V;
Vca = 499.42/120 V.
P 11.14 [a] Van = Vbn
120 = 150/15 V (rms);
Zy = Z /3 = 43 + j57 ⌦.
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Problems
11–13
The a-phase circuit is
150/15
= 2/ 38.13 A (rms).
45 + j60
[c] VAN = (43 + j57)IaA = 142.8/14.84 V (rms);
p
VAB = 3/ 30 VAN = 247.34/ 15.16 A (rms).
[b] IaA =
P 11.15 Zy = Z /3 = 4 + j3 ⌦.
The a-phase circuit is
IaA =
240/ 170
= 37.48/151.34 A (rms);
(1 + j1) + (4 + j3)
1
IAB = p /
3
30 IaA = 21.64/121.34 A (rms).
p
P 11.16 Van = 1/ 3/
208
30 Vab = p /20 V (rms);
3
Zy = Z /3 = 1
j3 ⌦.
The a-phase circuit is
Zeq = (4 + j3)k(1
j3) = 2.6
j1.8 ⌦;
!
2.6 j1.8
208
p /20 = 92.1/
VAN =
(1.4 + j0.8) + (2.6 j1.8)
3
p
VAB = 3/30 VAN = 159.5/29.34 V (rms).
0.66 V (rms);
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11–14
CHAPTER 11. Balanced Three-Phase Circuits
P 11.17 [a]
p
p
7650/ 3 7650/ 3
+
= 145.8/
IaA =
72 + j21
50
|IaA | = 145.8 A.
p
7650/ 3
= 58.89/
[b] IAN =
72 + j21
6.49 A;
16.26 A;
|IAN | = 58.89 A.
[c] IAB =
7650/30
= 51/30 A;
150
|IAB | = 51 A.
p
[d] Van = (145.8/ 6.49 )(j1) + 7650/ 3 = 4435.6/1.87 V;
p
|Vab | = 3(4435.6) = 7682.66 V.
P 11.18 [a]
[b] IaA = p
13,800
= 2917/
3(2.375 + j1.349)
29.6 A (rms);
|IaA | = 2917 A (rms).
[c] VAN = (2.352 + j1.139)(2917/ 29.6 ) = 7622.93/
p
|VAB | = 3|VAN | = 13,203.31 V (rms).
3.76 V (rms);
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Problems
[d] Van = (2.372 + j1.319)(2917/ 29.6 ) = 7616.93/
p
|Vab | = 3|Van | = 13,712.52 V (rms).
11–15
0.52 V (rms);
|IaA |
[e] |IAB | = p = 1684.13 A (rms).
3
[f ] |Iab | = |IAB | = 1684.13 A (rms).
P 11.19 [a] IAB =
4160/0
= 20.8/
160 + j120
36.87 A;
IBC = 20.8/83.13 A;
ICA = 20.8/ 156.87 A.
p
[b] IaA = 3/30 IAB = 36.03/
6.87 A;
IbB = 36.03/113.13 A;
IcC = 36.03/
126.87 A.
[c] Iba = IAB = 20.8/
36.87 A;
Icb = IBC = 20.8/83.13 A;
Iac = ICA = 20.8/
P 11.20 [a] IAB =
156.87 A.
480/0
= 192/16.26 A (rms);
2.4 j0.7
IBC =
ICA =
480/120
= 48/83.13 A (rms);
8 + j6
480/
[b] IaA = IAB
120
20
= 24/
120 A (rms).
ICA
= 210/20.79 ;
IbB = IBC
IAB
= 178.68/
IcC = ICA
= 70.7/
178.04 ;
IBC
104.53 .
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11–16
CHAPTER 11. Balanced Three-Phase Circuits
P 11.21 [a] Since the phase sequence is acb (negative) we have:
Van = 2399.47/30 V;
Vbn = 2399.47/150 V;
Vcn = 2399.47/
90 V;
1
ZY = Z = 0.9 + j4.5 ⌦/ .
3
p
[b] Vab = 2399.47/30
2399.47/150 = 2399.47 3/0 = 4156/0 V.
Since the phase sequence is negative, it follows that
Vbc = 4156/120 V;
Vca = 4156/
120 V.
[c]
Iba =
4156
= 301.87/
2.7 + j13.5
Iac = 301.87/
IaA = Iba
78.69 A;
198.69 A;
Iac = 522.86/
48.69 A.
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Problems
11–17
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 522.86/71.31 A;
IcC = 522.86/
168.69 A.
[d]
IaA =
2399.47/30
= 522.86/
0.9 + j4.5
48.69 A.
Since we have a balanced three-phase circuit and a negative phase
sequence we have:
IbB = 522.86/71.31 A;
IcC = 522.86/
168.69 A.
P 11.22 [a]
[b] IaA =
2399.47/30
= 1.2/46.15 A;
1920 j556
VAN = (1910 j636)(1.2/46.15 ) = 2416.54/27.73 V;
p
|VAB | = 3(2416.54) = 4185.57 V.
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11–18
CHAPTER 11. Balanced Three-Phase Circuits
1.2
[c] |Iab | = p = 0.69 A.
3
[d] Van = (1919.1 j564.5)(1.2/46.15 ) = 2400/29.76 V;
p
|Vab | = 3(2400) = 4156.92 V.
P 11.23 [a]
IaA =
1365/0
= 27.3/
30 + j40
53.13 A (rms);
IaA
ICA = p /150 = 15.76/96.87 A (rms).
3
[b] Sg/ =
1365I⇤aA =
22,358.75
j29,811.56 VA;
.·. Pdeveloped/phase = 22.359 kW.
Pabsorbed/phase = |IaA |2 28.5 = 21.241 kW;
% delivered =
21.241
(100) = 95%.
22.359
P 11.24 The complex power of the source per phase is
Ss = 20,000/(cos 1 0.6) = 20,000/53.13 = 12,000 + j16,000 kVA.
This complex power per phase must equal the sum of the per-phase complex
powers of the two loads:
Ss = S1 + S2
so
12,000 + j16,000 = 10,000 + S2 ;
.·. S2 = 2000 + j16,000 VA.
|Vrms |2
.
Z2⇤
Also,
S2 =
|Vrms | =
|Vload |
p
= 120 V (rms);
3
Thus,
Z2⇤ =
(120)2
|Vrms |2
= 0.11
=
S2
2000 + j16,000
j0.89 ⌦;
.·. Z2 = 0.11 + j0.89 ⌦.
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Problems
11–19
P 11.25 The a-phase of the circuit is shown below:
I1 =
120/20
= 12/
8 + j6
I⇤2 =
600/36
= 5/16 A (rms);
120/20
I = I1 + I2 = 12/
16.87 A (rms);
16.87 + 5/
16 = 17/
16.61 A (rms);
Sa = VI⇤ = (120/20 )(17/16.61 ) = 2040/36.61 VA;
ST = 3Sa = 6120/36.61 VA.
P 11.26 [a] I⇤aA =
(160 + j46.67)103
= 133.3 + j38.9;
1200
IaA = 133.3
j38.9 A;
Van = 1200 + (133.3
j38.9)(0.18 + j1.44) = 1280 + j184.95 V.
IC =
1280 + j184.95
=
j60
Ina =
(IaA + IC ) =
3.1 + j21.3 A;
130.2 + j17.6 = 131.4/172.3 A.
[b] Sg/ = (1280 + j184.95)( 130.2
SgT = 3Sg/ =
490.2
j17.6) =
163,400
j46,608.5 VA;
j139.8 kVA.
Therefore, the source is delivering 490.2 kW and 139.8 kvars.
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11–20
CHAPTER 11. Balanced Three-Phase Circuits
[c] Pdel = 490.2 kW;
Pabs = 3(160,000) + 3|IaA |2 (0.18)
= 490.4 kW ⇠
= Pdel
(roundo↵).
[d] Qdel = 3|IC |2 (60) + 139.8 ⇥ 103 = 223.2 kVAR;
Qabs = 3(46,666) + 3|IaA |2 (1.44)
= 223.3 kVAR = Qdel . (roundo↵)
P 11.27 [a] ST
= 14,000/41.41
9000/53.13 = 5.5/22 kVA;
S = ST /3 = 1833.46/22 VA.
3000/53.13
= 300 V (rms);
10/ 30
p
p
|Vline | = |Vab | = 3|Van | = 300 3 = 519.62 V (rms).
[b] |Van | =
P 11.28 [a] S1/ = 40,000(0.96)
j40,000(0.28) = 38,400
j11,200 VA;
S2/ = 60,000(0.8) + j60,000(0.6) = 48,000 + j36,000 VA;
S3/ = 33,600 + j5200 VA;
ST / = S1 + S2 + S3 = 120,000 + j30,000 VA.
.·. I⇤aA =
120,000 + j30,000
= 50 + j12.5;
2400
.·. IaA = 50
j12.5 A.
Van = 2400 + (50 j12.5)(1 + j8) = 2550 + j387.5 = 2579.27/8.64 V;
p
|Vab | = 3(2579.27) = 4467.43 V.
[b] Sg/ = (2550 + j387.5)(50 + j12.5) = 122,656.25 + j51,250 VA;
% efficiency =
120,000
(100) = 97.83%.
122,656.25
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Problems
11–21
P 11.29 [a] S1 = 10,200(0.87) + j10,200(0.493) = 8874 + j5029.13 VA;
S2 = 4200 + j1913.6 VA;
p
sin ✓3 = p
3VL IL sin ✓3 = 7250;
7250
= 0.517.
3(220)(36.8)
cos ✓3 = 0.856.
Therefore
Therefore
7250
P3 =
⇥ 0.856 = 12,003.9 W;
0.517
S3 = 12,003.9 + j7250 VA;
ST = S1 + S2 + S3 = 25.078 + j14.192 kVA;
1
ST / = ST = 8359.3 + j4730.7 VA;
3
220 ⇤
p IaA = (8359.3 + j4730.7);
3
IaA = 65.81
[b] pf = cos(0
j37.24 = 75.62/
I⇤aA = 65.81 + j37.24 A;
29.51 A,
so
|IaA | = 75.62 A.
29.51 ) = 0.87 lagging.
P 11.30 From the solution to Problem 11.20 we have:
SAB = (480/0 )(192/
16.26 ) = 88,473.7
SBC = (480/120 )(48/
83.13 ) = 18,431.98 + j13,824.03 VA;
SCA = (480/
j25,804.5 VA;
120 )(24/120 ) = 11,520 + j0 VA.
P 11.31 Let pa , pb , and pc represent the instantaneous power of phases a, b, and c,
respectively. Then assuming a positive phase sequence, we have
pa = van iaA = [Vm cos !t][Im cos(!t
pb = vbn ibB = [Vm cos(!t
✓ )];
120 )][Im cos(!t
pc = vcn icC = [Vm cos(!t + 120 )][Im cos(!t
✓
120 )];
✓ + 120 )].
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11–22
CHAPTER 11. Balanced Three-Phase Circuits
The total instantaneous power is pT = pa + pb + pc , so
pT = Vm Im [cos !t cos(!t
✓ ) + cos(!t
+ cos(!t + 120 ) cos(!t
120 ) cos(!t
✓
120 )
✓ + 120 )].
Now simplify using trigonometric identities. In simplifying, collect the
coefficients of cos(!t ✓ ) and sin(!t ✓ ). We get
pT = Vm Im [cos !t(1 + 2 cos2 120 ) cos(!t
+2 sin !t sin2 120 sin(!t
= 1.5Vm Im [cos !t cos(!t
✓ )
✓ )]
✓ ) + sin !t sin(!t
✓ )]
= 1.5Vm Im cos ✓ .
P 11.32 |Iline | =
1600
p = 11.547 A (rms);
240/ 3
p
240/ 3
|V |
|Zy | =
=
= 12;
|I|
11.547
Zy = 12/
50 ⌦;
Z = 3Zy = 36/
P 11.33 Assume a
50 = 23.14
j27.58 ⌦/ .
-connected load (series):
1
S = (96 ⇥ 103 )(0.8 + j0.6) = 25,600 + j19,200 VA;
3
Z⇤ =
Z
|480|2
= 5.76
25,600 + j19,200
j4.32 ⌦;
= 5.76 + 4.32 ⌦.
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Problems
11–23
Now assume a Y-connected load (series):
1
ZY = Z
3
= 1.92 + j1.44 ⌦.
Now assume a
P =
R
-connected load (parallel):
|480|2
;
R
|480|2
= 9 ⌦;
=
25,600
Q =
|480|2
;
X
X
=
|480|2
= 12 ⌦.
19,200
Now assume a Y-connected load (parallel):
1
RY = R
3
= 3 ⌦;
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11–24
CHAPTER 11. Balanced Three-Phase Circuits
1
XY = X
3
= 4 ⌦.
P 11.34 [a] POUT = 746 ⇥ 100 = 74,600 W;
PIN = 74,600/(0.97) = 76,907.22 W;
p
3VL IL cos ✓ = 76,907.22;
76,907.22
= 242.58 A (rms).
3(208)(0.88)
p
p
[b] Q = 3VL IL sin = 3(208)(242.58)(0.475) = 41,511.90 VAR.
IL = p
P 11.35
4000I⇤1 = (210 + j280)103 ;
I⇤1 =
280
210
+j
= 52.5 + j70 A (rms);
4
4
I1 = 52.5
j70 A (rms);
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Problems
I2 =
11–25
4000/0
= 240 + j70 A (rms);
15.36 j4.48
.·. IaA = I1 + I2 = 292.5 + j0 A (rms).
Van = 4000 + j0 + 292.5(0.1 + j0.8) = 4036.04/3.32 V (rms);
p
|Vab | = 3|Van | = 6990.62 V (rms).
P 11.36 [a]
p
24,000 3/0
= 66.5 j49.9 A (rms);
I1 =
400 + j300
p
24,000 3/0
= 33.3 + j24.9 A (rms);
I2 =
800 j600
I⇤3 =
57,600 + j734,400
p
= 1.4 + j17.7;
24,000 3
I3 = 1.4
j17.7 A (rms);
j42.7 A = 109.8/ 22.9 A (rms);
p
j42.7) + 24,000 3 = 42,454.8 + j1533.8 V (rms);
IaA = I1 + I2 + I3 = 101.2
Van = (2 + j16)(101.2
S = Van I⇤aA = (42,454.8 + j1533.8)(101.2 + j42.7)
= 4,230,932.5 + j1,968,040.5 VA;
ST = 3S = 12,692.8 + j5904.1 kVA.
p
[b] S1/ = 24,000 3(66.5 + j49.9) = 2764.4 + j2074.3 kVA;
p
S2/ = 24,000 3(33.3 j24.9) = 1384.3 j1035.1 kVA;
S3/ = 57.6 + j734.4 kVA;
S (load) = 4206.3 + j1773.6 kVA;
% delivered =
✓
◆
4206.3
(100) = 99.4%.
4230.9
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11–26
CHAPTER 11. Balanced Three-Phase Circuits
1
P 11.37 [a] Sg/ = (41.6)(0.707 + j0.707) ⇥ 103 = 9803.73 + j9803.73 VA;
3
I⇤aA =
9803.73 + j9803.73
p
= 70.76 + j70.76 A (rms);
240/ 3
IaA = 70.76
j70.76 A (rms).
240
(0.04 + j0.03)(70.76 j70.76)
VAN = p
3
= 133.61 + j0.71 = 133.61/0.30 V (rms).
|VAB | =
p
3(133.61) = 231.42 V (rms).
[b] SL/ = (133.61 + j0.71)(70.76 + j70.76) = 9404 + j9504.5 VA;
SL = 3SL/ = 28,212 + j28,513 VA.
Check:
Sg = 41,600(0.7071 + j0.7071) = 29,415 + j29,415 VA.
P` = 3|IaA |2 (0.04) = 1202 W;
Pg = PL + P` = 28,212 + 1202 = 29,414 W;
(checks)
Q` = 3|IaA |2 (0.03) = 901 VAR;
Qg = QL + Q` = 28,513 + 901 = 29,414 VAR. (checks)
P 11.38 [a]
1
Ss/ = (60)(0.96
3
j0.28) ⇥ 103 = 19.2
j5.6 kVA;
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Problems
11–27
1
S1/ = (45) = 15 + j0 kVA;
3
S2/ = Ss/
.·. I⇤2 =
S1/ = 4.2
j5.6 kVA;
4200 j5600
p
= 11.547
630/ 3
j15.396 A.
I2 = 11.547 + j15.396 A;
Zy = 11.34
j15.12 ⌦;
Z = 3Zy = 34.02
j45.36 ⌦.
p
(630/ 3)2
= 31.5 ⌦;
R = 3R = 94.5 ⌦;
[b] R =
4200
p
(630/ 3)2
= 23.625 ⌦;
X = 3XL =
XL =
5600
70.875 ⌦.
P 11.39 [a]
SL/ =
I⇤aA =

1
720
(0.6) 103 = 240,000 + j180,000 VA;
720 + j
3
0.8
240,000 + j180,000
= 100 + j75 A;
2400
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11–28
CHAPTER 11. Balanced Three-Phase Circuits
IaA = 100
j75 A;
Van = 2400 + (0.8 + j0.6)(100
j75)
= 2960 + j580 = 3016.29/11.09 V;
p
|Vab | = 3(3016.29) = 5224.37 V.
[b]
I1 = 100
S2 = 0
I⇤2 =
j75 A
(from part [a]);
1
j (576) ⇥ 103 =
3
j192,000
=
2400
j192,000 VAR;
j80 A;
.·. I2 = j80 A.
IaA = 100
j75 + j80 = 100 + j5 A;
Van = 2400 + (100 + j5)(0.8 + j6.4)
= 2448 + j644 = 2531.29/14.74 V;
p
|Vab | = 3(2531.29) = 4384.33 V.
[c] |IaA | = 125 A;
Ploss/ = (125)2 (0.8) = 12,500 W;
Pg/ = 240,000 + 12,500 = 252.5 kW;
%⌘ =
240
(100) = 95.05%.
252.5
[d] |IaA | = 100.125 A;
P`/ = (100.125)2 (0.8) = 8020 W;
%⌘ =
240,000
(100) = 96.77%.
248,200
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Problems
[e] Zcap/Y =
.·.
24002
=
192,000
j
1
= 30;
!C
C=
j30 ⌦;
1
= 88.42 µF.
(30)(120⇡)
P 11.40 [a] From Assessment Problem 11.9, IaA = (101.8
Therefore Icap = j135.7 A (rms)
Thus CY =
11–29
and
j135.7) A (rms).
p
2450/ 3
=
ZCY =
j135.7
j10.42 ⌦.
1
= 254.5 µF.
(10.42)(2⇡)(60)
ZC = ( j10.42)(3) =
Therefore C =
j31.26 ⌦;
254.5
= 84.84 µF.
3
[b] CY = 254.5 µF.
[c] |IaA | = 101.8 A (rms).
P 11.41 Wm1 = |VAB ||IaA | cos(/VAB
/IaA ) = (199.58)(2.4) cos(65.68 ) = 197.26 W;
Wm2 = |VCB ||IcC | cos(/VCB
/IcC ) = (199.58)(2.4) cos(5.68 ) = 476.64 W;
CHECK: W1 + W2 = 673.9 = (2.4)2 (39)(3) = 673.9 W.
p
3(W2 W1 )
= 0.75;
P 11.42 tan =
W 1 + W2
.·.
= 36.87 ;
p
.·. 2400 3|IL | cos 66.87 = 40,823.09.
|IL | = 25 A;
|Z| =
P 11.43 IaA =
2400
= 96 ⌦
25
.·. Z = 96/36.87 ⌦.
VAN
= |IL |/ ✓ A;
Z
Z = |Z|/✓ ,
VBC = |VL |/
Wm = |VL | |IL | cos[ 90
= |VL | |IL | cos(✓
90 V.
( ✓ )]
90 )
= |VL | |IL | sin ✓ .
Therefore
p
3Wm =
p
3|VL | |IL | sin ✓ = Qtotal .
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11–30
CHAPTER 11. Balanced Three-Phase Circuits
P 11.44 [a] Z = 16 + j12 = 20/36.87 ⌦;
.·. IaA = 34/ 36.87 A;
p
VCN = 680 3/ 90 V;
VAN = 680/0 V;
VBC = VBN
p
Wm = (680 3)(34) cos( 90 + 36.87 ) = 24,027.07 W;
p
3Wm = 41,616.1 W.
[b] Q = (342 )(12) = 13,872 VAR;
p
QT = 3Q = 41,616 VAR = 3Wm .
P 11.45 [a] W2
W1 = VL IL [cos(✓
30 )
cos(✓ + 30 )]
= VL IL [cos ✓ cos 30 + sin ✓ sin 30
cos ✓ cos 30 + sin ✓ sin 30 ]
= 2VL IL sin ✓ sin 30 = VL IL sin ✓.
Therefore
p
3(W2
[b] Z = (8 + j6) ⌦;
p
QT = 3[2476.25
W1 ) =
p
3VL IL sin ✓ = QT .
979.75] = 2592 VAR;
QT = 3(12)2 (6) = 2592 VAR;
(checks).
Z = (8 j6) ⌦;
p
QT = 3[979.75
2592 VAR;
2476.25] =
QT = 3(12)2 ( 6) = 2592 VAR;
(checks).
p
Z = 5(1 + j 3) ⌦;
p
QT = 3[2160 0] = 3741.23 VAR;
p
QT = 3(12)2 (5 3) = 3741.23 VAR;
(checks).
Z = 10/75 ⌦;
p
QT = 3[ 645.53
1763.63] =
QT = 3(12)2 [ 10 sin 75 ] =
4172.80 VAR;
4172.80 VAR;
(checks).
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Problems
11–31
VAN
;
IaA
P 11.46 Z = |Z|/✓ =
✓ = /VAN
/IaA ;
✓1 = /VAB
/IaA .
For a positive phase sequence,
/VAB = /VAN + 30 .
Thus,
/IaA = ✓ + 30 .
✓1 = /VAN + 30
Similarly,
VCN
;
IcC
Z = |Z|/✓ =
✓ = /VCN
/IcC ;
✓2 = /VCB
/IcC .
For a positive phase sequence,
/VCB = /VBA
120 = /VAB + 60 ;
/IcC = /IaA + 120 .
Thus,
✓2 = /VAB + 60
(/IaA + 120 ) = ✓1
= ✓ + 30
30 .
60 = ✓
60
P 11.47 [a] Z = 160 + j120 = 200/36.87 ⌦
S =
41602
= 69,222.3 + j51,916.9 VA
160 j120
ST = 3S = 207,667.2 + j155,750.4 VA
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11–32
CHAPTER 11. Balanced Three-Phase Circuits
[b] Wm1 = (4160)(36.03) cos(0 + 6.87 ) = 148,808.64 W
Wm2 = (4160)(36.03) cos( 60 + 126.87 ) = 58,877.55 W
Check:
PT = 207.7 kW = Wm1 + Wm2 .
P 11.48 From the solution to Prob. 11.20 we have
IaA = 210/20.79 A
and
[a] W1 = |Vac | |IaA | cos(✓ac
= 480(210) cos(60
[b] W2 = |Vbc | |IbB | cos(✓bc
IbB = 178.68/
178.04 A.
✓aA )
20.79 ) = 78,103.2 W.
✓bB )
= 480(178.68) cos(120 + 178.04 ) = 40,317.7 W.
[c] W1 + W2 = 118,421 W;
PAB = (192)2 (2.4) = 88,473.6 W;
PBC = (48)2 (8) = 18,432 W;
PCA = (24)2 (20) = 11,520 W;
PAB + PBC + PCA = 118,425.7.
therefore W1 + W2 ⇡ Ptotal .
P 11.49 [a] I⇤aA =
(round-o↵ di↵erences)
(432/3)(0.96 j0.28)103
= 20/
7200
VBN = 7200/
VBC = VBN
16.26 A;
VCN = 7200/120 V;
p
VCN = 7200 3/ 90 V;
120 V;
IbB = 20/
103.74 A;
p
Wm1 = (7200 3)(20) cos( 90 + 103.74 ) = 242,278.14 W.
[b] Current coil in line aA, measure IaA .
Voltage coil across AC, measure VAC .
[c] IaA = 20/16.76 A;
VCA = VAN
p
VCN = 7200 3/
p
Wm2 = (7200 3)(20) cos( 30
30 V;
16.26 ) = 172,441.86 W.
[d] Wm1 + Wm2 = 414.72kW;
PT = 432,000(0.96) = 414.72 kW = Wm1 + Wm2 .
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Problems
11–33
P 11.50 [a] W1 = |VBA ||IbB | cos ✓.
Negative phase sequence:
p
VBA = 240 3/150 V;
IaA =
240/0
= 18/30 A;
13.33/ 30
IbB = 18/150 A;
p
W1 = (18)(240) 3 cos 0 = 7482.46 W;
W2 = |VCA ||IcC | cos ✓;
p
VCA = 240 3/ 150 V;
IcC = 18/
90 A;
p
W2 = (18)(240) 3 cos( 60 ) = 3741.23 W.
[b] P = (18)2 (40/3) cos( 30 ) = 3741.23 W;
PT = 3P = 11,223.69 W;
W1 + W2 = 7482.46 + 3741.23 = 11,223.69 W;
.·. W1 + W2 = PT .
(checks)
1
P 11.51 [a] Z = Z = 4.48 + j15.36 = 16/73.74 ⌦;
3
IaA =
600/0
= 37.5/
16/73.74
73.74 A;
IbB = 37.5/ 193.74 A;
p
VAC = 600 3/ 30
p
VBC = 600 3/ 90 V;
p
W1 = (600 3)(37.5) cos( 30 + 73.74 ) = 28,156.15 W;
p
W2 = (600 3)(37.5) cos( 90 + 193.74 ) = 9256.15 W.
[b] W1 + W2 = 18,900 W;
PT = 3(37.5)2 (13.44/3) = 18,900 W.
p
[c] 3(W1 W2 ) = 64,800 VAR;
QT = 3(37.5)2 (46.08/3) = 64,800 VAR.
(checks)
(checks)
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11–34
CHAPTER 11. Balanced Three-Phase Circuits
P 11.52 [a] Negative phase sequence:
p
VAB = 240 3/ 30 V;
p
VBC = 240 3/90 V;
p
VCA = 240 3/ 150 V;
p
240 3/ 30
= 20.78/ 60 A;
IAB =
20/30
p
240 3/90
= 6.93/90 A;
IBC =
60/0
p
240 3/ 150
= 10.39/ 120 A;
ICA =
40/ 30
IaA = IAB + IAC = 18/
30 A;
IcC = ICB + ICA = ICA + IBC = 16.75/
108.06 ;
p
Wm1 = 240 3(18) cos( 30 + 30 ) = 7482.46 W;
p
Wm2 = 240 3(16.75) cos( 90 + 108.07 ) = 6621.23 W.
[b] Wm1 + Wm2 = 14,103.69 W.
p
PA = (12 3)2 (20 cos 30 ) = 7482.46 W;
p
PB = (4 3)2 (60) = 2880 W;
p
PC = (6 3)2 [40 cos( 30 )] = 3741.23 W;
PA + PB + PC = 14,103.69 = Wm1 + Wm2 .
P 11.53 [a]
[b]
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Problems
11–35
[c]
[d]
P 11.54 [a] Q =
|V|2
;
XC
.·. |XC | =
(13,800)2
= 158.70 ⌦;
1.2 ⇥ 106
1
1
= 158.70;
C=
= 16.71 µF.
!C
2⇡(60)(158.70)
p
(13,800/ 3)2
1
= (158.70);
[b] |XC | =
6
1.2 ⇥ 10
3
.·.
.·. C = 3(16.71) = 50.14 µF.
P 11.55 [a] The capacitor from Appendix H whose value is closest to 50.14 µF is 47 µF.
|XC | =
1
1
=
= 56.4 ⌦;
!C
2⇡(60)(47 ⇥ 10 6 )
|V |2
(13,800)2
Q=
= 1,124,775.6 VAR.
=
3XC
3(56.4)
[b] I⇤aA =
1,200,000 + j75,224
p
= 150.6 + j9.4 A;
13,800/ 3
Van =
13,800
p /0 + (0.6 + j4.8)(150.6
3
j9.4) = 8134.8/5.06 ;
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11–36
CHAPTER 11. Balanced Three-Phase Circuits
|Vab | =
p
3(8134.8) = 14,089.9 V.
This voltage falls within the allowable range of 13 kV to 14.6 kV.
P 11.56 [a] The capacitor from Appendix H whose value is closest to 16.71 µF is 22 µF.
|XC | =
Q=
[b] I⇤aA =
1
1
=
= 120.57 ⌦;
!C
2⇡(60)(22 ⇥ 10 6 )
|V |2
(13,800)2
= 1,579,497 VAR/ .
=
XC
120.57
1,200,000 j379,497
p
= 50.2
13,800/ 3
j15.9 A;
13,800
p /0 + (0.6 + j4.8)(50.2 + j15.9) = 7897.8/1.76 ;
3
p
|Vab | = 3(7897.8) = 13,679.4 V.
Van =
This voltage falls within the allowable range of 13 kV to 14.6 kV.
P 11.57 If the capacitors remain connected when the substation drops its load, the
expression for the line current becomes
13,800 ⇤
p IaA =
3
j1.2 ⇥ 106
I⇤aA =
j150.61 A
or
Hence
IaA = j150.61 A.
Now,
Van =
13,800
p /0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71 V.
3
The magnitude of the line-to-line voltage at the generating plant is
|Vab | =
p
3(7245.05) = 12,548.80 V.
This is a problem because the voltage is below the acceptable minimum of 13
kV. Thus when the load at the substation drops o↵, the capacitors must be
switched o↵.
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Problems
11–37
P 11.58 Before the capacitors are added the total line loss is
PL = 3|150.61 + j150.61|2 (0.6) = 81.66 kW.
After the capacitors are added the total line loss is
PL = 3|150.61|2 (0.6) = 40.83 kW.
Note that adding the capacitors to control the voltage level also reduces the
amount of power loss in the lines, which in this example is cut in half.
P 11.59 [a]
13,800 ⇤
p IaA = 80 ⇥ 103 + j200 ⇥ 103 j1200 ⇥ 103 ;
3
p
p
80
3
j1000
3
I⇤aA =
= 10.04 j125.51 A;
13.8
.·. IaA = 10.04 + j125.51 A.
13,800
p /0 + (0.6 + j4.8)(10.04 + j125.51)
3
= 7371.01 + j123.50 = 7372.04/0.96 V;
Van =
.·. |Vab | =
p
3(7372.04) = 12,768.75 V.
[b] Yes, the magnitude of the line-to-line voltage at the power plant is less
than the allowable minimum of 13 kV.
P 11.60 [a]
13,800 ⇤
p IaA = (80 + j200) ⇥ 103 ;
3
p
p
80 3 + j200 3
⇤
IaA =
= 10.04 + j25.1 A;
13.8
.·. IaA = 10.04 j25.1 A.
13,800
p /0 + (0.6 + j4.8)(10.04 j25.1)
3
= 8093.95 + j33.13 = 8094.02/0.23 V;
Van =
.·. |Vab | =
[b] Yes:
p
3(8094.02) = 14,019.25 V.
13 kV < 14,019.25 < 14.6 kV.
[c] Ploss = 3|10.04 + j125.51|2 (0.6) = 28.54 kW.
[d] Ploss = 3|10.04 + j25.1|2 (0.6) = 1.32 kW.
[e] Yes, the voltage at the generating plant is at an acceptable level and the
line loss is greatly reduced.
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Introduction to the Laplace
Transform
Assessment Problems
e t+e
2
Therefore,
t
.
AP 12.1 [a] cosh t =
1 Z 1 (s
L{cosh t} =
[e
2 0
1
=
2
"
1
=
2
[b] sinh t =
e t
Therefore,
e (s
(s
1
+ e (s
)t
1
) 0
1
+
s+
s
)t
e (s+ )t 1
+
(s + ) 0
!
=
1 Z 1 h (s
e
2 0
"
)t
2
.
1
2
"
s
1
s+
!
=
F (s) = L{te at } =
1
.
(s + a)2
L{tf (t)} =
dF (s)
.
ds
e (s
(s
1
)t
i
e (s+ )t dt
#1
1
=
2
s
s2
#
.
2
1
=
2
]dt
t
e
L{sinh t} =
)t
) 0
#1
e (s+ )t
(s + ) 0
(s2
2)
.
AP 12.2 [a] Let f (t) = te at :
Now,
12–1
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12–2
CHAPTER 12. Introduction to the Laplace Transform
So,
L{t · te
at
}=
"
#
2
d
1
=
.
2
ds (s + a)
(s + a)3
[b] Let f (t) = e at sinh t, then
L{f (t)} = F (s) =
(
df (t)
L
dt
)
(s + a)2
= sF (s)
2
f (0 ) =
;
s( )
(s + a)2
2
0=
s
(s + a)2
2
.
[c] Let f (t) = cos !t. Then
F (s) =
s
2
(s + ! 2 )
and
dF (s)
(s2 ! 2 )
= 2
.
ds
(s + ! 2 )2
dF (s)
s2 ! 2
= 2
.
ds
(s + ! 2 )2
Therefore L{t cos !t} =
AP 12.3
F (s) =
K1 =
K3 =
6s2 + 26s + 26
K1
K2
K3
=
+
+
;
(s + 1)(s + 2)(s + 3)
s+1 s+2 s+3
6
26 + 26
= 3;
(1)(2)
K2 =
24
52 + 26
= 2;
( 1)(1)
54 78 + 26
= 1;
( 2)( 1)
Therefore f (t) = [3e t + 2e 2t + e 3t ] u(t).
AP 12.4
7s2 + 63s + 134
K1
K2
K3
F (s) =
=
+
+
;
(s + 3)(s + 4)(s + 5)
s+3 s+4 s+5
K1 =
K3 =
63
189 134
= 4;
1(2)
175 315 + 134
=
( 2)( 1)
f (t) = [4e 3t + 6e 4t
K2 =
112
252 + 134
= 6;
( 1)(1)
3;
3e 5t ]u(t).
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Problems
12–3
AP 12.5 From Example 12.2,
V (s) =
K1
K2
+
.
s + 20,0000 s + 80,0000
K1 =
96 ⇥ 104
= 16;
s + 80,000 s= 20,0000
K2 =
96 ⇥ 104
=
s + 20,000 s= 80,0000
V (s) =
16
s + 20,0000
16.
16
;
s + 80,0000
Therefore,
v(t) = (16e 20,000t
AP 12.6
16e 80,000t )u(t) V.
10(s2 + 119)
;
(s + 5)(s2 + 10s + 169)
p
s1,2 = 5 ± 25 169 = 5 ± j12;
F (s) =
F (s) =
K2
K2⇤
K1
+
+
;
s + 5 s + 5 j12 s + 5 + j12
K1 =
10(25 + 119)
= 10;
25 50 + 169
K2 =
10[( 5 + j12)2 + 119]
= j4.17 = 4.17/90 ;
(j12)(j24)
Therefore
f (t) = [10e 5t + 8.33e 5t cos(12t + 90 )] u(t)
= [10e 5t
8.33e 5t sin 12t] u(t).
AP 12.7 [a] Using the expression for V (s) found for the circuit in Fig. 12.16,
V (s) =
=
=
Idc /C
2
s + (1/RC)s + (1/LC)
0.024/(25 ⇥ 10 9 )
s2 + [1/(625)(25 ⇥ 10 9 )]s + [1/(0.025)(25 ⇥ 10 9 )]
96 ⇥ 104
.
s2 + 64,000s + 16 ⇥ 108
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12–4
CHAPTER 12. Introduction to the Laplace Transform
[b] V (s) =
K1 =
K1
K1⇤
+
;
s + 32,000 j24,000 s + 32,000 + j24,000
96 ⇥ 104
= 20/
s + 32,000 + j24,000 s= 32,000+j24,000
90 .
Therefore,
v(t) = 40e 32,000t cos(24,000t
AP 12.8
F (s) =
K0 =
90 )u(t) V.
4s2 + 7s + 1
K1
K0
K2
+
;
=
+
s(s + 1)2
s
(s + 1)2 s + 1
1
= 1;
(1)2
K1 =
"
#
4
7+1
= 2;
1
d 4s2 + 7s + 1
s(8s + 7)
K2 =
=
ds
s
s= 1
=
(4s2 + 7s + 1)
s2
s= 1
1+2
= 3;
1
Therefore f (t) = [1 + 2te t + 3e t ] u(t).
AP 12.9 [a] Using the expression for V (s) found for the circuit in Fig. 12.16,
V (s) =
Idc /C
2
s + (1/RC)s + (1/LC)
0.024/(25 ⇥ 10 9 )
= 2
s + [1/(500)(25 ⇥ 10 9 )]s + [1/(0.025)(25 ⇥ 10 9 )]
=
[b] V (s) =
96 ⇥ 104
.
s2 + 80,000s + 16 ⇥ 108
K2
K1
+
2
(s + 40,000)
s + 40,000
=
K2 s + K1 + 40,000K2
;
(s + 40,000)2
So,
K1 = 96 ⇥ 104
and
K2 = 0.
Therefore,
v(t) = 96 ⇥ 104 te 40,000t u(t) V.
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Problems
AP 12.10
F (s) =
40
(s2 + 4s + 5)2
(s + 2
40
j1)2 (s + 2 + j1)2
K1⇤
K1
K2
+
+
(s + 2 j1)2 (s + 2 j1) (s + 2 + j1)2
=
+
K1 =
=
12–5
K2⇤
;
(s + 2 + j1)
40
=
(j2)2
and K1⇤ =
10 = 10/180
"
#
10;
80(j2)
=
(j2)4
d
40
=
K2 =
ds (s + 2 + j1)2 s= 2+j1
j10 = 10/
90 ;
K2⇤ = j10;
Therefore
f (t) = [20te 2t cos(t + 180 ) + 20e 2t cos(t
= 20e 2t [sin t
AP 12.11
F (s) =
90 )] u(t)
t cos t] u(t).
5s2 + 29s + 32
5s2 + 29s + 32
=
=5
(s + 2)(s + 4)
s2 + 6s + 8
s+8
;
(s + 2)(s + 4)
K1
K2
s+8
=
+
(s + 2)(s + 4)
s + 2 s + 4;
K1 =
2+8
= 3;
2
K2 =
4+8
=
2
2.
Therefore,
F (s) = 5
3
2
+
;
s+2 s+4
f (t) = 5 (t) + [ 3e 2t + 2e 4t ]u(t).
AP 12.12
2s3 + 8s2 + 2s
F (s) =
s2 + 5s + 4
f (t) = 2
d (t)
dt
4
= 2s
2+
4(s + 1)
= 2s
(s + 1)(s + 4)
2+
4
;
s+4
2 (t) + 4e 4t u(t).
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12–6
CHAPTER 12. Introduction to the Laplace Transform
AP 12.13 [a] Factoring the numerator,
20s2 + 80s + 100 = 20(s + 2 + j)(s + 2
There are zeros at
2
j and
j).
2 + j. Factoring the denominator,
s3 + 10s2 + 21s = s(s + 3)(s + 7).
There are poles at 0,
3 and
7.
[b] Factoring the numerator,
10s2 + 20s + 10 = 10(s + 1)2 .
There are two zeros at
1. Factoring the denominator,
2s3 + 28s2 + 258s + 712 = 2(s + 4)(s + 5 + j8)(s + 5
There are poles at
4,
5 + j8 and
5
j8).
j8.
[c] The numerator is
125s.
There is a zero at 0. Factoring the denominator,
5s4 + 90s3 + 670s2 + 2360s + 3400
= 5(s + 5 + j3)(s + 5
There are poles at
AP 12.14
5 + j3,
j3)(s + 4 + j2)(s + 4
j2).
5
4
j3,
4 + j2, and
j2.
#
"
7s3 [1 + (9/s) + (134/(7s2 ))]
lim sF (s) = lim 3
= 7;
s!1
s!1 s [1 + (3/s)][1 + (4/s)][1 + (5/s)]
.·. f (0+ ) = 7.
#
"
7s3 + 63s2 + 134s
lim sF (s) = lim
= 0;
s!0
s!0 (s + 3)(s + 4)(s + 5)
.·. f (1) = 0.
"
#
s3 [4 + (7/s) + (1/s)2 ]
lim sF (s) = lim
= 4;
s!1
s!1
s3 [1 + (1/s)]2
.·. f (0+ ) = 4.
"
#
4s2 + 7s + 1
lim sF (s) = lim
= 1;
s!0
s!0
(s + 1)2
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Problems
12–7
.·. f (1) = 1.
#
"
40s
lim sF (s) = lim 4
= 0;
s!1
s!1 s [1 + (4/s) + (5/s2 )]2
.·. f (0+ ) = 0.
"
#
40s
lim sF (s) = lim
= 0;
s!0
s!0 (s2 + 4s + 5)2
.·. f (1) = 0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–8
CHAPTER 12. Introduction to the Laplace Transform
Problems
P 12.1
[a] (10 + t)[u(t + 10)
u(t)] + (10
= (t + 10)u(t + 10)
[b] ( 24
8t)[u(t + 3)
+ 8[u(t
=
2tu(t) + (t
u(t + 2)]
u(t
u(t
10)]
10)u(t
10).
8[u(t + 2)
2)] + (24
8t)[u(t
u(t + 1)] + 8t[u(t + 1)
2)
u(t
2)u(t
[a] f (t) = 5t[u(t)
2) + 8(t
u(t
3)u(t
2)] + 10[u(t
( 5t + 40)[u(t
6)
u(t
[b] f (t) = 10 sin ⇡t[u(t)
u(t
2)].
[c] f (t) = 4t[u(t)
5)].
u(t
u(t
1)]
1)u(t
1)
3)]
8(t + 3)u(t + 3) + 8(t + 2)u(t + 2) + 8(t + 1)u(t + 1)
8(t
P 12.2
1)
t)[u(t)
8(t
3).
2)
u(t
6)]+
8)].
P 12.3
P 12.4
[a]
[b] f (t) =
20t[u(t)
u(t
1)]
20[u(t
1)
+20 cos(⇡t/2)[u(t
2)
u(t
4)]
+(100
4)
u(t
5)].
20t)[u(t
u(t
2)]
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 12.5
As " ! 0 the amplitude ! 1; the duration ! 0; and the area is independent
of ", i.e.,
A=
P 12.6
12–9
Z 1
" 1
dt = 1.
1 ⇡ "2 + t2
✓ ◆
✓ ◆
✓ ◆
1
1
1
[a] A =
bh =
(2")
= 1.
2
2
"
[b] 0.
[c] 1.
P 12.7
F (s) =
Z
"/2
"
✓
Z "/2
4 st
e dt +
"3
"/2
Therefore F (s) =
4 s"
[e
s"3
4
"3
◆
e
st
dt +
Z "
4 st
e dt.
"/2 "3
2es"/2 + 2e s"/2
e s" ];
L{ 00 (t)} = lim F (s).
"!0
After applying L’Hopital’s rule three times, we have

s s"/2
e
4
2s
ses"
"!0 3
lim
Therefore L{ 00 (t)} = s2 .
P 12.8
[a] I =
Z 3
1
3
✓
◆
s s"/2
2s 3s
e
+ se s" =
.
4
3 2
(t + 2) (t) dt +
Z 3
1
8(t3 + 2) (t
1) dt
= (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26.
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12–10
CHAPTER 12. Introduction to the Laplace Transform
[b] I =
Z 2
2
2
t (t) dt +
Z 2
2
2
t (t + 1.5) dt +
Z 2
2
t2 (t
3) dt
= 02 + ( 1.5)2 + 0 = 2.25.
P 12.9
F (s) =
Z "
es" e s"
1 st
e dt =
.
2"s
" 2"
"
#
1
1 2s
ses" + se s"
F (s) =
lim
·
= 1.
=
"!0
2s
1
2s 1
✓
1 Z 1 (4 + j!)
1
P 12.10 f (t) =
· ⇡ (!) · ejt! d! =
2⇡ 1 (9 + j!)
2⇡
(
dn f (t)
P 12.11 L
dtn
)
= sn F (s)
!
◆
2
4 + j0
⇡e jt0 = .
9 + j0
9
sn 2 f 0 (0 )
sn 1 f (0 )
···.
Therefore
L{ n (t)} = sn (1)
sn 1 (0 )
sn 2 0 (0 )
a) dt,
v = (t
P 12.12 [a] Let dv = 0 (t
sn 3 00 (0 )
a),
du = f 0 (t) dt.
u = f (t),
Therefore
Z 1
1
0
f (t) (t
Z 1
(t
d(e st )
=
dt
t=0
h
a) dt = f (t) (t
a)
1
1
1
Z 1
0
0
(t)e st dt =
P 12.13 [a] L{20e 500(t 10) u(t
a)f 0 (t) dt
f 0 (a).
=0
[b] L{ 0 (t)} =
· · · = sn .
10)} =
"
#
se st
i
t=0
= s.
20e 10s
.
(s + 500)
[b] First rewrite f (t) as
f (t) = (5t + 20)u(t + 4)
+(10t
20)u(t
= 5(t + 4)u(t + 4)
+10(t
.·. F (s) =
2)u(t
5[e4s
(10t + 20)u(t + 2)
2)
(5t
20)u(t
4)
10(t + 2)u(t + 2)
2)
5(t
4)u(t
2e2s + 2e 2s
s2
e 4s ]
4).
.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 12.14 [a] f (t) = 5t[u(t)
u(t
2)]
+(20
5t)[u(t
2)
u(t
6)]
+(5t
40)[u(t
6)
u(t
8)]
= 5tu(t)
10(t
2)u(t
2)
+10(t
6)u(t
6)
.·. F (s) =
5[1
5(t
8)u(t
2e 2s + 2e 6s
s2
e 8s ]
12–11
8).
.
[b]
f 0 (t) = 5[u(t)
u(t
2)]
5[u(t
+5[u(t
6)
u(t
8)]
= 5u(t)
L{f 0 (t)} =
5[1
10u(t
2) + 10u(t
2e 2s + 2e 6s
s
2)
u(t
6)
5u(t
e 8s ]
6)]
8).
.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–12
CHAPTER 12. Introduction to the Laplace Transform
[c]
f 00 (t) = 5 (t)
10 (t
L{f 00 (t)} = 5[1
P 12.15 L{e
at
f (t)} =
P 12.16 L{f (at)} =
Z 1
Z 1
0
Let u = at,
0
2) + 10 (t
2e 2s + 2e 6s
[e
at
f (t)]e
st
dt =
6)
5 (t
8).
e 8s ].
Z 1
0
f (t)e (s+a)t dt = F (s + a).
f (at)e st dt
u=0
when t = 0 ,
f (u)e (u/a)s
1
du
= F (s/a).
a
a
du = a dt,
and u = 1 when t = 1.
Therefore L{f (at)} =
P 12.17 [a] L{te at } =
Z 1
0
Z 1
0
te (s+a)t dt
e (s+a)t
=
(s + a)2

= 0+
1
.
(s + a)2
.·. L{te at } =
1
.
(s + a)2
(s + a)t
1
1
0
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Problems
(
)
(
)
s
d
[b] L
(te at )u(t) =
dt
(s + a)2
12–13
0.
s
d
L
(te at )u(t) =
.
dt
(s + a)2
[c]
d
(te at ) =
dt
ate at + e at .
L{ ate at + e at } =
a
a
1
s+a
=
+
+
.
(s + a)2 (s + a)
(s + a)2 (s + a)2
)
(
s
d
.·. L
(te at ) =
.
dt
(s + a)2
P 12.18 [a] L
[b]
⇢Z t
0
Z t
0
e ax dx =
e ax dx =
1
a
(
)
e at
a
1
L
a
P 12.19 [a]
Z t
0
x dx =
(
)
t2
L
2
CHECKS
1
F (s)
=
.
s
s(s + a)
e at
.
a
=

1 1
a s
1
1
.
=
s+a
s(s + a)
t2
.
2
1 Z 1 2 st
=
t e dt
2 0
"
1#
1 e st 2 2
=
(s t + 2st + 2)
2
s3
0
=
1
1
(2) = 3 .
3
2s
s
.·. L
⇢Z t
⇢Z t
x dx =
[b] L
0
0
.·. L
⇢Z t
P 12.20 [a] L{t} =
0
1
;
s2
x dx =
1
.
s3
1/s2
1
L{t}
=
= 3.
s
s
s
x dx =
1
.
s3
CHECKS
therefore L{te at } =
1
.
(s + a)2
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12–14
CHAPTER 12. Introduction to the Laplace Transform
[b] sin !t =
ej!t
e j!t
.
j2
Therefore
1
j2
L{sin !t} =
=
!
1
s + j!
1
s
!
s2 + ! 2
j!
!
=
1
j2
!✓
2j!
2
s + !2
◆
.
[c] sin(!t + ✓) = (sin !t cos ✓ + cos !t sin ✓).
Therefore
L{sin(!t + ✓)} = cos ✓L{sin !t} + sin ✓L{cos !t}
! cos ✓ + s sin ✓
.
=
s2 + ! 2
Z 1
1
e st
(
st
1)
=0
s2
0
0
[e] f (t) = cosh t cosh ✓ + sinh t sinh ✓.
From Assessment Problem 12.1(a)
s
.
L{cosh t} = 2
s
1
From Assessment Problem 12.1(b)
[d] L{t} =
te st dt =
L{sinh t} =
1
s2
1
=
Z 1
0
1
.
s2
f (t)e
"
s
(s2
1)
#
+ sinh ✓

1
s2
1
sinh ✓ + s[cosh ✓]
.
(s2 1)
st
Therefore L{tf (t)} =
Z 1
dt =
0
tf (t)e st dt
dF (s)
.
ds
d2 F (s) Z 1 2
=
t f (t)e st dt;
[b]
2
ds
0
d3 F (s) Z 1 3
=
t f (t)e st dt.
3
ds
0
Z 1
dn F (s)
n
= ( 1)
tn f (t)e st dt = ( 1)n L{tn f (t)}.
n
ds
0
Therefore
5
1) =
.
.·. L{cosh(t + ✓)} = cosh ✓
dF (s)
d
P 12.21 [a]
=
ds
ds
1
(0
s2
◆
4 ✓
120
1
4 d
1) 4 2 = 6 ;
ds s
s
4
[c] L{t } = L{t t} = (
1 d
L{t sin t} = ( 1)
ds
s2 +
2
!
=
2 s
(s2 +
2 )2
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Problems
12–15
L{te t cosh t}: .
From Assessment Problem 12.1(a),
s
F (s) = L{cosh t} = 2
;
s
1
(s2 1)1 s(2s)
dF
=
=
ds
(s2 1)2
s2 + 1
;
(s2 1)2
dF
s2 + 1
= 2
.
ds
(s
1)2
Therefore
Thus
L{t cosh t} =
s2 + 1
(s2 1)2
L{e t t cosh t} =
(
(s + 1)2 + 1
s2 + 2s + 2
=
.
[(s + 1)2 1]2
s2 (s + 2)2
)
s!
d sin !t
u(t) = 2
P 12.22 [a] L
dt
s + !2
sin(0) =
s2
d cos !t
[b] L
u(t) = 2
dt
s + !2
cos(0) =
(
(
)
)
✓
d3 (t2 )
2
[c] L
u(t) = s3 3
3
dt
s
[d]
◆
d sin !t
= (cos !t) · !,
dt
d cos !t
=
dt
.
s2
s2 + ! 2
s(0)
L{! cos !t} =
1=
!2
.
s2 + ! 2
2(0) = 2.
!s
s2 + ! 2
;
! sin !t
L{ ! sin !t} =
!2
s2 + ! 2
d3 (t2 u(t))
= 2 (t);
dt3
P 12.23 [a] L{f 0 (t)} =
s2 (0)
s!
s2 + ! 2
Z "
L{2 (t)} = 2.
Z 1
e st
dt +
" 2"
"
ae a(t ") e st dt
✓
◆
1 s"
a
(e
e s" )
e s" = F (s).
2s"
s+a
s
a
=
.
lim F (s) = 1
"!0
s+a
s+a
=
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12–16
CHAPTER 12. Introduction to the Laplace Transform
[b] L{e at } =
1
.
s+a
Therefore L{f 0 (t)} = sF (s)
P 12.24 [a] f1 (t) = e at cos !t;
F (s) = sF1 (s)
F1 (s) =
[c]
0=
s
.
s+a
s+a
;
(s + a)2 + ! 2
s(s + a)
a2 sa ! 2
1
=
.
(s + a)2 + ! 2
(s + a)2 + ! 2
!
F1 (s) =
;
(s + a)2 + ! 2
!
F1 (s)
=
.
s
s[(s + a)2 + ! 2 ]
d
[e at cos !t] =
dt
!e at sin !t
ae at cos !t;
Therefore F (s) =
! 2 a(s + a)
a2 sa ! 2
=
;
(s + a)2 + ! 2
(s + a)2 + ! 2
Z t
ae at sin !t !e at cos !t + !
.
a2 + ! 2
0
e ax sin !x dx =
Therefore
"
1
a!
F (s) = 2
a + ! 2 (s + a)2 + ! 2
=
P 12.25 [a]
s
s+a
f1 (0 ) =
[b] f1 (t) = e at sin !t;
F (s) =
f (0 ) =
Z 1
s
!(s + a)
!
+
(s + a)2 + ! 2
s
#
!
.
s[(s + a)2 + ! 2 ]
F (u)du =
Z 1 Z 1
s
0
=
Z 1
=
Z 1
[b] L{t sin t} =
(s2 +
0
0
f (t)e ut dt du =
Z 1
f (t)
f (t)
s
"
2 s
(
2 )2
t sin t
therefore L
t
e
ut
du dt =
#
Z 1 Z 1
0
s
"
Z 1
f (t)e ut du dt
#
e tu 1
f (t)
dt
t s
0
(
)
e st
f (t)
dt = L
.
t
t
;
)
=
Z 1"
s
#
2 u
du.
(u2 + 2 )2
Let ! = u2 + 2 , then ! = s2 + 2 when u = s, and ! = 1 when u = 1;
also d! = 2u du. Thus
(
)
#
✓
◆
Z 1 "
t sin t
1 1
d!
L
=
.
=
= 2
t
!
s + 2
s2 + 2 ! 2
s2 + 2
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Problems
P 12.26 Ig (s) =
1
= 1.6;
RC
1.2s
;
s2 + 1
V (s) 1 V (s)
+
+ C[sV (s)
R
L s
V (s)

1
= 1;
LC
12–17
1
= 1.6.
C
v(0 )] = Ig (s);
1
1
+
+ sC = Ig (s);
R Ls
V (s) =
(1.6)(1.2)s2
1.92s2
=
.
(s2 + 1.6s + 1)(s2 + 1)
(s2 + 1.6s + 1)(s2 + 1)
=
P 12.27 io = C
LsIg (s)
(1/C)sIg (s)
Ig (s)
=
= 2
2
(1/R) + (1/Ls) + sC
RLs + 1 + s LC
s + (R/C)s + (1/LC)
dvo
;
dt
.·. Io (s) = sCVo (s) =
sIdc
.
2
s + (1/RC)s + (1/LC)
0+ :
P 12.28 [a] For t
dvo
vo
+C
+ io = 0;
R
dt
vo = L
dvo
d 2 io
=L 2;
dt
dt
dio
;
dt
.·.
L dio
d2 io
+ LC 2 + io = 0,
R dt
dt
or
d2 io
1
1 dio
+
io = 0.
+
2
dt
RC dt
LC
[b] s2 Io (s)

sIdc
Io (s) s2 +
Io (s) =
0+
1
[sIo (s)
RC
Idc ] +
1
Io (s) = 0;
LC
1
1
s+
= Idc (s + 1/RC);
RC
LC
Idc [s + (1/RC)]
.
2
[s + (1/RC)s + (1/LC)]
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12–18
CHAPTER 12. Introduction to the Laplace Transform
0+ :
P 12.29 [a] For t
Rio + L
io = C
dio
+ vo = 0;
dt
dvo
dt
.·. RC
d 2 vo
dio
=C 2 ;
dt
dt
d2 vo
dvo
+ LC 2 + vo = 0
dt
dt
or
1
d2 vo R dvo
+
vo = 0.
+
2
dt
L dt
LC
1
R
[b] s2 Vo (s) sVdc 0 + [sVo (s) Vdc ] +
Vo (s) = 0;
L
LC

R
1
= Vdc (s + R/L);
Vo (s) s2 + s +
L
LC
Vo (s) =
Vdc [s + (R/L)]
.
2
[s + (R/L)s + (1/LC)]
1Zt
dvo
= 0;
vo dx + C
R
L 0
dt
RZ t
dvo
·
= Vdc .
vo dx + RC
..
vo +
L 0
dt
R Vo
Vdc
[b] Vo +
+ RCsVo =
;
L s
s
.·.
sLVo + RVo + RCLs2 Vo = LVdc ;
P 12.30 [a]
vo
Vdc
+
.·.
Vo (s) =
[c] io =
1Zt
vo dx;
L 0
Io (s) =
(1/RC)Vdc
.
2
s + (1/RC)s + (1/LC)
Vdc /RLC
Vo
=
.
2
sL
s[s + (1/RC)s + (1/LC)]
dv1 v1 v2
+
= ig ;
dt
R
1Zt
v2 v 1
=0
v2 d⌧ +
L 0
R
or
dv1 v1 v2
+
= ig ;
C
dt
R
R
1Zt
v1 v2
+
+
v2 d⌧ = 0.
R
R L 0
P 12.31 [a] C
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Problems
[b] CsV1 (s) +
12–19
V2 (s)
= Ig (s);
R
V1 (s)
R
V1 (s) V2 (s) V2 (s)
+
+
=0
R
R
sL
or
(RCs + 1)V1 (s)
V2 (s) = RIg (s);
sLV1 (s) + (R + sL)V2 (s) = 0.
Solving,
V2 (s) =
sIg (s)
.
2
C[s + (R/L)s + (1/LC)]
P 12.32 [a] 300 = 60i1 + 25
0=5
d
(i2
dt
d
di1
+ 10 (i2
dt
dt
i1 ) + 10
i1 ) + 5
d
(i1
dt
i2 )
10
di1
;
dt
di1
+ 40i2 .
dt
Simplifying the above equations gives:
300 = 60i1 + 10
0 = 40i2 + 5
[b]
di2
di1
+5 ;
dt
dt
di2
di1
+5 .
dt
dt
300
= (105s + 60)I1 (s) + 5sI2 (s);
s
0 = 5sI1 (s) + (5s + 40)I2 (s).
[c] Solving the equations in (b),
I1 (s) =
60(s + 8)
;
s(s + 4)(s + 24)
I2 (s) =
60
.
(s + 4)(s + 24)
P 12.33 From Problem 12.26:
V (s) =
1.92s2
;
(s2 + 1.6s + 1)(s2 + 1)
s2 + 1.6s + 1 = (s + 0.8 + j0.6)(s + 0.8
j0.6);
s2 + 1 = (s
j1)(s + j1).
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12–20
CHAPTER 12. Introduction to the Laplace Transform
Therefore
V (s) =
=
K1 =
1.92s2
(s + 0.8 + j0.6)(s + 0.8 j0.6)(s
j1)(s + j1)
K1
K1⇤
K2
K2⇤
+
+
+
.
s + 0.8 j0.6 s + 0.8 + j0.6 s j1 s + j1
1.92s2
= 1/
(s + 0.8 + j0.6)(s2 + 1) s= 0.8+j0.6
126.87 ;
1.92s2
K2 =
= 0.6/0 .
(s + j1)(s2 + 1.6s + 1) s= j1
Therefore
v(t) = [2e 0.8t cos(0.6t
P 12.34 [a]
126.87 ) + 1.2 cos(t)]u(t) V.
1
1
=
= 500;
3
RC
(1 ⇥ 10 )(2 ⇥ 10 6 )
1
1
=
= 40,000;
LC
(12.5)(2 ⇥ 10 6 )
Vo (s) =
500,000Idc
s + 500s + 40,000
=
500,000Idc
(s + 100)(s + 400)
=
15,000
(s + 100)(s + 400)
=
K1
K2
+
;
s + 100 s + 400
K1 =
15,000
= 50;
300
Vo (s) =
50
s + 100
vo (t) = [50e 100t
K2 =
50;
50
;
s + 400
50e 400t ]u(t) V.
[b] Io (s) =
0.03s
(s + 100)(s + 400)
=
K1
K2
+
;
s + 100 s + 400
K1 =
15,000
=
300
0.03( 100)
=
300
0.01;
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Problems
12–21
0.03( 400)
= 0.04;
300
0.04
0.01
Io (s) =
+
;
s + 100 s + 400
K2 =
io (t) = (40e 400t
10e 100t )u(t) mA.
[c] io (0) = 40 10 = 30 mA.
Yes. The initial inductor current is zero by hypothesis, the initial resistor
current is zero because the initial capacitor voltage is zero by hypothesis.
Thus at t = 0 the source current appears in the capacitor.
P 12.35
1
= 8000;
RC
1
= 16 ⇥ 106 ;
LC
0.005(s + 8000)
Io (s) =
s2 + 8000s + 16 ⇥ 106
;
s1,2 =
4000;
Io (s) =
0.005(s + 8000)
K1
K2
;
=
+
(s + 4000)2
(s + 4000)2 s + 4000
K1 = 0.005(s + 8000)
K2 =
= 20;
s= 4000
d
[0.005(s + 8000)]s= 4000 = 0.005;
ds
20
0.005
;
+
2
(s + 4000)
s + 4000
Io (s) =
io (t) = [20te 4000t + 0.005e 4000t ]u(t) V.
P 12.36
R
= 10,000;
L
Vo (s) =
=
K1 =
1
= 16 ⇥ 106 ;
LC
120(s + 10,000)
s2 + 10,000s + 16 ⇥ 106
K1
K2
120(s + 10,000)
=
+
;
(s + 2000)(s + 8000)
s + 2000 s + 8000
120(8000)
= 160 V;
6000
Vo (s) =
160
s + 2000
vo (t) = [160e 2000t
K2 =
120(2000)
=
6000
40 V;
40
;
s + 8000
40e 8000t ]u(t) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12–22
CHAPTER 12. Introduction to the Laplace Transform
P 12.37 [a]
1
1
=
= 50 ⇥ 106 ;
3
9
LC
(200 ⇥ 10 )(100 ⇥ 10 )
1
1
=
= 2000;
RC
(5000)(100 ⇥ 10 9 )
70,000
Vo (s) =
s2 + 2000s + 50 ⇥ 106
1000 ± j7000 rad/s;
s1,2 =
Vo (s) =
=
K1 =
;
(s + 1000
70,000
j7000)(s + 1000 + j7000)
K1
K1⇤
+
;
s + 1000 j7000 s + 1000 + j7000
70,000
= 5/
j14,000
90 ;
vo (t) = 10e 1000t cos(7000t
90 )]u(t) V
= 10e 1000t sin 7000tu(t) V.
[b] Io (s) =
=
s(s + 1000
35(10,000)
j7000)(s + 1000 + j7000)
K1
K2
K2⇤
+
+
;
s
s + 1000 j7000 s + 1000 + j7000
K1 =
35(10,000)
= 7 mA;
50 ⇥ 106
K2 =
35(10,000)
= 3.54/
(1000 + j7000)(j14,000)
io (t) = [7 + 7.07e 1000t cos(7000t
P 12.38
1
= 5 ⇥ 106 ;
C
1
= 25 ⇥ 106 ;
LC
V2 (s) =
(6 ⇥ 10 3 )(5 ⇥ 106 )
s2 + 8000s + 25 ⇥ 106
s1,2 =
4000 ± j3000;
V2 (s) =
(s + 4000
171.87 mA;
171.87 )]u(t) mA.
R
= 8000;
L
30,000
j3000)(s + 4000 + j3000)
K1
K1⇤
=
+
;
s + 4000 j3000 s + 4000 + j3000
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
K1 =
30,000
=
j6000
j5 = 5/
12–23
90 ;
v2 (t) = 10e 4000t cos(3000t
90 )
= [10e 4000t sin 3000t]u(t) V.
P 12.39 [a] I1 (s) =
K2
K3
K1
+
+
;
s
s + 4 s + 24
K1 =
(60)(8)
= 5;
(4)(24)
K2 =
K3 =
(60)( 16)
=
( 24)( 20)
2;
✓
3
s+4
5
I1 (s) =
s
3e 4t
i1 (t) = (5
I2 (s) =
K1 =
3;
◆
2
;
s + 24
2e 24t )u(t) A.
K2
K1
+
;
s + 4 s + 24
60
=
20
I2 (s) =
(60)(4)
=
( 4)(20)
✓
3;
K2 =
60
= 3;
20
◆
3
3
+
;
s + 4 s + 24
i2 (t) = (3e 24t
[b] i1 (1) = 5 A;
3e 4t )u(t) A.
i2 (1) = 0 A.
[c] Yes, at t = 1
300
= 5 A.
60
Since i1 is a dc current at t = 1 there is no voltage induced in the 10 H
inductor; hence, i2 = 0. Also note that i1 (0) = 0 and i2 (0) = 0. Thus our
solutions satisfy the condition of no initial energy stored in the circuit.
i1 =
P 12.40 [a] F (s) =
K1
K2
K3
+
+
;
s+1 s+2 s+4
K1 =
8s2 + 37s + 32
= 1;
(s + 2)(s + 4) s= 1
K2 =
8s2 + 37s + 32
= 5;
(s + 1)(s + 4) s= 2
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12–24
CHAPTER 12. Introduction to the Laplace Transform
K3 =
8s2 + 37s + 32
= 2;
(s + 1)(s + 2) s= 4
f (t) = [e t + 5e 2t + 2e 4t ]u(t).
[b] F (s) =
K2
K3
K4
K1
+
+
+
;
s
s+2 s+3 s+5
K1 =
8s3 + 89s2 + 311s + 300
= 10;
(s + 2)(s + 3)(s + 5) s=0
K2 =
8s3 + 89s2 + 311s + 300
= 5;
s(s + 3)(s + 5)
s= 2
K3 =
8s3 + 89s2 + 311s + 300
=
s(s + 2)(s + 5)
s= 3
8;
8s3 + 89s2 + 311s + 300
= 1;
K4 =
s(s + 2)(s + 3)
s= 5
f (t) = [10 + 5e 2t
[c] F (s) =
8e 3t + e 5t ]u(t).
K2
K2⇤
K1
+
+
;
s+1 s+2 j s+2+j
K1 =
22s2 + 60s + 58
= 10;
s2 + 4s + 5
s= 1
K2 =
22s2 + 60s + 58
= 10 + j8 = 10/53.13 ;
(s + 1)(s + 2 + j) s= 2+j
f (t) = [10e t + 20e 2t cos(t + 53.13 )]u(t).
[d] F (s) =
K2
K2⇤
K1
+
+
;
s
s+7 j s+7+j
K1 =
250(s + 7)(s + 14)
= 490;
s2 + 14s + 50
s=0
K2 =
250(s + 7)(s + 14)
= 125/
s(s + 7 + j)
s= 7+j
f (t) = [490 + 250e 7t cos(t
P 12.41 [a] F (s) =
K1 =
163.74 ;
163.74 )]u(t).
K1⇤
K1
+
;
s + 7 j14 s + 7 + j14
280
=
s + 7 + j14 s= 7+j14
f (t) = [20e 7t cos(14t
j10 = 10/
90 ;
90 )]u(t) = [20e 7t sin 14t]u(t).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[b] F (s) =
K1
K2
K2⇤
+
+
;
s
s + 5 j8 s + 5 + j8
K1 =
s2 + 52s + 445
= 5;
s2 + 10s + 89 s=0
K2 =
s2 + 52s + 445
=
s(s + 5 + j8) s= 5+j8
f (t) = [5 + 7.2e 5t cos(8t
[c] F (s) =
12–25
3
j2 = 3.6/
146.31 ;
146.31 )]u(t).
K1
K2
K2⇤
+
+
;
s + 6 s + 2 j4 s + 2 + j4
K1 =
14s2 + 56s + 152
= 10;
s2 + 4s + 20
s= 6
K2 =
14s2 + 56s + 152
= 2 + j2 = 2.83/45 ;
(s + 6)(s + 2 + j4) s= 2+j4
f (t) = [10e 6t + 5.66e 2t cos(4t + 45 )]u(t).
[d] F (s) =
K1
K1⇤
K2
K2⇤
+
+
+
;
s + 5 j3 s + 5 + j3 s + 4 j2 s + 4 + j2
8(s + 1)2
= 4.62/
K1 =
(s + 5 + j3)(s2 + 8s + 20) s= 5+j3
K2 =
8(s + 1)2
= 3.61/168.93 ;
(s + 4 + j2)(s2 + 10s + 34) s= 4+j2
f (t) = [9.25e 5t cos(3t
P 12.42 [a] F (s) =
K1 =
40.05 ) + 7.21e 4t cos(2t + 168.93 )]u(t).
K3
K1 K2
+
;
+
2
s
s
s+8
320
= 40;
s + 8 s=0
"

#
d 320
320
K2 =
=
=
ds s + 8
(s + 8)2 s=0
K3 =
40.04 ;
5;
320
= 5;
s2 s= 8
f (t) = [40t
5 + 5e 8t ]u(t).
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12–26
CHAPTER 12. Introduction to the Laplace Transform
[b] F (s) =
K1
K2
K3
+
;
+
2
s
(s + 2)
s+2
K1 =
80(s + 3)
= 60;
(s + 2)2 s=0
K2 =
80(s + 3)
=
s
s= 2
"
#
40;
"
d 80(s + 3)
80
=
K3 =
ds
s
s
f (t) = [60
[c] F (s) =
K1 =
40te 2t
#
80(s + 3)
=
s2
s= 2
60e 2t ]u(t).
K1
K3
K3⇤
K2
+
+
;
+
(s + 1)2 s + 1 s + 3 j4 s + 3 + j4
60(s + 5)
s2 + 6s + 25
= 12;
s= 1
"
#
"
d
60(s + 5)
60
= 2
K2 =
2
ds s + 6s + 25
s + 6s + 25
K3 =
60;
#
60(s + 5)(2s + 6)
= 0.6;
(s2 + 6s + 25)2 s= 1
60(s + 5)
= 1.68/100.305 ;
(s + 1)2 (s + 3 + j4) s= 3+j4
f (t) = [12te t + 0.6e t + 3.35e 3t cos(4t + 100.305 )]u(t).
[d] F (s) =
K1 =
K1 K2
K3
K4
+
;
+
+
2
2
s
s
(s + 5)
s+5
25(s + 4)2
= 16;
(s + 5)2 s=0
"
#
"
d 25(s + 4)2
25(2)(s + 4)
K2 =
=
2
ds (s + 5)
(s + 5)2
K3 =
#
25(2)(s + 4)2
= 1.6;
(s + 5)3
s=0
25(s + 4)2
= 1;
s2
s= 5
"
#
"
d 25(s + 4)2
25(2)(s + 4)
=
K4 =
2
ds
s
s2
f (t) = [16t + 1.6 + te 5t
#
25(2)(s + 4)2
=
s3
s= 5
1.6;
1.6e 5t ]u(t).
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Problems
K1
K2
K3
K4
+
;
+
+
3
2
s
(s + 3)
(s + 3)
s+3
P 12.43 [a] F (s) =
K1 =
135
= 5;
(s + 3)3 s=0
K2 =
135
=
s s= 3

d 135
K3 =
=
ds s
45;

135
=
s2 s= 3
1 d
K4 =
2 ds

f (t) = [5
22.5t2 e 3t

135
1
= ( 2)
2
s
2
✓
15te 3t
15;
135
s3
◆
=
5;
s= 3
5e 3t ]u(t).
K1
K2⇤
K1⇤
K2
+
;
+
+
(s + 1 j1)2 (s + 1 + j1)2 s + 1 j1 s + 1 + j1
[b] F (s) =
K1 =
12–27
10(s + 2)2
=
(s + 1 + j1)2 s= 1+j1
"
#
j5 = 5/
"
d
10(s + 2)2
10(2)(s + 2)
K2 =
=
2
ds (s + 1 + j1)
(s + 1 + j1)2
=
j5 = 5/
f (t) = [10te t cos(t
90 ;
#
10(2)(s + 2)2
(s + 1 + j1)3 s=0
90 ;
90 ) + 10e t cos(t
[c]
90 )]u(t).
25
F (s) = s2 + 15s + 54
25s2 + 395s + 1494
25s2 + 375s + 1350
20s + 144
F (s) = 25 +
20s + 144
s2 + 15s + 54
K1 =
20s + 144
=8
s+9
s= 6
K2 =
20s + 144
= 12
s+6
s= 9
= 25 +
K1
K2
+
s+6 s+9
f (t) = 25 (t) + [8e 6t + 12e 9t ]u(t)
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12–28
CHAPTER 12. Introduction to the Laplace Transform
[d]
F (s) = s2 + 7s + 10
5s3 + 20s2
5s
15
49s
108
5s3 + 35s2 + 50s
15s2
99s
108
15s2
105s
150
6s + 42
F (s) = 5s
15 +
K2
K1
+
;
s+2 s+5
K1 =
6s + 42
= 10;
s + 5 s= 2
K2 =
6s + 42
=
s + 2 s= 5
f (t) = 5 0 (t)
P 12.44 f (t) = L
1
(
4;
15 (t) + [10e 2t
4e 5t ]u(t).
K⇤
K
+
s+↵ j
s+↵+j
)
= Ke ↵t ej t + K ⇤ e ↵t e j t
= |K|e ↵t [ej✓ ej t + e j✓ e j t ]
= |K|e ↵t [ej( t+✓) + e j( t+✓) ]
= 2|K|e ↵t cos( t + ✓).
n
n
P 12.45 [a] L{t f (t)} = ( 1)
"
#
dn F (s)
.
dsn
1
then F (s) = ,
s
Let f (t) = 1,
n
n
Therefore L{t } = ( 1)
"
It follows that L{t(r 1) } =
and L{t(r 1) e at } =
Therefore
K
(r
dn F (s)
( 1)n n!
=
.
dsn
s(n+1)
#
n!
( 1)n n!
= (n+1) .
(n+1)
s
s
(r
1)!
sr
(r 1)!
.
(s + a)r
r 1
1)!
thus
L{t
e
at
(
)
K
Ktr 1 e at
}=
=
L
.
(s + a)r
(r 1)!
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
1
[b] f (t) = L
(
12–29
)
K⇤
K
+
.
(s + ↵ j )r (s + ↵ + j )r
Therefore
Ktr 1 (↵ j )t K ⇤ tr 1 (↵+j )t
f (t) =
e
e
+
(r 1)!
(r 1)!
=
i
|K|tr 1 e ↵t h j✓ j t
e e + e j✓ e j t
(r 1)!
#
"
2|K|tr 1 e ↵t
cos( t + ✓).
=
(r 1)!
P 12.46 F (s) =
80(s + 3)
;
s(s + 2)2
This function has a zero at
F (s) =
3 rad/s, a pole at 0, and two poles at
2 rad/s.
60(s + 5)
;
(s + 1)2 (s2 + 6s + 25)
This function has a zero at 5 rad/s, two poles at 1 rad/s, and complex
conjugate poles at 3 + j4 rad/s and 3 j4 rad/s.
P 12.47 F (s) =
135
;
s(s + 3)3
This function has no zeros, a pole at 0, and three poles at
F (s) =
3 rad/s.
10(s + 2)2
;
(s2 + 2s + 2)2
This function has two zeros at 2 rad/s and two pairs of complex conjugate
poles at 1 + j1 rad/s and 1 j1 rad/s.
"
#
1.92s3
P 12.48 [a] lim sV (s) = lim 4
= 0.
s!1
s!1 s [1 + (1.6/s) + (1/s2 )][1 + (1/s2 )]
Therefore v(0+ ) = 0.
[b] No, V has a pair of poles on the imaginary axis.
P 12.49 sVo (s) =
(Idc /C)s
2
s + (1/RC)s + (1/LC)
lim sVo (s) = 0,
.·. vo (1) = 0;
lim sVo (s) = 0,
.·. vo (0+ ) = 0;
s!0
s!1
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12–30
CHAPTER 12. Introduction to the Laplace Transform
sIo (s) =
s2 Idc
;
s2 + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s!0
.·. io (1) = 0;
.·. io (0+ ) = Idc .
lim sIo (s) = Idc ,
s!1
P 12.50 sIo (s) =
Idc s[s + (1/RC)]
;
2
s + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s!0
.·. io (1) = 0;
.·. io (0+ ) = Idc .
lim sIo (s) = Idc ,
s!1
P 12.51 sVo (s) =
sVdc /RC
;
2
s + (1/RC)s + (1/LC)
lim sVo (s) = 0,
.·. vo (1) = 0;
lim sVo (s) = 0,
.·. vo (0+ ) = 0;
s!0
s!1
sIo (s) =
Vdc /RLC
;
2
s + (1/RC)s + (1/LC)
lim sIo (s) =
s!0
Vdc
Vdc /RLC
=
,
1/LC
R
lim sIo (s) = 0,
s!1
P 12.52 [a] sF (s) =
Vdc
.·. io (1) =
;
R
.·. io (0+ ) = 0.
8s3 + 37s2 + 32s
;
(s + 1)(s + 2)(s + 4)
lim sF (s) = 0,
.·. f (1) = 0;
lim sF (s) = 8,
.·. f (0+ ) = 8.
s!0
s!1
[b] sF (s) =
8s3 + 89s2 + 311s + 300
;
(s + 2)(s2 + 8s + 15)
lim sF (s) = 10;
.·. f (1) = 10;
lim sF (s) = 8,
.·. f (0+ ) = 8.
s!0
s!1
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[c] sF (s) =
22s3 + 60s2 + 58s
;
(s + 1)(s2 + 4s + 5)
.·. f (1) = 0;
lim sF (s) = 0,
s!0
lim sF (s) = 22,
s!1
[d] sF (s) =
12–31
.·. f (0+ ) = 22.
250(s + 7)(s + 14)
;
(s2 + 14s + 50)
250(7)(14)
= 490,
.·. f (1) = 490;
s!0
50
lim sF (s) = 250,
.·. f (0+ ) = 250.
lim sF (s) =
s!1
P 12.53 [a] sF (s) =
280s
s2 + 14s + 245
;
lim sF (s) = 0,
.·. f (1) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
s2 + 52s + 445
;
[b] sF (s) = 2
s + 10s + 89
.·. f (1) = 5;
lim sF (s) = 5,
s!0
lim sF (s) =
s!1
[c] sF (s) =
1,
1.
14s3 + 56s2 + 152s
;
(s + 6)(s2 + 4s + 20)
lim sF (s) = 0,
s!0
lim sF (s) = 14,
s!1
[d] sF (s) =
.·. f (0+ ) =
.·. f (1) = 0;
.·. f (0+ ) = 14.
8s(s + 1)2
;
(s2 + 10s + 34)(s2 + 8s + 20)
lim sF (s) = 0,
.·. f (1) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
320
.
s(s + 8)
F (s) has a second-order pole at the origin so we cannot use the final
value theorem here.
P 12.54 [a] sF (s) =
lim sF (s) = 0,
s!1
.·. f (0+ ) = 0.
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12–32
CHAPTER 12. Introduction to the Laplace Transform
[b] sF (s) =
80(s + 3)
;
(s + 2)2
lim sF (s) = 60,
.·. f (1) = 60;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
[c] sF (s) =
60s(s + 5)
;
(s + 1)2 (s2 + 6s + 25)
lim sF (s) = 0,
.·. f (1) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
25(s + 4)2
.
s(s + 5)2
F (s) has a second-order pole at the origin so we cannot use the final
value theorem here.
[d] sF (s) =
lim sF (s) = 0,
s!1
P 12.55 [a] sF (s) =
.·. f (0+ ) = 0.
135
;
(s + 3)3
lim sF (s) = 5,
.·. f (1) = 5;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
[b] sF (s) =
10s(s + 2)2
;
(s2 + 2s + 2)2
lim sF (s) = 0,
.·. f (1) = 0;
lim sF (s) = 0,
.·. f (0+ ) = 0.
s!0
s!1
[c] This F (s) function is an improper rational function, and thus the
corresponding f (t) function contains impulses ( (t)). Neither the initial
value theorem nor the final value theorem may be applied to this F (s)
function!
[d] This F (s) function is an improper rational function, and thus the
corresponding f (t) function contains impulses ( (t)). Neither the initial
value theorem nor the final value theorem may be applied to this F (s)
function!
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 12.56 [a] ZL = j120⇡(0.01) = j3.77 ⌦;
ZC =
The phasor-transformed circuit is
IL =
j
=
120⇡(100 ⇥ 10 6 )
12–33
j26.526 ⌦.
1
= 36.69/56.61 mA;
15 + j3.77 j26.526
.·. iL(ss) (t) = 36.69 cos(120⇡t + 56.61 ) mA.
[b] The steady-state response is the second term in the expression for iL (t)
derived in the Practical Perspective. This matches the steady-state
response just derived in part (a).
P 12.57 The transient and steady-state components are both proportional to the
magnitude of the input voltage. Therefore,
K=
40
= 0.947.
42.26
So if we make the amplitude of the sinusoidal source 0.947 instead of 1, the
current will not exceed the 40 mA limit. A plot of the current through the
inductor is shown below with the amplitude of the sinusoidal source set at
0.947.
P 12.58 We begin by using the Laplace transform of the describing di↵erential
equation for the circuit in Fig. 12.19. Change the right-hand side so it is the
Laplace transform of Kte 100t to give:
15IL (s) + 0.01sIL (s) + 104
A
IL (s)
=
.
s
(s + 100)2
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12–34
CHAPTER 12. Introduction to the Laplace Transform
Solving for IL (s),
K1⇤
100Ks
K1
IL (s) = 2
+
=
(s + 1500s + 106 )(s + 100)2
s + 750 j661.44 s + 750 + j661.44
+
K1 =
K2 =
K2
K3
.
+
2
(s + 100)
s + 100
100Ks
= 87.9K /139.59 µA;
(s + 750 + j661.44)(s + 100)2 s= 750+j661.44
100Ks
(s2 + 1500s + 106 )
"
=
11.63K mA;
s= 100
#
d
100Ks
K3 =
= 133.86K µA.
2
ds (s + 1500s + 106 ) s= 100
Therefore,
iL (t) = K[0.176e 750t cos(661.44t + 139.59 )
11.63te 100t + 0.134e 100t ]u(t) mA.
Plot the expression above with K = 1:
The maximum value of the inductor current is 0.068K mA. Therefore,
K=
40
= 588.
0.068
So the inductor current rating will not be exceeded if the input to the RLC
circuit is 588te 100t V.
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13
The Laplace Transform in Circuit
Analysis
Assessment Problems
AP 13.1 [a] Z = 2000 +
4 ⇥ 107 s
1
= 2000 + 2
Y
s + 80,000s + 25 ⇥ 108
2000(s2 + 105 s + 25 ⇥ 108 )
2000(s + 50,000)2
=
.
s2 + 80,000s + 25 ⇥ 108
s2 + 80,000s + 25 ⇥ 108
z1 = z2 = 50,000 rad/s;
=
[b]
p1 =
40,000
p2 =
40,000 + j30,000 rad/s.
AP 13.2 [a] At t = 0 ,
j30,000 rad/s;
0.2v1 = (0.8)v2 ;
v1 = 4v2 ;
Therefore v1 (0 ) = 80V = v1 (0+ );
v1 + v2 = 100 V;
v2 (0 ) = 20V = v2 (0+ ).
20 ⇥ 10 3
(80/s) + (20/s)
=
;
I=
5000 + [(5 ⇥ 106 )/s] + (1.25 ⇥ 106 /s)
s + 1250
13–1
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13–2
CHAPTER 13. The Laplace Transform in Circuit Analysis
80
V1 =
s
5 ⇥ 106
s
20
V2 =
s
1.25 ⇥ 106
s
20 ⇥ 10 3
s + 1250
[b] i = 20e 1250t u(t) mA;
!
=
80
;
s + 1250
!
=
20 ⇥ 10 3
s + 1250
20
.
s + 1250
v1 = 80e 1250t u(t) V;
v2 = 20e 1250t u(t) V.
AP 13.3 [a]
I=
Vdc /L
Vdc /s
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
Vdc
= 40;
L
I=
(s + 0.6
K1 =
40
=
j1.6
R
1
= 1.2;
= 1.0;
L
LC
40
K1
K1⇤
=
+
;
j0.8)(s + 0.6 + j0.8)
s + 0.6 j0.8 s + 0.6 + j0.8
j25 = 25/
90 ;
K1⇤ = 25/90 .
[b] i = 50e 0.6t cos(0.8t
90 ) = [50e 0.6t sin 0.8t]u(t) A.
160s
[c] V = sLI =
(s + 0.6 j0.8)(s + 0.6 + j0.8)
=
K1 =
K1
K1⇤
+
;
s + 0.6 j0.8 s + 0.6 + j0.8
160( 0.6 + j0.8)
= 100/36.87 .
j1.6
[d] v(t) = [200e 0.6t cos(0.8t + 36.87 )]u(t) V.
AP 13.4 Transforming the circuit into the s domain for t > 0:
✓
40
1
s
+
Zeq =
+
6
40 ⇥ 10
625
s
◆ 1
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Problems
=
40 ⇥ 106 s
;
s2 + 64,000s + 16 ⇥ 108
Zeq
I=
s/40
=
=
K1 =
13–3
24s
2
s + 40,0002
!
384 ⇥ 106 s
(s2 + 40,0002 )(s + 32,000 j24,000)(s + 32,000 + j24,000)
s
K1⇤
K2
K2⇤
K1
+
+
+
;
j40,000 s + j40,000 s + 32,000 j24,000 s + 32,000 + j24,000
384 ⇥ 106 s
=
(s + j40,000)(s2 + 64,000s + 16 ⇥ 108 ) s=j40,000
j0.075 = 0.075/
90 ;
384 ⇥ 106 s
= j0.125 = 0.125/90 .
K2 = 2
(s + j40,0002 )(s + 32,000 + j24,000) s= 32,000+j24,000
Therefore,
i(t) = (0.15 sin 40,000t
0.25e 32,000t sin 24,000t)u(t) A.
AP 13.5 [a]
The two node voltage equations are
V2 V2 V1 V2
5
and
+
+
s
s
3
s
Solving for V1 and V2 yields
V1
V2
V1 =
5(s + 3)
,
s(s2 + 2.5s + 1)
(15/s)
= 0.
15
+ V1 s =
V2 =
2.5(s2 + 6)
.
s(s2 + 2.5s + 1)
[b] The partial fraction expansions of V1 and V2 are
50/3
5/3
15
15
+
and V2 =
s
s + 0.5 s + 2
s
It follows that

50 0.5t 5 2t
v1 (t) = 15
e
+ e
u(t) V and
3
3

125 0.5t 25 2t
v2 (t) = 15
e
+ e
u(t) V.
6
3
V1 =
125/6
25/3
+
.
s + 0.5 s + 2
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13–4
CHAPTER 13. The Laplace Transform in Circuit Analysis
50 5
+ = 0;
3
3
[c] v1 (0+ ) = 15
v2 (0+ ) = 15
125 25
+
= 2.5 V.
6
3
[d] v1 (1) = 15 V;
v2 (1) = 15 V.
AP 13.6 [a]
With no load across terminals a


1 20
2 s
VTh s + 1.2
Therefore VTh =
Vx = 5IT
✓
◆
20
s
b Vx = 20/s:
VTh = 0;
20(s + 2.4)
.
s(s + 2)
and ZTh =
VT
.
IT
Solving for IT gives
IT =
(VT
5IT )s
+ VT
2
6IT .
Therefore
14IT = VT s + 5sIT + 2VT ;
therefore ZTh =
5(s + 2.8)
.
s+2
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Problems
13–5
[b]
I=
20(s + 2.4)
VTh
=
.
ZTh + 2 + s
s(s + 3)(s + 6)
AP 13.7 [a] i2 = 1.25e t
Therefore
1.25e 3t ;
so
di2
=
dt
1.25e t + 3.75e 3t .
di2
= 0 when
dt
1.25e t = 3.75e 3t
or e2t = 3,
i2 (max) = 1.25[e 0.549
t = 0.5(ln 3) = 549.31 ms.
e 3(0.549) ] = 481.13 mA.
[b] Solving the mesh current equations from Example 13.7 are
(3 + 2s)I1 + 2sI2 = 10;
and
2sI1 + (12 + 8s)I2 = 10.
Solving for I1 ,
I1 =
5(s + 2)
.
(s + 1)(s + 3)
A partial fraction expansion leads to the expression
2.5
2.5
+
.
s+1 s+3
Therefore we get
I1 =
i1 = 2.5[e t + e 3t ]u(t) A.
di1 (0.54931)
di1
= 2.5[e t + 3e 3t ];
=
dt
dt
[d] When i2 is at its peak value,
2.89 A/s.
[c]
di2
= 0.
dt
Therefore L2
[e] i2 (max) =
di2
dt
!
= 0 and i2 =
✓
M
12
◆
!
di1
.
dt
2( 2.89)
= 481.13 mA. (checks)
12
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13–6
CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V0
V 0s
(20/s)
+
+
= 0;
2
1.25s
20
100/3 100/3
200
V0 =
=
;
(s + 2)(s + 8)
s+2
s+8
V0
100 2t
[e
e 8t ]u(t) V.
3
[b] With the current source acting alone,
v0 =
V 00
V 00 s
5
V 00
+
+
= ;
2
1.25s
20
s
50/3
100
V 00 =
=
(s + 2)(s + 8)
s+2
v 00 =
50 2t
[e
3
e 8t ]u(t) V.
[c] v = v 0 + v 00 = [50e 2t
Vo
Vo s
+
= Ig ;
s+2
10
[b] z1 = 2 rad/s;
AP 13.9 [a]
AP 13.10 [a]
Vo =
50/3
;
s+8
50e 8t ]u(t) V.
Vo
10(s + 2)
.
= H(s) = 2
Ig
s + 2s + 10
1 + j3 rad/s;
p2 = 1 j3 rad/s.
therefore
p1 =
10(s + 2)
1
Ko
K1
K1⇤
·
=
+
+
;
s2 + 2s + 10 s
s
s + 1 j3 s + 1 + j3
Ko = 2;
K1 = 5/3/
vo = [2 + (10/3)e t cos(3t
126.87 ;
K1⇤ = 5/3/126.87 ;
126.87 )]u(t) V.
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Problems
[b] Vo =
13–7
K2
K2⇤
10(s + 2)
·
1
=
+
;
s2 + 2s + 10
s + 1 j3 s + 1 + j3
K2 = 5.27/
K2⇤ = 5.27/18.43 ;
18.43 ;
vo = [10.54e t cos(3t
18.43 )]u(t) V.
AP 13.11 [a]
H(s) = L{h(t)} = L{vo (t)}
vo (t) = 10,000 cos ✓e 70t cos 240t
= 9600e 70t cos 240t
Therefore H(s) =
=
[b] Vo (s) = H(s) ·
2800e 70t sin 240t.
2800(240)
(s + 70)2 + (240)2
9600(s + 70)
(s + 70)2 + (240)2
9600s
s2 + 140s + 62,500
.
9600
1
= 2
s
s + 140s + 62,500
=
K1 =
10,000 sin ✓e 70t sin 240t
9600
=
j480
K1
K1⇤
+
s + 70 j240 s + 70 + j240
j20 = 20/
90 .
Therefore
vo (t) = [40e 70t cos(240t
90 )]u(t) V = [40e 70t sin 240t]u(t) V.
AP 13.12 From Assessment Problem 13.9:
H(s) =
10(s + 2)
.
s2 + 2s + 10
Therefore H(j4) =
10(2 + j4)
= 4.47/
10 16 + j8
63.43 .
Thus,
vo = (10)(4.47) cos(4t
63.43 ) = 44.7 cos(4t
63.43 ) V.
AP 13.13 [a]
Let R1 = 10 k⌦,
R2 = 50 k⌦,
Then V1 = V2 =
Vg R2
.
R2 + (1/sC)
C = 400 pF,
R2 C = 2 ⇥ 10 5 .
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13–8
CHAPTER 13. The Laplace Transform in Circuit Analysis
Also
V1
Vg
R1
+
V1
therefore Vo = 2V1
Vo
R1
= 0,
Vg .
Now solving for Vo /Vg , we get H(s) =
It follows that H(j50,000) =
R2 Cs 1
.
R2 Cs + 1
j 1
= j1 = 1/90.
j+1
Therefore vo = 10 cos(50,000t + 90 ) V.
[b] Replacing R2 by Rx gives us H(s) =
Rx Cs 1
.
Rx Cs + 1
Therefore
H(j50,000) =
Thus,
j20 ⇥ 10 6 Rx 1
Rx + j50,000
=
.
6
j20 ⇥ 10 Rx + 1
Rx j50,000
50,000
= tan 60 = 1.7321,
Rx
Rx = 28,867.51 ⌦.
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Problems
13–9
Problems
P 13.1
1Zt
i=
vd⌧ + I0 ;
L 0
P 13.2
IN =
✓
1
therefore I =
L
◆✓
V
s
◆
+
V
I0
I0
=
+ .
s
sL
s
I0
LI0
=
;
ZN = sL.
sL
s
Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
✓
P 13.3
VTh = Vab = CV0
P 13.4
[a] Z = R + sL +
1
sC
◆
=
V0
;
s
ZTh =
1
.
sC
L[s2 + (R/L)s + (1/LC)]
1
=
sC
s
0.025[s2 + 16 ⇥ 104 s + 1010 ]
.
s
[b] Zeros at 80,000 + j60,000 rad/s and
Pole at 0.
=
P 13.5
j60,000 rad/s;
1
1
C[s2 + (1/RC)s + (1/LC)]
[a] Y = +
+ sC =
;
R sL
s
Z=
1
s/C
4 ⇥ 106 s
= 2
= 2
.
Y
s + (1/RC)s + (1/LC)
s + 2000s + 64 ⇥ 104
[b] zero at z1 = 0,
poles at p1 =
P 13.6
80,000
400 rad/s and
p2 =
1600 rad/s.
[a]
Z=
(1/C)(s + R/L)
(R + sL)(1/sC)
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
250
R
=
= 3125;
L
0.08
1
1
=
= 25 ⇥ 106 ;
6
LC
(0.08)(0.5 ⇥ 10 )
2 ⇥ 106 (s + 3125)
Z= 2
.
s + 3125s + 25 ⇥ 106
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13–10
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Z =
P 13.7
(s + 1562.5
2 ⇥ 106 (s + 3125)
;
j4749.6)(s + 1562.5 + j4749.6)
z1 =
3125 rad/s;
p2 =
1562.5
p1 =
1562.5 + j4749.6 rad/s;
j4749.6 rad/s.
Transform the Y-connection of the two resistors and the capacitor into the
equivalent delta-connection:
where
Za =
(1/s)(1) + (1)(1/s) + (1)(1)
= s + 2;
(1/s)
Zb = Zc =
s+2
(1/s)(1) + (1)(1/s) + (1)(1)
=
.
1
s
Then
Zab = Za k[(skZc ) + (skZb )] = Za k2(skZb );
skZb =
s(s + 2)
s[(s + 2)/s]
= 2
;
s + [(s + 2)/s]
s +s+2
2s(s + 2)2
2s(s + 2)
s2 + s + 2
Zab = (s + 2)k 2
=
2s(s + 2)
s +s+2
s+2+ 2
s +s+2
2s(s + 2)2
2s(s + 2)
2s
=
= 2
=
.
2
(s + 2)(s + s + 2) + 2s(s + 2)
s + 3s + 2
s+1
Zero at 0; pole at
P 13.8
Z1 =
1 rad/s.
16
4s
4(s2 + 4s + 16)
16
+ sk4 =
+
=
;
s
s
s+4
s(s + 4)
Zab = 4k
16(s2 + 4s + 16)
4(s2 + 4s + 16)
=
s(s + 4)
8s2 + 32s + 64
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Problems
=
2(s2 + 4s + 16)
2(s + 2 + j3.46)(s + 2 j3.46)
=
.
2
s + 4s + 8
(s + 2 + j2)(s + 2 j2)
Zeros at 2 + j3.46 rad/s and
2 j2 rad/s.
P 13.9
13–11
2
j3.46 rad/s; poles at
2 + j2 rad/s and
[a] For t > 0:
2.5s
[b] Vo =
(16 ⇥ 105 )/s + 5000 + 2.5s
=
✓
150
s
◆
150s
s2 + 2000s + 64 ⇥ 104
150s
.
(s + 400)(s + 1600)
K2
K1
+
.
[c] Vo =
s + 400 s + 1600
=
K1 =
150s
= 50;
s + 1600 s= 400
K2 =
150s
=
s + 400 s= 1600
Vo =
50
s + 400
200;
200
.
s + 1600
vo (t) = (50e 400t
200e 1600t )u(t) V.
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13–12
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.10 [a] io (0 ) =
20
= 5 mA.
4000
sC(20/s + L⇢)
20/s + L⇢
= 2
R + sL + 1/sC
s LC + RsC + 1
Io =
=
Vo =
=
20/L + s⇢
s2 + sR/L + 1/LC
L⇢ + sLIo =
40 + s(0.005)
s2 + 8000s + 16 ⇥ 106
0.0025 +
40,000
.
(s + 4000)2
;
0.0025s(s + 8000)
s2 + 8000s + 16 ⇥ 106
40,000te 4000t u(t) V.
vo (t) =
[b] Io =
=
0.005(s + 8000)
s2 + 8000s + 16 ⇥ 106
=
K1
K2
.
+
2
(s + 4000)
s + 4000
K1 = 20
K2 = 0.005;
io (t) = [20te 4000t + 0.005e 4000t ]u(t) A.
P 13.11
✓
5 ⇥ 106 /s
240
Vo =
6
1120 + 0.8s + 5 ⇥ 10 /s s
=
◆
12 ⇥ 108
s(0.8s2 + 1120s + 5 ⇥ 106 )
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Problems
=
15 ⇥ 108
s(s2 + 1400s + 625 ⇥ 104 )
=
K1
K2
K2⇤
+
+
.
s
s + 700 j2400 s + 700 + j2400
K1 = 240;
13–13
K2 = 125/163.74 ;
vo (t) = [240 + 250e 700t cos(2400t + 163.74 )]u(t) V.
P 13.12
Vo s
Vo 240/s
+
+ 14.4 ⇥ 10 6 = 0;
1120 + 0.8s 5 ⇥ 106
Vo
✓
s
1
+
1120 + 0.8s 5 ⇥ 106
=
240/s
0.8s + 1120
14.4 ⇥ 10 6 ;
72s2 100,800s + 15 ⇥ 108
s(s2 + 1400s + 625 ⇥ 104 )
Vo =
=
.·.
◆
162.5/163.74
162.5/ 163.74
240
+
+
.
s
s + 700 j2400
s + 700 + j2400
vo (t) = [240 + 325e 700t cos(2400t + 163.74 )]u(t) V.
P 13.13 [a]
Vo =
Ig /C
(1/sC)(sL)(Ig /s)
= 2
;
R + sL + (1/sC)
s + (R/L)s + (1/LC)
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13–14
CHAPTER 13. The Laplace Transform in Circuit Analysis
15
Ig
=
= 150;
C
0.1
1
= 10;
LC
R
= 7;
L
Vo =
[b] sVo =
150
s2 + 7s + 10
150s
s2 + 7s + 10
;
lim sVo = 0;
.·. vo (1) = 0.
lim sVo = 0;
.·. vo (0+ ) = 0.
s!0
s!1
[c] Vo =
150
50
50
=
+
;
(s + 2)(s + 5)
s+2 s+5
vo = [50e 2t
P 13.14 IL =
.
Ig
s
15
IL =
s
50e 5t ]u(t) V.
Vo
Ig
=
1/sC
s
sCVo ;
15s
15
=
(s + 2)(s + 5)
s
iL (t) = [15 + 10e 2t

25
10
+
;
s+2 s+5
25e 5t ]u(t) A.
Check:
iL (0+ ) = 0 (ok);
iL (1) = 15 A. (ok)
P 13.15 [a] For t < 0:
1
1
1
1
+
= 0.1875;
= +
Re
8 80 20
Re = 5.33 ⌦;
v1 = (9)(5.33) = 48 V;
iL (0 ) =
48
= 2.4 A;
20
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Problems
vC (0 ) =
v1 =
13–15
48 V.
For t = 0+ :
s-domain circuit:
where
[b]
R = 20 ⌦;
C = 6.25 µF;
=
L = 6.4 mH;
and
⇢=
2.4 A.
Vo
sL
⇢
= 0;
s
Vo
+ Vo sC
R
.·. Vo =
C+
[s + (⇢/ C)]
s2 + (1/RC)s + (1/LC)
48 V;
.
2.4
⇢
=
= 8000;
C
( 48)(6.25 ⇥ 10 6 )
1
1
=
= 8000;
RC
(20)(6.25 ⇥ 10 6 )
1
1
=
= 25 ⇥ 106 ;
3
LC
(6.4 ⇥ 10 )(6.25 ⇥ 10 6 )
Vo =
[c] IL =
48(s + 8000)
s2 + 8000s + 25 ⇥ 106
Vo
sL
=
.
⇢
Vo
2.4
=
+
s
0.0064s
s
7500(s + 8000)
s(s2 + 8000s + 25 ⇥ 106 )
2.4
2.4(s + 4875)
= 2
.
s
(s + 8000s + 25 ⇥ 106 )
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13–16
CHAPTER 13. The Laplace Transform in Circuit Analysis
[d] Vo =
=
48(s + 8000)
s2 + 8000s + 25 ⇥ 106
K1
s + +4000
K1 =
j3000
+
K1⇤
.
s + 4000 + j3000
48(s + 8000)
= 40/126.87 ;
s + 4000 + j3000 s= 4000+j3000
vo (t) = [80e 4000t cos(3000t + 126.87 ]u(t) V.
[e] IL =
=
2.4(s + 4875)
s2 + 8000s + 25 ⇥ 106
K1⇤
K1
+
.
s + 4000 j3000 s + 4000 + j3000
K1 =
2.4(s + 4875)
= 1.25/
s + 4000 + j3000 s= 4000+j3000
iL (t) = [2.5e 4000t cos(3000t
16.26 ;
16.26 )]u(t) A.
P 13.16 [a] For t < 0:
Vc
Vc 275
100
+
+
= 0;
80
240
100
✓
◆
1
1
100 275
1
Vc
+
+
+
;
=
80 240 100
80
100
Vc
Vc = 150 V;
275 150
= 1.25 A.
100
For t > 0:
iL (0 ) =
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[b] Vo =
150
12,500
I+
;
s
s
275
12,500
150
+ 20I +
I+
s
s
s
0=
✓
20 ⇥ 10 3 + 0.016sI;
◆
I 20 +
12,500
125
+ 0.016s =
+ 20 ⇥ 10 3 ;
s
s
.·. I =
7812.5 + 1.25s
.
s2 + 1250s + 781,250
12,500
Vo =
s
=
[c] Vo =
13–17
!
150
7812.5 + 1.25s
+
s2 + 1250s + 781,250
s
150s2 + 203,125s + 214,843,750
.
s(s2 + 1250s + 781,250)
K2
K2⇤
K1
+
+
.
s
s + 625 j625 s + 625 + j625
K1 =
150s2 + 203,125s + 214,843,750
= 275;
s2 + 1250s + 781,250
s=0
K2 =
150s2 + 203,125s + 214,843,750
= 80.04/141.34 ;
s(s + 625 + j625)
s= 625+j625
vo (t) = [275 + 160.08e 625t cos(625t + 141.34 )]u(t) V.
P 13.17 For t < 0:
vo (0 ) 500 vo (0 ) vo (0 )
+
+
= 0;
5
25
100
25vo (0 ) = 10,000
iL (0 ) =
.·.
vo (0 ) = 400 V;
400
vo (0 )
=
= 16 A.
25
25
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13–18
CHAPTER 13. The Laplace Transform in Circuit Analysis
For t > 0 :
Vo
Vo (400/s)
Vo + 400
+
+
= 0;
25 + 25s 100
100/s
Vo
✓
◆
1
s
1
+
+
=4
25 + 25s 100 100
.·.
Io =
Vo =
Vo
400
;
25 + 25s
400(s 3)
.
s2 + 2s + 5
20s 20
(400/s)
= 2
100/s
s + 2s + 5
K1⇤
K1
+
.
=
s + 1 j2 s + 1 + j2
K1 =
20(s + 1)
=
s + 1 + j2 s= 1+j2
10;
io (t) = [ 20e t cos 2t]u(t) A.
P 13.18 [a] For t < 0:
iL (0 ) =
40
40
=
= 0.25 A;
32 + 80k320 + 64
160
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
i1 =
80
( 0.25) =
400
13–19
0.05 A;
vC (0 ) = 80( 0.05) + 32( 0.25) + 40 = 28 V.
For t > 0:
[b] (160 + 0.2s + 40,000/s)I = 0.05 +
.·.
I=
28
;
s
0.25(s + 560)
s2 + 800s + 200,000
K1⇤
K1
+
.
=
s + 400 j200 s + 400 + j200
K1 =
0.25(s + 560)
= 0.16/
s + 400 + j200 s= 400+j200
[c] io (t) = 0.32e 400t cos(200t
38.66 .
38.66 )u(t) A.
P 13.19 [a]
0 = 0.5s(I1
30/s) +
2500
(I1
s
I2 ) + 100I1 ;
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13–20
CHAPTER 13. The Laplace Transform in Circuit Analysis
2500
375
=
(I2
s
s
I1 ) + 50(I2
30/s)
or
(s2 + 200s + 5000)I1
5000I2 = 30s;
50I1 + (s + 50)I2 = 22.5.
=
[b] sI1 =
50
(s + 50)
5000
= 30(s2 + 50s + 3750);
N1
=
30(s2 + 50s + 3750)
.
s(s + 100)(s + 150)
(s2 + 200s + 5000) 30s
50
N2
=
= 22.5s2 + 6000s + 112,500;
22.5
22.5s2 + 6000s + 112,500
.
s(s + 100)(s + 150)
30(s2 + 50s + 3750)
(s + 100)(s + 150)
lim sI1 = i1 (0+ ) = 30 A;
lim sI1 = i1 (1) = 7.5 A.
s!1
sI2 =
s!0
22.5s2 + 6000s + 112,500
;
(s + 100)(s + 150)
lim sI2 = i2 (0+ ) = 22.5 A;
s!1
[c] I1 =
= s(s + 100)(s + 150);
22.5 (s + 50)
N2 =
I2 =
5000
30s
N1 =
I1 =
(s2 + 200s + 5000)
lim sI2 = i2 (1) = 7.5 A.
s!0
K1
K2
K3
30(s2 + 50s + 3750)
=
+
+
s(s + 100)(s + 150)
s
s + 100 s + 150
K1 = 7.5;
K2 =
i1 (t) = [7.5
52.5e 100t + 75e 150t ]u(t) A.
I2 =
52.5;
K3 = 75
K1
K2
K3
22.5s2 + 6000s + 112,500
=
+
+
s(s + 100)(s + 150)
s
s + 100 s + 150
K1 = 7.5;
K2 = 52.5;
i2 (t) = [7.5 + 52.5e 100t
K3 =
37.5
37.5e 250t ]u(t) A
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–21
P 13.20 [a]
200i1 + 0.4s(I1
I2 ) =
200
;
s
200I2 +
105
I2 + 0.4s(I2
s
I1 ) = 0.
Solving the second equation for I1 :
s2 + 500s + 25 ⇥ 104
I2 .
s2
Substituting into the first equation and solving for I2 :
I1 =
(0.4s + 200)
.·.
I2 =
.·.
I1 =
=
Io = I1
0.5s
s2 + 500s + 125,000
;
0.5(s2 + 500s + 25 ⇥ 104 )
s(s2 + 500s + 125,000)
0.5s
s2 + 500s + 125,000
250(s + 500)
s(s2 + 500s + 125,000)
K1
K2
K2⇤
+
+
.
s
s + 250 j250 s + 250 + j250
K2 = 0.5/
io (t) = [1
[b] Vo = 0.4sIo =
K1 = 70.71/
.·.
200
;
s
0.5(s2 + 500s + 25 ⇥ 104 )
;
s(s2 + 500s + 125,000)
K1 = 1;
.·.
0.4s =
s2 + 500s + 25 ⇥ 104
0.5s
· 2
2
s
s + 500s + 125,000
I2 =
=
=
s2 + 500s + 25 ⇥ 104
s2
180 =
0.5;
1e 250t cos 250t]u(t) A.
K1
K1⇤
100(s + 500)
=
+
;
s2 + 500s + 125,000
s + 250 j250 s + 250 + j250
45 ;
vo (t) = 141.42e 250t cos(250t
45 )u(t) V.
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13–22
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] At t = 0+ the circuit is
.·. vo (0+ ) = 100 V = 141.42 cos( 45 );
Io (0+ ) = 0.
Both values agree with our solutions for vo and io .
At t = 1 the circuit is
.·. vo (1) = 0;
io (1) = 1 A.
Both values agree with our solutions for vo and io .
P 13.21 [a]
V1 V1 50/s V1 Vo
+
+
= 0;
10
25/s
4s
5 Vo V1 Vo 50/s
+
+
= 0.
s
4s
30
Simplfying,
(4s2 + 10s + 25)V1
25Vo = 200s;
15V1 + (2s + 15)Vo = 400.
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Problems
=
(4s2 + 10s + 25)
25
15
(2s + 15)
(4s2 + 10s + 25) 200s
No =
Vo =
K1 =
15
No
= 8s(s + 5)2 ;
= 200(8s2 + 35s + 50);
400
200(8s2 + 35s + 50)
K2
K1
K3
+
.
=
+
8s(s + 5)2
s
(s + 5)2 s + 5
=
(25)(50)
= 50;
25
"
K2 =
25(200
#
175 + 50)
=
5
"
d 8s2 + 35s + 50
s(16s + 35)
= 25
K3 = 25
ds
s
s= 5
=
5( 45)
375;
#
(8s2 + 35s + 50)
s2
s= 5
75 = 150;
375
150
;
+
2
(s + 5)
s+5
50
.·. Vo =
s
[b] vo (t) = [50
13–23
375te 5t + 150e 5t ]u(t) V.
[c] At t = 0+ :
vo (0+ ) = 50 + 150 = 200 V.(checks)
At t = 1:
vo (1)
10
5+
vo (1) 50
= 0;
30
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13–24
CHAPTER 13. The Laplace Transform in Circuit Analysis
.·. 3vo (1)
150 + vo (1)
.·. 4vo (1) = 200;
50 = 0;
.·.
vo (1) = 50 V.(checks)
V1
V1
V1 V2
325/s
+
+
= 0;
250
500
250
P 13.22 [a]
V2 V1 (V2 325/s)s
V2
+
+
= 0.
0.625s
250
125 ⇥ 104
Thus,
5V1
2V2 =
650
;
s
5000sV1 + (s2 + 5000s + 2 ⇥ 106 )V2 = 325s;
=
5
5000s s2 + 5000s + 2 ⇥ 106
5
N2 =
V2 =
2
650/s
= 5(s + 1000)(s + 2000; )
= 1625(s + 2000);
5000s 325s
N2
=
1625(s + 2000)
325
=
;
5(s + 1000)(s + 2000)
s + 1000
325
325,000
=
;
s + 1000
s(s + 1000)
Vo =
325
s
Io =
520
0.52
V2
=
=
0.625s
s(s + 1000)
s
[b] vo (t) = (325
io (t) = (520
0.52
.
s + 1000
325e 1000t )u(t) V;
520e 1000t )u(t) mA.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–25
[c] At t = 0+ the circuit is
vo (0+ ) = 0;
io (0+ ) = 0.
Checks
At t = 1 the circuit is
vo (1) = 325 V;
io (1) =
500
325
·
= 0.52 A. Checks
250 + (500k250) 750
P 13.23 Begin by transforming the circuit from the time domain to the s domain:
Io =
=
5
s2 + 502
400
4 + 0.24s +
s
=
20.833s
(s2 + 502 )(s2 + 16.67s + 1666.67)
K1
K1⇤
K2
K2⇤
+
+
+
.
s j50 s + j50 s + 8.33 j39.965 s + 8.33 + j39.965
K1 =
K2 =
20.833s
= 0.00884/
(s + j50)(s2 + 16.67s + 1666.67) s=j50
20.833s
(s2 + 502 )(s + 8.33 + j39.965)
135 ;
= 0.00903/46.194 .
s= 8.33+j39.965
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13–26
CHAPTER 13. The Laplace Transform in Circuit Analysis
Therefore,
io (t) = [17.68 cos(50t
135 ) + 18.06e 8.33t cos(39.965t + 46.194 )] mA
P 13.24 Begin by transforming the circuit from the time domain to the s domain:
30k0.0012s =
30s
.
s + 25,000
Use voltage division to find Vo (s):
30s
s + 25,000
Vo = 6
30s
10
+
s
s + 25,000
2 ⇥ 106
s2 + 50,0002
!
s2
= 2
(s + 33,333.33s + 0.833 ⇥ 109 )
=
!
K1⇤
K1
+
s + 16,666.67 j23,570.226 s + 16,666.67 + j23,570.226
+
K1 =
2 ⇥ 106
s2 + 50,0002
s
K2
K2⇤
+
.
j50,000 s + j50,000
2 ⇥ 106 s2
(s + 16,666.67 + j23,570.226)(s2 + 50,0002 ) s= 16,666.67+j23,570.226
= 15/180 ;
K2 =
2 ⇥ 106 s2
(s2 + 33,333.33s + 0.833 ⇥ 109 )(s + j50,000) s=j50,000
= 21.21/
45 .
Therefore,
vo (t) = [ 30e 16,666.67t cos(23,570.226t) + 42.43 cos(50,000t
45 )] V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–27
P 13.25 [a]
Vo 8I
35/s
+ 0.4V +
= 0;
2
s + (250/s)
Vo
"
#
Vo 8I
V =
s;
s + (250/s)
I =
(35/s)
2
Vo
.
Solving for Vo yields:
Vo =
29.4s2 + 56s + 1750
29.4s2 + 56s + 1750
=
;
s(s2 + 2s + 50)
s(s + 1 j7)(s + 1 + j7)
Vo =
K2
K2⇤
K1
+
+
.
s
s + 1 j7 s + 1 + j7
K1 =
29.4s2 + 56s + 1750
= 35;
s2 + 2s + 50
s=0
K2 =
29.4s2 + 56s + 1750
s(s + 1 + j7)
s= 1+j7
=
2.8 + j0.6 = 2.86/167.91 .
.·. vo (t) = [35 + 5.73e t cos(7t + 167.91 )]u(t) V.
[b] At t = 0+
vo
35
2
vo = 35 + 5.73 cos(167.91 ) = 29.4 V.
+ 0.4v = 0;
vo
35 + 0.8v = 0;
vo = v + 8i = v + 8(0.4v ) = 4.2 V;
vo + (0.8)
vo
= 35;
4.2
.·. vo (0+ ) = 29.4 V.(checks)
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13–28
CHAPTER 13. The Laplace Transform in Circuit Analysis
At t = 1, the circuit is
v = 0,
i =0
.·. vo = 35 V.(checks)
P 13.26
Vo
Vo
5 ⇥ 10 3
=
+ 3.75 ⇥ 10 3 V +
;
6
s
200 + 4 ⇥ 10 /s
0.04s
V =
4 ⇥ 106 /s
4 ⇥ 106 Vo
V
=
;
o
200 + 4 ⇥ 106 /s
200s + 4 ⇥ 106
.·.
5 ⇥ 10 3
Vo s
15,000Vo
25Vo
=
;
+
+
s
200s + 4 ⇥ 106 200s + 4 ⇥ 106
s
.·. Vo =
s + 20,000
s2 + 20,000s + 108
K1 = 10,000;
Vo =
=
K1
K2
.
+
2
(s + 10,000)
s + 10,000
K2 = 1;
10,000
1
;
+
2
(s + 10,000)
s + 10,000
vo (t) = [10,000te 10,000t + e 10,000t ]u(t) V.
P 13.27 [a]
120
= 50(I1
s
0.05V ) +
250
(I1
s
I2 );
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
✓
◆
250
2.5
(I2
s
250
= 50I1
s
I1 ) +
250
I1
s
13–29
250
I2 .
s
Simplfying,
(50s + 875)I1
1)I1 + (20s2 + 450s + 250)I2 = 0.
250(s
(50s + 875)
=
250(s
875
= 1200(2s2 + 45s + 25);
0 (20s + 450s + 250)
(50s + 875) 120
250(s
N1
N2
I = I2
I2
= 1000s(s2 + 40s + 625);
1) (20s + 450s + 250)
2
N2 =
I2 =
875
2
120
N1 =
I1 =
875I2 = 120;
1)
=
30,000(s
=
1200(2s2 + 45s + 25)
;
s(s2 + 40s + 625)
=
30,000(s 1)
;
s(s2 + 40s + 625)
0.05V = I2
0.05
2400(s + 35)
I1 =
1);
0
s(s2 + 40s + 625)

250
(I2
s
I1 ) .
.
600,000(s + 35)
;
s(s2 + 40s + 625)
250
(I2
s
I1 ) =
.·.
30,000(s + 35)
1080
30,000(s 1)
+
=
.
s(s2 + 40s + 625) s(s2 + 40s + 625)
s(s2 + 40s + 625)
[b] sI =
I=
1080
(s2 + 40s + 625)
;
i(0+ ) = lim sI = 0;
s!1
i(1) = lim sI =
s!0
1080
= 1728 mA.
625
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13–30
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] At t = 0+ the circuit is
From this circuit, i(0+ ) = 0. At t = 1 the circuit is
120 = 50(ia
i1 ) + 700ia
= 50(ia
v =
700ia
0.05v ) + 700ia = 750ia
.·.
120 = 750ia + 1750ia = 2500ia ;
ia =
120
= 48 mA;
2500
v =
700ia =
33.60 V;
i(1) = 48 ⇥ 10 3
[d] I =
2.5v ;
0.05( 33.60) = 48 ⇥ 10 3 + 1.68 = 1728 mA. (checks)
1080
K1
K2
K2⇤
=
+
+
.
s(s2 + 40s + 625)
s
s + 20 j15 s + 20 + j15
K1 =
1080
= 1.728;
625
K2 =
1080
= 1.44/126.87 ;
( 20 + j15)(j30)
i(t) = [1728 + 2880e 20t cos(15t + 126.87 )]u(t) mA.
Check:
i(0+ ) = 0 mA;
i(1) = 1728 mA.
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Problems
13–31
P 13.28 vo (0 ) = vo (0+ ) = 0.
Vo
Vo
0.05
+
+
s
1000 25
Vo
.·.
✓
s
20
+ 7
1000 10
Vo =
◆
21
=
Vo
Vo
+ 7 = 0;
1000 10 /s
0.05
;
s
500,000
=
s(s + 200,000)
vo (t) = [2.5e 200,000t
P 13.29 [a] iL (0 ) = iL (0+ ) =
"
2.5
2.5
+
;
s
s + 200,000
2.5]u(t) V.
24
= 8 A.
3
directed upward
#
✓
◆
25IT (10/s)
200
20(10/s)
+
VT = 25I +
IT =
IT ;
20 + (10/s)
20 + (10/s)
10 + 20s
45
250 + 200
VT
=
;
=Z=
IT
20s + 10
2s + 1
Vo
8
Vo Vo (2s + 1)
+
+
= ;
5
45
5.625s
s
8
[9s + (2s + 1)s + 8]Vo
= ;
45s
s
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13–32
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo [2s2 + 10s + 8] = 360;
180
360
=
.
2s2 + 10s + 8
s2 + 5s + 4
K1
K2
180
[b] Vo =
=
+
.
(s + 1)(s + 4)
s+1 s+4
Vo =
180
180
= 60;
K2 =
=
3
3
60
60
Vo =
;
s+1 s+4
K1 =
vo (t) = [60e t
60e 4t ]u(t) V.
20I + 25s(Io
I ) + 25(Io
60;
P 13.30 [a]
50
I + 5I1 + 25(I1
s
100
.·.
I1 =
I
s
Simplifying,
( 25s
I1 ) = 0;
Io ) + 25s(I
I1 = I
5)I + (25s + 25)Io =
(50/s + 25s + 30)I + ( 25s
Io ) = 0;
100
.
s
2500/s;
25)Io = 3000/s.
5(5s + 1)
25(s + 1)
= 5
=
(5s2 + 6s + 10) 25(s + 1)
s
5(5s + 1)
2500/s
N2 = 5
=
(5s2 + 6s + 10) 3000/s
s
20(s2 4.8s 10)
N2
=
.
Io =
s(s + 1)(s + 2)
625(s + 1)(1 + 2/s);
12,500
s2
4.8s
s2
10
;
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Problems
13–33
[b] io (0+ ) = lim sIo = 20 A;
s!1
200
=
2
io (1) = lim sIo =
s!0
100 A.
[c] At t = 0+ the circuit is
20I + 5I1 = 0;
I
I1 = 100;
.·. 20I + 5(I
100) = 0;
25I = 500;
.·. I = Io (0+ ) = 20 A.(checks)
At t = 1 the circuit is
Io (1) =
100 A.(checks)
K1
K2
K3
20(s2 4.8s 10)
[d] Io =
=
+
+
.
s(s + 1)(s + 2)
s
s+1 s+2
K1 =
200
=
(1)(2)
100;
K2 =
20(1 + 4.8 10)
= 84;
( 1)(1)
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13–34
CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 =
20(4 + 9.6 10)
= 36;
( 2)( 1)
Io =
84
36
100
+
+
.
s
s+1 s+2
io (t) = ( 100 + 84e t + 36e 2t )u(t) A.
io (1) =
100 A.(checks)
io (0+ ) =
100 + 84 + 36 = 20 A.(checks)
P 13.31 [a]
[b] I1 =
0.01
25/s
=
;
2500 + (125,000/s)
s + 50
V1 =
1000
(100,000/s)(25/s)
=
;
2500 + (125,000/s)
s(s + 50)
V2 =
250
(25,000/s)(25/s)
=
.
2500 + (125,000/s)
s(s + 50)
[c] i1 (t) = 10e 50t u(t) mA;
20
20
.·.
s
s + 50
5
5
.·.
V2 = =
s
s + 50
V1 =
20e 50t )u(t) V;
v1 (t) = (20
5e 50t )u(t) V.
v2 (t) = (5
[d] i1 (0+ ) = 10 mA;
i1 (0+ ) =
25
= 10 mA.(checks)
2.5 ⇥ 10 3
v1 (0+ ) = 0;
v1 (1) = 20 V;
v2 (0+ ) = 0.
(checks)
v2 (1) = 5 V.
v1 (1) + v2 (1) = 25 V.
(checks)
(checks)
(10 ⇥ 10 6 )v1 (1) = 200 µC;
(40 ⇥ 10 6 )v2 (1) = 200 µC.
(checks)
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Problems
13–35
P 13.32 [a] For t < 0:
V2 =
[b] V1 =
10
(450) = 90 V.
10 + 40
25(450/s)
(125,000/s) + 25 + 1.25 ⇥ 10 3 s
=
9 ⇥ 106
9 ⇥ 106
=
;
s2 + 20, 000s + 108
(s + 10,000)2
v1 (t) = (9 ⇥ 106 te 10,000t )u(t) V.
[c] V2 =
90
s
=
=
(25,000/s)(450/s)
(125,000/s) + 1.25 ⇥ 10 3 s + 25
90(s + 20,000)
s2 + 20,000s + 108
900,000
90
;
+
2
(s + 10,000)
s + 10,000
v2 (t) = [9 ⇥ 105 te 10,000t + 90e 10,000t ]u(t) V.
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13–36
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.33 [a]
10
10
I1 + (I1 I2 ) + 10(I1 9/s) = 0;
s
s
10
10
(I2 9/s) + (I2 I1 ) + 10I2 = 0.
s
s
Simplifying,
(s + 2)I1
I2 = 9;
9
I1 + (s + 2)I2 = .
s
=
(s + 2)
1
1
(s + 2)
9
N1 =
I1 =
N1
"
#
(s + 2) 9
1
N2
9
s
9
9s2 + 18s + 9
= (s + 1)2 ;
s
s
9
9(s + 1)
(s + 1)2
=
.
=
s (s + 1)(s + 3)
s(s + 3)
=
I a = I1 =
Ib =
=
9/s (s + 2)
N2 =
I2 =
1
= s2 + 4s + 3 = (s + 1)(s + 3);
9/s
=
18
(s + 1);
s
18(s + 1)
18
=
.
s(s + 1)(s + 3)
s(s + 3)
3
6
9(s + 1)
= +
;
s(s + 3)
s s+3
I1 =
9
s
9(s + 1)
6
=
s(s + 3)
s
6
.
s+3
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Problems
13–37
[b] ia (t) = 3(1 + 2e 3t )u(t) A;
e 3t )u(t) A.
ib (t) = 6(1
✓
6
10 3
10
[c] Va = Ib =
+
s
s s s+3
=
◆
30 20
30
60
= 2 +
+
2
s
s(s + 3)
s
s
✓
20
;
s+3
◆
✓
Vb =
10
(I2
s
=
10 3
s s
12
30
= 2
s+3
s
40
40
+
;
s
s+3
Vc =
10
(9/s
s
I2 ) =
✓
6
6
+
s s+3
=
30 20
+
s2
s
20
.
s+3

I1 ) =
[d] va (t) = [30t + 20
vb (t) = [30t
10
s
6
s
10 9
s s
6
s+3
6
3
+
s s+3
◆
◆
20e 3t ]u(t) V;
40 + 40e 3t ]u(t) V;
vc (t) = [30t + 20
20e 3t ]u(t) V.
[e] Calculating the time when the capacitor voltage drop first reaches 1000 V:
30t + 20
20e 3t = 1000 or 30t
40 + 40e 3t = 1000.
Note that in either of these expressions the exponential tem is negligible
when compared to the other terms. Thus,
30t + 20 = 1000 or 30t
40 = 1000.
Thus,
980
1040
t=
= 32.67 s or t =
= 34.67 s.
30
30
Therefore, the breakdown will occur at t = 32.67 s.
P 13.34 [a] The s-domain equivalent circuit is
I=
Vg /L
Vg
=
,
R + sL
s + (R/L)
Vg =
Vm (! cos + s sin )
;
s2 + ! 2
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13–38
CHAPTER 13. The Laplace Transform in Circuit Analysis
I=
K0
K1
K1⇤
+
+
.
s + R/L s j! s + j!
K0 =
Vm (!L cos
R sin )
,
2
2
R + ! L2
K1 =
✓(!)
90
Vm /
p
2 R 2 + ! 2 L2
where tan ✓(!) = !L/R. Therefore, we have
i(t) =
Vm (!L cos
R sin ) (R/L)t Vm sin[!t +
✓(!)]
p
e
+
.
2
2
2
R +! L
R 2 + ! 2 L2
[b] iss (t) = p
Vm
R 2 + ! 2 L2
sin[!t +
✓(!)].
Vm (!L cos
R sin ) (R/L)t
e
.
R 2 + ! 2 L2
Vg
[d] I =
,
Vg = Vm /
90 .
R + j!L
[c] itr =
Therefore I = p
Vm /
90
Vm
/
p
=
R2 + ! 2 L2 /✓(!)
R 2 + ! 2 L2
Therefore iss = p
Vm
R 2 + ! 2 L2
sin[!t +
✓(!)
90 .
✓(!)].
[e] The transient component vanishes when
!L cos
= R sin
P 13.35 vC = 12 ⇥ 105 te 5000t V,
dvC
iC = C
dt
!
or
tan
=
!L
R
C = 5 µF;
= 6e 5000t (1
or
= ✓(!).
therefore
5000t) A.
iC > 0 when 1 > 5000t or iC > 0 when 0 < t < 200 µs
and iC < 0 when t > 200 µs.
iC = 0 when 1
5000t = 0,
dvC
= 12 ⇥ 105 e 5000t [1
dt
.·. iC = 0 when
or t = 200 µs.
5000t];
dvC
= 0.
dt
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Problems
13–39
P 13.36 [a]
100k5s =
100s
500s
=
;
5s + 100
s + 20
"
#
100s
5000s
50
=
;
Vo =
2
s + 20 (s + 25)
(s + 20)(s + 25)2
Io =
50s
Vo
=
;
100
(s + 20)(s + 25)2
IL =
1000
Vo
=
.
5s
(s + 20)(s + 25)2
[b] Vo =
K2
K3
K1
+
.
+
s + 20 (s + 25)2 s + 25
K1 =
5000s
=
(s + 25)2 s= 20
K2 =
5000s
= 25,000;
(s + 20) s= 25

4000;
"
d 5000s
5000
K3 =
=
ds s + 20 s= 25
s + 20
#
5000s
= 4000;
(s + 20)2 s= 25
vo (t) = [ 4000e 20t + 25,000te 25t + 4000e 25t ]u(t) V.
Io =
K2
K3
K1
+
.
+
2
s + 20 (s + 25)
s + 25
K1 =
50s
=
(s + 25)2 s= 20
K2 =
50s
= 250;
(s + 20) s= 25

40;
"
d
50s
50
K3 =
=
ds s + 20 s= 25
s + 20
#
50s
= 40;
(s + 20)2 s= 25
io (t) = [ 40e 20t + 250te 25t + 40e 25t ]u(t) V.
IL =
K2
K3
K1
+
.
+
2
s + 20 (s + 25)
s + 25
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13–40
CHAPTER 13. The Laplace Transform in Circuit Analysis
K1 =
1000
= 40;
(s + 25)2 s= 20
K2 =
1000
=
(s + 20) s= 25

200;
d 1000
K3 =
=
ds s + 20 s= 25
iL (t) = [40e 20t
"
200te 25t
#
1000
=
(s + 20)2 s= 25
40;
40e 25t ]u(t) V.
P 13.37
40
400
40
10s
·
=
=
;
10s + 1000 s
10s + 1000
s + 100
VTh =
ZTh = 1000 + 1000k10s = 1000 +
I=
2000(s + 50)
10,000s
=
.
10s + 1000
s + 100
40/(s + 100)
(5 ⇥ 105 )/s + 2000(s + 50)/(s + 100)
=
K1 =
=
40s
2000s2 + 600,000s + 5 ⇥ 107
K1
K1⇤
0.02s
=
+
.
s2 + 300s + 25,000
s + 150 j50 s + 150 + j50
0.02s
= 31.62 ⇥ 10 3 /71.57 ;
s + 150 + j50 s= 150+j50
i(t) = 63.25e 150t cos(50t + 71.57 )u(t) mA.
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Problems
13–41
P 13.38 [a]
180
= (100 + 15s)I1 + 10sI2 ;
s
0 = 10sI1 + (20s + 200)I2 .
=
N2 =
I2 =
15s + 100
10s
10s
20s + 200
15s + 100 180/s
10s
N2
=
=
1800;
0
9
;
(s + 5)(s + 20)
Vo = 160I2 =
[b] sVo =
= 200(s + 5)(s + 20);
1440
.
(s + 5)(s + 20)
1440s
.
(s + 5)(s + 20)
lim sVo = vo (1) = 0 V;
s!0
lim sVo = vo (0+ ) = 0 V.
s!1
[c] Vo =
96
96
+
;
s + 5 s + 20
vo (t) = [ 96e 5t + 96e 20t ]u(t) V.
P 13.39
180
= (100 + 15s)I1
s
10sI2 ;
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13–42
CHAPTER 13. The Laplace Transform in Circuit Analysis
0=
10sI1 + (20s + 200)I2 .
=
N2 =
I2 =
15s + 100
10s
10s
20s + 200
15s + 100 180/s
10s
N2
=
= 200(s + 5)(s + 20; )
= 1800;
0
9
;
(s + 5)(s + 20)
Vo = 160I2 =
96
1440
=
(s + 5)(s + 20)
s+5
vo (t) = [96e 5t
96
;
s + 20
96e 20t ]u(t) V.
1
1
P 13.40 [a] W = L1 i21 + L2 i22 + M i1 i2 ;
2
2
W = 4(15)2 + 9(100) + 150(6) = 2700 J.
[b] 120i1 + 8
di1
dt
270i2 + 18
6
di2
= 0;
dt
di2
dt
6
di1
= 0.
dt
LaPlace transform the equations to get
120I1 + 8(sI1
15)
6(sI2 + 10) = 0;
270I2 + 18(sI2 + 10)
6(sI1
15) = 0.
In standard form,
(8s + 120)I1
6sI2 = 180;
6sI1 + (18s + 270)I2 =
=
N1 =
N2 =
8s + 120
6s
6s
18s + 270
180
6s
270.
= 108(s + 10)(s + 30);
= 1620(s + 30);
270 18s + 270
8s + 120 180
6s
=
1080(s + 30);
270
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Problems
I1 =
I2 =
N1
N2
=
1620(s + 30)
15
=
;
108(s + 10)(s + 30)
s + 10
=
1080(s + 30)
10
=
.
108(s + 10)(s + 30)
s + 10
[c] i1 (t) = 15e 10t u(t) A;
[d] W120⌦ =
Z 1
0
W270⌦ =
10e 10t u(t) A.
i2 (t) =
(225e 20t )(120) dt = 27,000
Z 1
0
13–43
e 20t 1
= 1350 J;
20 0
(100e 20t )(270) dt = 27,000
e 20t 1
= 1350 J;
20 0
W120⌦ + W270⌦ = 2700 J.
1
1
[e] W = L1 i21 + L2 i22 M i1 i2 = 900 + 900 900 = 900 J.
2
2
With the dot reversed the s-domain equations are
(8s + 120)I1 + 6sI2 = 60;
6sI1 + (18s + 270)I2 =
As before,
N2 =
I1 =
= 108(s + 10)(s + 30). Now,
60
N1 =
90.
6s
= 1620(s + 10);
90 18s + 270
8s + 120 60
N1
6s
=
=
15
;
s + 30
i1 (t) = 15e 30t u(t) A;
W270⌦ =
W120⌦ =
Z 1
0
Z 1
0
1080(s + 10);
90
I2 =
N2
=
i2 (t) =
10
;
s + 30
10e 30t u(t) A.
(100e 60t )(270) dt = 450 J;
(225e 60t )(120) dt = 450 J;
W120⌦ + W270⌦ = 900 J.
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13–44
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.41 [a] s-domain equivalent circuit is
[b]
20
=
10
i2 (0+ ) =
Note:
2 A.
24
= (120 + 3s)I1 + 3sI2 + 6
s
0 = 6 + 3sI1 + (360 + 15s)I2 + 36.
In standard form,
(s + 40)I1 + sI2 = (8/s)
sI1 + (5s + 120)I2 =
=
[c] sI1 =
s
s
5s + 120
(8/s)
N1 =
I1 =
s + 40
10
N1
=
2
2;
10.
= 4(s + 20)(s + 60);
s
=
5s + 120
200(s
s
4.8)
;
50(s 4.8)
.
s(s + 20)(s + 60)
50(s 4.8)
;
(s + 20)(s + 60)
lim sI1 = i1 (0+ ) = 0 A;
s!1
lim sI1 = i1 (1) =
s!0
[d] I1 =
( 50)( 4.8)
= 0.2 A.
(20)(60)
K2
K3
K1
+
+
.
s
s + 20 s + 60
K1 =
K3 =
240
= 0.2;
1200
K2 =
50( 20) + 240
=
( 20)(40)
1.55;
50( 60) + 240
= 1.35;
( 60)( 40)
i1 (t) = [0.2
1.55e 20t + 1.35e 60t ]u(t) A.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–45
P 13.42 For t < 0:
For t > 0+ :
Note that because of the dot locations on the coils, the sign of the mutual
inductance is negative! (See Example C.1 in Appendix C.)
L1
M = 3 + 1 = 4 H;
L2
M = 2 + 1 = 3 H;
18 ⇥ 4 = 72;
18 ⇥ 3 = 54.
V 72
+
4s + 20
V
V + 54
+
= 0;
s + 10
3s
V
✓
"
1
+
4s + 20
◆
1
1
72
+
=
s + 10 3s
4s + 20
54
;
3s
#
72(3s) 54(4s + 20)
3s( s + 10) + 3s(4s + 20) + (4s + 20)( s + 10)
;
=
V
3s( s + 10)(4s + 20)
3s(4s + 20)
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13–46
CHAPTER 13. The Laplace Transform in Circuit Analysis
[72(3s)
V =
54(4s + 20)]( s + 10)
5s2 + 110s + 200
;
216
12
12
V
=
=
+
;
s + 10
(s + 2)(s + 20)
s + 2 s + 20
Io =
io (t) = 12[e 20t
e 2t ]u(t) A.
P 13.43 The s-domain equivalent circuit is
V1 + 9.6
V1
V1 48/s
+
+
= 0;
4 + (100/s)
0.8s
0.8s + 20
1200
V1 =
s2 + 10s + 125
;
30,000
20
V1 =
0.8s + 20
(s + 25)(s + 5 j10)(s + 5 + j10)
Vo =
=
K1 =
K2 =
K2
K2⇤
K1
+
+
.
s + 25 s + 5 j10 s + 5 + j10
30,000
s2 + 10s + 125
=
60;
s= 25
30,000
= 67.08/63.43 ;
(s + 25)(s + 5 + j10) s= 5+j10
vo (t) = [ 60e 25t + 134.16e 5t cos(10t + 63.43 )]u(t) V.
P 13.44 [a] Voltage source acting alone:
Vo1
V01
60/s V01 s
+
+
= 0;
10
80
20 + 10s
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
.·. V01 =
13–47
480(s + 2)
.
s(s + 4)(s + 6)
Vo2 V02 s V02 30/s
+
+
= 0;
10
80
10(s + 2)
.·. V02 =
240
.
s(s + 4)(s + 6)
Vo = Vo1 + Vo2 =
[b] Vo =
480(s + 2.5)
480(s + 2) + 240
=
.
s(s + 4)(s + 6)
s(s + 4)(s + 6)
K2
K3
K1
+
+
.
s
s+4 s+6
K1 =
(480)(2.5)
= 50;
(4)(6)
vo (t) = [50 + 90e 4t
K2 =
480( 1.5)
= 90;
( 4)(2)
K3 =
480( 3.5)
=
( 6)( 2)
140;
140e 6t ]u(t) V.
P 13.45 [a] Let va be the voltage across the 500 nF capacitor, positive at the upper
terminal.
Let vb be the voltage across the 100 k⌦ resistor, positive at the upper
terminal.
Also note
106
2 ⇥ 106
=
0.5s
s
and
106
4 ⇥ 106
=
;
0.25s
s
Vg =
0.5
.
s
Va
Va (0.5/s)
sVa
+
=0
+
6
2 ⇥ 10
200,000
200,000
sVa + 10Va
Va =
5
+ 10Va = 0;
s
5
;
s(s + 20)
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13–48
CHAPTER 13. The Laplace Transform in Circuit Analysis
(0 Vb )s
0 Va
+
= 0.
200,000
4 ⇥ 106
.·.
Vb =
100
20
Va = 2
.
s
s (s + 20)
(Vb 0)s (Vb Vo )s
Vb
+
+
= 0;
100,000
4 ⇥ 106
4 ⇥ 106
40Vb + sVb + sVb = sVo ;
.·.
2(s + 20)Vb
Vo =
;
s
[b] vo (t) =
100t2 u(t) V.
[c]
100t2 =
4;
Vo = 2
✓
◆
100
=
s3
200
.
s3
t = 0.2 s = 200 ms.
Zf
Vg ;
Zi
P 13.46 [a] Vo =
Zf =
107
1010 /s
1010
107
k1000 = 7
=
=
;
s
10 /s + 1000
1000s + 107
s + 104
Zi =
400s + 2 ⇥ 106
400
2 ⇥ 106
+ 400 =
=
(s + 5000);
s
s
s
Vg =
20,000
;
s2
.·. Vo =
[b] Vo =
107 /(s + 104 ) 20,000
5 ⇥ 108
·
.
=
(400/s)(s + 5000)
s2
s(s + 5000)(s + 10,000)
K2
K3
K1
+
+
.
s
s + 5000 s + 10,000
K1 =
5 ⇥ 108
=
(s + 5000)(s + 10,000) s=0
K2 =
5 ⇥ 108
= 20;
s(s + 10,000) s= 5000
K3 =
5 ⇥ 108
=
s(s + 5000) s= 10,000
.·. vo (t) = [ 10 + 20e 5000t
10;
10;
10e 10,000t ]u(t) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[c]
10 + 20e 5000ts 10e 10,000ts =
Let x = e 5000ts . Then
10x2
13–49
5.
20x + 5 = 0.
Solving,
x = 0.292893.
e 5000t = 0.292893
[d] vg = m tu(t);
Vo =
=
.·.
Vg =
t = 245.6 µs.
m
;
s2
m
108 s
· 2
4000(s + 5000)(s + 10,000) s
25,000m
.
s(s + 5000)(s + 10,000)
K1 =
25,000m
=
(5000)(10,000)
.·.
5=
Vp =
50/s
50
Vg2 =
Vg2 ;
5 + 50/s
5s + 50
5 ⇥ 10 4 m;
.·. m = 10,000 V/s.
5 ⇥ 10 4 m
P 13.47 [a]
Vp
Vp
40/s Vp Vo Vp Vo
+
+
= 0;
20
5
100/s
✓
1
s
1
+ +
20 5 100
✓
◆
◆
50
16
s + 25
100
5s + 50 s
Vo
✓
◆
s
2
1
+
= ;
5 100
s
✓
◆
✓
◆
2
s
1
s + 20
= Vo
+
= Vo
;
s
5 100
100
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13–50
CHAPTER 13. The Laplace Transform in Circuit Analysis
"
#
2
40s + 2000
=
s
s(s + 10)(s + 20)
100
16(s + 25)
Vo =
s + 20 10(s + 10)(s)
=
K2
K3
K1
+
+
.
s
s + 10 s + 20
K1 = 10;
.·.
[b] 10
K2 =
24;
K3 = 14;
24e 10t + 14e 20t ]u(t) V.
vo (t) = [10
24x + 14x2 = 5;
14x2
24x + 5 = 0;
x = 0 or 0.242691;
e 10t = 0.242691
.·.
t = 141.60 ms.
P 13.48
Va
0
Va
Va Vo
4.8/s
+
+
= 0;
0.8
1/s
1/s
Va 0 Vo
+
= 0;
1/s
1.25
Va =
Vo
;
1.25s
Va (2s + 1.25)
sVo = 6/s;
Vo
"
(2s + 1.25)
+ 3 = 6/s;
1.25s
Vo
"
125s2 + 2s + 1.25
= 6/s;
1.25s
#
#
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
7.5
Vo =
1.25s2 + 2s + 1.25
=
13–51
6
s2 + 1.6s + 1
K1
K1⇤
=
+
.
s + 0.8 j0.6 s + 0.8 + j0.6
K1 =
6
= 5/90 ;
s + 0.8 + j0.6 s= 0.8+j0.6
vo (t) = 10e 0.8t cos(0.6t + 90 )u(t) V.
RsC
s
R
=
=
.
R + 1/sC
RsC + 1
s + 1/RC
There is a single zero at 0 rad/sec, and a single pole at
sL
s
[b]
=
.
R + sL
s + R/L
There is a single zero at 0 rad/sec, and a single pole at
P 13.49 [a]
1/RC rad/sec.
R/L rad/sec.
[c] There are several possible solutions. One is
R = 100 ⌦;
L = 10 mH;
C = 1 µF.
1/sC
1
1/RC
=
=
.
R + 1/sC
RsC + 1
s + 1/RC
There are no zeros, and a single pole at
R
R/L
[b]
=
.
R + sL
s + R/L
There are no zeros, and a single pole at
P 13.50 [a]
1/RC rad/sec.
R/L rad/sec.
[c] There are several possible solutions. One is
R = 10 ⌦;
P 13.51 [a]
C = 100 µF.
Vo
1
1/sC
=
;
=
Vi
R + 1/sC
RCs + 1
H(s) =
[b]
L = 10 mH;
(1/RC)
200
=
;
s + (1/RC)
s + 200
p1 =
200 rad/s.
RCs
s
R
Vo
=
=
=
Vi
R + 1/sC
RCs + 1
s + (1/RC)
s
;
z1 = 0,
p1 = 200 rad/s.
s + 200
Vo
s
s
sL
[c]
=
=
;
=
Vi
R + sL
s + R/L
s + 8000
=
z1 = 0;
p1 =
8000 rad/s.
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13–52
CHAPTER 13. The Laplace Transform in Circuit Analysis
[d]
R/L
8000
R
Vo
=
=
;
=
Vi
R + sL
s + (R/L)
s + 8000
p1 =
8000 rad/s.
[e]
Vo Vi
Vo
Vo s
+
= 0;
+
6
4 ⇥ 10
10,000
40,000
sVo + 400Vo + 100Vo = 100Vi ;
P 13.52 [a]
H(s) =
Vo
100
;
=
Vi
s + 500
p1 =
500 rad/s.
(R/L)s
R
= 2
.
1/sC + sL + R
s + (R/L)s + 1/LC
There is a single zero at 0 rad/sec, and two poles:
p1 =
(R/2L) +
p2 =
(R/2L)
q
(R/2L)2
q
(1/LC)
(R/2L)2
(1/LC).
[b] There are several possible solutions. One is
R = 250 ⌦;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
p1 =
5000 rad/sec
and
p2 =
20,000 rad/sec.
[c] There are several possible solutions. One is
R = 200 ⌦;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
p1 =
10,000 rad/sec
and
p2 =
10,000 rad/sec.
[d] There are several possible solutions. One is
R = 120 ⌦;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
p1 =
6000 + j8000 rad/sec
and
p2 =
6000
j8000 rad/sec.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 13.53 [a]
13–53
1/LC
Rk1/sC
Vo
= 2
.
=
Vi
(Rk1/sC) + sL
s + (1/RC)s + 1/LC
There are no zeros and two poles:
p1 =
(1/2RC) +
p2 =
(1/2RC)
q
(1/2RC)2
q
(1/2RC)2
(1/LC);
(1/LC).
[b] There are several
qpossible solutions. Any selection of R, L, and C such
that 1/2RC < 1/LC will yield two real, distinct poles. When L = 10
mH and C = 1 µF, R must be less than 50 ⌦. For example, let
R = 47 ⌦;
L = 10 mH;
C = 1 µF.
These component values yield the following poles:
p1 =
7008.8 rad/s
and
p2 =
[c] To get repeated real poles, 1/2RC =
R=
1
= 50 ⌦.
2(10,000)(10 6 )
14,267.8 rad/s.
q
1/LC, so choose
We can create a resistance of 50 ⌦ using two parallel-connected 100 ⌦
resistors. Then
R = 50 ⌦;
L = 10 mH;
C = 1 µF,
These component values yield the following poles:
p1 =
10,000 rad/s
and
p2 =
10,000 rad/s,
[d] There are several possible solutions. Any resistor with a value greater
than 50 ⌦ will yield two complex conjugate poles. For example, let
R = 100 ⌦;
L = 10 mH;
C = 1 µF,
These component values yield the following poles:
p1 =
5000 + j8660.25 rad/s
and
p2 =
5000
j8660.25 rad/s,
P 13.54 [a]
Va =
s
500,000Vi
=
Vi ;
500,000 + [(20 ⇥ 106 )/s]
s + 40
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13–54
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2Vi = Vo + Va ;
.·.
s
Vi .
s + 40
Vo = 0.2Vi
0.2(s + 40)
Vo
=
Vi
s + 40
[b]
s
=
0.8s + 8
=
s + 40
0.8(s 10)
.
s + 40
z1 = 10 rad/s;
p1 =
40 rad/s.
P 13.55 [a]
sVa
(Va Vo )s
Va Vg
+
+
= 0;
6
1000
5 ⇥ 10
5 ⇥ 106
5000Va
5000Vg + 2sVa
sVo = 0;
sVo = 5000Vg ;
(5000 + 2s)Va
(0 Va )s 0 Vo
=0
+
5 ⇥ 106
5000
sVa
.·.
1000Vo = 0;
(2s + 5000)
✓
◆
1000
Vo
s
Va =
1000
Vo ,
s
sVo = 5000Vg ;
1000Vo (2s + 5000) + s2 Vo =
5000sVg ;
Vo (s2 + 2000s + 5 ⇥ 106 ) =
5000sVg ;
5000s
Vo
= 2
;
Vg
s + 2000s + 5 ⇥ 106
p
s1,2 = 1000 ± 106 5 ⇥ 106 =
Vo
=
Vg
(s + 1000
1000 ± j2000;
5000s
.
j2000)(s + 1000 + j2000)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[b] z1 = 0;
p1 =
1000 + j2000;
p2 =
1000
13–55
j2000.
[c] If vi (t) = (t), Vi (s) = 1. Then,
Vo =
(s + 1000
K1⇤
K1
+
.
s + 1000 j2000 s + 1000 + j2000
=
K1 =
5000s
j2000)(s + 1000 + j2000)
5000s
= 2795.085/
s + 1000 + j2000 s= 1000+j2000
153.43 .
Therefore,
vo (t) = 5590.17e 1000t cos(2000t
153.43 ) V.
Note that the op amp will saturate once the output voltage exceeds the
power supply voltages.
P 13.56 [a] Let R1 = 250 k⌦; R2 = 125 k⌦; C2 = 1.6 nF;
Then
(R2 + 1/sC2 )1/sCf
(s + 1/R2 C2 )
⌘ =
⇣
⌘.
Zf = ⇣
C +C
1
1
R2 + sC2 + sCf
Cf s s + C22Cf Rf2
and
Cf = 0.4 nF.
1
= 2.5 ⇥ 109 ;
Cf
62.5 ⇥ 107
1
=
= 5000 rad/s;
R2 C2
125 ⇥ 103
2 ⇥ 10 9
C2 + Cf
= 25,000 rad/s;
=
C2 Cf R2
(0.64 ⇥ 10 18 )(125 ⇥ 103 )
.·. Zf =
2.5 ⇥ 109 (s + 5000)
⌦.
s(s + 25,000)
Zi = R1 = 250 ⇥ 103 ⌦;
[b]
H(s) =
Vo
Zf
104 (s + 5000)
.
=
=
Vg
Zi
s(s + 25,000)
z1 =
5000 rad/s;
p1 = 0;
p2 =
25,000 rad/s.
[c] If vg (t) = u(t) then Vg (s) = 1/s. Then,
Vo =
104 (s + 5000)
=
s2 (s + 25,000)
2000
+
s2
0.32
0.32
+
.
s
s + 25,000
Thus,
vo = ( 2000t
0.32 + 0.32e 25,000t )u(t) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–56
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.57 [a] Zi = 1000 +
1000(s + 5000)
5 ⇥ 106
=
;
s
s
40 ⇥ 106
40 ⇥ 106
k40,000 =
;
s
s + 1000
Zf =
H(s) =
40,000s
40 ⇥ 106 /(s + 1000)
=
.
1000(s + 5000)/s
(s + 1000)(s + 5000)
Zf
=
Zi
[b] Zero at z1 = 0;
Poles at
p1 =
1000 rad/s and
p2 =
5000 rad/s.
[c] If vg (t) = u(t) then Vg (s) = 1/s. Then,
Vo =
10
10
40,000
=
+
.
(s + 1000)(s + 5000)
s + 1000 s + 5000
Thus,
vo = ( 10e 1000t + 10e 5000t )u(t) V.
P 13.58 [a]
Vo
Vo
+
+ Vo (10 7 )s = Ig ;
5000 0.2s
.·. Vo =
Ig =
10 ⇥ 106 s
· Ig ;
s2 + 2000s + 50 ⇥ 106
0.1s
;
s2 + 108
.·. H(s) =
[b] Io =
s2
s2 + 2000s + 50 ⇥ 106 .
(s + 1000
Io =
Io = 10 7 sVo ;
(s2 )(0.1s)
;
j7000)(s + 1000 + j7000)(s2 + 108 )
(s + 1000
0.1s3
j7000)(s + 1000 + j7000)(s + j104 )(s
j104 )
.
[c] Damped sinusoid of the form
M e 1000t cos(7000t + ✓1 ).
[d] Steady-state sinusoid of the form
N cos(104 t + ✓2 ).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[e] Io =
13–57
K1⇤
K2
K1
K2⇤
+
+
+
.
s + 1000 j7000 s + 1000 + j7000 s j104 s + j104
K1 =
0.1( 1000 + j7000)3
= 46.9 ⇥ 10 3 /
(j14,000)( 1000 j5000)( 1000 + j17,000)
K2 =
0.1(j104 )3
= 92.85 ⇥ 10 3 /21.8 ;
(j20,000)(1000 + j3000)(1000 + j17,000)
io (t) = [93.8e 1000t cos(7000t
140.54 ;
140.54 ) + 185.7 cos(104 t + 21.8 )] mA.
Test:
Z=
1
;
Y
.·. Z =
Y =
1
1
+
+
5000 j2000
10,000
= 1856.95/
2 + j5
1
2 + j5
=
;
j1000
10,000
68.2 ⌦.
Vo = Ig Z = (0.1/0 )(1856.95/
68.2 ) = 185.695/
68.2 V;
Io = (10 7 )(j104 )Vo = 185.7/21.8 mA;
ioss = 185.7 cos(104 t + 21.8 ) mA.(checks)
P 13.59 [a]
2000(Io
Ig ) + 8000Io + µ(Ig
.·. Io =
1000(1 µ)
Ig ;
s + 1000(5 µ)
.·. H(s) =
Io )(2000) + 2sIo = 0;
1000(1 µ)
.
s + 1000(5 µ)
[b] µ < 5.
[c]
µ
H(s)
Io
3 4000/(s + 8000)
20,000/s(s + 8000)
0 1000/(s + 5000)
5000/s(s + 5000)
4
3000/(s + 1000)
15,000/s(s + 1000)
5
4000/s
20,000/s2
6
5000/(s
1000)
25,000/s(s
1000)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–58
CHAPTER 13. The Laplace Transform in Circuit Analysis
µ=
3:
Io =
2.5
s
2.5
;
(s + 8000)
io = [2.5
2.5e 8000t ]u(t) A.
µ = 0:
1
;
s + 5000
1
s
µ = 4:
Io =
15
s
Io =
io = [1
15
;
s + 1000
e 5000t ]u(t) A.
io = [ 15 + 15e 1000t ]u(t) A.
µ = 5:
20,000
;
s2
Io =
io =
20,000t u(t) A.
µ = 6:
Io =
25
;
s 1000
25
s
e1000t ]u(t) A.
io = 25[1
P 13.60
Vg = 25sI1
0=
=
N2 =
I2 =
35sI2 ;
!
16 ⇥ 106
35sI1 + 50s + 10,000 +
I2 ;
s
25s
35s
35s 50s + 10,000 + 16 ⇥ 106 /s
25s Vg
35s 0
N2
=
= 25(s + 2000)(s + 8000);
= 35sVg ;
35sVg
;
25(s + 2000)(s + 8000)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
H(s) =
.·.
I2
1.4s
;
=
Vg
(s + 2000)(s + 8000)
z1 = 0;
p1 =
Vo
1
;
=
Vi
s+1
For 0  t  1:
Z t
0
e
d = (1
2000 rad/s;
p2 =
8000 rad/s.
h(t) = e t .
P 13.61 H(s) =
vo =
13–59
e t ) V.
For 1  t  1:
vo =
Z t
t 1
P 13.62 H(s) =
e
d = (e
Vo
s
=1
=
Vi
s+1
h( ) = ( )
e
1)e t V.
1
;
s+1
h(t) = (t)
e t;
.
For 0  t  1:
vo =
Z t
0
[ ( )
e
] d = [1 + e
] |t0 = e t V.
For 1  t  1:
vo =
Z t
t 1
t
( e
)d = e
= (1
e)e t V.
t 1
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13–60
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.63 [a] 0  t  40:
y(t) =
Z t
0
t
(10)(1)(d ) = 10
= 10t.
0
40  t  80:
y(t) =
t
Z 40
t 40
80 :
40
(10)(1)(d ) = 10
= 10(80
t).
t 40
y(t) = 0.
[b] 0  t  10:
y(t) =
Z t
0
t
40 d = 40
= 40t.
0
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Problems
13–61
10  t  40:
y(t) =
Z t
t 10
t
40 d = 40
= 400.
t 10
40  t  50:
y(t) =
t
Z 40
t 10
50 :
40
40 d = 40
= 40(50
t).
t 10
y(t) = 0.
[c] The expressions are
Z t
0t1:
y(t) =
1  t  40 :
y(t) =
0
t
400 d = 400
= 400t;
0
Z t
t 1
t
400 d = 400
Z 40
40  t  41 :
y(t) =
41  t < 1 :
y(t) = 0.
t 1
= 400;
t 1
40
400 d = 400
= 400(41
t);
t 1
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13–62
CHAPTER 13. The Laplace Transform in Circuit Analysis
[d]
[e] Yes, note that h(t) is approaching 40 (t), therefore y(t) must approach
40x(t), i.e.
Z t
y(t) =
0
)x( ) d !
h(t
Z t
0
40 (t
)x( ) d
! 40x(t).
This can be seen in the plot, e.g., in part (c), y(t) ⇠
= 40x(t).
P 13.64 [a] h( ) =
2
5
0
✓
2
5
h( ) = 4
 5;
◆
5
 10.
0  t  5:
vo = 10
Z t
vo = 10
Z 5
2
d = 2t2 ;
0 5
5  t  10:
=
=
✓
Z t
2
d + 10
4
0 5
5
4 2 5
+40
2 0
100 + 40t
t
5
2
5
◆
d
4 2 t
2 5
2t2 ;
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
10  t  1:
Z 5
Z 10 ✓
2
vo = 10
d + 10
4
0 5
5
=
4 2 5
+40
2 0
= 50 + 200
vo = 2t2 V
vo = 40t
vo = 100 V
10
5
2
5
◆
13–63
d
4 2 10
2 5
150 = 100.
0  t  5;
100
2t2 V
5  t  10;
10  t  1.
[b]
1
[c] Area = (10)(2) = 10 .·.
2
5
h( ) =
,
0   2;
2
✓
◆
5
h( ) = 10
,
2
2
1
(4)h = 10 so h = 5;
2
 4.
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13–64
CHAPTER 13. The Laplace Transform in Circuit Analysis
0  t  2:
Z t
5
d = 12.5t2 ;
0 2
2  t  4:
Z 2
Z t✓
5
d + 10
10
vo = 10
0 2
2
vo = 10
=
25 2 2
+100
2 0
=
2
✓
Z 4
5
d + 10
10
vo = 10
2
0 2
Z 2
25 2 2
+100
2 0
5
2
◆
d
25 2 4
2 2
4
2
150 = 100.
vo = 12.5t2 V
vo = 100t
d
12.5t2 ;
100 + 100t
= 50 + 200
◆
25 2 t
2 2
t
4  t  1:
=
5
2
0  t  2;
12.5t2 V
100
2  t  4;
4  t  1.
vo = 100 V
[d] The waveform in part (c) is closer to replicating the input waveform
because in part (c) h( ) is closer to being an ideal impulse response.
That is, the area was preserved as the base was shortened.
P 13.65 [a]
y(t) = 0
0  t  10 :
t < 0;
y(t) =
Z t
0
625 d = 625t;
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Problems
Z 10
10  t  20 :
y(t) =
20  t < 1 :
y(t) = 0.
t 10
625 d = 625(10
t + 10) = 625(20
13–65
t);
[b]
y(t) = 0
t < 0;
Z t
0  t  10 :
y(t) =
10  t  20 :
y(t) =
20  t  30 :
y(t) =
30  t < 1 :
y(t) = 0.
0
312.5 d = 312.5t;
Z 10
0
Z 10
312.5 d = 3125;
t 20
312.5 d = 312.5(30
t);
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13–66
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c]
y(t) = 0
t < 0;
0t1:
y(t) =
1  t  10 :
Z t
y(t) =
0
625 d = 625t;
Z 1
0
625 d = 625;
Z 1
10  t  11 :
y(t) =
11  t < 1 :
y(t) = 0.
t 10
625 d = 625(11
t);
P 13.66 [a] From Problem 13.51(a)
H(s) =
200
.
s + 200
h( ) = 200e 200 .
0  t  5 ms:
vo =
Z t
0
20(200)e 200 d = 20(1
e 200t ) V.
5 ms  t  1:
vo =
Z t
t 5⇥10
3
20(200)e 200 d = 20(e1
1)e 200t V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–67
[b]
2000
s + 2000
0  t  5 ms:
P 13.67 [a] H(s) =
vo =
Z t
0
.·. h( ) = 2000e 2000 .
20(2000)e 2000 d = 20(1
e 2000t ) V.
5 ms  t  1:
vo =
Z t
t 5⇥10
3
20(2000)e 2000 d = 20(e10
1)e 2000t V.
[b] Decrease.
[c] The circuit with R = 5 k⌦.
P 13.68 [a]
1  t  4:
vo =
Z t+1
0
10 d = 5 2
Z t+1
t 4
10 d = 5 2
Z 10
t 4
t+1
= 50t
75 V;
t 4
9  t  14:
vo = 10
= 5t2 + 10t + 5 V;
0
4  t  9:
vo =
t+1
d + 10
Z t+1
10
10 d
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13–68
CHAPTER 13. The Laplace Transform in Circuit Analysis
10
=5 2
t+1
+100
=
t 4
5t2 + 140t
480 V;
10
14  t  19:
Z t+1
vo = 100
t 4
d = 500 V;
19  t  24:
vo =
Z 20
t 4
100 d +
Z t+2
20
= 100
=
10(30
)d
t+1
t+2
20
+300
t 2
20
5t2 + 190t
1305 V;
5 2
20
24  t  29:
vo = 10
Z t+1
t 4
= 1575
t+1
(30
) d = 300
t 4
vo = 10
t 4
= 5t2
t 4
50t V;
29  t  34:
Z 30
t+1
5 2
30
(30
5 2
) d = 300
t 4
30
t 2
340t + 5780 V.
Summary:
1t
vo = 0
vo = 5t2 + 10t + 5 V
vo = 50t
vo =
5t2 + 190t
vo = 1575
vo = 5t2
vo = 0
9  t  14;
480 V
14  t  19;
vo = 500 V
vo =
1  t  4;
4  t  9;
75 V
5t2 + 140t
1;
19  t  24;
1305 V
50t V
24  t  29;
340t + 5780 V
29  t  34;
34  t  1.
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Problems
13–69
[b]
✓
◆
0.8s
16s
=
= 0.8 1
P 13.69 H(s) =
40 + 4s + 16s
s+2
h( ) = 0.8 ( )
vo =
Z t
0
120
1.6
;
s+2
1.6e 2 u( );
75[0.8 ( )
= 60
2
= 0.8
s+2
1.6e 2 ] d =
e 2
2
t
Z t
0
= 60 + 60(e 2t
60 ( ) d
120
Z t
0
e 2 d
1)
0
= 60e 2t u(t) V.
P 13.70 [a] H(s) =
=
1/LC
Vo
= 2
Vi
s + (R/L)s + (1/LC)
100
s2 + 20s + 100
=
100
;
(s + 10)2
h( ) = 100 e 10 u( ).
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13–70
CHAPTER 13. The Laplace Transform in Circuit Analysis
0  t  0.5:
Z t
vo = 500
0
e 10 d
(
e 10
( 10
= 500
100
t
1)
0
)
e 10t (10t + 1)];
= 5[1
0.5  t  1:
Z t
vo = 500
t 0.5
e 10 d
(
e 10
( 10
= 500
100
1)
= 5e 10t [e5 (10t
10t
4)
t
t 0.5
)
1].
[b]
P 13.71 [a] Vo =
.·.
16
Vg ;
20
H(s) =
Vo
4
= .
Vg
5
h( ) = 0.8 ( ).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–71
[b]
0 < t < 0.5 s :
vo =
Z t
0
75[0.8 ( )] d = 60 V.
0.5 s  t  1.0 s:
vo =
Z t 0.5
0
75[0.8 ( )] d =
1s < t < 1 :
60 V.
vo = 0.
[c]
Yes, because the circuit has no memory.
P 13.72 [a]
Vo
Vg
5
+
Vo s Vo
+
= 0;
4
20
(5s + 5)Vo = 4Vg ;
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13–72
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo
0.8
;
=
Vg
s+1
H(s) =
h( ) = 0.8e
u( ).
[b]
0  t  0.5 s;
vo =
Z t
0
75(0.8e
) d = 60
60e t V,
vo = 60
t
e
1
;
0
0  t  0.5 s.
0.5 s  t  1 s:
vo =
Z t 0.5
=
60
0
( 75)(0.8e
t 0.5
e
1
+60
)d +
= 120e (t 0.5)
60e t
t 0.5
75(0.8e
)d
t
e
0
Z t
1
t 0.5
60 V,
0.5 s  t  1 s;
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Problems
1 s  t  1:
vo =
Z t 0.5
=
60
t 1
( 75)(0.8e
t 0.5
e
1
+60
t 1
= 120e (t 0.5)
Z t
)d +
t 0.5
75(0.8e
13–73
)d
t
e
1
60e (t 1)
t 0.5
60e t V,
1 s  t  1.
[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.73 [a]
vo =
Z t
0
10(10e 4 ) d
= 100
e 4
4
= 25(1
t
=
25[e 4t
1]
0
e 4t ) V,
0  t  1.
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13–74
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
0  t  0.5:
vo =
Z t
0
100(1
2 ) d = 100(
2
t
) = 100t(1
0.5  t  1:
vo =
Z 0.5
0
t);
0
100(1
2 ) d = 100(
2
0.5
)
= 25.
0
[c]
P 13.74 [a] Ig =
Vo
Vo s
Vo (s + 50)
+
=
;
105 5 ⇥ 106
5 ⇥ 106
5 ⇥ 106
Vo
;
= H(s) =
Ig
s + 50
h( ) = 5 ⇥ 106 e 50 u( ).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
0  t  0.1 s:
vo =
Z t
0
(50 ⇥ 10 6 )(5 ⇥ 106 )e 50 d = 250
e 50
50
t
= 5(1
13–75
e 50t ) V.
0
0.1 s  t  0.2 s:
vo =
Z t 0.1
0
+
=
Z t
( 50 ⇥ 10 6 )(5 ⇥ 106 e 50 d )
t 0.1
250
h
(50 ⇥ 10 6 )(5 ⇥ 106 e 50 d )
e 50
50
t 0.1
+250
0
= 5 e 50(t 0.1)
vo = [10e 50(t 0.1)
i
1
h
e 50
50
5 e 50t
5e 50t
5] V.
t
t 0.1
i
e 50(t 0.1) ;
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13–76
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2 s  t  1:
vo =
Z t 0.1
t 0.2
"
= 5e 50
250e 50 d +
Z t
t 0.1
t
vo = [10e 50(t 0.1)
[b] Io =
5e 50
t 0.2
t 0.1
t 0.1
#
5e 50(t 0.2)
250e 50 d
;
5e 50t ] V.
s
5 ⇥ 106 Ig
Vo s
;
=
·
5 ⇥ 106
5 ⇥ 106
s + 50
50
;
s + 50
s
Io
=1
= H(s) =
Ig
s + 50
h( ) = ( )
50e 50 .
0 < t < 0.1 s:
io =
Z t
0
(50 ⇥ 10 6 )[ ( )
= 50 ⇥ 10 6
50e 50 ] d
25 ⇥ 10 3
e 50
50
= 50 ⇥ 10 6 + 50 ⇥ 10 6 [e 50t
t
0
1] = 50e 50t µA.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
13–77
0.1 s < t < 0.2 s:
io =
Z t 0.1
0
+
=
=
( 50 ⇥ 10 6 )[ ( )
Z t
t 0.1
(50 ⇥ 10 6 )( 50e 50 ) d
50 ⇥ 10 6 + 2.5 ⇥ 10 3
50 ⇥ 10 6
e 50
50
t 0.1
0
2.5 ⇥ 10 3
e 50
50
t
t 0.1
50 ⇥ 10 6 e 50(t 0.1) + 50 ⇥ 10 6
+50 ⇥ 10 6 e 50t
= 50e 50t
50e 50 ] d
50 ⇥ 10 6 e 50(t 0.1)
100e 50(t 0.1) µA.
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13–78
CHAPTER 13. The Laplace Transform in Circuit Analysis
0.2 s < t < 1:
io =
Z t 0.1
t 0.2
+
( 50 ⇥ 10 6 )( 50e 50 ) d
Z t
t 0.1
(50 ⇥ 10 6 )( 50e 50 ) d
= 50e 50t
100e 50(t 0.1) + 50e 50(t 0.2) µA.
[c] At t = 0.1 :
vo = 5(1
e 5 ) = 4.97 V;
.·. io = 50
i100k⌦ =
4.97
= 49.66 µA;
0.1
49.66 = 0.34 µA.
From the solution for io we have
io (0.1 ) = 50e 5 = 0.34 µA. (checks)
At t = 0.1+ :
vo (0.1+ ) = vo (0.1 ) = 4.97 V;
i100k⌦ = 49.66 µA;
.·. io (0.1+ ) =
(50 + 49.66) =
99.66 µA.
From the solution for io we have
io (0.1+ ) = 50e 5
100 =
99.66 µA. (checks)
At t = 0.2 :
vo = 10e 5
5e 10
5=
4.93 V;
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Problems
13–79
i100k⌦ = 49.33 µA;
io =
50 + 49.33 =
0.67 µA.
From the solution for io ,
vo (0.2 ) = 50e 10
100e 5 =
0.67 µA. (checks)
At t = 0.2+ :
vo (0.2+ ) = vo (0.2 ) =
4.93 V;
i100k⌦ =
49.33 µA;
io = 0 + 49.33 = 49.33 µA.
From the solution for io ,
io (0.2+ ) = 50e 10
P 13.75 H(s) =
100e 5 + 50 = 49.33 µA.(checks)
2
5
Vo
=
;
=
Vi
5 + 2.5s
s+2
h( ) = 2e 2 ;
⇡
T
= ;
2
2
h(t
T = ⇡ s;
vi ( ) = 20 sin 2 [u( )
) = 2e 2(t
f=
1
Hz;
⇡
u(
⇡/2)].
)
= 2e 2t e2 ;
(⇡/2) s  t  1:
vo =
Z ⇡/2
0
(2e 2t e2 )(20 sin 2 ) d = 40e 2t
Z ⇡/2
0
e2 sin 2 d
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13–80
CHAPTER 13. The Laplace Transform in Circuit Analysis
2t
= 40e
"
e2
(2 sin 2
8
2 cos 2
#⇡/2
= 10e 2t [e⇡ (sin ⇡
cos ⇡)
1(0
1)]
0
= 10e 2t (e⇡ + 1) = 10(e⇡ + 1)e 2t V.
vo (2.2) = 241.41e 4.4 = 2.96 V.
P 13.76
20 ⇥ 103
(5 ⇥ 103 Ig );
Vo =
3
6
25 ⇥ 10 + 2.5 ⇥ 10 /s
4000s
Vo
;
= H(s) =
Ig
s + 100

H(s) = 4000 1
100
= 4000
s + 100
h(t) = 4000 (t)
4 ⇥ 105 e 100t ;
vo =
Z 10 3
0
+
( 20 ⇥ 10 3 )[4000 ( )
Z 5⇥10 3
10
3
4 ⇥ 105 e 100 ] d
(10 ⇥ 10 3 )[ 4 ⇥ 105 e 100 ] d
Z 10 3
=
80 + 8000
=
80
= 40e 0.5
4 ⇥ 105
;
s + 100
0
80(e 0.1
e
100
d
4000
1) + 40(e 0.5
120e 0.1 =
Z 5⇥10 3
10
3
e 100 d
e 0.1 )
84.32 V.
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Problems
13–81
Alternate:
Ig =
=
Z 4⇥10 3

3
st
Z 6⇥10 3
( 20 ⇥ 10 3 )e st dt
0
(10 ⇥ 10 )e
10
s
30 4⇥10 3 s 20 6⇥10 3 s
e
+ e
⇥ 10 3 ;
s
s
Vo = Ig H(s) =
40
[1
s + 100
40
=
s + 100
120e 4⇥10
s + 100
vo (t) = 40e 100t
3)
vo (5 ⇥ 10 3 ) = 40e 0.5
Z 1
Y (s) =
=
=
0
Z 1
0
e
st
0
Z 1
0
Z 1
0
Z 1Z 1
0
3s
]
3
80e 6⇥10 s
+
];
s + 100
3)
u(t
6 ⇥ 10 3 ).
u(t
4 ⇥ 10 3 )
h( )x(t
)d
84.32 V. (checks)
dt
e st h( )x(t
) d dt
Z 1
) dt d .
h( )
0
e st x(t
) = 0 when t < .
Let u = t
=
+ 2e 6⇥10
120e 0.1 + 80(0) =
Therefore Y (s) =
=
3s
3s
3
y(t)e st dt.
But x(t
Y (s) =
4⇥10
3e 4⇥10
120e 100(t 4⇥10
+80e 100(t 6⇥10
P 13.77 [a] Y (s) =
dt +
Z 1
0
Z 1
0
Z 1
0
;
Z 1
0
h( )
du = dt;
Z 1
h( )
h( )e
0
s
Z 1
e st x(t
) dt d .
u = 0 when t = ;
u = 1 when t = 1.
e s(u+ ) x(u) du d
Z 1
0
e su x(u) du d
h( )e s X(s) d = H(s) X(s).
Note on x(t
) = 0,
t< .
We are using one-sided Laplace transforms; therefore h(t) and x(t) are
assumed zero for t < 0.
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13–82
CHAPTER 13. The Laplace Transform in Circuit Analysis
a
a
1
= ·
= H(s)X(s);
2
s(s + a)
s (s + a)2
[b] F (s) =
.·. h(t) = u(t),
.·. f (t) =
Z t
0
x(t) = at e at u(t).
(1)a e
a
"
e a
( a
d =a
a2
1
1( 1)] = [1
a
=
1 at
[e ( at
a
1)
=

te at u(t).
1
a
1 at
e
a
t
1)
0
e at
ate at ]
Check:
F (s) =
K1
K0
K2
a
+
;
=
+
2
2
s(s + a)
s
(s + a)
s+a
1
K0 = ;
a
f (t) =
P 13.78 H(j500) =

1
a
✓ ◆
d a
K2 =
ds s
K1 =
1;
te at
1 at
e
u(t).
a
=
s= a
1
;
a
125(400 + j500)
= 0.6157 ⇥ 10 3 /163.96 ;
(j500)(104 5002 + j105 )
.·. io (t) = 49.3 cos(500t + 163.96 ) mA.
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Problems
P 13.79 H(j3) =
13–83
4(3 + j3)
= 0.42/8.13 ;
9 + j24 + 41
.·. vo (t) = 16.97 cos(3t + 8.13 ) V.
P 13.80 Vo =
50
s + 8000
20
30(s + 3000)
=
;
s + 5000
(s + 5000)(s + 8000)
✓
◆
30
;
Vo = H(s)Vg = H(s)
s
.·. H(s) =
s(s + 3000)
.
(s + 5000)(s + 8000)
H(j6000) =
(j6000)(3000 + j6000)
= 0.52/66.37 ;
(5000 + j6000)(8000 + j6000)
.·. vo (t) = 61.84 cos(6000t + 66.37 ) V.
P 13.81 [a]
Vp =
0.01s
s
Vg =
Vg ;
80 + 0.01s
s + 8000
Vn = Vp ;
Vn Vo
Vn
+
+ (Vn
5000
25,000
5Vn + Vn
Vo + (Vn
Vo )8 ⇥ 10 9 s = 0;
Vo )2 ⇥ 10 4 s = 0;
6Vn + 2 ⇥ 10 4 sVn = Vo + 2 ⇥ 10 4 sVo ;
2 ⇥ 10 4 Vn (s + 30,000) = 2 ⇥ 10 4 Vo (s + 5000);
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13–84
CHAPTER 13. The Laplace Transform in Circuit Analysis
✓
s + 30,000
s + 30,000
Vo =
Vn =
s + 5000
s + 5000
H(s) =
◆✓
◆
sVg
;
s + 8000
s(s + 30,000)
Vo
.
=
Vg
(s + 5000)(s + 8000)
[b] vg = 0.6u(t);
Vg =
0.6
;
s
Vo =
K1
K2
0.6(s + 30,000)
=
+
;
(s + 5000)(s + 8000)
s + 5000 s + 8000
K1 =
0.6(25,000)
= 5;
3000
.·. vo (t) = (5e 5000t
K2 =
0.6(22,000)
=
3000
4.4;
4.4e 8000t )u(t) V.
[c] Vg = 2 cos 10,000t V;
H(j!) =
.·.
j10,000(30,000 + j10,000)
= 2.21/
(5000 + j10,000)(8000 + j10,000)
vo = 4.42 cos(10,000t
P 13.82 [a] H(s) =
6.34 ;
6.34 ) V.
Zf
;
Zi
108
(1/Cf )
Zf =
=
;
s + (1/Rf Cf )
s + 1000
Zi =
10,000(s + 400)
Ri [s + (1/Ri Ci )]
=
;
s
s
H(s) =
104 s
.
(s + 400)(s + 1000)
[b] H(j400) =
104 (j400)
= 6.565/
(400 + j400)(1000 + j400)
vo (t) = 13.13 cos(400t
156.8 ;
156.8 ) V.
P 13.83 Original charge on C1 ; q1 = V0 C1 .
The charge transferred to C2 ; q2 = V0 Ce =
The charge remaining on C1 ;
Therefore V2 =
q10 = q1
V0 C1
q2
=
C2
C1 + C2
V0 C1 C2
.
C1 + C2
V0 C12
q2 =
.
C1 + C2
and V1 =
q10
V0 C1
=
.
C1
C1 + C2
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Problems
13–85
P 13.84 [a] After making a source transformation, the circuit is as shown. The
impulse current will pass through the capacitive branch since it appears
as a short circuit to the impulsive current,
+
So vo (0 ) = 10
6
Z 0+ "
0
#
(t)
dt = 1000 V;
1000
Therefore wC = (0.5)Cv 2 = 0.5 J.
[b] iL (0+ ) = 0;
therefore wL = 0 J.
Vo
Vo
+
= 10 3 ;
[c] Vo (10 6 )s +
250 + 0.05s 1000
Therefore
1000(s + 5000)
Vo = 2
s + 6000s + 25 ⇥ 106
=
K1⇤
K1
+
.
s + 3000 j4000 s + 3000 + j4000
K1 = 559.02/
26.57 ;
K1⇤ = 559.02/26.57 ;
vo = [1118.03e 3000t cos(4000t
26.57 )]u(t) V.
[d] The s-domain circuit is
Vo
Vo
Vo s
+
= 10 3 .
+
6
10
250 + 0.05s 1000
Note that this equation is identical to that derived in part [c], therefore
the solution for Vo will be the same.
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13–86
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.85 [a]
V1
V1
= 10 5 ;
+
4
5
4
10
[(2 ⇥ 10 )/s)] + [(5 ⇥ 10 )/s]
sV1
V1
+
= 10 5 ;
4
10
25 ⇥ 104
25 + sV1 = 2.5;
V1 =
2.5
;
s + 25
✓
sV1
Vo =
25 ⇥ 104
.·.
Vo =
◆
5 ⇥ 104
s
0.5
;
s + 25
!
1
= V1 ;
5
vo = 0.5e 25t u(t) V.
[b] vo (0+ ) = 0.5 V;
vo (0+ ) =
Ce =
106 Z 0+
10 ⇥ 10 6 (x) dx = 0.5 V. (checks)
20 0
(5)(20)
= 4 µF;
25
⌧ = RCe = (10 ⇥ 103 )(4 ⇥ 10 6 ) = 4 ⇥ 10 2 s;
100
1
=
= 25. (checks)
⌧
4
Yes, the impulsive current establishes an instantaneous charge on each
capacitor. After the impulsive current vanishes the capacitors discharge
exponentially to zero volts.
P 13.86 [a] The s-domain circuit is
The node-voltage equation is
V
V
V
⇢
+ +
=
sL1 R sL2
s
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Problems
So V =
⇢R
s + (R/Le )
where Le =
13–87
L 1 L2
;
L1 + L2
Therefore v = ⇢Re (R/Le )t u(t) V.
[b] I1 =
V
K0
K1
V
⇢[s + (R/L2 )]
+
=
+
.
=
R sL2
s[s + (R/Le )]
s
s + (R/Le )
K0 =
⇢L1
;
L1 + L2
K1 =
Thus we have i1 =
[c] I2 =
⇢
[L1 + L2 e (R/Le )t ]u(t) A.
L1 + L2
K2
K3
V
(⇢R/L2 )
=
+
.
=
sL2
s[s + (R/Le )]
s
s + (R/Le )
⇢L1
;
L1 + L2
K3 =
Therefore i2 =
⇢L1
[1
L1 + L2
K2 =
[d]
⇢L2
;
L1 + L2
⇢L1
;
L1 + L2
e (R/Le )t ]u(t).
(t) = L1 i1 + L2 i2 = ⇢L1 .
P 13.87 [a] As R ! 1, v(t) ! ⇢Le (t) since the area under the impulse generating
function is ⇢Le .
i1 (t) !
⇢L1
u(t) A as R ! 1;
L1 + L2
i2 (t) !
⇢L1
u(t) A as R ! 1.
L1 + L2
[b] The s-domain circuit is
V
⇢
V
+
= ;
sL1 sL2
s
so V =
⇢L1 L2
= ⇢Le ;
L1 + L2
Therefore v(t) = ⇢Le (t)
✓
V
⇢L1
=
I1 = I2 =
sL2
L1 + L2
Thus i1 = i2 =
◆✓ ◆
1
;
s
⇢L1
u(t) A.
L1 + L2
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13–88
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.88 [a]
20 = sI1
0.5sI2 ;
✓
◆
3
I2 ;
0.5sI1 + s +
s
0=
s
=
0.5s
= s2 + 3
0.25s2 = 0.75(s2 + 4);
0.5s (s + 3/s)
20
N1 =
0.5s
= 20s +
0 (s + 3/s)
N1
I1 =
=
20(s2 + 3)
80
s2 + 3
=
·
s(0.75)(s2 + 4)
3 s(s2 + 4)
K0
K1
K1⇤
+
+
.
s
s j2 s + j2
=
✓ ◆
80 3
= 20;
K0 =
3 4

.·. i1 = 20 +
s
[b] N2 =
I2 =
"
#
10
80
4+3
= /0 ;
K1 =
3 (j2)(j4)
3
20
cos 2t u(t) A.
3
20
= 10s;
0.5s 0
N2
40
K1 =
3
i2 =
20s2 + 60
20(s2 + 3)
60
=
=
;
s
s
s

✓
◆
10s
40
K1⇤
K1
s
=
+
.
=
0.75(s2 + 4)
3 s2 + 4
s j2 s + j2
=
j2
j4
!
=
20
/0 ;
3
40
cos 2t u(t) A.
3
✓ ◆
✓
◆
3
K1
K1⇤
40
3 40
s
[c] V0 = I2 =
=
+
.
=
s
s 3 s2 + 4
s2 + 4
s j2 s + j2
K1 =
40
=
j4
j10 = 10/90 ;
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Problems
vo = 20 cos(2t
13–89
90 ) = 20 sin 2t;
vo = [20 sin 2t]u(t) V.
[d] Let us begin by noting i1 jumps from 0 to (80/3) A between 0 and 0+
and in this same interval i2 jumps from 0 to (40/3) A. Therefore in the
derivatives of i1 and i2 there will be impulses of (80/3) (t) and
(40/3) (t), respectively. Thus
di1
80
40
=
(t)
sin 2t A/s;
dt
3
3
40
80
di2
=
(t)
sin 2t A/s.
dt
3
3
From the circuit diagram we have
di2
di1
0.5
20 (t) = 1
dt
dt
=
40
sin 2t
3
80
(t)
3
20 (t) 40
+
sin 2t
3
3
= 20 (t).
Thus our solutions for i1 and i2 are in agreement with known circuit
behavior.
Let us also note the impulsive voltage will impart energy into the circuit.
Since there is no resistance in the circuit, the energy will not dissipate.
Thus the fact that i1 , i2 , and vo exist for all time is consistent with
known circuit behavior.
Also note that although i1 has a dc component, i2 does not. This follows
from known transformer behavior.
Finally we note the flux linkage prior to the appearance of the impulsive
voltage is zero. Now since v = d /dt, the impulsive voltage source must
be matched to an instantaneous change in flux linkage at t = 0+ of 20.
For the given polarity dots and reference directions of i1 and i2 we have
(0+ ) = L1 i1 (0+ ) + M i1 (0+ )
(0+ ) = 1
=
✓
◆
✓
80
80
+ 0.5
3
3
120
3
◆
L2 i2 (0+ )
1
✓
40
3
◆
M i2 (0+ );
0.5
✓
40
3
◆
60
= 20. (checks)
3
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13–90
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.89 [a]
106
0.5
·
Vo =
50,000 + 5 ⇥ 106 /s s
=
500,000
10
;
=
6
50,000s + 5 ⇥ 10
s + 100
vo = 10e 100t u(t) V.
[b] At t = 0 the current in the 1 µF capacitor is 10 (t) µA;
.·.
+
vo (0 ) = 10
6
Z 0+
0
10 ⇥ 10 6 (t) dt = 10 V.
After the impulsive current has charged the 1 µF capacitor to 10 V it
discharges through the 50 k⌦ resistor.
Ce =
0.25
C1 C2
= 0.2 µF;
=
C1 + C2
1.25
⌧ = (50,000)(0.2 ⇥ 10 6 ) = 10 2 ;
1
= 100. (checks)
⌧
Note – after the impulsive current passes the circuit becomes
The solution for vo in this circuit is also
vo = 10e 100t u(t) V.
P 13.90 [a] For t < 0,
0.5v1 = 2v2 ;
v1 + v2 = 100;
therefore v1 = 4v2 ;
therefore v1 (0 ) = 80 V.
[b] v2 (0 ) = 20 V.
[c] v3 (0 ) = 0 V.
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Problems
13–91
[d] For t > 0:
I=
100/s
⇥ 10 6 = 32 ⇥ 10 6 ;
3.125/s
i(t) = 32 (t) µA.
106 Z 0+
32 ⇥ 10 6 (t) dt + 80 = 64 + 80 = 16 V.
0.5 0
106 Z 0+
32 ⇥ 10 6 (t) dt + 20 = 16 + 20 = 4 V.
[f ] v2 (0+ ) =
2 0
0.625 ⇥ 106
20
· 32 ⇥ 10 6 =
[g] V3 =
s
s
[e] v1 (0+ ) =
v3 (t) = 20u(t) V;
v3 (0+ ) = 20 V.
Check: v1 (0+ ) + v2 (0+ ) = v3 (0+ ).
P 13.91 [a] For t < 0:
Req = 0.8 k⌦k4 k⌦k16 k⌦ = 0.64 k⌦;
i1 (0 ) =
3200
= 0.8 A;
4000
i2 (0 ) =
v = 5(640) = 3200 V;
3200
= 0.2 A.
1600
[b] For t > 0:
i1 + i2 = 0;
8( i1 ) = 2( i2 );
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13–92
CHAPTER 13. The Laplace Transform in Circuit Analysis
i1 (0 ) +
i2 =
i1 + i2 (0 ) +
0.8 A;
i2 = 0;
therefore
i1 (0+ ) = 0.8
0.2 = 0.6 A.
i1 =
0.2 A;
[c] i2 (0 ) = 0.2 A.
[d] i2 (0+ ) = 0.2
0.8 =
0.6 A.
[e] The s-domain equivalent circuit for t > 0 is
0.6
0.006
=
;
0.01s + 20,000
s + 2 ⇥ 106
I1 =
6
i1 (t) = 0.6e 2⇥10 t u(t) A.
[f ] i2 (t) =
[g] V =
i1 (t) =
6
0.6e 2⇥10 t u(t) A.
0.0064 + (0.008s + 4000)I1 =
1.6 ⇥ 10 3
=
7200
;
s + 2 ⇥ 106
v(t) = [ 1.6 ⇥ 10 3 (t)]
P 13.92 [a] Z1 =
0.0016(s + 6.5 ⇥ 106 )
s + 2 ⇥ 106
6
[7200e 2⇥10 t u(t)] V.
1/C1
25 ⇥ 1010
=
⌦;
s + 1/R1 C1
s + 20 ⇥ 104
Z2 =
1/C2
6.25 ⇥ 1010
⌦;
=
s + 1/R2 C2
s + 12,500
V0 V0 10/s
+
= 0;
Z2
Z1
10 (s + 20 ⇥ 104 )
V0 (s + 12,500) V0 (s + 20 ⇥ 104 )
+
=
;
6.25 ⇥ 1010
25 ⇥ 1010
s
25 ⇥ 1010
V0 =
K1
K2
2(s + 200,000)
=
+
.
s(s + 50,000)
s
s + 50,000
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Problems
K1 =
2(200,000)
= 8;
50,000
K2 =
2(150,000)
=
50,000
.·. vo = [8
[b] I0 =
13–93
6;
6e 50,000t ]u(t) V.
V0
2(s + 200,000)(s + 12,500)
=
Z2
s(s + 50,000)6.25 ⇥ 1010
= 32 ⇥ 10
12
"
162,500s + 25 ⇥ 108
1+
s(s + 50,000)
= 32 ⇥ 10
12
"
K2
K1
+
1+
.
s
s + 50,000
K1 = 50,000;
#
#
K2 = 112,500;
io = 32 (t) + [1.6 ⇥ 106 + 3.6 ⇥ 106 e 50,000t ]u(t) pA.
[c] When
C1 = 64 pF
Z1 =
156.25 ⇥ 108
⌦
s + 12,500
10 (s + 12,500)
V0 (s + 12,500) V0 (s + 12,500)
+
=
625 ⇥ 108
156.25 ⇥ 108
s 156.25 ⇥ 108
.·. V0 + 4V0 =
40
.
s
8
V0 = ;
s
vo = 8u(t) V.
I0 =

V0
8 (s + 12,500)
12,500
12
=
=
128
⇥
10
1
+
;
Z2
s 6.25 ⇥ 1010
s
io (t) = 128 (t) + 1.6 ⇥ 106 u(t) pA.
P 13.93 Let a =
1
1
=
.
R1 C1
R2 C2
Then Z1 =
1
C1 (s + a)
and
Z2 =
1
.
C2 (s + a)
Vo
10/s
Vo
+
=
;
Z2 Z1
Z1
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13–94
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo C2 (s + a) + V0 C1 (s + a) = (10/s)C1 (s + a);
✓
◆
10
C1
Vo =
;
s C1 + C2
Thus, vo is the input scaled by the factor
C1
.
C1 + C2
P 13.94 [a] The circuit parameters are
288
1202
1202
1202
= 12 ⌦
Rb =
= 8⌦
Xa =
=
⌦.
Ra =
1200
1800
350
7
The branch currents are
35
35
120/0
120/0
= 10/0 A(rms)
= j
= / 90 A(rms);
I2 =
I1 =
12
j1440/35
12
12
I3 =
120/0
= 15/0 A(rms);
8
.·. Io = I1 + I2 + I3 = 25
j
35
= 25.17/
12
6.65 A(rms).
and
p
io = 25.17 2 cos(!t
Therefore,
✓
◆
35 p
i2 =
2 cos(!t
12
Thus,
i2 (0 ) = i2 (0+ ) = 0 A
90 ) A
and
6.65 ) A.
p
io (0 ) = io (0+ ) = 25 2 A.
[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for V0
symbolically. Write a single node voltage equation:
V0
V0
V0
(Vg + L` Io )
+
+
= 0;
sL`
Ra sLa
.·. V0 =
(Ra /L` )Vg + Io Ra
,
s + [Ra (La + L` )]/La L`
p
where L` = 1/120⇡ H, La = 12/35⇡ H, Ra = 12 ⌦, and I0 Ra = 300 2 V.
Thus,
p
p
p
300 2
1440⇡(122.92 2s 3000⇡ 2)
V0 =
+
(s + 1475⇡)(s2 + 14,400⇡ 2 )
s + 1475⇡
p
K1
K2
K2⇤
300 2
=
+
+
+
.
s + 1475⇡ s j120⇡ s + j120⇡ s + 1475⇡
The coefficients are
p
K1 = 121.18 2 V
p
K2 = 61.03 2/6.85 V
p
K2⇤ = 61.03 2/
6.85 .
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Problems
13–95
p
p
Note that K1 + 300 2 = 178.82 2 V. Thus, the inverse transform of V0 is
p
p
v0 = 178.82 2e 1475⇡t + 122.06 2 cos(120⇡t + 6.85 ) V.
Initially,
p
p
p
v0 (0+ ) = 178.82 2 + 122.06 2 cos 6.85 = 300 2 V.
p
Note that at t = 0+ the initial value of iL , which is p
25 2 A, existspin the
12 ⌦ resistor Ra . Thus, the initial value of V0 is (25 2)(12) = 300 2 V.
[c] The phasor domain equivalent circuit has a j1 ⌦ inductive impedance in
series with the parallel combination of a 12 ⌦ resistive impedance and a
j1440/35 ⌦ inductive impedance (remember that ! = 120⇡ rad/s). Note
that Vg = 120/0 + (25.17/ 6.65 )(j1) = 125.43/11.50 V(rms). The
node voltage equation in the phasor domain circuit is
V0
V0 35V0
125.43/11.50
+
+
= 0;
j1
12
1440
.·. V0 = 122.06/6.85 V(rms).
p
Therefore, v0 = 122.06 2 cos(120⇡t + 6.85 ) V, agreeing with the
steady-state component of the result in part (b).
[d] A plot of v0 , generated in Excel, is shown below.
P 13.95 [a] At t = 0 the phasor domain equivalent circuit is
I1 =
j120
=
12
I2 =
j120(35)
=
j1440
I3 =
j120
=
8
j10 = 10/
90 A (rms);
35
35
= /180 A (rms);
12
12
j15 = 15/
90 A (rms);
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13–96
CHAPTER 13. The Laplace Transform in Circuit Analysis
35
12
IL = I1 + I2 + I3 =
j25 = 25.17/
96.65 A (rms);
p
iL = 25.17 2 cos(120⇡t
96.65 )A.
p
2.92 2A;
iL (0 ) = iL (0+ ) =
35 p
2 cos(120⇡t + 180 )A.
12
p
35 p
i2 (0 ) = i2 (0+ ) =
2 = 2.92 2A;
12
i2 =
Vg = Vo + j1IL ;
35
12
= 25 j122.92 = 125.43/ 78.50 V (rms);
p
vg = 125.43 2 cos(120⇡t 78.50 )V
p
= 125.43 2[cos 120⇡t cos 78.50 + sin 120⇡t sin 78.50 ]
p
p
= 25 2 cos 120⇡t + 122.92 2 sin 120⇡t;
Vg =
j120 + 25
j
p
p
25 2s + 122.92 2(120⇡)
·
.
. . Vg =
s2 + (120⇡)2
s-domain circuit:
where
Ll =
1
H;
120⇡
iL (0) =
La =
p
2.92 2 A;
12
H;
35⇡
Ra = 12 ⌦;
i2 (0) =
p
2.92 2 A.
The node voltage equation is
0=
Vo
Vo
Vo + i2 (0)La
(Vg + iL (0)Ll )
+
+
.
sLl
Ra
sLa
Solving for Vo yields
Vo =
Ra [iL (0) i2 (0)]
Vg Ra /Ll
+
.
[s + Ra (Ll + La )/La Ll ] [s + Ra (Ll + La )/Ll La ]
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Problems
13–97
Ra
= 1440⇡;
Ll
1
12
+ 35⇡
)
12( 120⇡
Ra (Ll + La )
=
= 1475⇡;
1
12
Ll La
( 35⇡ )( 120⇡ )
p
p
iL (0) i2 (0) = 2.92 2 + 2.92 2 = 0;
p
p
1440⇡[25
2s
+
122.92
2(120⇡)]
.·. Vo =
2
(s + 1475⇡)[s + (120⇡)2 ]
K2
K2⇤
K1
+
+
;
=
s + 1475⇡ s j120⇡ s + j120⇡
p
p
K1 = 14.55 2
K2 = 61.03 2/ 83.15 ;
p
p
.·. vo (t) = 14.55 2e 1475⇡t + 122.06 2 cos(120⇡t
83.15 )V.
Check:
p
vo (0) = ( 14.55 + 14.55) 2 = 0.
[b]
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13–98
CHAPTER 13. The Laplace Transform in Circuit Analysis
p
[c] In Problem 13.94, the line-to-neutral voltage spikes at 300 2 V. Here the
line-to-neutral voltage has no spike. Thus the amount of voltage
disturbance depends on what part of the cycle the sinusoidal steady-state
voltage is switched.
P 13.96 [a] First find Vg before Rb is disconnected. The phasor domain circuit is
120/✓
120/✓
120/✓
+
+
Ra
Rb
jXa
120/✓
[(Ra + Rb )Xa = jRa Rb ].
=
Ra Rb Xa
IL =
Since Xl = 1 ⌦ we have
Vg = 120/✓ +
120/✓
[Ra Rb + j(Ra + Rb )Xa ];
Ra Rb Xa
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Problems
Ra = 12 ⌦;
Vg =
Rb = 8 ⌦;
Xa =
13–99
1440
⌦;
35
120/✓
(1475 + j300)
1400
25
/✓ (59 + j12) = 125.43/(✓ + 11.50) ;
12
p
vg = 125.43 2 cos(120⇡t + ✓ + 11.50 )V.
=
Let
= ✓ + 11.50 . Then
p
vg = 125.43 2(cos 120⇡t cos
sin 120⇡t sin )V.
Therefore
p
125.43 2(s cos
120⇡ sin )
Vg =
.
2
s + (120⇡)2
The s-domain circuit becomes
where ⇢1 = iL (0+ ) and ⇢2 = i2 (0+ ).
The s-domain node voltage equation is
Vo
Vo
Vo + ⇢2 La
(Vg + ⇢1 Ll )
+
+
= 0.
sLl
Ra
sLa
Solving for Vo yields
Vo =
Vg Ra /Ll + (⇢1
⇢2 )Ra
.
(La +Ll )Ra
[s + La Ll ]
Substituting the numerical values
Ll =
1
H;
120⇡
La =
12
H;
35⇡
Ra = 12 ⌦;
Rb = 8 ⌦;
gives
Vo =
1440⇡Vg + 12(⇢1 ⇢2 )
.
(s + 1475⇡)
Now determine the values of ⇢1 and ⇢2 .
⇢1 = iL (0+ )
and
⇢2 = i2 (0+ );
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13–100
CHAPTER 13. The Laplace Transform in Circuit Analysis
120/✓
[(Ra + Rb )Xa
Ra Rb Xa
IL =
jRa Rb ]
"
120/✓
(20)(1440)
=
96(1440/35)
35
= 25.17/(✓
#
j96
6.65) A(rms);
p
.·. iL = 25.17 2 cos(120⇡t + ✓ 6.65 )A.
p
iL (0+ ) = ⇢1 = 25.17 2 cos(✓ 6.65 )A;
p
p
.·. ⇢1 = 25 2 cos ✓ + 2.92 2 sin ✓A.
I2 =
35
120/✓
= /(✓
j(1440/35)
12
90) ;
35 p
2 cos(120⇡t + ✓ 90 )A;
12
p
35 p
⇢2 = i2 (0+ ) =
2 sin ✓ = 2.92 2 sin ✓A;
12
p
.·. ⇢1 = ⇢2 = 25 2 cos ✓.
p
(⇢1 ⇢2 )Ra = 300 2 cos ✓.
p
300
1440⇡
2 cos ✓
.·. Vo =
· Vg +
s + 1475⇡
s + 1475⇡
p
p
"
#
300 2 cos ✓
1440⇡
125.43 2(s cos
120⇡ sin )
+
=
s + 1475⇡
s2 + 14,400⇡ 2
s + 1475⇡
p
K2
K2⇤
K1 + 300 2 cos ✓
+
+
.
=
s + 1475⇡
s j120⇡ s + j120⇡
i2 =
Now
p
(1440⇡)(125.43 2)[ 1475⇡ cos
120⇡ sin ]
K1 =
2
2
2
1475 ⇡ + 14,400⇡
p
1440(125.43 2)[1475 cos + 120 sin ]
.
=
14752 + 14,000
Since
K1 =
= ✓ + 11.50 , K1 reduces to
p
p
121.18 2 cos ✓ + 14.55 2 sin ✓.
From the partial fraction expansionpfor Vo we see vo (t) will go directly
into steady state when K1 = 300 2 cos ✓. It follows that
p
p
14.55 2 sin ✓ = 178.82 2 cos ✓
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Problems
or
tan ✓ =
13–101
12.29;
Therefore,
✓=
[b] When ✓ =
85.35 ,
85.35 .
=
73.85 .
p
1440⇡(125.43
2)[ 120⇡ sin( 73.85 ) + j120⇡ cos( 73.85 )
.·. K2 =
(1475⇡ + j120⇡)(j240⇡)
p
720 2(120.48 + j34.88)
=
120 + j1475
p
= 61.03 2/ 78.50 ;
p
.·. vo = 122.06 2 cos(120⇡t
= 172.61 cos(120⇡t
78.50 ) V t > 0
78.50 ) V t > 0.
[c] vo1 = 169.71 cos(120⇡t
85.35 )V
t < 0;
vo2 = 172.61 cos(120⇡t
78.50 )V
t > 0.
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13–102
CHAPTER 13. The Laplace Transform in Circuit Analysis
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14
Introduction to
Frequency-Selective Circuits
Assessment Problems
AP 14.1
fc = 8 kHz,
!c =
1
;
RC
!c = 2⇡fc = 16⇡ krad/s;
R = 10 k⌦;
1
1
.·. C =
=
= 1.99 nF.
!c R
(16⇡ ⇥ 103 )(104 )
AP 14.2 [a] !c = 2⇡fc = 2⇡(2000) = 4⇡ krad/s;
L=
R
5000
= 0.40 H.
=
!c
4000⇡
[b] H(j!) =
!c
4000⇡
=
.
!c + j!
4000⇡ + j!
When ! = 2⇡f = 2⇡(50,000) = 100,000⇡ rad/s,
H(j100,000⇡) =
1
4000⇡
=
= 0.04/87.71 ;
4000⇡ + j100,000⇡
1 + j25
.·. |H(j100,000⇡)| = 0.04.
[c] .·. ✓(100,000⇡) =
AP 14.3
!c =
87.71 .
5000
R
=
= 1.43 Mrad/s.
L
3.5 ⇥ 10 3
14–1
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14–2
CHAPTER 14. Introduction to Frequency-Selective Circuits
106
106
1
=
=
= 10 krad/s.
RC
R
100
106
= 200 rad/s.
[b] !c =
5000
106
= 33.33 rad/s.
[c] !c =
3 ⇥ 104
AP 14.4 [a] !c =
AP 14.5 Let Z represent the parallel combination of (1/SC) and RL . Then
RL
.
(RL Cs + 1)
Z=
Thus H(s) =
=
where K =
AP 14.6
!o2 =
1
LC
!o
Q=
=
RL
Z
=
R+Z
R(RL Cs + 1) + RL
(1/RC)
L
s + R+R
RL
⇣
1
RC
⌘ =
(1/RC)
s + K1
⇣
1
RC
⌘,
RL
.
R + RL
so L =
!o
R/L
1
!o2 C
=
so R =
1
(24⇡ ⇥ 103 )2 (0.1 ⇥ 10 6 )
= 1.76 mH;
!o L
(24⇡ ⇥ 103 )(1.76 ⇥ 10 3 )
=
= 22.10 ⌦.
Q
6
AP 14.7
!o = 2⇡(2000) = 4000⇡ rad/s;
= 2⇡(500) = 1000⇡ rad/s;
R = 250 ⌦;
1
RC
so C =
1
1
=
= 1.27 µF;
R
(1000⇡)(250)
!o2 =
1
LC
so L =
1
106
=
= 4.97 mH.
!o2 C
(4000⇡)2 (1.27)
!o2 =
1
LC
so L =
1
RC
so R =
=
AP 14.8
=
1
!o2 C
=
1
(104 ⇡)2 (0.2 ⇥ 10 6 )
= 5.07 mH;
1
1
=
= 3.98 k⌦.
C
400⇡(0.2 ⇥ 10 6 )
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Problems
AP 14.9
!o2 =
Q=
1
LC
fo
so L =
=
1
!o2 C
=
1
(400⇡)2 (0.2 ⇥ 10 6 )
14–3
= 31.66 mH;
5 ⇥ 103
= 25 = !o RC;
200
25
Q
=
= 9.95 k⌦.
.·. R =
!o C
(400⇡)(0.2 ⇥ 10 6 )
AP 14.10
!o = 8000⇡ rad/s;
C = 500 nF;
!o2 =
Q=
1
LC
!o
so L =
=
1
!o2 C
= 3.17 mH;
!o L
1
=
;
R
!o CR
1
1
=
= 15.92 ⌦.
!o CQ
(8000⇡)(500)(5 ⇥ 10 9 )
.·. R =
AP 14.11
!o = 2⇡fo = 2⇡(20,000) = 40⇡ krad/s;
Q=
!o2 =
!o
=
1
LC
!o
(R/L)
so L =
so C =
R = 100 ⌦;
Q = 5;
QR
5(100)
= 3.98 mH;
=
!o
(40⇡ ⇥ 103 )
1
1
=
= 15.92 nF.
2
3
2
!o L
(40⇡ ⇥ 10 ) (3.98 ⇥ 10 3 )
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14–4
CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems
P 14.1
[a] !c =
127
R
=
= 12.7 krad/s;
L
10 ⇥ 10 3
12,700
!c
=
= 2021.27 Hz.
.·. fc =
2⇡
2⇡
!c
12,700
[b] H(s) =
;
=
s + !c
s + 12,700
H(j!) =
12,700
;
12,700 + j!
H(j!c ) =
12,700
= 0.7071/
12,700 + j12,700
12,700
= 0.981/
12,700 + j2540
11.31 ;
12,700
= 0.196/
12,700 + j63,500
78.69 .
H(j0.2!c ) =
H(j5!c ) =
[c] vo (t)|!c = 7.07 cos(12,700t
P 14.2
45;
45 ) V;
vo (t)|0.2!c = 9.81 cos(2540t
11.31 ) V;
vo (t)|5!c = 1.96 cos(63,500t
78.69 ) V.
Vo
R
[a] H(U L) (s) =
=
=
Vi
R + sL
R
L
.
R
s+
L
✓
◆
RL
R
Vo
RkRL
L R + RL
[b] H(L) (s) =
=
=
✓
◆.
R
RL
Vi
RkRL + sL
s+
L R + RL
✓
◆
R
R
RL
!c(L) =
so the cuto↵ frequencies are di↵erent.
[c] !c(U L) = ;
L
L R + RL
H(0)(U L) = 1;
H(0)(L) = 1
so the passband gains are the same.
330
= 33,000 rad/s.
0.01
[e] !c(L) = 33,000 0.05(33,000) = 31,350 rad/s;
[d] !c(U L) =
✓
330
RL
31,350 =
0.01 330 + RL
◆
.·. 0.05RL = 313.5 so RL
so
RL
= 0.95;
330 + RL
6270 ⌦.
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Problems
P 14.3
[a] !o =
14–5
R
= 2000⇡ rad/s;
L
R = L!o = (0.005)(2000⇡) = 31.42 ⌦.
[b] Re = 31.42k270 = 28.14 ⌦;
Re
= 5628 rad/s;
L
!loaded
.·. floaded =
= 895.72 Hz.
2⇡
!loaded =
[c] Use a 33 ⌦ resistor, which gives
!o = R/L = 33/0.005 = 6600 rad/s = 1050.4 Hz.
P 14.4
R
(R/L)
Vo
.
=
=
Vi
sL + R + Rl
s + (R + Rl )/L
(R/L)
[b] H(j!) = ⇣ R+R ⌘
;
l
+ j!
L
[a] H(s) =
|H(j!)| = r⇣
(R/L)
⌘
R+Rl 2
+ !2
L
;
|H(j!)|max occurs when ! = 0.
R
.
R + Rl
R/L
R
[d] |H(j!c )| = p
= r⇣
⌘
2(R + Rl )
R+Rl 2
[c] |H(j!)|max =
L
.·. !c2 =
[e] !c =
✓
R + Rl
L
◆2
;
;
+ !c2
.·. !c = (R + Rl )/L.
127 + 75
= 20,200 rad/s;
0.01
H(j!) =
12,700
;
20,200 + j!
H(j0) = 0.6287;
H(j20,200) =
H(j4040) =
0.6287
p /
2
45 = 0.4446/
12,700
= 0.6165/
20,200 + j4040
H(j101,000) =
45;
11.31 ;
12,700
= 0.1233/
20,200 + j101,000
78.69 .
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14–6
P 14.5
CHAPTER 14. Introduction to Frequency-Selective Circuits
R
R
250
so L =
= 31.25 mH.
=
L
!c
8000
[b] Req = 1000k250 = 200 ⌦;
[a] !c =
!c =
P 14.6
200
Req
=
= 6400 rad/s.
L
0.03125
[a] !c = 0.9(8000) = 7200 =
Then,
225 = 250kRL =
Req
0.03125
so Req = 7200(0.03125) = 225 ⌦.
250RL
.
250 + RL
Solving,
RL = 2250 ⌦.
7200
!c
;
=
[b] H(s) =
s + !c
s + 7200
|H(j0)| =
P 14.7
7200
= 1.
7200
1
1
=
= 10 krad/s;
3
RC
(10 )(100 ⇥ 10 9 )
!c
= 1591.55 Hz.
fc =
2⇡
10,000
!c
;
=
[b] H(j!) =
s + !c
s + 10,000
[a] !c =
H(j!) =
10,000
;
10,000 + j!
H(j!c ) =
10,000
= 0.7071/
10,000 + j10,000
45 ;
H(j0.1!c ) =
10,000
= 0.9950/
10,000 + j1000
H(j10!c ) =
10,000
= 0.0995/
10,000 + j100,000
[c] vo (t)|!c = 0.2(0.7071) cos(10,000t
= 141.42 cos(10,000t
45 ) mV.
5.71 )
5.71 ) mV.
vo (t)|10!c = 0.2(0.0995) cos(100,000t
= 19.90 cos(100,000t
84.29 .
45 )
vo (t)|0.1!c = 0.2(0.9950) cos(1000t
= 199.01 cos(1000t
5.71 ;
84.29 )
84.29 ) mV.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 14.8
[a] Let Z =
14–7
RL
RL (1/SC)
=
;
RL + 1/SC
RL Cs + 1
Then H(s) =
=
=
Z
Z +R
RL
RRL Cs + R + RL
(1/RC)
✓
◆.
R + RL
s+
RRL C
(1/RC)
[b] |H(j!)| = q
! 2 + [(R + RL )/RRL C]2
;
|H(j!)| is maximum at ! = 0.
RL
.
R + RL
(1/RC)
RL
=q
;
[d] |H(j!c )| = p
2(R + RL )
!C2 + [(R + RL )/RRL C]2
[c] |H(j!)|max =
.·. !c =
1
R + RL
=
(1 + (R/RL )) .
RRL C
RC
1
[1 + (103 /104 )] = 10,000(1 + 0.1) = 11,000 rad/s;
(103 )(10 7 )
[e] !c =
H(j0) =
10,000
= 0.9091/0 ;
11,000
H(j!c ) =
P 14.9
45 ;
H(j0.1!c ) =
10,000
= 0.9046/
11,000 + j1100
H(j10!c ) =
10,000
= 0.0905/
11,000 + j110,000
[a] fc =
[b]
10,000
= 0.6428/
11,000 + j11,000
5.71 ;
84.29 .
50,000
50
!c
=
=
⇥ 103 = 7957.75 Hz.
2⇡
2⇡
2⇡
1
= 50 ⇥ 103 ;
RC
R=
1
(50 ⇥ 103 )(0.5 ⇥ 10 6 )
= 40 ⌦.
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14–8
CHAPTER 14. Introduction to Frequency-Selective Circuits
[c] With a load resistor added in parallel with the capacitor the transfer
function becomes
RL k(1/sC)
RL /sC
H(s) =
=
R + RL k(1/sC)
R[RL + (1/sC)] + RL /sC
=
1/RC
RL
.
=
RRL sC + R + RL
s + [(R + RL )/RRL C]
This transfer function is in the form of a low-pass filter, with a cuto↵
frequency equal to the quantity added to s in the denominator.
Therefore,
!c =
.·.
✓
◆
1
R + RL
R
=
1+
;
RRL C
RC
RL
R
= 0.05
RL
[d] H(j0) =
.·. RL = 20R = 800 ⌦.
RL
800
= 0.9524.
=
R + RL
840
P 14.10 [a] !c = 2⇡(100) = 628.32 rad/s.
1
1
1
[b] !c =
so R =
=
= 338.63 ⌦.
RC
!c C
(628.32)(4.7 ⇥ 10 6 )
[c]
1/RC
628.32
Vo
1/sC
=
=
.
=
Vi
R + 1/sC
s + 1/RC
s + 628.32
Vo
628.32
(1/sC)kRL
1/RC
[e] H(s) =
=
.
=
=
✓
◆
R + RL
Vi
R + (1/sC)kRL
s + 2(628.32)
s+
1/RC
RL
[f ] !c = 2(628.32) = 1256.64 rad/s.
[d] H(s) =
[g] H(0) = 1/2.
P 14.11 [a] H(s) =
s
s
sL
=
=
.
R + sL
s + R/L
s + 15,000
R
= 15,000 rad/s.
L
1
jR/L
j
[c] |H(jR/L)| =
=
=p .
jR/L + R/L
j+1
2
[b] !c =
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
✓
14–9
◆
RL
s
RL ksL
Vo
R + RL
=
=
P 14.12 [a] H(s) =
✓
◆
R
RL
Vi
R + RL ksL
s+
L R + RL
=
1
s
2
1
s + (15,000)
2
✓
.
◆
R
1
RL
[b] !c =
= (15,000) = 7500 rad/s.
L R + RL
2
1
[c] !c(L) = !c(U L) .
2
[d] The gain in the passband is also reduced by a factor of 1/2 for the loaded
filter.
P 14.13 [a] !c =
R
L
so R = !c L = (25 ⇥ 103 )(5 ⇥ 10 3 ) = 125 ⌦.
[b] !c (loaded) =
.·.
RL
R
·
= 24,000;
L R + RL
24,000
!c (loaded)
RL
=
= 0.96.
=
R + RL
!c (unloaded)
25,000
RL = 0.96(R + RL )
.·.
0.04RL = 0.96R = (0.96)(125);
(0.96)(125)
= 3 k⌦.
.·. RL =
0.04
P 14.14 [a] !c = 2⇡(500) = 3141.59 rad/s.
1
1
1
[b] !c =
so R =
=
= 1.45 M⌦.
RC
!c C
(3141.59)(220 ⇥ 10 12 )
[c]
s
s
Vo
R
=
=
.
=
Vi
R + 1/sC
s + 1/RC
s + 3141.59
s
s
RkRL
Vo
=
=
.
=
[e] H(s) =
✓
◆
R + RL
Vi
RkRL + (1/sC)
s + 2(3141.59)
s+
1/RC
RL
[f ] !c = 2(3141.59) = 6283.19 rad/s.
[d] H(s) =
[g] H(1) = 1.
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14–10
P 14.15 [a]
CHAPTER 14. Introduction to Frequency-Selective Circuits
1
1
=
= 4000 rad/s;
3
RC
(50 ⇥ 10 )(5 ⇥ 10 9 )
fc =
4000
= 636.62 Hz.
2⇡
[b] H(s) =
s
s + !c
.·.
H(j!c ) = H(j4000) =
H(j!) =
j!
;
4000 + j!
j4000
= 0.7071/45 ;
4000 + j4000
H(j0.2!c ) = H(j800) =
j800
= 0.1961/78.69 ;
4000 + j800
H(j5!c ) = H(j20,000!c ) =
j20,000
= 0.9806/11.31 .
4000 + j20,000
[c] vo (t)|!c = (0.7071)(0.5) cos(4000t + 45 )
= 353.33 cos(4000t + 45 ) mV;
vo (t)|0.2!c = (0.1961)(0.5) cos(800t + 78.60 )
= 98.06 cos(800t + 78.69 ) mV;
vo (t)|5!c = (0.9806)(0.5) cos(20,000t + 11.31 )
= 490.29 cos(20,000t + 11.31 ) mV.
P 14.16 [a] H(s) =
R
Vo
=
Vi
R + Rc + (1/sC)
R
s
.
·
R + Rc [s + (1/(R + Rc )C)]
j!
R
;
·
[b] H(j!) =
R + Rc j! + (1/(R + Rc )C)
=
|H(j!)| =
R
!
·q
R + Rc
!2 +
1
(R+Rc )2 C 2
.
The magnitude will be maximum when ! = 1
R
[c] |H(j!)|max =
.
R + Rc
R!c
q
.
[d] |H(j!c )| =
(R + Rc ) !c2 + [1/(R + Rc )C]2
.·. |H(j!)| = p
R
2(R + Rc )
when
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
1
(R + Rc )2 C 2
.·. !c2 =
or !c =
[e] !c =
14–11
1
.
(R + Rc )C
1
(62.5 ⇥ 103 )(5 ⇥ 10 9 )
= 3200 rad/s;
50
R
= 0.8;
=
R + Rc
62.5
.·.
H(j!) =
H(j!c ) =
0.8j!
.
3200 + j!
(0.8)j3200
= 0.5657/45 ;
3200 + j3200
H(j0.2!c ) =
H(j5!c ) =
(0.8)j640
= 0.1569/78.69 ;
3200 + j640
(0.8)j16,000
= 0.7845/11.31 .
3200 + j16,000
1
= 2⇡(300) = 600⇡ rad/s;
RC
P 14.17 [a] !c =
.·. R =
1
1
=
= 5305.16 ⌦ = 5.305 k⌦.
!c C
(600⇡)(100 ⇥ 10 9 )
[b] Re = 5305.16k47,000 = 4767.08 ⌦;
P 14.18 [a]
!c =
1
1
=
= 2097.7 rad/s;
Re C
(4767.08)(100 ⇥ 10 9 )
fc =
2097.7
!c
=
= 333.86 Hz.
2⇡
2⇡
= !c2
Then,
2
+
=
=
!c1
v
u
u
t
2
!2
!c1
+ !o2 =
!c2
2
!c1
!c2
2
!o2 = !c1 !c2 .
and
+
+
s
s
2
!c2
!c2
!c1
2
+
s
✓
!c2
!c1
2
◆2
+ !c1 !c2
2
2!c1 !c2 + !c1
+ 4!c1 !c2
4
2
2
!c2
+ !c1
+ 2!c1 !c2
4
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–12
CHAPTER 14. Introduction to Frequency-Selective Circuits
s
(!c1 + !c2 )2
2
4
!c1 !c2 !c1 + !c2
=
+
= !c1 .
2
2
Similarly,
=
2
+
v
u
u
t
=
=
2
!c2
!2
+
+ !o2 =
!c2
!c1
2
!c2
!c1
2
+
+
!c2
!c1
2
s
2
!c2
s
+
s
✓
!c2
!c1
+
v
u
!2
u
1
1
6
+ t1 +
[b] !o 4
2Q
2Q
But Q = !o / . Thus
v
u
=
2
+
= !c1
v
u
u
t
2
!2
6 1
+
!o 4
2Q
!2
v
u
u
t
v
u
u
t
=
1
1+
2Q
!2
!o2 +
!o
2Q
v
u
u
t
!2
2
+
= !c2
+ !c1 !c2
3
v
u
!o u
!o
7
+ t!o2 +
5=
2Q
2Q
!2
.
v
u
!2
+ !o2
from part (a).
But Q = !o / . Thus
!o
+
2Q
2
◆2
u
!o
!o
+ t!o2 +
=
2(!o / )
2(!o / )
Similarly,
2
!c1
2
2
!c2
+ !c1
+ 2!c1 !c2
4
s
!o u
!o
+ t!o2 +
2Q
2Q
!c2
2
2!c1 !c2 + !c1
+ 4!c1 !c2
4
(!c1 + !c2 )2
2
4
!c2 !c1 !c1 + !c2
=
+
= !c2 .
2
2
=
2
!c1
2
!2
3
v
u
!o u
!o
7
+ t!o2 +
5=
2Q
2Q
!o
+
=
2(!o / )
v
u
u
t
!o2 +
!2
.
!o
2(!o / )
!2
+ !o2
from part (a).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 14.19
=
!o
50,000
=
= 12.5 krad/s;
Q
4
2
s
1
!c2 = 50,000 4 +
8
fc2 =
1
+
!c1 = 50,000 4
8
P 14.20 !o =
fo =
3
5 = 56.64 krad/s;
s
✓ ◆2
1
1+
8
3
5 = 44.14 krad/s;
44.14 k
= 7.02 kHz.
2⇡
p
!c1 !c2 =
q
(121)(100) = 110 krad/s;
!o
= 17.51 kHz;
2⇡
= 121
Q=
1
1+
8
12,500
= 1.99 kHz
2⇡
56.64 k
= 9.01 kHz;
2⇡
2
fc1 =
✓ ◆2
14–13
!o
100 = 21 krad/s or 2.79 kHz;
=
P 14.21 [a] !o =
Q=
=
110
= 5.24.
21
q
1/LC
!o
R
L
so L =
so
=
=
1
= 79.16 mH.
[8000(2⇡)]2 (5 ⇥ 10 9 )
!o
8000
=
= 4 kHz.
Q
2
so R = L = (79.16 ⇥ 10 3 )(8000⇡) = 1.99 k⌦.
[b] From part (a),
fc1,2 = ±
1
!o2 C
2
+
= 4 kHz. Then,
v
u
u
t
2
!2
!c1 = 39,246.1 rad/s
4000
+
+ fo2 = ±
2
s
✓
4000
2
◆2
+ 80002 = ±2000 + 8246.2 Hz.
!c2 = 64,378.8 rad/s
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14–14
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.22 H(j!) =
j!(8000⇡)
.
(16,000⇡)2 ! 2 + j!(8000⇡)
[a] H(j16,000⇡) =
.·. vo (t) = 20 cos 16,000⇡t V.
Vo = (1)Vi
[b] H(j39,246.1) =
(16,000⇡)2
1
Vo = p /45 Vi
2
[c] H(j64,378.8) =
1
Vo = p /
2
(16,000⇡)2
H(j!) =
45 ;
45 ) V.
(j1600⇡)(8000⇡)
= 0.0504/87.1 ;
(1600⇡)2 + (j1600⇡)(8000⇡)
.·. vo (t) = 1.01 cos(1600⇡t + 87.1 ) V.
Vo = 0.0504/87.1 Vi
Vo = 0.0504/
1
j64,378.8(8000⇡)
=p /
2
64,378.8 + j64,378.8(8000⇡)
2
.·. vo (t) = 14.14 cos(64,378.8t
(16,000⇡)2
[e] H(j160,000) =
1
j39,246.1(8000⇡)
/45 ;
p
=
39,246.12 + j39,246.1(8000⇡)
2
.·. vo (t) = 14.14 cos(39,246.1t + 45 ) V.
45 Vi
[d] H(j1600⇡) =
P 14.23 H(s) = 1
(j16,000⇡)(8000⇡)
= 1;
(16,000⇡)2 + (j16,000⇡)(8000⇡)
(16,000⇡)2
(16,000⇡)2
87.1 Vi
(j160,000⇡)(8000⇡)
= 0.0504/
(160,000⇡)2 + (j160,000⇡)(8000⇡)
.·. vo (t) = 1.01 cos(160,000⇡t
87.1 ;
87.1 ) V.
s2 + (1/LC)
(R/L)s
=
;
s2 + (R/L)s + (1/LC)
s2 + (R/L)s + (1/LC)
(16,000⇡)2 ! 2
.
(16,000⇡)2 ! 2 + j!(8000⇡)
(16,000⇡)2 (16,000⇡)2
= 0;
[a] H(j16,000⇡) =
(16,000⇡)2 (16,000⇡)2 + (j16,000⇡)(8000⇡)
Vo = (0)Vi
.·. vo (t) = 0 V.
(16,000⇡)2 39,246.12
1
[b] H(j39,246.1) =
=p /
2
2
(16,000⇡)
39,246.1 + j39,246.1(8000⇡)
2
1
Vo = p /
2
45 Vi
.·. vo (t) = 14.14 cos(39,246.1t
45 ;
45 ) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
(16,000⇡)2 64,378.82
1
= p /45 ;
2
2
(16,000⇡)
64,378.8 + j64,378.8(8000⇡)
2
[c] H(j64,378.8) =
1
Vo = p /45 Vi
2
[d] H(j1600⇡) =
.·. vo (t) = 14.14 cos(64,378.8t + 45 ) V.
(16,000⇡)2 (1600⇡)2
= 0.9987/
(16,000⇡)2 (1600⇡)2 + (j1600⇡)(8000⇡)
Vo = 0.9987/
[e] H(j160,000) =
14–15
.·. vo (t) = 19.97 cos(1600⇡t
2.89 Vi
2.89 ;
2.89 ) V.
(16,000⇡)2 (160,000⇡)2
= 0.9987/2.89 ;
(16,000⇡)2 (160,000⇡)2 + (j160,000⇡)(8000⇡)
Vo = 0.9987/2.89 Vi
.·. vo (t) = 19.97 cos(160,000⇡t + 2.89 ) V.
1
1
=
= 625 ⇥ 106 ;
3
LC
(40 ⇥ 10 )(40 ⇥ 10 9 )
P 14.24 [a] !o2 =
!o = 25 ⇥ 103 rad/s = 25 krad/s;
fo =
[b] Q =
25,000
= 3978.87 Hz.
2⇡
!o L
(25 ⇥ 103 )(40 ⇥ 10 3 )
= 5.
=
R + Ri
200
2
1
[c] !c1 = !o 6
+
4
2Q
v
u
u
t
1
1+
2Q
!2
3
2
1
7
+
5 = 25,000 4
10
s
3
1 5
1+
100
= 22.625 krad/s or 3.60 kHz.
2
1
[d] !c2 = !o 6
+
4
2Q
v
u
u
t
1
1+
2Q
!2
3
2
1
7
+
5 = 25,000 4
10
s
3
1 5
1+
100
= 27.625 krad/s or 4.4 kHz.
[e]
= !c2 !c1 = 27.62 22.62 = 5 krad/s
or
25,000
!o
=
= 5 krad/s or 795.77 Hz.
=
Q
5
P 14.25 [a] H(s) =
(R/L)s
1
i)
s2 + (R+R
s + LC
L
.
For the numerical values in Problem 14.24 we have
4500s
H(s) = 2
;
s + 5000s + 625 ⇥ 106
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14–16
CHAPTER 14. Introduction to Frequency-Selective Circuits
.·. H(j!) =
H(j!o ) =
4500j!
.
! 2 ) + j5000!
(625 ⇥ 106
j4500(25 ⇥ 103 )
= 0.9/0 ;
j5000(25 ⇥ 103 )
.·. vo (t) = 0.5(0.9) cos 25,000 = 450 cos 25,000t mV.
[b] From the solution to Problem 14.24,
!c1 = 22.62 krad/s;
H(j22.62 k) =
j4500(22.62 ⇥ 103 )
= 0.6364/45 ;
(113.12 + j113.12) ⇥ 106
.·. vo (t) = 0.5(0.6364) cos(22,620t + 45 ) = 318.2 cos(22,620t + 45 ) mV.
[c] From the solution to Problem 14.24,
!c2 = 27.62 krad/s;
H(j27.62 k) =
j4500(27.62 ⇥ 103 )
= 0.6364/
( 138.12 + j138.12) ⇥ 106
.·. vo (t) = 0.5(0.6364) cos(27,620t
P 14.26 [a] L =
45 ) mV.
Q
5
=
= 5 k⌦.
3
!o C
(20 ⇥ 10 )(50 ⇥ 10 9 )
2
1
[b] !c2 = !o 6
+
4
2Q
v
u
u
t
1
1+
2Q
.·.
= 22.10 krad/s
2
1
!c1 = !o 6
+
4
2Q
v
u
u
t
= 18.10 krad/s
=
45 ) = 318.2 cos(27,620t
1
1
=
= 50 mH;
!o2 C
(50 ⇥ 10 9 )(20 ⇥ 103 )2
R=
[c]
45 ;
!2
3
2
3
s
1
1
7
5
+ 1+
5 = 20,000 4
fc2 =
1
1+
2Q
.·.
10
!2
3
!c2
= 3.52 kHz;
2⇡
2
1
7
+
5 = 20,000 4
10
fc1 =
100
!c1
= 2.88 kHz.
2⇡
s
3
1 5
1+
100
!o
20,000
=
= 4000 rad/s or 636.62 Hz.
Q
5
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Problems
14–17
P 14.27 [a] We need !c = 20,000 rad/s. There are several possible approaches – this
one starts by choosing L = 1 mH. Then,
1
C=
20,0002 (0.001)
= 2.5 µF.
Use the closest value from Appendix H, which is 2.2 µF to give
!c =
s
1
= 21,320 rad/s.
(0.001)(2.2 ⇥ 10 6 )
Then, R =
5
Q
=
= 106.6 ⌦.
!o C
(21320)(2.2 ⇥ 10 6 )
Use the closest value from Appendix H, which is 100 ⌦ to give
Q = 100(21,320)(2.2 ⇥ 10 6 ) = 4.69.
[b] % error in !c =
% error in Q =
P 14.28 [a] !o2 =
1
LC
R=
21,320 20,000
(100) = 6.6%;
20,000
4.69 5
(100) =
5
so L =
1
= 79.16 mH;
[8000(2⇡)]2 (5 ⇥ 10 9 )
!o L
8000(2⇡)(79.16 ⇥ 10 3 )
=
= 1.99 k⌦.
Q
2
2
1
+
[b] fc1 = 8000 4
4
2
1
[c] fc2 = 8000 4 +
4
[d]
6.2%.
s
s
3
1 5
1+
= 6.25 kHz.
16
3
1 5
= 10.25 kHz.
1+
16
= fc2 fc1 = 4 kHz
or
fo
8000
=
=
= 4 kHz.
Q
2
P 14.29 [a] We need !c close to 2⇡(8000) = 50,265.48 rad/s. There are several
possible approaches – this one starts by choosing L = 10 mH. Then,
C=
1
= 39.58 nF.
[2⇡(8000)]2 (0.01)
Use the closest value from Appendix H, which is 0.047 µF to give
!c =
s
1
= 46,126.56 rad/s or fc = 7341.27 Hz.
(0.01)(47 ⇥ 10 9 )
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14–18
CHAPTER 14. Introduction to Frequency-Selective Circuits
Then, R =
(46,126.56)(0.01)
!o L
=
= 230 ⌦.
Q
2
Use the closest value from Appendix H, which is 220 ⌦ to give
(46,126.56)(0.01)
= 2.1.
220
7341.27 8000
(100) =
[b] % error in fc =
8000
Q=
% error in Q =
P 14.30 [a] !o2 =
8.23%;
2.1 2
(100) = 5%.
2
1
1
=
= 1010 ;
3
LC
(10 ⇥ 10 )(10 ⇥ 10 9 )
!o = 105 rad/s = 100 krad/s.
!o
105
=
= 15.9 kHz.
2⇡
2⇡
[c] Q = !o RC = (100 ⇥ 103 )(8000)(10 ⇥ 10 9 ) = 8.
[b] fo =
2
v
u
u
t
1
6
+
[d] !c1 = !o 4
2Q
[e] .·. fc1 =
1
1+
2Q
!2
!c1
= 14.95 kHz.
2⇡
2
6 1
+
[f ] !c2 = !o 4
2Q
v
u
u
t
1
1+
2Q
!2
3
2
1
7
5
+
5 = 10 4
16
3
3
s
1 5
1+
= 93.95 krad/s.
256
2
3
s
1
1
7
5
5 = 106.45 krad/s.
+ 1+
5 = 10 4
16
256
!c2
= 16.94 kHz.
[g] .·. fc2 =
2⇡
105
!o
=
= 12.5 krad/s or 1.99 kHz.
[h] =
Q
8
P 14.31 [a] !o2 =
1
1
=
= 1012 ;
3
LC
(5 ⇥ 10 )(200 ⇥ 10 12 )
!o = 1 Mrad/s.
[b]
R + RL 1
=
=
·
RL
RC
500 ⇥ 103
400 ⇥ 103
106
= 16.
62.5 ⇥ 103
RL
= 0.8/0 ;
[d] H(j!o ) =
R + RL
[c] Q =
!o
!
1
3
(100 ⇥ 10 )(200 ⇥ 10 12 )
!
= 62.5 krad/s.
=
.·. vo (t) = 0.25(0.8) cos(106 t) = 200 cos 106 t mV.
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Problems
[e]
✓
R
= 1+
RL
◆
✓
14–19
◆
100
1
= 1+
(50 ⇥ 103 ) rad/s
RC
RL
!o = 106 rad/s;
Q=
!o
20
1 + (100/RL )
=
where RL is in kilohms.
[f ]
P 14.32 [a]
[b] L =
1
!o2 C
R=
=
1
(50 ⇥ 103 )2 (20 ⇥ 10 4 )
= 20 mH;
!o L
(50 ⇥ 103 )(20 ⇥ 10 3 )
=
= 160 ⌦.
Q
6.25
[c] Re = 160k480 = 120 ⌦;
Re + Ri = 120 + 80 = 200 ⌦;
Qsystem =
[d]
(50 ⇥ 103 )(20 ⇥ 10 3 )
!o L
= 5.
=
R e + Ri
200
50 ⇥ 103
= 10 krad/s;
=
system =
Qsystem
5
!o
system (Hz) =
10,000
= 1591.55 Hz.
2⇡
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14–20
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.33 [a]
Vo
1
Z
where Z =
=
Vi
Z +R
Y
1
1
LCRL s2 + sL + RL
and Y = sC +
+
.
=
sL RL
RL Ls
H(s) =
=
RL
RL Ls
2
RLCs + (R + R
s2 +
⇣
h⇣
(1/RC)s
R+RL
RL
⌘⇣
⌘⇣
1
RC
⌘i
⌘⇣
L )Ls + RRL
1
s + LC
⌘
R+RL
RL
1
s
R+RL
RL
RC
h⇣
⌘⇣
⌘i
=
1
1
L
s2 + R+R
s + LC
RL
RC
=
[b]
[c]
K s
,
s + !o2
s2 +
✓
◆
R + RL
RL
1
;
U =
RC
=
.·.
[d] Q =
L =
!o
✓
K=
RL
,
R + RL
✓
R
RL
=
1
.
(RkRL )C
1
.
RC
R + RL
RL
◆
U = 1+
◆
U.
!o RC
= ⇣ R+R ⌘ .
L
RL
[e] QU = !o RC;
.·. QL =
[f ] H(j!) =
✓
!o2
◆
1
RL
QU .
QU =
R + RL
[1 + (R/RL )]
Kj!
;
! 2 + j!
H(j!o ) = K.
Let !c represent a corner frequency. Then
K!c
K
;
|H(j!c )| = p = q
2
(!o2 !c2 )2 + !c2 2
1
.·. p = q
2
(!o2
!c
!c2 )2 + !c2 2
.
Squaring both sides leads to
(!o2
!c2 )2 = !c2 2 or (!o2
!c2 ) = ±!c ;
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Problems
.·. !c2 ± !c
or
!c = ⌥
±
2
14–21
!o2 = 0
s
2
+ !o2 .
4
The two positive roots are
!c1 =
2
+
s
4
◆
1
1
and !o2 =
.
RC
LC
where
✓
= 1+
P 14.34 !o2 =
R
RL
2
+ !o2
and !c2 =
2
+
s
2
4
+ !o2
1
1
=
= 1016 ;
6
12
LC
(2 ⇥ 10 )(50 ⇥ 10 )
!o = 100 Mrad/s;
QU = !o RC = (100 ⇥ 106 )(2.4 ⇥ 103 )(50 ⇥ 10 12 ) = 12;
.·.
✓
◆
RL
12 = 7.5;
R + RL
P 14.35 H(s) =
.·. RL =
7.5
R = 4 k⌦.
4.5
1
s2 + LC
1 .
s2 + R
s + LC
L
[a] !o =
s
[b] fo =
1
=
LC
s
1
= 1000 rad/s.
(0.05)(20 ⇥ 10 6 )
!o
= 159.155 Hz.
2⇡
s
s
L
0.05
[c] Q =
=
= 2.
2
2
R C
(25) (20 ⇥ 10 6 )
R
25
[d] =
=
= 500 rad/s.
L
0.05
(Hertz) =
[e] !c1 =
2
=
+
500
= 79.5775 Hz.
2⇡
v
u
u
t
500
+
2
2
!2
s
✓
+ !o2
500
2
◆2
+ 10002 = 780.7764 rad/s.
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14–22
CHAPTER 14. Introduction to Frequency-Selective Circuits
[f ] fc1 =
[g] !c2 =
780.7764
= 124.264 Hz.
2⇡
2
+
v
u
u
t
2
500
+
=
2
[h] fc2 =
!2
s
✓
+ !o2
500
2
◆2
+ 10002 = 1280.7764 rad/s.
1280.7764
= 203.842 Hz.
2⇡
P 14.36 [a] H(j!) =
1
LC
1
LC
!2
!2 + j
R
!
L
=
10002 ! 2
.
10002 ! 2 + j500!
!o = 1000 rad/s :
H(j!o ) =
10002 10002
= 0.
10002 10002 + j500(1000)
!c1 = 780.7764 rad/s :
H(j!c1 ) =
10002
10002 780.77642
= 0.7071/
780.77642 + j500(780.7764)
45 .
!c2 = 1280.7764 rad/s :
H(j!c2 ) =
10002
10002 1280.77642
= 0.7071/45 .,
1280.77642 + j500(1280.7764)
0.125!o = 125 rad/s :
H(j0.125!o ) =
10002 1252
= 0.998/
10002 1252 + j500(125)
3.633 .
8!o = 8000 rad/s :
10002 80002
H(j8!o ) =
= 0.998/3.633 .
10002 80002 + j500(8000)
[b] ! = !o = 1000 rad/s :
Vo = H(j1000)Vi = 0Vi ;
vo (t) = 0.
! = !c1 = 780.7764 rad/s :
Vo = H(j780.7764)Vi = (0.7071/
vo (t) = 3.54 cos(780.7764t
45 )(5) = 3.54/
45 ;
45 ) V.
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Problems
14–23
! = !c2 = 1280.7764 rad/s :
Vo = H(j1280.7764)Vi = (0.7071/45 )(5) = 3.54/45 ;
vo (t) = 3.54 cos(1280.7764t + 45 ) V.
! = 0.125!o = 125 rad/s :
Vo = H(j125)Vi = (0.998/
vo (t) = 4.99 cos(125t
3.633 )(5) = 4.99/
3.633 ;
3.633 ) V.
! = 8!o = 8000 rad/s :
Vo = H(j8000)Vi = (0.998/3.633 )(5) = 4.99/3.633 ;
vo (t) = 4.99 cos(8000t + 3.3633 ) V.
P 14.37 [a] !o =
Q=
=
q
1/LC
!o
R
L
so
so L =
=
1
!o2 C
=
1
(20,000)2 (50 ⇥ 10 9 )
= 50 mH.
!o
20,000
=
= 4000 rad/s.
Q
5
so R = L = (50 ⇥ 10 3 )(4000) = 200 ⌦.
[b] From part (a),
!c1,2 = ±
2
+
= 4000 rad/s.
v
!
u
u beta 2
t
2
4000
+
+ !o2 = ±
2
s
✓
4000
2
◆2
+ 20,0002
= ±2000 + 20,099.75;
!c1 = 18,099.75 rad/s;
!o2 ! 2
P 14.38 H(j!) = 2
!o ! 2 + j!
!c2 = 22,099.75 rad/s.
20,0002 ! 2
=
.
20,0002 ! 2 + j!(4000)
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14–24
CHAPTER 14. Introduction to Frequency-Selective Circuits
[a] H(j20,000) =
20,0002
.·. vo (t) = 0 V.
Vo = (0)Vi
[b] H(j18,099.75) =
1
Vo = p /
2
20,0002
.·. vo (t) = 35.4 cos(22,099.75t + 45 ) V.
20,0002
j!
! 2 + j!
[a] H(j20,000) =
Vo = (1)Vi
1.157 ) V.
.·. vo (t) = 49.99 cos(200,000t + 1.157 ) V.
=
j!(4000)
.
20,0002 ! 2 + j!(4000)
j(20,000)(4000)
= 1;
20,0002 + j(20,000)(4000)
20,0002
.·. vo (t) = 50 cos 20,000t V.
[b] H(j18,099.75) =
20,0002
1
Vo = p /45 Vi
2
[c] H(j22,099.75) =
1
Vo = p /
2
1.157 ;
20,0002 200,0002
= 0.9998/1.157 ;
200,0002 + j(200,000)(4000)
Vo = 0.9998/1.157 Vi
!o2
.·. vo (t) = 49.99 cos(2000t
1.157 Vi
[e] H(j200,000) =
P 14.39 H(j!) =
45 ) V.
20,0002 20002
= 0.9998/
20,0002 20002 + j(2000)(4000)
Vo = 0.9998/
1
j(18,099.75)(4000)
= p /45 ;
2
18,099.75 + j(18,099.75)(4000)
2
.·. vo (t) = 35.4 cos(18,099.75t + 45 ) V.
20,0002
j(22,099.75)(4000)
1
=p /
2
22,099.75 + j(22,099.75)(4000)
2
45 Vi
.·. vo (t) = 35.4 cos(22,099.75t
20,0002
j(2000)(4000)
= 0.0202/88.84 ;
20002 + j(2000)(4000)
Vo = 0.0202/88.84 Vi
.·. vo (t) = 1.01 cos(2000t + 88.84 ) V.
[d] H(j2000) =
45 ;
1
20,0002 22,099.752
= p /45 ;
2
22,099.75 + j(22,099.75)(4000)
2
20,0002
1
Vo = p /45 Vi
2
20,0002 18,099.752
1
=p /
2
18,099.75 + j(18,099.75)(4000)
2
.·. vo (t) = 35.4 cos(18,099.75t
45 Vi
[c] H(j22,099.75) =
[d] H(j2000) =
20,0002 20,0002
= 0;
20,0002 + j(20,000)(4000)
45 ;
45 ) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
[e] H(j200,000) =
20,0002
Vo = 0.0202/
Problems
14–25
j(200,000)(4000)
= 0.0202/
200,0002 + j(200,000)(4000)
88.84 ;
.·. vo (t) = 1.01 cos(200,000t
88.84 Vi
88.84 ) V.
P 14.40 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage
and we seek the steady-state nature of the output voltage vo .
At zero frequency the inductor provides a direct connection between the
input and the output, hence vo = vi when ! = 0.
At infinite frequency the capacitor provides the direct connection, hence
vo = vi when ! = 1.
At the resonant frequency of the parallel combination of L and C the
impedance of the combination is infinite and hence the output voltage
will be zero when ! = !o .
At frequencies on either side of !o the amplitude of the output voltage
will be nonzero but less than the amplitude of the input voltage.
Thus the circuit behaves like a band-reject filter.
[b] Let Z represent the impedance of the parallel branches L and C, thus
Z=
sL(1/sC)
sL
= 2
.
sL + 1/sC
s LC + 1
Then
H(s) =
=
H(s) =
R(s2 LC + 1)
R
Vo
=
=
Vi
Z +R
sL + R(s2 LC + 1)
[s2 + (1/LC)]
s2 +
⇣
1
RC
⌘
s+
s2 + !o2
.
s2 + s + !o2
⇣
1
LC
⌘;
[c] From part (b) we have
H(j!) =
!o2 ! 2
.
!o2 ! 2 + j!
It follows that H(j!) = 0 when ! = !o .
.·. !o = p
1
.
LC
!o2
[d] |H(j!)| = q
(!o2
!2
! 2 )2 + ! 2 2
;
1
|H(j!)| = p when ! 2 2 = (!o2
2
or ± ! = !o2
! 2 )2
! 2 , thus
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14–26
CHAPTER 14. Introduction to Frequency-Selective Circuits
!2 ± !
!o2 = 0.
The two positive roots of this quadratic are
!c1 =
!c2 =
+
2
2
+
v
u
u
t
v
u
u
t
2
2
!2
!2
+ !o2 ;
+ !o2 .
Also note that since
2
!c1 = !o 6
4
1
+
2Q
!c2 = !o 6
4
1
+
2Q
2
= !o /Q
v
u
u
t
1+
v
u
u
t
1+
1
2Q
!2
1
2Q
!2
3
7
5;
3
7
5.
[e] It follows from the equations derived in part (d) that
= !c2
!c1 = 1/RC.
[f ] By definition Q = !o / = !o RC =
P 14.41 [a] !o2 =
q
R2 C/L.
1
1
=
= 1012 ;
6
LC
(50 ⇥ 10 )(20 ⇥ 10 9 )
.·. !o = 1 Mrad/s.
!o
= 159.15 kHz.
2⇡
[c] Q = !o RC = (106 )(750)(20 ⇥ 10 9 ) = 15.
[b] fo =
2
v
u
u
t
1
[d] !c1 = !o 6
+
4
2Q
1
1+
2Q
!2
= 967.22 krad/s.
[e] fc1 =
!c1
= 153.94 kHz.
2⇡
2
1
[f ] !c2 = !o 6
+
4
2Q
v
u
u
t
1
1+
2Q
= 1.03 Mrad/s.
[g] fc2 =
!2
3
3
2
1
7
6
+
5 = 10 4
30
2
1
7
6
+
5 = 10 4
30
s
3
s
1 5
1+
900
3
1 5
1+
900
!c1
= 164.55 kHz.
2⇡
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[h]
= fc2
14–27
fc1 = 10.61 kHz.
P 14.42 [a] !o = 2⇡fo = 8⇡ krad/s;
L=
1
1
=
= 3.17 mH;
!o2 C
(8000⇡)2 (0.5 ⇥ 10 6 )
R=
Q
5
=
= 397.89 ⌦.
!o C
(8000⇡)(0.5 ⇥ 10 6 )
2
1
[b] fc2 = fo 6
+
4
2Q
v
u
u
t
1
1+
2Q
!2
= 4.42 kHz;
2
1
+
2Q
6
fc1 = fo 4
v
u
u
t
3
2
3
s
1
1
7
5
+ 1+
5 = 4000 4
1
2Q
1+
!2
= 3.62 kHz.
[c]
10
3
2
7
5 = 4000 4
100
1
+
10
s
1+
3
1 5
100
= fc2 fc1 = 800 Hz
or
4000
fo
=
= 800 Hz.
=
Q
5
P 14.43 [a] Re = 397.89k1000 = 284.63 ⌦
Q = !o Re C = (8000⇡)(284.63)(0.5 ⇥ 10 6 ) = 3.58
[b]
=
fo
4000
=
= 1.12 kHz
Q
3.58
2
[c] fc2 = 4000 4
1
+
7.16
2
s
1
[d] fc1 = 4000 4
+
7.16
1+
s
3
1 5
= 4.60 kHz
7.162
3
1 5
= 3.48 kHz
1+
7.162
P 14.44 [a] We need !c = 2⇡(4000) = 25,132.74 rad/s. There are several possible
approaches – this one starts by choosing L = 100 µH. Then,
C=
1
[2⇡(4000)]2 (100 ⇥ 10 6 )
= 15.83 µF.
Use the closest value from Appendix H, which is 22 µF, to give
!c =
s
1
= 21,320.07 rad/s so fc = 3393.19 Hz.
100 ⇥ 10 6 )(22 ⇥ 10 6 )
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14–28
CHAPTER 14. Introduction to Frequency-Selective Circuits
Then, R =
Q
5
=
= 10.66 ⌦.
!o C
(21320.07)(22 ⇥ 10 6 )
Use the closest value from Appendix H, which is 10 ⌦, to give
Q = 10(21,320.07)(22 ⇥ 10 6 ) = 4.69.
[b] % error in fc =
% error in Q =
P 14.45 [a] Let Z =
Z=
3393.19 4000
(100) =
4000
4.69 5
(100) =
5
15.2%;
6.2%.
RL (sL + (1/sC))
;
RL + sL + (1/sC)
RL (s2 LC + 1)
.
s2 LC + RL Cs + 1
Then H(s) =
s2 RL CL + RL
Vo
=
.
Vi
(R + RL )LCs2 + RRL Cs + R + RL
Therefore
H(s) =
=
✓
◆
[s2 + (1/LC)]
RL
⇣
⌘
i
·h
RRL
s
1
R + RL
s2 + R+R
+
L
LC
L
K(s2 + !o2 )
,
s2 + s + !o2
RL
;
where K =
R + RL
!o2 =
1
;
LC
✓
RRL
=
R + RL
◆
1
.
L
1
.
LC
✓
◆
RRL
1
.
[c] =
R + RL L
!o
!o L
[d] Q =
=
.
[RRL /(R + RL )]
K(!o2 ! 2 )
[e] H(j!) = 2
(!o ! 2 ) + j !
[b] !o = p
H(j!o ) = 0.
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Problems
14–29
K!o2
RL
=K=
.
2
!o
R + RL
[f ] H(j0) =
h
i
K (!o /!)2
[g] H(j!) = nh
(!o /!)2
1
i
1 + j /!
o;
RL
K
=K=
.
1
R + RL
H(j1) =
K(!o2 ! 2 )
;
(!o2 ! 2 ) + j !
Let !c represent a corner frequency. Then
[h] H(j!) =
K
|H(j!c )| = p ;
2
K
K(!o2 !c2 )
.·. p = q
.
2
(!o2 !c2 )2 + !c2 2
Squaring both sides leads to
(!o2
!c2 )2 = !c2 2 or (!o2
.·. !c2 ± !c
or
!c = ⌥
2
±
!c2 ) = ±!c ;
!o2 = 0
s
2
+ !o2 .
4
The two positive roots are
!c1 =
2
+
s
2
4
+ !o2
and !c2 =
2
+
s
2
4
+ !o2 ,
where
=
P 14.46 [a] !o2 =
RRL
1
1
.
· and !o2 =
R + RL L
LC
1
1
=
= 0.25 ⇥ 1018 = 25 ⇥ 1016 ;
6
LC
(10 )(4 ⇥ 10 12 )
!o = 5 ⇥ 108 = 500 Mrad/s;
=
Q=
(30)(150)
1
1
RRL
·
· =
= 25 Mrad/s = 3.98 MHz;
R + RL L
180
10 6
!o
=
500 M
= 20.
25 M
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14–30
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] H(j0) =
RL
150
= 0.8333;
=
R + RL
180
H(j1) =
2
RL
= 0.8333.
R + RL
250 4 1
[c] fc2 =
+
⇡ 40
2
3
s
1 5
1+
= 81.59 MHz;
1600
250 4 1
fc1 =
+
⇡
40
3
s
Check:
= fc2
!o
500 ⇥ 106
[d] Q =
= RRL 1
·
R+RL L
1 5
1+
= 77.61 MHz.
1600
fc1 = 3.98 MHz.
✓
50
500(R + RL )
30
=
=
1+
RRL
3
RL
where RL is in ohms.
◆
[e]
1
= 625 ⇥ 106 ;
LC
1
.·. L =
= 64 mH.
6
(625 ⇥ 10 )(25 ⇥ 10 9 )
P 14.47 [a] !o2 =
[b]
RL
= 0.9;
R + RL
.·. 0.1RL = 0.9R;
.·.
.·. R =
=
RL = 9R
✓
Q=
◆
500
= 55.6 ⌦.
9
1
RL
R · = 781.25 rad/s;
R + RL
L
!o
=
25,000
= 32.
781.25
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Problems
14–31
10 ⇥ 1010
P 14.48 [a] |H(j!)| = q
= 1;
(10 ⇥ 1010 ! 2 )2 + (50,000!)2
.·.
0=
100 ⇥ 1020 + (10 ⇥ 1010
! 2 )2 + (50,000!)2
20 ⇥ 1010 ! 2 + ! 4 + 25 ⇥ 108 ! 2 .
=
From the above equation it is obvious that ! = 0 is one solution. The
other is as follows:
! 2 = 1975 ⇥ 108
so
! = 444.41 krad/s.
[b] From the equation for |H(j!)| in part (a), the frequency for which the
magnitude is maximum is the frequency for which the denominator is
minimum. This occurs when 10 ⇥ 1010 ! 2 = 0:
p
p
! 2 = 10 ⇥ 1010
so
! = 10 ⇥ 1010 = 10 ⇥ 105 rad/s.
p
[c] |H(j 10 ⇥ 105 )| =
P 14.49 [a] H(s) =
10 ⇥ 1010
p
= 6.3
50,000( 10 ⇥ 105 )
sL
s2
s2 LC
=
=
1
1 .
RsC + s2 LC + 1
R + sL + sC
s2 + R
s + LC
L
[b] When s = j! is very small (think of ! approaching 0),
H(s) ⇡
s2
1
LC
= 0.
[c] When s = j! is very large (think of ! approaching 1),
s2
= 1.
s2
[d] The magnitude of H(s) approaches 0 as the frequency approaches 0, and
approaches 1 as the frequency approaches 1. Therefore, this circuit is
behaving like a high pass filter when the output is the voltage across the
inductor.
!c2
1
[e] |H(j!c )| = q
=p ;
2
(106 !c2 )2 + (1500!c )2
H(s) ⇡
2!c4 = (106
!c2 )2 + (1500!c )2 = 1012
Simplifying,
!c4
25 ⇥ 104 !c2
2 ⇥ 106 !c2 + !c4 + 225 ⇥ 104 !c2 .
1012 = 0.
Solve for !c2 and then !c :
!c2 = 1,132,782.22
so
!c = 1064.322 rad/s.
f = 169.4 Hz
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14–32
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.50 [a] H(s) =
1
sC
=
1
R + sL + sC
1
1
LC
=
1 .
RsC + s2 LC + 1
s + LC
s2 + R
L
[b] When s = j! is very small (think of ! approaching 0),
H(s) ⇡
1
LC
1
LC
= 1.
[c] When s = j! is very large (think of ! approaching 1),
H(s) ⇡
1
LC
s2
= 0.
[d] The magnitude of H(s) approaches 1 as the frequency approaches 0, and
approaches 0 as the frequency approaches 1. Therefore, this circuit is
behaving like a low pass filter when the output is the voltage across the
capacitor.
1
106
[e] |H(j!c )| = q
=p ;
2
(106 !c2 )2 + (1500!c )2
2 ⇥ 1012 = (106
!c2 )2 + (1500!c )2 = 1012
!c4 + (15002
Simplifying,
2 ⇥ 106 )!c2
2 ⇥ 106 !c2 + !c4 + 225 ⇥ 104 !c2 .
1012 = 0
Solve for !c2 and then !c :
!c2 = 882,782.22
so
!c = 939.565 rad/s
f = 149.54 Hz
P 14.51 [a] Use the cuto↵ frequencies to calculate the bandwidth:
!c1 = 2⇡(697) = 4379.38 rad/s
Thus
= !c2
!c2 = 2⇡(941) = 5912.48 rad/s;
!c1 = 1533.10 rad/s.
Calculate inductance using Eq. 14.16 and capacitance using Eq. 14.13:
L=
C=
R
=
600
= 0.39 H;
1533.10
1
L!c1 !c2
=
1
= 0.10 µF.
(0.39)(4379.38)(5912.48)
[b] At the outermost two frequencies in the low-frequency group (687 Hz and
941 Hz) the amplitudes are
|V697Hz | = |V941Hz | =
|Vpeak |
p
= 0.707|Vpeak |
2
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Problems
14–33
because these are cuto↵ frequencies. We calculate the amplitudes at the
other two low frequencies:
|V | = (|Vpeak |)(|H(j!)|) = |Vpeak | q
(!o2
!
! 2 )2 + (! )2
.
Therefore
(4838.05)(1533.10)
|V770Hz | = |Vpeak | = q
(5088.522
4838.052 )2 + [(4838.05)(1533.10)]2
= 0.948|Vpeak |,
and
(5353.27)(1533.10)
|V852Hz | = |Vpeak | = q
(5088.522
5353.272 )2 + [(5353.27)(1533.10)]2
= 0.948|Vpeak |.
It is not a coincidence that these two magnitudes are the same. The
frequencies in both bands of the DTMF system were carefully chosen to
produce this type of predictable behavior with linear filters. In other
words, the frequencies were chosen to be equally far apart with respect to
the response produced by a linear filter. Most musical scales consist of
tones designed with this dame property – note intervals are selected to
place the notes equally far apart. That is why the DTMF tones remind
us of musical notes! Unlike musical scales, DTMF frequencies were
selected to be harmonically unrelated, to lower the risk of misidentifying
a tone’s frequency if the circuit elements are not perfectly linear.
[c] The high-band frequency closest to the low-frequency band is 1209 Hz.
The amplitude of a tone with this frequency is
(7596.37)(1533.10)
|V1209Hz | = |Vpeak | = q
(5088.522
7596.372 )2 + [(7596.37)(1533.10)]2
= 0.344|Vpeak |.
This is less than one half the amplitude of the signals with the low-band
cuto↵ frequencies, ensuring adequate separation of the bands.
P 14.52 The cuto↵ frequencies and bandwidth are
!c1 = 2⇡(1209) = 7596 rad/;s
!c2 = 2⇡(1633) = 10.26 krad/s;
= ! c2
!c1 = 2664 rad/s.
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14–34
CHAPTER 14. Introduction to Frequency-Selective Circuits
Telephone circuits always have R = 600 ⌦. Therefore, the filters inductance
and capacitance values are
R
L=
C=
=
600
= 0.225 H;
2664
1
= 0.057 µF.
!c1 !c2 L
At the highest of the low-band frequencies, 941 Hz, the amplitude is
|V! | = |Vpeak | q
where
!
(!o2
! 2 )2 + ! 2 2
p
!c1 !c2 . Thus,
!o =
|V! | = q
[(8828)2
|Vpeak |(5912)(2664)
(5912)2 ]2 + [(5912)(2664)]2
= 0.344 |Vpeak |.
Again it is not coincidental that this result is the same as the response of the
low-band filter to the lowest of the high-band frequencies.
P 14.53 From Problem 14.51 the response to the largest of the DTMF low-band tones
is 0.948|Vpeak |. The response to the 20 Hz tone is
|V20Hz | =
[(50892
|Vpeak |(125.6)(1533)
125.62 )2 + [(125.6)(1533)]2 ]1/2
= 0.00744|Vpeak |;
.·.
|V20Hz |
0.00744|Vpeak |
|V20Hz |
=
=
= 0.5.
|V770Hz |
|V852Hz |
0.948|Vpeak |
.·. |V20Hz | = 63.7|V770Hz |.
Thus, the 20Hz signal can be 63.7 times as large as the DTMF tones.
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Active Filter Circuits
15
Assessment Problems
AP 15.1
H(s) =
(R2 /R1 )s
;
s + (1/R1 C)
1
= 1 rad/s;
R1 C
R2
= 1,
R1
.·.
R1 = 1 ⌦,
.·. R2 = R1 = 1 ⌦;
Hprototype (s) =
AP 15.2
H(s) =
.·. C = 1 F.
s
.
s+1
20,000
(1/R1 C)
=
.
s + (1/R2 C)
s + 5000
1
= 20,000;
R1 C
.·. R1 =
C = 5 µF;
1
= 10 ⌦.
(20,000)(5 ⇥ 10 6 )
1
= 5000;
R2 C
.·. R2 =
1
= 40 ⌦.
(5000)(5 ⇥ 10 6 )
15–1
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15–2
CHAPTER 15. Active Filter Circuits
AP 15.3
!c = 2⇡fc = 2⇡ ⇥ 104 = 20,000⇡ rad/s;
.·. kf = 20,000⇡ = 62,831.85.
C0 =
C
kf km
.·. km =
.·.
0.5 ⇥ 10 6 =
1
;
kf km
1
= 31.83.
(0.5 ⇥ 10 6 )(62,831.85)
AP 15.4 For a 2nd order Butterworth high pass filter
H(s) =
s2
p
.
s2 + 2s + 1
For the circuit in Fig. 15.25
H(s) =
s2 +
⇣
2
R2 C
⌘
s2
s+
⇣
1
R1 R2 C 2
⌘.
Equate the transfer functions. For C = 1F,
p
2
= 2,
R2 C
.·. R2 =
1
= 1,
R1 R2 C 2
p
2 = 1.414 ⌦;
1
.·. R1 = p = 0.707 ⌦.
2
AP 15.5
Q = 8, K = 5, !o = 1000 rad/s, C = 1 µF.
For the circuit in Fig 15.26
✓
◆
R3 =
2
;
C
1
s
R1 C
!
H(s) =
✓
◆
R
+
R
2
2
1
s2 +
s+
R3 C
R1 R2 R3 C 2
K s
= 2
.
s + s + !o2
=
2
,
R3 C
=
!o
1000
=
= 125 rad/s;
Q
8
.·.
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Problems
.·. R3 =
K =
15–3
2 ⇥ 106
= 16 k⌦.
(125)(1)
1
;
R1 C
.·. R1 =
1
K C
=
1
= 1.6 k⌦.
5(125)(1 ⇥ 10 6 )
!o2 =
R 1 + R2
;
R1 R2 R3 C 2
106 =
(1600 + R2 )
.
(1600)(R2 )(16,000)(10 6 )2
Solving for R2 ,
R2 =
(1600 + R2 )106
,
256 ⇥ 105
246R2 = 16,000,
R2 = 65.04 ⌦.
AP 15.6
!o = 1000 rad/s;
Q = 4;
C = 2 µF
s2 + (1/R2 C 2 )
#
H(s) =
✓
◆
1
4(1
)
2
s+
s +
RC
R2 C 2
2
2
s + !o
1
;
= 2
;
!o =
2
s + s + !o
RC
"
R=
=
=
4(1
)
;
RC
1
1
=
= 500 ⌦;
!o C
(1000)(2 ⇥ 10 6 )
!o
1000
=
= 250;
Q
4
.·.
4(1
)
= 250.
RC
4(1
) = 250RC = 250(500)(2 ⇥ 10 6 ) = 0.25;
1
=
0.25
= 0.0625;
4
.·.
= 0.9375.
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15–4
CHAPTER 15. Active Filter Circuits
Problems
P 15.1
[a] K = 10(5/20) = 1.778 =
R2
;
R1
R2 =
1
1
=
= 318.31 ⌦;
!c C
(2⇡)(104 )(50 ⇥ 10 9 )
R1 =
318.31
R2
=
= 179 ⌦.
K
1.778
[b]
P 15.2
[a]
1
= 2⇡(10,000) so Rf C = 1.5915 ⇥ 10 5 .
Rf C
There are several possible approaches. Here, choose C = 47 nF. Then
1.5915 ⇥ 10 5
= 338.63 ⌦.
Rf =
47 ⇥ 10 9
Choose Rf = 330 ⌦. This gives
!c =
1
= 64,474.5 rad/s so fc = 10,261.44 Hz.
(330)(47 ⇥ 10 9 )
To get a passband gain of 5 dB,
330
Rf
=
= 185.6 ⌦.
1.778
1.778
Choose Ri = 180 ⌦ to give K = 20 log10 (330/180) = 5.265 dB.
Ri =
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Problems
15–5
The resulting circuit is
[b] % error in fc =
10,261.44 10,000
(100) = 2.6%;
10,000
% error in passband gain =
P 15.3
[a] !c =
1
R2 C
so R2 =
R2
R1
so R1 =
K=
5.265
5
5
(100) = 5.3%.
1
1
=
= 63.66 ⌦;
!c C
2⇡(500)(5 ⇥ 10 6 )
63.66
R2
=
= 7.96 ⌦.
K
8
[b] From part (a), R2 a↵ects the cuto↵ frequency and the passband gain, so
both are changed.
P 15.4
[a] 8(2) = 16 V so Vcc 16 V.
8(2⇡)(500)
[b] H(j!) =
.
j! + 2⇡(500)
H(j1000⇡) =
8(1000⇡)
=
1000⇡ + j1000⇡
8
Vo = p /135 Vi
2
8
4 + j4 = p /135 ;
2
so vo (t) = 11.3 cos(1000⇡t + 135 ) V.
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15–6
CHAPTER 15. Active Filter Circuits
8(1000⇡)
= 7.96/174.3 ;
1000⇡ + j100⇡
[c] H(j100⇡) =
Vo = 7.96/174.3 Vi
8(1000⇡)
= 0.796/95.7 ;
1000⇡ + j10,000⇡
[d] H(j10,000⇡) =
Vo = 0.796/95.7 Vi
P 15.5
so vo (t) = 1.59 cos(10,000⇡t + 95.7 ) V.
Summing the currents at the inverting input node yields
0
Vi
Zi
+
0
Vo
Zf
= 0;
Vo
=
Zf
Vi
;
Zi
.·. H(s) =
Vo
=
Vi
.·.
P 15.6
so vo (t) = 15.92 cos(100⇡t + 174.3 ) V.
[a] Zf =
Zf
.
Zi
R2
R2 (1/sC2 )
=
[R2 + (1/sC2 )]
R2 C2 s + 1
(1/C2 )
.
s + (1/R2 C2 )
Likewise
(1/C1 )
.
Zi =
s + (1/R1 C1 )
=
.·. H(s) =
(1/C2 )[s + (1/R1 C1 )]
[s + (1/R2 C2 )](1/C1 )
=
C1 [s + (1/R1 C1 )]
.
C2 [s + (1/R2 C2 )]
"
#
C1 j! + (1/R1 C1 )
[b] H(j!) =
;
C2 j! + (1/R2 C2 )
C1
H(j0) =
C2
[c] H(j1) =
C1
C2
✓
R2 C2
R1 C1
j
j
!
=
◆
=
R2
.
R1
C1
.
C2
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Problems
15–7
[d] As ! ! 0 the two capacitor branches become open and the circuit reduces
to a resistive inverting amplifier having a gain of R2 /R1 .
As ! ! 1 the two capacitor branches approach a short circuit and in
this case we encounter an indeterminate situation; namely vn ! vi but
vn = 0 because of the ideal op amp. At the same time the gain of the
ideal op amp is infinite so we have the indeterminate form 0 · 1.
Although ! = 1 is indeterminate we can reason that for finite large
values of ! H(j!) will approach C1 /C2 in value. In other words, the
circuit approaches a purely capacitive inverting amplifier with a gain of
( 1/j!C2 )/(1/j!C1 ) or C1 /C2 .
P 15.7
[a] Zf =
(1/C2 )
;
s + (1/R2 C2 )
Zi = R 1 +
1
R1
[s + (1/R1 C1 )];
=
sC1
s
H(s) =
s
(1/C2 )
·
[s + (1/R2 C2 )] R1 [s + (1/R1 C1 )]
=
[b] H(j!) =
1
s
.
R1 C2 [s + (1/R1 C1 )][s + (1/R2 C2 )]
1
⇣
R1 C2 j! +
j!
1
R1 C1
H(j0) = 0.
⌘⇣
j! + R21C2
⌘;
[c] H(j1) = 0.
[d] As ! ! 0 the capacitor C1 disconnects vi from the circuit. Therefore
vo = vn = 0.
As ! ! 1 the capacitor short circuits the feedback network, thus
ZF = 0 and therefore vo = 0.
P 15.8
[a] R1 =
1
1
=
= 1273.24 ⌦;
!c C
(2⇡)(2500)(50 ⇥ 10 9 )
K = 10(10/20) = 3.16 =
R2
;
R1
.·. R2 = 3.16R1 = 4026.337 ⌦.
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15–8
CHAPTER 15. Active Filter Circuits
[b]
P 15.9
[a]
1
= 2⇡(2500) so R1 C = 6.366 ⇥ 10 5 .
R1 C
There are several possible approaches. Here, choose C = 0.047 µF. Then
6.366 ⇥ 10 5
= 1354.5 ⌦.
0.047 ⇥ 10 6
Choose R1 = 1500 ⌦. This gives
R1 =
!c =
1
= 14.18 krad/s so fc = 2257.5 Hz.
(0.047 ⇥ 10 6 )(1500)
To get a passband gain of 10 dB, choose
R2 = 3.16R1 = 3.16(1500) = 4740 ⌦.
Choose R2 = 4700 ⌦ to give a passband gain of
20 log10 (4700/1500) = 9.92 dB. The resulting circuit is
[b] % error in fc =
2257.5 2500
(100) =
2500
% error in passband gain =
P 15.10 [a] !c =
9.7%;
9.92 10
(100) =
10
0.8%.
1
1
=
= 212.2 ⌦.
!c C
2⇡(10,000)(75 ⇥ 10 9 )
1
R1 C
so R1 =
R2
R1
so R2 = KR1 = (4)(212.2) = 848.8 ⌦.
K=
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Problems
15–9
[b] From part (a), R2 a↵ects the passband gain but not the cuto↵ frequency,
so only the passband gain is changed.
P 15.11 [a] 4(4) = 16 V so Vcc 16 V.
4j!
[b] H(j!) =
.
j! + 20,000⇡
H(j20,000⇡) =
4
Vo = p /
2
4(j20,000⇡)
4
=p /
20,000⇡ + j20,000⇡
2
135 Vi
[c] H(j2000⇡) =
95.7 Vi
[d] H(j200,000⇡) =
Vo = 3.98/
so vo (t) = 11.3 cos(20,000⇡t
4(j2000⇡)
= 0.398/
20,000⇡ + j2000⇡
Vo = 0.398/
135 ;
95.7 ;
so vo (t) = 1.6 cos(2000⇡t
4(j200,000⇡)
= 3.98/
20,000⇡ + j200,000⇡
174.3 Vi
135 ) V.
95.7 ) V.
174.3 ;
so vo (t) = 15.9 cos(200,000⇡t
174.3 ) V.
P 15.12 For the RC circuit
H(s) =
s
Vo
;
=
Vi
s + (1/RC)
C
;
km kf
R0 = km R;
C0 =
.·. R0 C 0 =
RC
1
= ;
kf
kf
H 0 (s) =
1
= kf .
R0 C 0
s
s
(s/kf )
=
.
=
0
0
s + (1/R C )
s + kf
(s/kf ) + 1
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15–10
CHAPTER 15. Active Filter Circuits
For the RL circuit
s
H(s) =
;
s + (R/L)
R0 = km R;
✓
km L
;
kf
L0 =
◆
R
R0
= kf
= kf ;
0
L
L
H 0 (s) =
(s/kf )
s
.
=
s + kf
(s/kf ) + 1
P 15.13 For the RC circuit
H(s) =
Vo
(1/RC)
;
=
Vi
s + (1/RC)
R0 = km R;
C0 =
.·. R0 C 0 = km R
C
;
km kf
C
1
1
= RC = .
km kf
kf
kf
1
= kf .
R0 C 0
kf
(1/R0 C 0 )
H (s) =
=
;
0
0
s + (1/R C )
s + kf
0
H 0 (s) =
1
.
(s/kf ) + 1
For the RL circuit
R0 = km R;
H(s) =
L0 =
km
L;
kf
✓
◆
R/L
s + R/L
so
km R
R
R0
= km = k f
= kf ;
0
L
L
L
kf
H 0 (s) =
kf
(R0 /L0 )
=
;
0
0
s + (R /L )
s + kf
H 0 (s) =
1
.
(s/kf ) + 1
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Problems
P 15.14 H(s) =
(R/L)s
s2 + (R/L)s + (1/LC)
=
s2 +
For the prototype circuit !o = 1 and
For the scaled circuit
15–11
s
.
s + !o2
= !o /Q = 1/Q.
(R0 /L0 )s
H (s) = 2
s + (R0 /L0 )s + (1/L0 C 0 )
0
where R0 = km R; L0 =
.·.
C
km
L; and C 0 =
;
kf
kf km
✓
◆
R0
km R
R
= km = k f
= kf .
0
L
L
L
kf
kf2
k f km
1
= kf2 ;
= km
=
0
0
LC
LC
LC
kf
Q0 =
!o0
=
0
kf !o
= Q.
kf
therefore the Q of the scaled circuit is the same as the Q of the unscaled
circuit. Also note 0 = kf .
.·.
⇣
⌘
⇣ ⌘⇣
s
kf
kf
s
Q
0
⇣ ⌘
;
H (s) =
k
s2 + Qf s + kf2
H 0 (s) = ⇣ ⌘2
s
kf
P 15.15 [a] L = 1 H;
R=
1
Q
+ Q1
⇣
s
kf
⌘
.
+1
C = 1 F;
1
1
=
= 0.1 ⌦.
Q
10
!o0
= 75,000;
!o
Thus,
[b] kf =
⌘
km =
2500
R0
=
= 25,000.
R
0.1
R0 = km R = (0.1)(25,000) = 2.5 k⌦;
L0 =
km
25,000
(1) = 333.33 mH;
L=
kf
75,000
C0 =
C
1
= 533.33 pF.
=
km kf
(25,000)(75,000)
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15–12
CHAPTER 15. Active Filter Circuits
[c]
P 15.16 [a] Since !o2 = 1/LC and !o = 1 rad/s,
C=
1
1
= F.
L
Q
[b] H(s) =
(R/L)s
s2 + (R/L)s + (1/LC)
H(s) =
(1/Q)s
s2 + (1/Q)s + 1
;
.
[c] In the prototype circuit
R = 1 ⌦;
L = 12 H;
km =
and
C=
R0
= 50,000;
R
1
= 0.0833 F;
L
kf =
!o0
= 40,000.
!o
Thus
R0 = km R = 50 k⌦;
L0 =
km
50,000
(12) = 15 H;
L=
kf
40,000
C0 =
C
0.0833
= 41.67 pF.
=
km kf
(50,000)(40,000)
[d]
[e] H 0 (s) =
H 0 (s) =
1
12
s
40,000
!2
s
40,000
1
+
12
!
!
s
+1
40,000
3333.33s
.
s2 + 3333.33s + 16 ⇥ 108
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Problems
15–13
P 15.17 [a] Using the first prototype
!o = 1 rad/s;
C = 1 F;
L = 1 H;
10,000
R0
=
= 666.67;
R
15
km =
kf =
R = 15 ⌦.
!o0
= 100,000.
!o
Thus,
R0 = km R = 10 k⌦;
L0 =
km
666.67
(1) = 6.67 mH;
L=
kf
100,000
C
1
= 15 nF.
=
km kf
(666.67)(100,000)
C0 =
Using the second prototype
!o = 1 rad/s;
L=
C = 15 F;
1
= 66.67 mH;
15
R = 1⌦
R0
= 10,000;
R
kf =
!o0
= 100,000.
!o
R0 = km R = 10 k⌦;
L0 =
10,000
km
(0.06667) = 6.67 mH;
L=
kf
100,000
km =
Thus,
C0 =
C
15
= 15 nF.
=
km kf
(10,000)(100,000)
[b]
P 15.18 [a] For the circuit in Fig. P15.18(a)
s+
1
s
s2 + 1
Vo
⇣ ⌘
=
.
=
H(s) =
1
1
2+ 1 s+1
Vi
s
+s+
Q
Q
s
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15–14
CHAPTER 15. Active Filter Circuits
For the circuit in Fig. P15.18(b)
Qs + Qs
Vo
=
Vi
1 + Qs + Qs
H(s) =
Q(s2 + 1)
;
Qs2 + s + Q
=
s2 + 1
H(s) =
[b] H 0 (s) = ⇣
=
s
104
⇣
⌘2
s
104
⌘2
+ 18
2
⇣ ⌘
1
Q
s2 +
s+1
.
+1
⇣
s
104
8
s + 10
⌘
+1
s2 + 1250s + 108
.
P 15.19 For the scaled circuit
s2 +
H 0 (s) =
L0 =
.·.
⇣
s2 +
R0
L0
km
L;
kf
⇣
⌘
1
L0 C 0
s+
C0 =
⇣
1
L0 C 0
⌘;
C
km kf ;
kf2
1
;
=
L0 C 0
LC
✓
⌘
R0 = km R;
◆
R0
R
.·.
= kf
.
0
L
L
It follows then that
2
H 0 (s) =
s +
s2 +
✓
⇣ ⌘
R
L
= ⇣ ⌘ 2
s
kf
⇣
◆
k2
kf s + LCf
s
kf
+
kf2
LC
⌘2
+
⇣ ⌘⇣
R
L
⇣
1
LC
s
kf
= H(s)|s=s/kf .
⌘
⌘
+
⇣
1
LC
⌘
P 15.20 For the circuit at the bottom of Fig. 14.31
H(s) =
s2 +
⇣
1
LC
s
+
s2 + RC
⌘
⇣
1
LC
⌘.
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Problems
15–15
It follows that
s2 + L01C 0
s2 + R0sC 0 + L01C 0
H 0 (s) =
where R0 = km R;
C0 =
.·.
L0 =
km
L;
kf
C
;
km kf
kf2
1
.
=
L0 C 0
LC
kf
1
=
0
0
RC
RC.
s2 +
H 0 (s) =
✓
⇣
kf
RC
= ⇣ ⌘2
+
s2 +
s
kf
⇣
kf2
LC
◆
⌘
s + LCf
⇣
⌘⇣
s
kf
⌘2
1
RC
= H(s)|s=s/kf .
k2
1
+ LC
s
kf
⌘
1
+ LC
P 15.21 For prototype circuit (a):
H(s) =
Vo
Q
Q
=
;
1 =
Vi
Q + s2s+1
Q + s+ 1
s
H(s) =
s2 + 1
Q(s2 + 1)
⇣ ⌘
.
=
Q(s2 + 1) + s
s2 + 1 s + 1
Q
For prototype circuit (b):
H(s) =
=
Vo
1
=
Vi
1 + (s(s/Q)
2 +1)
s2 + 1
s2 +
⇣ ⌘
1
Q
s+1
.
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15–16
CHAPTER 15. Active Filter Circuits
500
= 5;
100
P 15.22 [a] km =
km
(160 ⇥ 10 6 ) = 0.002
kf
C0 =
so
kf =
(160 ⇥ 10 6 )(5)
= 0.4.
0.002
25 ⇥ 10 9
= 12.5 nF.
(5)(0.4)
[b] Zab =
1
j!RL
1
+ Rkj!L =
+
j!C
j!C R + j!L
=
(R + j!L ! 2 RLC)( ! 2 LC j!RC)
( ! 2 LC + j!RC)( ! 2 LC j!RC)
The denominator is purely real, so set the imaginary part of the
numerator equal to 0 and solve for !:
!R2 C
.·.
! 3 L2 C + ! 3 R2 LC 2 = 0
!2 =
Thus,
(500)2
R2
=
R2 LC L2
(500)2 (0.002(12.5 ⇥ 10 9 )
(0.002)2
= 111.11 ⇥ 109 .
! = 333.33 krad/s.
[c] In the original, unscaled circuit, the frequency at which the impedance Zab
is purely real is
2
=
!us
(100)2
(100)2 (160 ⇥ 10 6 )(25 ⇥ 10 9 )
Thus,
!us = 833.33 krad/s.
(160 ⇥ 10 6 )2
= 694.44 ⇥ 109 .
333.33
!scaled
= 0.4 = kf .
=
!us
833.33
P 15.23 [a] kf =
500
= 0.025;
20,000
125 ⇥ 10 9
= 5 µF;
0.025
0.06
= 2.4 H.
L0 =
0.025
C0 =
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Problems
[b] Io =
(200
15–17
j400)k(60 + j1200)
(0.4) = 0.57/15.14 A;
200 j400
io = 0.57 cos(500t + 15.14 ) A.
The magnitude and phase angle of the output current are the same as in
the unscaled circuit.
P 15.24 From the solution to Problem 14.30, !o = 100 krad/s and
Compute the two scale factors:
kf =
2⇡(200 ⇥ 103 )
!o0
=
= 4⇡;
!o
100 ⇥ 103
km =
1 C
1 10 ⇥ 10 9
1
=
= .
0
9
kf C
4⇡ 2.5 ⇥ 10
⇡
= 12.5 krad/s.
Thus,
R 0 = km R =
8000
= 2546.45 ⌦
⇡
L0 =
km
1/⇡
(10 ⇥ 10 3 ) = 253.3 µH.
L=
kf
4⇡
Calculate the cuto↵ frequencies:
0
!c1
= kf !c1 = 4⇡(93.95 ⇥ 103 ) = 1180.6 krad/s;
0
!c2
= kf !c2 = 4⇡(106.45 ⇥ 103 ) = 1337.7 krad/s.
To check, calculate the bandwidth:
0
0
= !c2
0
!c1
= 157.1 krad/s = 4⇡ . (checks!)
P 15.25 From the solution to Problem 14.41, !o = 106 rad/s and
Calculate the scale factors:
= 2⇡(10.61) krad/s.
50 ⇥ 103
!o0
kf =
=
= 0.05;
!o
106
km =
0.05(200 ⇥ 10 6 )
k f L0
=
= 0.2.
L
50 ⇥ 10 6
Thus,
R0 = km R = (0.2)(750) = 150 ⌦
C0 =
C
20 ⇥ 10 9
= 2 µF.
=
km kf
(0.2)(0.05)
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15–18
CHAPTER 15. Active Filter Circuits
Calculate the bandwidth:
0
= (0.05)[2⇡(10.6 ⇥ 103 )] = 3330 rad/s.
= kf
To check, calculate the quality factor:
Q=
Q0 =
=
106
= 15;
2⇡(10.61 ⇥ 103 )
=
0
50 ⇥ 103
= 15. (checks)
3330
!o
!o0
P 15.26 [a] From Eq 15.1 we have
H(s) =
K!c
s + !c
where K =
R2
,
R1
.·. H 0 (s) =
K 0 !c0
s + !c0
where K 0 =
R20
R10
!c =
1
;
R2 C
!c0 =
1
R20 C 0
.
By hypothesis R10 = km R1 ;
and C 0 =
R20 = km R2 ,
C
. It follows that
kf km
K 0 = K and !c0 = kf !c , therefore
H 0 (s) =
K!c
Kkf !c
=⇣s⌘
.
s + kf !c
+ !c
kf
[b] H(s) =
K
.
s+1
K
[c] H 0 (s) = ⇣ s ⌘
kf
+1
=
Kkf
.
s + kf
P 15.27 [a] From Eq. 15.4
H(s) =
!c =
Ks
R2
where K =
and
s + !c
R1
1
;
R1 C
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Problems
.·. H 0 (s) =
and !c0 =
15–19
K 0s
R20
0
where
K
=
s + !c0
R10
1
R10 C 0
.
By hypothesis
R10 = km R1 ;
R20 = km R2 ;
C0 =
C
.
km kf
It follows that
K 0 = K and !c0 = kf !c ;
.·. H 0 (s) =
[b] H(s) =
Ks
.
s+1
K(s/kf )
[c] H 0 (s) = ⇣ s
kf
P 15.28 [a] Hhp =
K(s/kf )
Ks
= ⇣s⌘
.
s + kf !c
+
!
c
kf
s
;
s+1
0
=
.·. Hhp
⌘ =
+1
Ks
.
s + kf
1000(2⇡)
!o0
=
= 2000⇡;
!
1
kf =
s
.
s + 2000⇡
1
= 2000⇡;
RH CH
.·. RH =
1
= 1.59 k⌦.
(2000⇡)(0.1 ⇥ 10 6 )
5000(2⇡)
1
!0
;
kf = o =
= 10,000⇡;
s+1
!
1
10,000⇡
0
.
=
.·. Hlp
s + 10,000⇡
Hlp =
1
= 10,000⇡;
RL CL
.·. RL =
1
= 318.3 ⌦.
(10,000⇡)(0.1 ⇥ 10 6 )
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15–20
CHAPTER 15. Active Filter Circuits
[b] H 0 (s) =
=
10,000⇡
s
·
s + 2000⇡ s + 10,000⇡
10,000⇡s
.
(s + 2000⇡)(s + 10,000⇡)
q
p
(2000⇡)(10,000⇡) = 1000⇡ 20 rad/s;
p
(10,000⇡)(j1000⇡
20)
p
p
H 0 (j!o ) =
(2000⇡ + j1000⇡ 20)(10,000⇡ + j1000⇡ 20)
p
j10 20
p
p
= 0.8333/0 .
=
(2 + j 20)(10 + j 20)
[c] !o =
p
!c1 !c1 =
[d] G = 20 log10 (0.8333) =
1.58 dB.
[e]
P 15.29 [a] For the high-pass section:
4000(2⇡)
!o0
=
= 8000⇡;
!
1
s
H 0 (s) =
;
s + 8000⇡
kf =
.·.
1
= 8000⇡;
R1 (10 ⇥ 10 9 )
R1 = 3.98 k⌦
.·.
R2 = 3.98 k⌦.
R2 = 39.8 k⌦
.·.
R1 = 39.8 k⌦.
For the low-pass section:
kf =
400(2⇡)
!o0
=
== 800⇡;
!
1
H 0 (s) =
.·.
800⇡
;
s + 800⇡
1
= 800⇡;
R2 (10 ⇥ 10 9 )
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Problems
15–21
0 dB gain corresponds to K = 1. In the summing amplifier we are free to
choose Rf and Ri so long as Rf /Ri = 1. To keep from having many
di↵erent resistance values in the circuit we opt for Rf = Ri = 39.8 k⌦.
[b]
[c] H 0 (s) =
800⇡
s
+
s + 8000⇡ s + 800⇡
s2 + 1600⇡s + 64 ⇥ 105 ⇡ 2
.
(s + 800⇡)(s + 8000⇡)
q
p
[d] !o = (8000⇡)(800⇡) = 800⇡ 10;
p
p
p
(800⇡ 10)2 + 1600⇡(j800⇡ 10) + 64 ⇥ 105 ⇡ 2
0
p
p
H (j800⇡ 10) =
(800⇡ + j800⇡ 10)(8000⇡ + j800⇡ 10)
p
j128 ⇥ 104 10⇡ 2
p
p
=
(800⇡)2 (1 + j 10)(10 + j 10)
p
j2 10
p
p
=
(1 + j 10)(10 + j 10)
=
= 0.1818/0 .
[e] G = 20 log10 0.1818 =
14.81 dB.
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15–22
CHAPTER 15. Active Filter Circuits
[f ]
P 15.30 !o = 2⇡fo = 1000⇡ rad/s;
= 2⇡(4000) = 8000⇡ rad/s;
.·. !c2
p
!c1 = 8000⇡.
!c1 !c2 = !o = 1000⇡.
Solve for the cuto↵ frequencies:
!c1 !c2 = 106 ⇡ 2 ;
!c2 =
106 ⇡ 2
;
!c1
106 ⇡ 2
·
..
!c1
!c1 = 8000⇡
or !c21 + 8000⇡!c1
106 ⇡ 2 = 0.
Solving,
!c1 = 386.75 rad/s;
.·. !c2 = 8000⇡ + 386.75 = 25,519.5 rad/s.
Thus, fc1 = 61.553 Hz
and
fc2 = 4061.553 Hz.
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Problems
Check:
= fc2
15–23
fc1 = 400 Hz.
!c2 =
1
= 25,519.5 rad/s;
RL CL
RL =
1
= 156.7 ⌦;
(25,519.5)(250 ⇥ 10 9 )
!c1 =
1
= 386.75 rad/s;
RH CH
RH =
1
= 10.34 k⌦.
(386.75)(250 ⇥ 10 9 )
P 15.31 !o = 250 rad/s;
= 2000 rad/s;
= ! c2
!o =
p
C = 1 µF.
!c1 = 2000;
!c1 !c2 = 250.
Solve for the cuto↵ frequencies:
.·. !c21 + 2000!c1
2502 = 0;
!c1 = 30.7764 rad/s;
!c2 = 2000 + !c1 = 2030.7764 rad/s.
!c1 =
1
;
RL CL
.·. RL =
1
(1 ⇥ 10
6 )(30.7764)
= 32.49 k⌦;
1
= !c2 = 2030.7764;
RH CH
RH =
1
(1 ⇥ 10
6 )(2030.7764)
= 492.4 ⌦.
Rf
= 1;
Ri
If Ri = 1 k⌦
Rf = Ri = 1 k⌦.
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15–24
CHAPTER 15. Active Filter Circuits
P 15.32 H(s) =
Zf =
Vo
Zf
=
;
Vi
Zi
1
(1/C2 )
;
kR2 =
sC2
s + (1/R2 C2 )
Zi = R1 +
1
sR1 C1 + 1
=
;
sC1
sC1
1/C2
(1/R1 C2 )s
s + (1/R2 C2 )
=
.·. H(s) =
s + (1/R1 C1 )
[s + (1/R1 C1 )][s + (1/R2 C2 )]
s/R1
=
K s
.
s2 + s + !o2
250s
250s
3.57(70s)
p
= 2
= 2
;
(s + 50)(s + 20)
s + 70s + 1000
s + 70s + ( 1000)2
p
!o = 1000 = 31.6 rad/s;
[a] H(s) =
= 70 rad/s;
K = 3.57.
!o
[b] Q =
= 0.45;
!c1,2 = ±
2
+
v
u
u
t
!c1 = 12.17 rad/s
P 15.33 [a] H(s) =
2
!2
+ !o2 = ±35 +
p
352 + 1000 = ±35 + 47.17;
!c2 = 82.17 rad/s.
(1/sC)
(1/RC)
=
;
R + (1/sC)
s + (1/RC)
H(j!) =
(1/RC)
;
j! + (1/RC)
(1/RC)
;
|H(j!)| = q
! 2 + (1/RC)2
|H(j!)|2 =
(1/RC)2
.
! 2 + (1/RC)2
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Problems
15–25
[b] Let Va be the voltage across the capacitor, positive at the upper terminal.
Then
Va
Va Vin
= 0.
+ sCVa +
R1
R2 + sL
Solving for Va yields
(R2 + sL)Vin
.
2
R1 LCs + (R1 R2 C + L)s + (R1 + R2 )
Va =
But
vo =
sLVa
.
R2 + sL
Therefore
Vo =
sLVin
;
2
R1 LCs + (L + R1 R2 C)s + (R1 + R2 )
H(s) =
sL
R1 LCs2 + (L + R1 R2 C)s + (R1 + R2 )
H(j!) =
[(R1 + R2 )
|H(j!)| = q
[R1 + R2
|H(j!)|2 =
=
(R1 + R2
;
j!L
;
R1 LC! 2 ] + j!(L + R1 R2 C)
!L
R1 LC! 2 ]2 + ! 2 (L + R1 R2 C)2
;
! 2 L2
R1 LC! 2 )2 + ! 2 (L + R1 R2 C)2
! 2 L2
.
R12 L2 C 2 ! 4 + (L2 + R12 R22 C 2 2R12 LC)! 2 + (R1 + R2 )2
[c] Let Va be the voltage across R2 positive at the upper terminal. Then
Va
Vin
R1
(0
+
Va
+ Va sC + Va sC = 0;
R2
Va )sC + (0
.·. Va =
and Va =
Va )sC +
0
Vo
R3
= 0;
R2 Vin
2R1 R2 Cs + R1 + R2
Vo
.
2R3 Cs
It follows directly that
H(s) =
2R2 R3 Cs
Vo
;
=
Vin
2R1 R2 Cs + (R1 + R2 )
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15–26
CHAPTER 15. Active Filter Circuits
H(j!) =
2R2 R3 C(j!)
;
(R1 + R2 ) + j!(2R1 R2 C)
2R2 R3 C!
|H(j!)| = q
;
(R1 + R2 )2 + ! 2 4R12 R22 C 2
|H(j!)|2 =
4R22 R32 C 2 ! 2
.
(R1 + R2 )2 + 4R12 R22 C 2 ! 2
( 0.05)( 25) ⇠
P 15.34 [a] n ⇠
=
= 2.526;
log10 (2500/800)
.·. n = 3.
1
[b] Gain = 20 log10 q
1 + (2500/800)6
=
29.7 dB.
P 15.35 For the scaled circuit
H 0 (s) =
1/(R0 )2 C10 C20
,
s2 + R02C 0 s + (R0 )21C 0 C 0
1
1
2
where
R0 = km R;
C10 = C1 /kf km ;
C20 = C2 /kf km .
It follows that
kf2
1
=
;
(R0 )2 C10 C20
R2 C1 C2
2
R0 C10
=
2kf
.
RC1
.·. H 0 (s) =
kf2 /RC1 C2
= ⇣ ⌘2
s
kf
P 15.36 [a] H(s) =
k2
2k
s2 + RCf1 s + R2 Cf1 C2
1/RC1 C2
2
+ RC
1
⇣
s
kf
⌘
+ R2 C11 C2
.
1
(s + 1)(s2 + s + 1)
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Problems
[b] fc = 800 Hz;
H 0 (s) =
!c = 1600⇡ rad/s;
15–27
kf = 1600⇡.
1
( ksf + 1)[( ksf )2 + ksf + 1]
=
kf3
(s + kf )(s2 + kf s + kf2 )
=
(1600⇡)3
.
(s + 1600⇡)[s2 + 1600⇡s + (1600⇡)2 ]
[c] H 0 (j5000⇡) = 0.03275/127.366 ;
Gain = 20 log10 (0.03275) =
29.7 dB.
P 15.37 [a] In the first-order circuit R = 1 ⌦ and C = 1 F.
km =
2700
R0
=
= 2700;
R
1
R0 = km R = 2700 ⌦;
kf =
C0 =
!c0
2⇡(800)
= 1600⇡;
=
!c
1
C
1
= 73.68 nF.
=
km kf
(2700)(1600⇡)
In the second-order circuit R = 1 ⌦, 2/C1 = 1 so C1 = 2 F, and
C2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit
R0 = km R = 2700 ⌦;
C20 =
C10 =
2
C1
= 147.37 nF;
=
km kf
(2700)(1600⇡)
0.5
C2
= 36.84 nF.
=
km kf
(2700)(1600⇡)
[b]
1
= 10 log10 (1 + ! 2n ).
1 + ! 2n
From the laws of logarithms we have
P 15.38 [a] y = 20 log10 p
✓
◆
10
ln(1 + ! 2n ).
y=
ln 10
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15–28
CHAPTER 15. Active Filter Circuits
Thus
✓
◆
dy
10 2n! 2n 1
=
;
d!
ln 10 (1 + ! 2n )
x = log10 ! =
ln !
;
ln 10
.·. ln ! = x ln 10.
d!
= ! ln 10;
dx
1 d!
= ln 10,
! dx
dy
=
dx
dy
d!
!
d!
dx
!
20n! 2n
dB/decade.
1 + ! 2n
=
At ! = !c = 1 rad/s
dy
=
dx
10n dB/decade.
1
[b] y = 20 log10 p
=
[ 1 + ! 2 ]n
=
10n log10 (1 + ! 2 )
10n
ln(1 + ! 2 );
ln 10
✓
◆
d!
= !(ln 10);
dx
.·.
At the corner !c =
p
10n
20n!
1
dy
=
.
2!
=
d!
ln 10 1 + ! 2
(ln 10)(1 + ! 2 )
As before
dy
=
dx
20n[21/n
21/n
1]
dy
20n! 2
=
.
dx
(1 + ! 2 )
21/n
.·. !c2 = 21/n
1.
dB/decade.
[c] For the Butterworth Filter
n
1
dy/dx (dB/decade)
For the cascade of identical sections
n
dy/dx (dB/decade)
1
10
1
10
2
20
2
11.72
3
30
3
12.38
4
40
4
12.73
1
1
1
13.86
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Problems
15–29
[d] It is apparent from the calculations in part (c) that as n increases the
amplitude characteristic at the cuto↵ frequency decreases at a much
faster rate for the Butterworth filter.
Hence the transition region of the Butterworth filter will be much
narrower than that of the cascaded sections.
P 15.39 n = 5: 1 + ( 1)5 s10 = 0;
s10 = 1;
s10 = 1/(0 + 360k)
so
k sk+1
k sk+1
0 1/0
5 1/180
1 1/36
6 1/216
2 1/72
7 1/252
3 1/108
8 1/288
4 1/144
9 1/324
s = 1/36k .
Group by conjugate pairs to form denominator polynomial.
(s + 1)[s
· [(s
(cos 108 + j sin 108 )][(s
(cos 144 + j sin 144 )][(s
= (s + 1)(s + 0.309
(cos 252 + j sin 252 )]
(cos 216 + j sin 216 )]
j0.951)(s + 0.309 + j0.951)·
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15–30
CHAPTER 15. Active Filter Circuits
(s + 0.809
j0.588)(s + 0.809 + j0.588)
which reduces to
(s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1).
n = 6: 1 + ( 1)6 s12 = 0
s12 =
1;
s12 = 1/180 + 360k.
k sk+1
k sk+1
0 1/15
6 1/195
1 1/45
7 1/225
2 1/75
8 1/255
3 1/105
9 1/285
4 1/135
10 1/315
5 1/165
11 1/345
Grouping by conjugate pairs yields
(s + 0.2588
j0.9659)(s + 0.2588 + j0.9659)⇥
(s + 0.7071
(s + 0.9659
j0.7071)(s + 0.7071 + j0.7071)⇥
j0.2588)(s + 0.9659 + j0.2588)
or (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9318s + 1).
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Problems
s2
P 15.40 H 0 (s) =
s2 +
.
s2
H 0 (s) =
s2 +
=
1
2
s+
2 k2 )
km R2 (C/km kf )
km R1 km R2 (C 2 /km
f
15–31
kf2
2kf
s+
R2 C
R1 R2 C 2
(s/kf )2
!
.
1
2
s
2
+
(s/kf ) +
R2 C kf
R1 R2 C 2
( 0.05)( 48)
= 3.99 .·.
n = 4.
log10 (32/8)
From Table 15.1 the transfer function is
1
.
H(s) = 2
(s + 0.765s + 1)(s2 + 1.848s + 1)
P 15.41 [a] n =
The capacitor values for the first stage prototype circuit are
2
= 0.765
C1
C2 =
.·.
C1 = 2.61 F;
1
= 0.38 F.
C1
The values for the second stage prototype circuit are
2
= 1.848
C1
C2 =
.·.
C1 = 1.08 F;
1
= 0.92 F.
C1
The scaling factors are
km =
R0
= 1000;
R
kf =
!o0
= 16,000⇡.
!o
Therefore the scaled values for the components in the first stage are
R1 = R2 = R = 1000 ⌦;
C1 =
2.61
= 52.01 nF;
(16,000⇡)(1000)
C2 =
0.38
= 7.61 nF.
(16,000⇡)(1000)
The scaled values for the second stage are
R1 = R2 = R = 1000 ⌦;
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15–32
CHAPTER 15. Active Filter Circuits
C1 =
1.08
= 21.53 nF;
(16,000⇡)(1000)
C2 =
0.92
= 18.38 nF.
(16,000⇡)(1000)
[b]
( 0.05)( 48)
= 3.99 .·. n = 4.
log10 (2000/500)
From Table 15.1 the transfer function of the first section is
P 15.42 [a] n =
s2
.
s2 + 0.765s + 1
For the prototype circuit
H1 (s) =
2
= 0.765;
R2
R2 = 2.61 ⌦;
R1 =
1
= 0.383 ⌦.
R2
The transfer function of the second section is
s2
.
s2 + 1.848s + 1
For the prototype circuit
H2 (s) =
2
= 1.848;
R2
R2 = 1.082 ⌦;
R1 =
1
= 0.9240 ⌦.
R2
The scaling factors are:
kf =
!o0
2⇡(2000)
= 4000⇡;
=
!o
1
C0 =
C
km kf
.·.
.·.
km =
1
= 7957.75.
4000⇡(10 ⇥ 10 9 )
10 ⇥ 10 9 =
1
;
4000⇡km
Therefore in the first section
R10 = km R1 = 3.05 k⌦;
R20 = km R2 = 20.77 k⌦.
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Problems
15–33
In the second section
R10 = km R1 = 7.35 k⌦;
R20 = km R2 = 8.61 k⌦.
[b]
P 15.43 [a] The cascade connection is a bandpass filter.
[b] The cuto↵ frequencies areq2 kHz and 8 kHz.
The center frequency is (2)(8) = 4 kHz.
The Q is 4/(8 2) = 2/3 = 0.67.
[c] For the high pass section kf = 4000⇡. The prototype transfer function is
Hhp (s) =
.·.
s4
;
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
0
Hhp
(s) =
·
(s/4000⇡)4
[(s/4000⇡)2 + 0.765(s/4000⇡) + 1]
1
[(s/4000⇡)2 + 1.848(s/4000⇡) + 1]
s4
= 2
.
(s + 3060⇡s + 16 ⇥ 106 ⇡ 2 )(s2 + 7392⇡s + 16 ⇥ 106 ⇡ 2 )
For the low pass section kf = 16,000⇡:
Hlp (s) =
.·.
1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
0
Hlp
(s) =
·
=
;
1
[(s/16,000⇡)2 + 0.765(s/16,000⇡) + 1]
1
[(s/16,000⇡)2 + 1.848(s/16,000⇡) + 1]
(16,000⇡)4
.
([s2 + 12,240⇡s + (16,000⇡)2 )][s2 + 29,568⇡s + (16,000⇡)2 ]
The cascaded transfer function is
0
0
H 0 (s) = Hhp
(s)Hlp
(s).
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15–34
CHAPTER 15. Active Filter Circuits
For convenience let
D1 = s2 + 3060⇡s + 16 ⇥ 106 ⇡ 2 ;
D2 = s2 + 7392⇡s + 16 ⇥ 106 ⇡ 2 ;
D3 = s2 + 12,240⇡s + 256 ⇥ 106 ⇡ 2 ;
D4 = s2 + 29,568⇡s + 256x106 ⇡ 2 .
Then
H 0 (s) =
65,536 ⇥ 1012 ⇡ 4 s4
.
D1 D2 D3 D4
[d] !o = 2⇡(4000) = 8000⇡ rad/s;
s = j8000⇡;
s2 =
64 ⇥ 106 ⇡ 2 ;
D1 = (16 ⇥ 106 ⇡ 2
64 ⇥ 106 ⇡ 2 ) + j(8000⇡)(3060⇡)
= 106 ⇡ 2 ( 48
j24.48) = 106 ⇡ 2 (53.88/152.98 ).
D2 = (16 ⇥ 106 ⇡ 2
64 ⇥ 106 ⇡ 2 ) + j(8000⇡)(7392⇡)
= 106 ⇡ 2 ( 48 + j59.136) = 106 ⇡ 2 (76.16/129.07 ).
D3 = (256 ⇥ 106 ⇡ 2
64 ⇥ 106 ⇡ 2 ) + j(8000⇡)(12,240⇡)
= 106 ⇡ 2 (192 + j97.92) = 106 ⇡ 2 (215.53/27.02 ).
D4 = (256 ⇥ 106 ⇡ 2
64 ⇥ 106 ⇡ 2 ) + j(8000⇡)(29,568⇡)
= 106 ⇡ 2 (192 + j236.544) = 106 ⇡ 2 (304.66/50.93 ).
H 0 (j!o ) =
(65,536)(4096)⇡ 8 ⇥ 1024
(⇡ 8 ⇥ 1024 )[(53.88)(76.16)(215.53)(304.66)/360 ]
= 0.996/
360 = 0.996/0 .
P 15.44 [a] First we will design a unity gain filter and then provide the passband gain
with an inverting amplifier. For the high pass section the cut-o↵
frequency is 500 Hz. The order of the Butterworth is
n=
( 0.05)( 20)
= 2.51;
log10 (500/200)
.·. n = 3.
Hhp (s) =
s3
.
(s + 1)(s2 + s + 1)
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Problems
15–35
For the prototype first-order section
R1 = R2 = 1 ⌦,
C = 1 F.
For the prototype second-order section
R1 = 0.5 ⌦,
R2 = 2 ⌦,
C = 1 F.
The scaling factors are
kf =
!o0
= 2⇡(500) = 1000⇡;
!o
km =
106
C
1
=
.
=
C 0 kf
(15 ⇥ 10 9 )(1000⇡)
15⇡
In the scaled first-order section
106
R10 = R20 = km R1 =
(1) = 21.22 k⌦;
15⇡
C 0 = 15 nF.
In the scaled second-order section
R10 = 0.5km = 10.61 k⌦;
R20 = 2km = 42.44 k⌦;
C 0 = 15 nF.
For the low-pass section the cut-o↵ frequency is 4500 Hz. The order of
the Butterworth filter is
( 0.05)( 20)
n=
= 2.51;
.·. n = 3;
log10 (11,250/4500)
Hlp (s) =
1
.
(s + 1)(s2 + s + 1)
For the prototype first-order section
R1 = R2 = 1 ⌦,
C = 1 F.
For the prototype second-order section
R1 = R2 = 1 ⌦;
C1 = 2 F;
C2 = 0.5 F.
The low-pass scaling factors are
R0
= 104 ;
km =
R
!o0
kf =
= (4500)(2⇡) = 9000⇡.
!o
For the scaled first-order section
1
C
= 3.54 nF.
=
C0 =
R10 = R20 = 10 k⌦;
kf km
(9000⇡)(104 )
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15–36
CHAPTER 15. Active Filter Circuits
For the scaled second-order section
R10 = R20 = 10 k⌦;
C10 =
2
C1
= 7.07 nF;
=
kf km
(9000⇡)(104 )
C20 =
0.5
C2
= 1.77 nF.
=
kf km
(9000⇡)(104 )
Gain Amplifier:
.·. K = 10.
20 log10 K = 20 dB,
Since we are using 10 k⌦ resistors in the low-pass stage, we will use
Rf = 100 k⌦ and Ri = 10 k⌦ in the inverting amplifier stage.
[b]
P 15.45 [a] Unscaled high-pass stage
Hhp (s) =
s3
.
(s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (!o0 /!o ) = 1000⇡. Therefore the
scaled transfer function is
0
Hhp
(s) = ⇣
=
(s/1000⇡)3
⌘ ⇣
s
+1
1000⇡
s
1000⇡
3
⌘3
s
+ 1000⇡
+1
s
.
(s + 1000⇡)[s2 + 1000⇡s + 106 ⇡ 2 ]
Unscaled low-pass stage
Hlp (s) =
1
.
(s + 1)(s2 + s + 1)
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Problems
15–37
The frequency scaling factor is kf = (!o0 /!o ) = 9000⇡. Therefore the
scaled transfer function is
1
Hlp0 (s) = ⇣
⌘ ⇣
⌘2 ⇣
⌘
s
s
s
+
1
+
+1
9000⇡
9000⇡
9000⇡
=
(9000⇡)3
.
(s + 9000⇡)(s2 + 9000⇡s + 81 ⇥ 106 ⇡ 2 )
Thus the transfer function for the filter is
729 ⇥ 1010 ⇡ 3 s3
0
H 0 (s) = 10Hhp
(s)Hlp0 (s) =
,
D1 D2 D3 D4
where
D1 = s + 1000⇡;
D2 = s + 9000⇡;
D3 = s2 + 1000⇡s + 106 ⇡ 2 ;
D4 = s2 + 9000⇡s + 81 ⇥ 106 ⇡ 2 .
[b] At 200 Hz
! = 400⇡ rad/s:
D1 (j400⇡) = 400⇡(2.5 + j1);
D2 (j400⇡) = 400⇡(22.5 + j1);
D3 (j400⇡) = 4 ⇥ 105 ⇡ 2 (2.1 + j1.0);
D4 (j400⇡) = 4 ⇥ 105 ⇡ 2 (202.1 + j9).
Therefore
D1 D2 D3 D4 (j400⇡) = 256⇡ 6 1014 (28,534.82/52.36 );
H 0 (j400⇡) =
(729⇡ 3 ⇥ 1010 )(64 ⇥ 106 ⇡ 3 )
256⇡ 6 ⇥ 1014 (28,534.82/52.36 )
= 0.639/
52.36 ;
.·. 20 log10 |H 0 (j400⇡)| = 20 log10 (0.639) =
At f = 1500 Hz,
3.89 dB.
! = 3000⇡ rad/s.
Then
D1 (j3000⇡) = 1000⇡(1 + j3);
D2 (j3000⇡) = 3000⇡(3 + j1);
D3 (j3000⇡) = 106 ⇡ 2 ( 8 + j3);
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15–38
CHAPTER 15. Active Filter Circuits
D4 (j3000⇡) = 106 ⇡ 2 (8 + j3.
H 0 (j3000⇡) =
(729 ⇥ ⇡ 3 ⇥ 1010 )(27 ⇥ 109 ⇡ 3 )
27 ⇥ 1018 ⇡ 6 (730/270 )
= 9.99/90 ;
.·. 20 log10 |H 0 (j3000⇡)| = 19.99 dB.
[c] From the transfer function the gain is down 19.99 + 3.89 or 23.88 dB at
200 Hz. Because the upper cut-o↵ frequency is nine times the lower
cut-o↵ frequency we would expect the high-pass stage of the filter to
predict the loss in gain at 200 Hz. For a 3nd order Butterworth
1
GAIN = 20 log10 q
1 + (500/200)6
=
23.89 dB.
1500 Hz is in the passband for this bandpass filter. Hence we expect the
gain at 1500 Hz to nearly equal 20 dB as specified in Problem 15.44.
Thus our scaled transfer function confirms that the filter meets the
specifications.
P 15.46 [a] From Table 15.1
Hlp (s) =
1
p
;
(s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
Hhp (s) = ⇣ 1
s2
+ 0.518
⇣ ⌘
1
s
⌘⇣
+1
1
⇣ ⌘
⌘⇣
⇣ ⌘
⌘;
p
1
1
1
1
2
+
+
1
+
1.932
+
1
2
2
s
s
s
s
s6
p
.
(s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
Hhp (s) =
P 15.47 [a] kf = 25,000;
(s/25,000)6
[(s/25,000)2 + 0.518(s/25,000) + 1]
0
Hhp
(s) =
·
=
1
[(s/25,000)2 + 1.414s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 + 1]
s6
(s2 + 12,950s + 625 ⇥ 106 )(s2 + 35,350s + 625 ⇥ 106 )
·
1
(s2 + 48,300s + 625 ⇥ 106 )
.
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Problems
[b] H 0 (j25,000) =
=
15–39
(25,000)6
[12,900(j25,000)][35,350(j25,000)][48,300(j25,000)]
(25,000)3
(12,900)(25,350)(48,300)j 3
= 0.7094/
90 ;
20 log10 |H 0 (j25,000)| =
Should be
2.98 dB.
3.01; di↵erence due to round o↵.
P 15.48 [a] At very low frequencies the two capacitor branches are open and because
the op amp is ideal the current in R3 is zero. Therefore at low frequencies
the circuit behaves as an inverting amplifier with a gain of R2 /R1 . At
very high frequencies the capacitor branches are short circuits and hence
the output voltage is zero.
[b] Let the node where R1 , R2 , R3 , and C2 join be denoted as a, then
(Va
Vi )G1 + Va sC2 + (Va
Va G3
Vo )G2 + Va G3 = 0;
Vo sC1 = 0.
or
(G1 + G2 + G3 + sC2 )Va
G2 Vo = G1 Vi ;
sC1
Vo .
G3
Va =
Solving for Vo /Vi yields
H(s) =
=
=
G1 G3
(G1 + G2 + G3 + sC2 )sC1 + G2 G3
G1 G3
2
s C1 C2 + (G1 + G2 + G3 )C1 s + G2 G3
h
G1 G3 /C1 C2
i
(G1 +G2 +G3 )
2 G3
s+ G
C2
C1 C2
G1 G2 G3
G2 C1 C2
h
i
=
(G1 +G2 +G3 )
2 G3
2
s +
s+ G
C2
C1 C2
=
s2 +
Kbo
,
2
s + b1 s + b o
where K =
and b1 =
G1
;
G2
bo =
G 2 G3
;
C1 C2
G 1 + G2 + G3
.
C2
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15–40
CHAPTER 15. Active Filter Circuits
[c] Rearranging we see that
G1 = KG2 ;
bo C1
bo C1 C2
;
=
G2
G2 .
G3 =
Since by hypothesis C2 = 1 F
b1 =
G 1 + G2 + G3
= G 1 + G2 + G3 ;
C2
bo C1
.·. b1 = KG2 + G2 +
;
G2
bo C1
.
G2
b1 = G2 (1 + K) +
Solving this quadratic equation for G2 we get
b1
G2 =
±
2(1 + K)
=
b1 ±
q
b21
s
b21
bo C1 4(1 + K)
4(1 + K)2
4bo (1 + K)C1
2(1 + K)
.
For G2 to be realizable
b21
C1 <
.
4bo (1 + K)
[d] 1. Select C2 = 1 F;
2. Select C1 such that C1 <
b21
;
4bo (1 + K)
3. Calculate G2 (R2 );
4. Calculate G1 (R1 ); G1 = KG2 ;
5. Calculate G3 (R3 ); G3 = bo C1 /G2 .
P 15.49 [a] In the second order section of a third order Butterworth filter bo = b1 = 1
Therefore,
C1 
1
b21
=
= 0.05 F;
4bo (1 + K)
(4)(1)(5)
.·.
C1 = 0.05 F. (limiting value)
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Problems
[b] G2 =
15–41
1
= 0.1 S;
2(1 + 4)
1
(0.05) = 0.5 S;
0.1
G1 = 4(0.1) = 0.4 S.
G3 =
Therefore,
1
1
R1 =
= 2.5 ⌦;
R2 =
= 10 ⌦;
G1
G2
!0
[c] kf = o = 2⇡(2500) = 5000⇡;
!o
km =
C10 =
R3 =
1
= 2 ⌦.
G3
C2
1
=
= 6366.2;
0
C2 kf
(10 ⇥ 10 9 )kf
0.05
= 0.5 ⇥ 10 9 = 500 pF;
kf km
R10 = (2.5)(6366.2) = 15.92 k⌦;
R20 = (10)(6366.2) = 63.66 k⌦;
R30 = (2)(6366.2) = 12.73 k⌦.
;
[d] R10 = R20 = (6366.2)(1) = 6.37 k⌦
C0 =
1
C
= 8 = 10 nF.
kf km
10
[e]
P 15.50 [a] By hypothesis the circuit becomes:
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15–42
CHAPTER 15. Active Filter Circuits
For very small frequencies the capacitors behave as open circuits and
therefore vo is zero. As the frequency increases, the capacitive branch
impedances become small compared to the resistive branches. When this
happens the circuit becomes an inverting amplifier with the capacitor C2
dominating the feedback path. Hence the gain of the amplifier
approaches (1/j!C2 )/(1/j!C1 ) or C1 /C2 . Therefore the circuit behaves
like a high-pass filter with a passband gain of C1 /C2 .
[b] Summing the currents away from the upper terminal of R2 yields
Va G2 + (Va
Vi )sC1 + (Va
Vo )sC2 + Va sC3 = 0
or
Va [G2 + s(C1 + C2 + C3 )]
Vo sC2 = sC1 Vi .
Summing the currents away from the inverting input terminal gives
(0
Va )sC3 + (0
Vo )G1 = 0
or
sC3 Va =
G1 Vo ;
Va =
G1 Vo
.
sC3
Therefore we can write
G1 Vo
[G2 + s(C1 + C2 + C3 )]
sC3
sC2 Vo = sC1 Vi .
Solving for Vo /Vi gives
H(s) =
Vo
C1 C3 s2
=
Vi
C2 C3 s2 + G1 (C1 + C2 + C3 )s + G1 G2 ]
= h
=
C1 2
s
C2
G
G G
s2 + C2 C1 3 (C1 + C2 + C3 )s + C12 C32
Ks2
.
s 2 + b1 s + b o
i
Therefore the circuit implements a second-order high-pass filter with a
passband gain of C1 /C2 .
[c] C1 = K:
b1 =
G1
(K + 2) = G1 (K + 2);
(1)(1)
.·. G1 =
bo =
b1
;
K +2
✓
◆
K +2
R1 =
.
b1
G 1 G2
= G 1 G2 ;
(1)(1)
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Problems
15–43
bo
bo
.·. G2 =
= (K + 2);
G1
b1
.·. R2 =
b1
.
bo (K + 2)
[d] From Table 15.1 the transfer function of the second-order section of a
third-order high-pass Butterworth filter is
H(s) =
Ks2
.
s2 + s + 1
Therefore b1 = bo = 1.
Thus
C1 = K = 8 F;
R1 =
8+2
= 10 ⌦;
1
R2 =
1
= 0.1 ⌦.
1(8 + 2)
P 15.51 [a] Low-pass filter:
n=
( 0.05)( 30)
= 3.77;
log10 (1000/400)
.·. n = 4.
In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F.
C1 
(0.765)2
b21

 0.0732.
4bo (1 + K)
(4)(2)
choose C1 = 0.03 F;
G2 =
0.765 ±
q
(0.765)2
4(2)(0.03)
4
=
0.765 ± 0.588
.
4
Arbitrarily select the larger value for G2 , then
G2 = 0.338 S;
.·.
G1 = KG2 = 0.338 S;
G3 =
R2 =
.·.
1
= 2.96 ⌦;
G2
R1 =
bo C1
(1)(0.03)
= 0.089
=
G2
0.338
.·.
1
= 2.96 ⌦;
G1
R3 = 1/G3 = 11.3 ⌦.
Therefore in the first second-order prototype circuit
R1 = R2 = 2.96 ⌦;
R3 = 11.3 ⌦;
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15–44
CHAPTER 15. Active Filter Circuits
C1 = 0.03 F;
C2 = 1 F.
In the second second-order prototype circuit:
b1 = 1.848, b0 = 1, C2 = 1 F;
.·. C1 
(1.848)2
 0.427.
8
choose C1 = 0.30 F;
G2 =
1.848 ±
q
(1.848)2
8(0.3)
4
=
1.848 ± 1.008
.
4
Arbitrarily select the larger value, then
G2 = 0.7139 S;
.·.
R2 =
G1 = KG2 = 0.7139 S;
G3 =
.·.
1
= 1.4008 ⌦;
G2
R1 =
bo C1
(1)(0.30)
= 0.4202 S
=
G2
0.7139
1
= 1.4008 ⌦;
G1
.·.
R3 = 1/G3 = 2.3796 ⌦.
In the low-pass section of the filter
kf =
!o0
= 2⇡(400) = 800⇡;
!o
km =
C
1
125,000
.
=
=
0
9
C kf
(10 ⇥ 10 )kf
⇡
Therefore in the first scaled second-order section
R10 = R20 = 2.96km = 118 k⌦;
R30 = 11.3km = 450 k⌦;
C10 =
0.03
= 300 pF;
kf km
C20 = 10 nF.
In the second scaled second-order section
R10 = R20 = 1.4008km = 55.74 k⌦;
R30 = 2.38km = 94.68 k⌦;
C10 =
0.3
= 3 nF;
kf km
C20 = 10 nF.
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Problems
15–45
High-pass filter section
n=
( 0.05)( 30)
= 3.77;
log10 (6400/2560)
n = 4.
In the first prototype second-order section:
b1 = 0.765; bo = 1; C2 = C3 = 1 F;
C1 = K = 1 F;
R1 =
3
K +2
= 3.92 ⌦;
=
b1
0.765
R2 =
0.765
b1
=
= 0.255 ⌦.
bo (K + 2)
3
In the second prototype second-order section: b1 = 1.848; bo = 1;
C2 = C3 = 1 F;
C1 = K = 10 F;
R1 =
3
K +2
= 1.623 ⌦;
=
b1
1.848
R2 =
1.848
b1
=
= 0.616 ⌦.
bo (K + 2)
3
In the high-pass section of the filter
kf =
!o0
= 2⇡(6400) = 12,800⇡;
!o
km =
7812.5
C
1
=
.
=
0
9
C kf
(10 ⇥ 10 )(12,800⇡)
⇡
In the first scaled second-order section
R10 = 3.92km = 9.75 k⌦;
R20 = 0.255km = 634 ⌦;
C10 = C20 = C30 = 10 nF.
In the second scaled second-order section
R10 = 1.623km = 4.04 k⌦;
R20 = 0.616km = 1.53 k⌦;
C10 = C20 = C30 = 10 nF.
In the gain section, let Ri = 10 k⌦ and Rf = 10 k⌦.
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15–46
CHAPTER 15. Active Filter Circuits
[b]
P 15.52 [a] The prototype low-pass transfer function is
Hlp (s) =
1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
.
The low-pass frequency scaling factor is
kflp = 2⇡(400) = 800⇡.
The scaled transfer function for the low-pass filter is
Hlp0 (s) = ⇣
=
s
800⇡
⌘2
1
+ 0.765s
+1
800⇡
⇣
⌘2
s
+ 1.848s
+1
800⇡
800⇡
8 4
4096 ⇥ 10 ⇡
[s2 + 612⇡s + (800⇡)2 ] [s2 + 1478.4⇡s + (800⇡)2 ]
.
The prototype high-pass transfer function is
Hhp (s) =
s4
.
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The high-pass frequency scaling factor is
kfhp = 2⇡(6400) = 12,800⇡.
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Problems
15–47
The scaled transfer function for the high-pass filter is
0
Hhp
(s) = ⇣
=
s
12,800⇡
⌘2
(s/12,800⇡)4
0.765s
+ 12,800⇡
+1
⇣
s
12,800⇡
4
s
⌘2
1.848s
+ 12,800⇡
+1
[s2 + 9792⇡s + (12,800⇡)2 ][s2 + 23,654.4⇡s + (12,800⇡)2 ]
.
The transfer function for the filter is
h
i
0
H 0 (s) = Hlp0 (s) + Hhp
(s) .
[b] fo =
q
fc1 fc2 =
q
400)(6400) = 1600 Hz;
!o = 2⇡fo = 3200⇡ rad/s;
(j!o )2 =
1024 ⇥ 104 ⇡ 2 ;
(j!o )4 = 1,048,576 ⇥ 108 ⇡ 4 ;
Hlp0 (j!o ) =
4096 ⇥ 108 ⇡ 4
⇥
[ 960 ⇥ 104 ⇡ 2 + j612(3200⇡ 2 )]
1
[
=
960 ⇥ 104 ⇡ 2 + j1478.4(3200⇡ 2 )]
40,000
( 3000 + j612)( 3000 + j1478.4)
= 3906.2 ⇥ 10 6 / 322.24 .
1,048,576 ⇥ 108 ⇡ 4
0
(j!o ) =
Hhp
[15,360 ⇥ 104 ⇡ 2 + j9792(3200⇡ 2 )]
1
[15,360 ⇥ 104 ⇡ 2 + j23,654.4(3200⇡ 2 )]
=
10.24 ⇥ 106
(48,000 + j9792)(48,000 + j23,654.4)
= 3906.2 ⇥ 10 6 /
.·. H 0 (j!o ) =
=
37.76 .
3906.2 ⇥ 10 6 (1/
3906.2 ⇥ 10 6 (1.58/0 ) =
322.24 + 1/
37.76 )
6176.35 ⇥ 10 6 /0 .
G = 20 log10 |H 0 (j!o )| = 20 log10 (6176.35 ⇥ 10 6 ) =
44.19 dB.
P 15.53 [a] At low frequencies the capacitor branches are open; vo = vi . At high
frequencies the capacitor branches are short circuits and the output
voltage is zero. Hence the circuit behaves like a unity-gain low-pass filter.
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15–48
CHAPTER 15. Active Filter Circuits
[b] Let va represent the voltage-to-ground at the right-hand terminal of R1 .
Observe this will also be the voltage at the left-hand terminal of R2 . The
s-domain equations are
(Va
Vi )G1 + (Va
(Vo
Va )G2 + sC2 Vo = 0
Vo )sC1 = 0;
or
(G1 + sC1 )Va
sC1 Vo = G1 Vi
G2 Va + (G2 + sC2 )Vo = 0.
.·. Va =
.·.
"
G2 + sC2 Vo
;
G2
(G2 + sC2 )
(G1 + sC1 )
G2
#
sC1 Vo = G1 Vi ;
Vo
G1 G 2
.·.
=
Vi
(G1 + sC1 )(G2 + sC2 )
C1 G2 s
;
which reduces to
G1 G2 /C1 C2
bo
Vo
= 2 G1
.
G1 G2 = 2
Vi
s + b1 s + b o
s + C1 s + C1 C2
[c] There are four circuit components and two restraints imposed by H(s);
therefore there are two free choices.
G1 ·
[d] b1 =
. . G1 = b1 C1 ;
C1
bo =
G 1 G2 ·
bo
. . G2 = C2 .
C1 C2
b1
[e] No, all physically realizeable capacitors will yield physically realizeable
resistors.
[f ] From Table 15.1 we know the transfer function of the prototype 4th order
Butterworth filter is
1
H(s) = 2
.
(s + 0.765s + 1)(s2 + 1.848s + 1)
In the first section bo = 1,
b1 = 0.765;
.·. G1 = (0.765)(1) = 0.765 S.
R1 = 1/G1 = 1.307 ⌦;
G2 =
1
(1) = 1.307 S;
0.765
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Problems
15–49
R2 = 1/G2 = 0.765 ⌦.
In the second section bo = 1,
b1 = 1.848;
.·. G1 = 1.848 S.
R1 = 1/G1 = 0.541 ⌦;
G2 =
✓
◆
1
(1) = 0.541 S;
1.848
R2 = 1/G2 = 1.848 ⌦.
P 15.54 [a] kf =
!o0
= 2⇡(3000) = 6000⇡;
!o
km =
106
C
1
=
.
=
C 0 kf
(4.7 ⇥ 10 9 )(6000⇡)
28.2⇡
In the first section
R10 = 1.307km = 14.75 k⌦;
R20 = 0.765km = 8.64 k⌦.
In the second section
R10 = 0.541km = 6.1 k⌦;
R20 = 1.848km = 20.86 k⌦.
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15–50
CHAPTER 15. Active Filter Circuits
[b]
P 15.55 [a] Interchanging the Rs and Cs yields the following circuit.
At low frequencies the capacitors appear as open circuits and hence the
output voltage is zero. As the frequency increases the capacitor branches
approach short circuits and va = vi = vo . Thus the circuit is a unity-gain,
high-pass filter.
[b] The s-domain equations are
(Va
Vi )sC1 + (Va
(Vo
Va )sC2 + Vo G2 = 0.
Vo )G1 = 0;
It follows that
Va (G1 + sC1 )
and Va =
Thus
("
G1 Vo = sC1 Vi
(G2 + sC2 )Vo
.
sC2
#
(G2 + sC2 )
(G1 + sC1 )
sC2
)
G1 Vo = sC1 Vi ;
Vo {s2 C1 C2 + sC1 G2 + G1 G2 } = s2 C1 C2 Vi ;
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Problems
15–51
Vo
s2
=✓
◆
G 1 G2
G2
Vi
2
s+
s +
C2
C1 C2
s2
Vo
= 2
.
=
Vi
s + b1 s + b o
H(s) =
[c] There are 4 circuit components: R1 , R2 , C1 and C2 .
There are two transfer function constraints: b1 and bo .
Therefore there are two free choices.
G 1 G2
G2
[d] bo =
;
b1 =
;
C1 C2
C2
.·. G2 = b1 C2 ;
G1 =
R2 =
1
.
b1 C2
bo
b1
C1 .·. R1 =
.
b1
bo C1
[e] No, all realizeable capacitors will produce realizeable resistors.
[f ] The second-order section in a 3rd-order Butterworth high-pass filter is
s2 /(s2 + s + 1). Therefore bo = b1 = 1 and
R1 =
1
= 1 ⌦;
(1)(1)
R2 =
1
= 1 ⌦.
(1)(1)
P 15.56 [a] kf =
!o0
= 104 ⇡;
!o
105
1
C
km = 0 =
=
;
C kf
(75 ⇥ 10 9 )(104 ⇡)
75⇡
C1 = C2 = 75 nF;
[b] R = 424.4 ⌦;
R10 = R20 = km R = 424.4 ⌦.
C = 75 nF.
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15–52
CHAPTER 15. Active Filter Circuits
[c]
[d] Hhp (s) =
s3
;
(s + 1)(s2 + s + 1)
0
Hhp
(s) =
=
[e]
(s/104 ⇡)3
[(s/104 ⇡) + 1][(s/104 ⇡)2 + (s/104 ⇡) + 1]
s3
.
(s + 104 ⇡)(s2 + 104 ⇡s + 108 ⇡ 2 )
0
Hhp
(j104 ⇡) =
.·.
(j104 ⇡)3
= 0.7071/135 ;
(j104 ⇡ + 104 ⇡)[(j104 ⇡)2 + 104 ⇡(j104 ⇡) + 108 ⇡ 2 ]
0
|Hhp
| = 0.7071 =
3 dB.
P 15.57 From Eq 15.19 we can write
✓
◆
1
s
R1 C
H(s) =
R 1 + R2
2
s+
s2 +
R3 C
R1 R2 R3 C 2
or
✓
◆✓
◆
R3
2
s
2R1
R3 C
.
H(s) =
R 1 + R2
2
s+
s2 +
R3 C
R1 R2 R3 C 2
Therefore
2
=
R3 C
=
!o
;
Q
R1 + R2
= !o2 ;
R1 R2 R3 C 2
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Problems
and K =
15–53
R3
.
2R1
By hypothesis C = 1 F and !o = 1 rad/s;
.·.
2
1
or R3 = 2Q;
=
R3
Q
R1 =
Q
R3
= ;
2K
K
R 1 + R2
= 1;
R1 R2 R3
✓
◆
Q
Q
+ R2 =
(2Q)R2 ;
K
K
.·. R2 =
Q
2Q2
K
.
P 15.58 [a] From the statement of the problem, K = 10 ( = 20 dB). Therefore for the
prototype bandpass circuit
R1 =
R2 =
16
Q
=
= 1.6 ⌦;
K
10
Q
2Q2
K
=
16
⌦;
502
R3 = 2Q = 32 ⌦.
The scaling factors are
kf =
!o0
= 2⇡(6400) = 12,800⇡;
!o
km =
C
1
= 1243.40.
=
0
9
C kf
(20 ⇥ 10 )(12,800⇡)
Therefore,
R10 = km R1 = (1.6)(1243.30) = 1.99 k⌦;
R20 = km R2 = (16/502)(1243.40) = 39.63 ⌦;
R30 = km R3 = 32(1243.40) = 39.79 k⌦.
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15–54
CHAPTER 15. Active Filter Circuits
[b]
P 15.59 [a] It follows directly from Eq. 15.21 that
s2 + 1
H(s) = 2
.
s + 4(1
)s + 1
Now note from Eq 15.22 that (1
H(s) =
) equals 1/4Q, hence
s2 + 1
.
s2 + Q1 s + 1
[b] For Example 15.14 !o = 5000 rad/s and Q = 5. Therefore kf = 5000 and
(s/5000)2 + 1
✓
◆
1
s
2
(s/5000) +
+1
5 5000
2
6
s + 25 ⇥ 10
.
= 2
s + 1000s + 25 ⇥ 106
H 0 (s) =
P 15.60 [a] !o = 2000⇡ rad/s;
!o0
= 2000⇡;
!o
.·.
kf =
km =
105
1
C
=
;
=
C 0 kf
(15 ⇥ 10 9 )(2000⇡)
3⇡
R 0 = km R =
=1
105
(1) = 10,610 ⌦;
3⇡
1
=1
4Q
R0 = 10,478 ⌦;
1
= 0.9875;
4(20)
(1
)R0 = 133 ⌦;
C 0 = 15 nF;
2C 0 = 30 nF.
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Problems
15–55
[b]
[c] kf = 2000⇡;
H(s) =
=
(s/2000⇡)2 + 1
1
(s/2000⇡)2 + 20
(s/2000⇡) + 1
s2 + 4 ⇥ 106 ⇡ 2
.
s2 + 100⇡s + 4 ⇥ 106 ⇡ 2
P 15.61 To satisfy the gain specification of 20 dB at ! = 0 and ↵ = 1 requires
R 1 + R2
= 10
R1
or
R2 = 9R1 .
Choose a standard resistor of 11.1 k⌦ for R1 and a 100 k⌦ potentiometer for
R2 . Since (R1 + R2 )/R1
1 the value of C1 is
C1 =
1
= 39.79 nF.
2⇡(40)(105 )
Choose a standard capacitor value of 39 nF. Using the selected values of R1
and R2 the maximum gain for ↵ = 1 is
20 log10
✓
111.1
11.1
◆
= 20.01 dB.
↵=1
When C1 = 39 nF the frequency 1/R2 C1 is
109
1
= 256.41 rad/s = 40.81 Hz.
= 5
R2 C1
10 (39)
The magnitude of the transfer function at 256.41 rad/s is
|H(j256.41)|↵=1 =
|111.1 ⇥ 103 + j256.41(11.1)(100)(39)10 3 |
= 7.11.
11.1 ⇥ 103 + j256.41(11.1)(100)(39)10 3 |
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15–56
CHAPTER 15. Active Filter Circuits
Therefore the gain at 40.81 Hz is
20 log10 (7.11)↵=1 = 17.04 dB.
P 15.62 20 log10
.·.
✓
R 1 + R2
R1
◆
= 13.98;
R 1 + R2
= 5;
R1
Choose
.·. R2 = 4R1 .
R1 = 100 k⌦. Then
1
= 100⇡ rad/s;
R2 C1
P 15.63 [a] |H(j0)| =
R2 = 400 k⌦;
.·. C1 =
1
= 7.96 nF.
(100⇡)(400 ⇥ 103 )
R1 + ↵R2
11.1 + ↵(100)
.
=
R1 + (1 ↵)R2
11.1 + (1 ↵)100
P 15.64 [a] Combine the impedances of the capacitors in series in Fig. P15.64(b) to
get
1 ↵
↵
1
1
=
+
=
,
sCeq
sC1
sC1
sC1
which is identical to the impedance of the capacitor in Fig. P15.64(a).
[b]
Vx =
(1
↵/sC1
V = ↵V ;
↵)/sC1 + ↵/sC1
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Problems
Vy =
(1
15–57
↵R2
= ↵V = Vx .
↵)R2 + ↵R2
[c] Since x and y are both at the same potential, they can be shorted together,
and the circuit in Fig. 15.34 can thus be drawn as shown in Fig. 15.64(c).
[d] The feedback path between Vo and Vs containing the resistance R4 + 2R3
has no e↵ect on the ratio Vo /Vs , as this feedback path is not involved in
the nodal equation that defines the voltage ratio. Thus, the circuit in
Fig. P15.64(c) can be simplified into the form of Fig. 15.2, where the
input impedance is the equivalent impedance of R1 in series with the
parallel combination of (1 ↵)/sC1 and (1 ↵)R2 , and the feedback
impedance is the equivalent impedance of R1 in series with the parallel
combination of ↵/sC1 and ↵R2 :
Zi = R1 +
=
(1 ↵)
· (1
sC1
(1
R1 + (1
↵)R2
↵)R2 + (1sC↵)
1
↵)R2 + R1 R2 C1 s
;
1 + R2 C1 s
↵
· ↵R2
sC1
Zf = R1 +
↵R2 + sC↵1
=
R1 + ↵R2 + R1 R2 C1 s
.
1 + R2 C1 s
P 15.65 As ! ! 0
|H(j!)| !
2R3 + R4
= 1.
2R3 + R4
Therefore the circuit would have no e↵ect on low frequency signals. As ! ! 1
|H(j!)| !
When
[(1
[(1
=1
|H(j1)| =1 =
If R4
)R4 + Ro ]( R4 + R3 )
.
)R4 + R3 ]( R4 + Ro )
Ro (R4 + R3 )
.
R3 (R4 + Ro )
Ro
Ro
|H(j1)| =1 ⇠
> 1.
=
R3
Thus, when
= 1 we have amplification or “boost”. When
|H(j1)| =0 =
R3 (R4 + Ro )
.
Ro (R4 + R3 )
=0
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15–58
CHAPTER 15. Active Filter Circuits
If R4
Ro
R3
|H(j1)| =0 ⇠
< 1.
=
R0
Thus, when = 0 we have attenuation or “cut”.
Also note that when = 0.5
|H(j!)| =0.5 =
(0.5R4 + Ro )(0.5R4 + R3 )
= 1.
(0.5R4 + R3 )(0.5R4 + Ro )
Thus, the transition from amplification to attenuation occurs at = 0.5. If
> 0.5 we have amplification, and if < 0.5 we have attenuation.
Also note the amplification an attenuation are symmetric about = 0.5. i.e.
|H(j!)| =0.6 =
1
.
|H(j!)| =0.4
Yes, the circuit can be used as a treble volume control because
• The circuit has no e↵ect on low frequency signals;
• Depending on the circuit can either amplify ( > 0.5) or attenuate
( < 0.5) signals in the treble range;
• The amplification (boost) and attenuation (cut) are symmetric around
= 0.5. When = 0.5 the circuit has no e↵ect on signals in the treble
frequency range.
P 15.66 [a] |H(j1)| =1 =
.·.
maximum boost = 20 log10 9.99 = 19.99 dB.
[b] |H(j1)| =0 =
.·.
(65.9)(505.9)
Ro (R4 + R3 )
=
= 9.99;
R3 (R4 + Ro )
(5.9)(565.9)
R3 (R4 + R3 )
;
Ro (R4 + Ro )
maximum cut =
[c] R4 = 500 k⌦;
19.99 dB.
Ro = R1 + R3 + 2R2 = 65.9 k⌦;
.·. R4 = 7.59Ro .
Yes, R4 is significantly greater than Ro .
[d] |H(j/R3 C2 )| =1 =
=
o
(R4 + R3 )
(2R3 + R4 ) + j R
R3
(2R3 + R4 ) + j(R4 + Ro )
(505.9)
511.8 + j 65.9
5.9
511.8 + j565.9
= 7.44.
20 log10 |H(j/R3 C2 )| =1 = 20 log10 7.44 = 17.43 dB.
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Problems
[e] When
15–59
=0
|H(j/R3 C2 )| =0 =
(2R3 + R4 ) + j(R4 + Ro )
.
Ro
(2R3 + R4 ) + j (R4 + R3 )
R3
Note this is the reciprocal of |H(j/R3 C2 )| =1 .
.·. 20 log10 |H(j/R3 C2 )| +0 =
17.43 dB.
[f ] The frequency 1/R3 C2 is very nearly where the gain is 3 dB o↵ from its
maximum boost or cut. Therefore for frequencies higher than 1/R3 C2 the
circuit designer knows that gain or cut will be within 3 dB of the
maximum.
P 15.67 |H(j1)| =
[(1
[(1
)R4 + Ro ][ R4 + R3 ]
R4 + R3 ][ R4 + Ro ]
=
[(1
[(1
)500 + 65.9][ 500 + 5.9]
.
)500 + 5.9][ 500 + 65.9]
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16
Fourier Series
Assessment Problems
AP 16.1
av =
✓
1 Z 2T /3
Vm
1ZT
Vm dt +
T 0
T 2T /3 3
◆
7
dt = Vm = 7⇡ V;
9
"
#
✓
◆
Z T
2 Z 2T /3
Vm
ak =
Vm cos k!0 t dt +
cos k!0 t dt
T
3
0
2T /3
=
✓
4Vm
3k!0 T
✓
4Vm
3k!0 T
◆
4k⇡
sin
3
!
✓ ◆
4k⇡
6
=
sin
k
3
!
V;
"
#
◆
✓
Z T
Vm
2 Z 2T /3
bk =
Vm sin k!0 t dt +
sin k!0 t dt
T
3
0
2T /3
=
◆"
1
4k⇡
cos
3
!#
✓ ◆"
6
=
k
1
4k⇡
cos
3
!#
V.
AP 16.2 [a] av = 7⇡ = 21.99 V.
[b] a1 =
5.196; a2 = 2.598; a3 = 0; a4 =
b1 = 9;
✓
b2 = 4.5;
b3 = 0;
1.299; a5 = 1.039.
b4 = 2.25;
b5 = 1.8.
◆
2⇡
= 50 rad/s.
T
[d] f3 = 3f0 = 23.87 Hz.
[c] w0 =
[e] v(t) = 21.99
5.2 cos 50t + 9 sin 50t + 2.6 sin 100t + 4.5 cos 100t
1.3 sin 200t + 2.25 cos 200t + 1.04 sin 250t + 1.8 cos 250t + · · · V.
AP 16.3 Odd function with both half- and quarter-wave symmetry.
✓
◆
6Vm
vg (t) =
t,
T
0  t  T /6;
av = 0,
ak = 0 for all k;
16–1
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16–2
CHAPTER 16. Fourier Series
bk = 0 for k even;
bk =
8 Z T /4
f (t) sin k!0 t dt,
T 0
✓
◆
k odd
8 Z T /6 6Vm
8 Z T /4
=
t sin k!0 t dt +
Vm sin k!0 t dt
T 0
T
T T /6
=
✓
vg (t) =
◆
!
12Vm
k⇡
sin
;
k2⇡2
3
1
n⇡
12Vm X
1
sin n!0 t V.
sin
⇡ 2 n=1,3,5,... n2
3
AP 16.4 [a] From the results of Assessment Problem 16.1:
av = 21.99 V;
4⇡
=
a1 = 6 sin
3
5.2;
✓
b2 = 3 1
a3 = 2 sin 4⇡ = 0;
4⇡
cos
3
b1 = 6 1
8⇡
a2 = 3 sin
= 2.6;
3
16⇡
a4 = 1.5 sin
=
3
✓
b3 = 2 (1
1.3;
8⇡
cos
3
16⇡
cos
3
b5 = 1.2 1
20⇡
cos
3
✓
120 ;
A3 = 0;
A4 =
j2.25 = 2.6/
A5 = 1.04
j1.8 = 2.1/
5.2
1.3
◆
◆
= 2.25;
= 1.8.
j4.5 = 5.2/
A2 = 2.6
60 ;
120 ;
60 .
✓1 =
120 ;
✓2 =
60 ;
✓4 =
120 ;
✓5 =
60 .
[b] !0 =
= 4.5;
✓
j9 = 10.4/
A1 =
= 9;
cos 4⇡) = 0;
b4 = 1.5 1
20⇡
a5 = 1.2 sin
= 1.04;
3
◆
◆
✓3 not defined;
2⇡
2⇡
=
= 50 rad/s.
T
0.12566
v(t) = 21.99 + 10.4 cos(50t
+ 2.6 cos(200t
120 ) + 5.2 cos(100t
120 ) + 2.1 cos(250t
60 )
60 ) + · · · V.
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Problems
16–3
AP 16.5 The Fourier series for the input voltage is
✓
◆
1
n⇡
8A X
1
vi = 2
sin
sin n!0 (t + T /4)
2
⇡ n=1,3,5,... n
2
✓
◆
1
8A X
n⇡
1
sin2
cos n!0 t
= 2
2
⇡ n=1,3,5,... n
2
=
1
8A X
1
cos n!0 t;
2
⇡ n=1,3,5,... n2
8(281.25⇡ 2 )
8A
=
= 2250 mV;
⇡2
⇡2
!0 =
2⇡
2⇡
=
⇥ 103 = 10;
T
200⇡
.·.
vi = 2250
1
X
1
cos 10nt mV.
2
n=1,3,5,... n
From the circuit we have
Vo =
1
Vi
Vi
·
=
;
R + (1/j!C) j!C
1 + j!RC
Vo =
1/RC
100
Vi =
Vi ;
1/RC + j!
100 + j!
Vi1 = 2250/0 mV;
!0 = 10 rad/s;
Vi3 =
2250
/0 = 250/0 mV;
9
3!0 = 30 rad/s;
Vi5 =
2250
/0 = 90/0 mV;
25
5!0 = 50 rad/s;
Vo1 =
100
(2250/0 ) = 2238.83/
100 + j10
Vo3 =
100
(250/0 ) = 239.46/
100 + j30
Vo5 =
100
(90/0 ) = 80.50/
100 + j50
.·.
vo = 2238.33 cos(10t
+80.50 cos(50t
5.71 mV;
16.70 mV;
26.57 mV;
5.71 ) + 239.46 cos(30t
16.70 )
26.57 ) + . . . mV.
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16–4
CHAPTER 16. Fourier Series
AP 16.6 [a] !o =
2⇡
2⇡
=
(103 ) = 104 rad/s;
T
0.2⇡
vg (t) = 840
1
X
n⇡
1
sin
cos n10,000t V
2
n=1,3,5,... n
= 840 cos 10,000t
280 cos 30,000t + 168 cos 50,000t
120 cos 70,000t + · · · V;
Vg1 = 840/0 V;
Vg3 = 280/180 V;
Vg5 = 168/0 V;
Vg7 = 120/180 V;
H(s) =
=
s
Vo
= 2
;
Vg
s + s + !c2
109
1
= 4
= 5000 rad/s;
RC
10 (20)
!c2 =
(109 )(103 )
1
=
= 25 ⇥ 108 ;
LC
400
H(s) =
5000s
s2 + 5000s + 25 ⇥ 108
H(j!) =
;
j5000!
;
! 2 + j5000!
25 ⇥ 108
H1 =
j5 ⇥ 107
= 0.02/88.81 ;
24 ⇥ 108 + j5 ⇥ 107
H3 =
j15 ⇥ 107
= 0.09/84.64 ;
16 ⇥ 108 + j15 ⇥ 107
H5 =
j25 ⇥ 107
= 1/0 ;
25 ⇥ 107
H7 =
j35 ⇥ 107
= 0.14/
24 ⇥ 108 + j35 ⇥ 107
81.70 ;
Vo1 = Vg1 H1 = 17.50/88.81 V;
Vo3 = Vg3 H3 = 26.14/
95.36 V;
Vo5 = Vg5 H5 = 168/0 V;
Vo7 = Vg7 H7 = 17.32/98.30 V;
vo = 17.50 cos(10,000t + 88.81 ) + 26.14 cos(30,000t
95.36 )
+ 168 cos(50,000t) + 17.32 cos(70,000t + 98.30 ) + · · · V.
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Problems
16–5
[b] The 5th harmonic because the circuit is a passive bandpass filter with a Q
of 10 and a center frequency of 50 krad/s.
AP 16.7
w0 =
2⇡ ⇥ 103
= 3 rad/s.
2094.4
j!0 k = j3k;
VR =
2
2sVg
(Vg ) = 2
;
2 + s + 1/s
s + 2s + 1
VR
Vg
H(s) =
!
=
2s
s2 + 2s + 1
H(j!0 k) = H(j3k) =
H(j3) =
j6k
;
9k 2 ) + j6k
(1
vg1 = 25.98 sin !0 t V;
;
Vg1 = 25.98/0 V;
j6
= 0.6/
8 + j6
53.13 ;
VR1 = 15.588/
53.13 V;
p
(15.588/ 2)2
= 60.75 W;
P1 =
2
vg3 = 0,
vg5 =
therefore P3 = 0 W;
1.04 sin 5!0 t V;
H(j15) =
Vg5 = 1.04/180 ;
j30
= 0.1327/
224 + j30
VR5 = (1.04/180 )(0.1327/
82.37 ;
82.37 ) = 138/97.63 mV;
p
(0.1396/ 2)2
P5 =
= 4.76 mW;
2
therefore P ⇠
= P1 ⇠
= 60.75 W.
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16–6
CHAPTER 16. Fourier Series
AP 16.8 Odd function with half- and quarter-wave symmetry, therefore av = 0, ak = 0
for all k, bk = 0 for k even; for k odd we have
8 Z T /8
8 Z T /4
bk =
2 sin k!0 t dt +
8 sin k!0 t dt
T 0
T T /8
=
✓
8
⇡k
◆"
Therefore Cn =
AP 16.9 [a] Irms =
[b] C1 =
C7 =
!#
k⇡
1 + 3 cos
4
s
✓
,
k odd.
j4
n⇡
◆
✓
✓
◆
✓

n⇡
1 + 3 cos
4
◆
,
n odd.
◆
T
2
T
3T
(2)2
(2) + (8)2
T
8
8
8
j1.5
j0.9
j12.5
; C3 =
; C5 =
;
⇡
⇡
⇡
j1.8
;
⇡
j1.4
;
⇡
C9 =
v
u
u
Irms = tI 2 + 2
dc
1
X
n=1,3,5,...
|Cn |2 ⇠
=
C11 =
s
=
p
34 = 5.7683 A.
j0.4
;
⇡
2
(12.52 + 1.52 + 0.92 + 1.82 + 1.42 + 0.42 )
⇡2
⇠
= 5.777 A.
5.777 5.831
⇥ 100 =
5.831
[d] Using just the terms C1 – C9 ,
[c] % Error =
v
u
u
Irms = tI 2 + 2
dc
1
X
n=1,3,5,...
|Cn |2 ⇠
=
s
0.93%.
2
(12.52 + 1.52 + 0.92 + 1.82 + 1.42 )
⇡2
⇠
= 5.774 A;
% Error =
5.774 5.831
⇥ 100 =
5.831
0.98%.
Thus, the % error is still less than 1%.
AP 16.10 T = 32 ms, therefore 8 ms requires shifting the function T /4 to the right.
i=
1
X
n= 1
n(odd)
1
X
✓
◆
4
n⇡ jn!0 (t T /4)
j
1 + 3 cos
e
n⇡
4
✓
◆
n⇡
4
1
1 + 3 cos
e j(n+1)(⇡/2) ejn!0 t .
=
⇡ n= 1 n
4
n(odd)
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Problems
16–7
Problems
P 16.1
2⇡
= 31, 415.93 rad/s;
200 ⇥ 10 6
[a] !oa =
2⇡
= 3978.87 rad/s.
40 ⇥ 10 6
!ob =
1
1
1
=
= 5000 Hz;
fob =
= 25,000 Hz.
6
T
200 ⇥ 10
40 ⇥ 10 6
100(10 ⇥ 10 6 )
avb =
= 25 V.
[c] ava = 0;
40 ⇥ 10 6
[d] The periodic function in Fig. P16.1(a):
[b] foa =
ava = 0.
aka =
2 Z T /2
2 Z T /4
2⇡kt
2⇡kt
dt +
dt
40 cos
80 cos
T 0
T
T T /4
T
2 Z 3T /4
2ZT
2⇡kt
2⇡kt
+
dt +
dt
40 cos
80 cos
T T /2
T
T 3T /4
T
=
80 T
2⇡kt T /4 160 T
2⇡kt T /2
sin
sin
+
T 2⇡k
T 0
T 2⇡k
T T /4
+
2⇡kt 3T /4
2⇡kt T
160 T
80 T
sin
sin
+
T 2⇡k
T T /2
T 2⇡k
T 3T /4
⇡k
80
sin
,
⇡k
2
=
k odd;
aka = 0,
k even.
2 Z T /2
2 Z T /4
2⇡kt
2⇡kt
dt +
dt
40 sin
80 sin
bka =
T 0
T
T T /4
T
+
=
80 T
2⇡kt T /4
cos
T 2⇡k
T 0
+
=
2 Z 3T /4
2ZT
2⇡kt
2⇡kt
dt +
dt
40 sin
80 sin
T T /2
T
T 3T /4
T
2⇡kt T /2
160 T
cos
T 2⇡k
T T /4
2⇡kt 3T /4 160 T
2⇡kt T
80 T
cos
cos
+
T 2⇡k
T T /2
T 2⇡k
T 3T /4
240
,
⇡k
k odd;
bka = 0,
k even.
The periodic function in Fig. P16.1(b):
avb = 25 V
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16–8
CHAPTER 16. Fourier Series
akb =
200 T
2⇡k T /8
⇡k
2 Z T /8
2⇡kt
200
dt =
sin
t
sin
.
100 cos
=
T T /8
T
T 2⇡k
T
⇡k
4
T /8
bkb =
2 Z T /8
2⇡kt
dt =
100 sin
T T /8
T
200 T
2⇡k T /8
cos
t
= 0.
T 2⇡k
T
T /8
[e] For the periodic function in Fig. P16.1(a),
1
80 X
v(t) =
⇡ n=1,3,5
✓
◆
n⇡
3
1
sin
cos n!o t + sin n!o t V.
n
2
n
For the periodic function in Fig. P16.1(b),
◆
✓
1
n⇡
200 X
1
sin
cos n!o t V.
v(t) = 25 +
⇡ n=1 n
4
P 16.2
In studying the periodic function in Fig. P16.2 note that it can be visualized
as the combination of two half-wave rectified sine waves, as shown in the figure
below. Hence we can use the Fourier series for a half-wave rectified sine wave
with a magnitude of Vm , which is derived first.
av =
✓
◆
1 Z T /2
Vm
2⇡
;
Vm sin
t dt =
T 0
T
⇡
2 Z T /2
Vm
2⇡
ak =
Vm sin t cos k!0 t dt =
T 0
T
⇡
Note: ak = 0 for k-odd, ak =
bk =
!
1 + cos k⇡
;
1 k2
2Vm
⇡(1 k 2 )
for k even.
2 Z T /2
2⇡
Vm sin t sin k!0 t dt = 0 for k = 2, 3, 4, . . . .
T 0
T
For k = 1, we have b1 =
v(t) =
Vm
;
2
therefore
1
V m Vm
2Vm X
1
+
sin !0 t +
cos n!0 t V.
⇡
2
⇡ n=2,4,6,... 1 n2
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Problems
v1 (t) =
1
100
200 X
cos n!o t
+ 50 sin !o t +
V;
⇡
⇡ n=2,4,6 (n2 1)
v2 (t) =
60
+ 30 sin !o (t
⇡
T /2) +
16–9
1
120 X
cos n!o (t T /2)
V.
⇡ n=2,4,6
(n2 1)
Observe the following:
✓
2⇡ T
T 2
sin !o (t
T /2) = sin !o t
cos n!o (t
T /2) = cos n!o t
✓
◆
= sin(!o t
2⇡n T
T 2
◆
⇡) =
= cos(n!o t
sin !o t;
n⇡) = cos n!o t.
Using the observations above and that fact that n is even,
60
v2 (t) =
⇡
30 sin !o t
1
120 X
cos(n!o t)
V.
⇡ n=2,4,6 (n2 1)
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16–10
CHAPTER 16. Fourier Series
Thus,
1
320 X
cos(n!o t)
V.
⇡ n=2,4,6 (n2 1)
160
+ 20 sin !o t
v(t) = v1 (t) + v2 (t) =
⇡
P 16.3
[a] Odd function with half- and quarter-wave symmetry, av = 0, ak = 0 for all
k, bk = 0 for even k; for k odd we have
8 Z T /4
4Vm
,
Vm sin k!0 t dt =
bk =
T 0
k⇡
and v(t) =
k odd
1
4Vm X
1
sin n!0 t V.
⇡ n=1,3,5,... n
[b] Even function: bk = 0 for k
av =
ak =
=
2Vm
2 Z T /2
⇡
;
Vm sin t dt =
T 0
T
⇡
4 Z T /2
2Vm
⇡
Vm sin t cos k!0 t dt =
T 0
T
⇡
✓
1
1
2k
+
1
1 + 2k
◆
4Vm /⇡
1 4k 2
"
#
1
X
2Vm
1
and v(t) =
cos n!0 t V.
1+2
⇡
4n2
n=1 1
✓
◆
1 Z T /2
Vm
2⇡
;
[c] av =
Vm sin
t dt =
T 0
T
⇡
2 Z T /2
Vm
2⇡
ak =
Vm sin t cos k!0 t dt =
T 0
T
⇡
Note: ak = 0 for k-odd, ak =
bk =
!
1 + cos k⇡
;
1 k2
2Vm
⇡(1 k 2 )
for k even.
2 Z T /2
2⇡
Vm sin t sin k!0 t dt = 0 for k = 2, 3, 4, . . . .
T 0
T
For k = 1, we have b1 =
v(t) =
Vm
;
2
therefore
1
V m Vm
2Vm X
1
+
sin !0 t +
cos n!0 t V.
⇡
2
⇡ n=2,4,6,... 1 n2
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Problems
P 16.4
16–11
5
1 Z T /4
1 Z T Vm
av =
dt = Vm = 62.5⇡ V;
Vm dt +
T 0
T T /4 2
8
"
#
Z T
2 Z T /4
Vm
cos k!0 t dt
Vm cos k!0 t dt +
ak =
T 0
T /4 2
=
k⇡
50
k⇡
Vm
sin
=
sin
;
k!0 T
2
k
2
#
"
Z T
2 Z T /4
Vm
bk =
sin k!0 t dt
Vm sin k!0 t dt +
T 0
T /4 2
"
#
Vm
=
1
k!0 T
P 16.5
[a] I1 =
Z to +T
to
"
k⇡
50
cos
=
1
2
k
to +T
1
cos m!0 t
m!0
to
sin m!0 t dt =
=
1
[cos m!0 (to + T )
m!0
=
1
[cos m!0 to cos m!0 T
m!0
=
1
[cos m!0 to
m!0
I2 =
Z to +T
to
0
#
k⇡
cos
.
2
cos m!0 to ]
sin m!0 to sin m!0 T
cos m!0 to ]
cos m!0 to ] = 0 for all m;
cos m!0 to dt =
=
1
[sin m!0 (to + T )
m!0
=
1
[sin m!0 to
m!0
to +T
1
[sin m!0 t]
m!0
to
sin m!0 to ]
sin m!0 to ] = 0 for all m.
Z to +T
1 Z to +T
cos m!0 t sin n!0 t dt =
[b] I3 =
[sin(m + n)!0 t sin(m n)!0 t] dt.
2 to
to
But (m + n) and (m n) are integers, therefore from I1 above, I3 = 0 for
all m, n.
Z to +T
1 Z to +T
[c] I4 =
[cos(m n)!0 t cos(m + n)!0 t] dt.
sin m!0 t sin n!0 t dt =
2 to
to
If m 6= n, both integrals are zero (I2 above). If m = n, we get
I4 =
1 Z to +T
dt
2 to
1 Z to +T
T
cos 2m!0 t dt =
2 to
2
0=
T
.
2
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16–12
CHAPTER 16. Fourier Series
[d] I5 =
Z to +T
=
to
cos m!0 t cos n!0 t dt
1 Z to +T
[cos(m
2 to
n)!0 t + cos(m + n)!0 t] dt.
If m 6= n, both integrals are zero (I2 above). If m = n, we have
T
1 Z to +T
T
1 Z to +T
dt +
cos 2m!0 t dt = + 0 = .
I5 =
2 to
2 to
2
2
P 16.6
f (t) sin k!0 t = av sin k!0 t +
1
X
an cos n!0 t sin k!0 t +
n=1
1
X
bn sin n!0 t sin k!0 t.
n=1
Now integrate both sides from to to to + T. All the integrals on the right-hand
side reduce to zero except in the last summation when n = k, therefore we
have
Z to +T
to
P 16.7
f (t) sin k!0 t dt = 0 + 0 + bk
1 Z to +T
1
av =
f (t) dt =
T to
T
Let t =
x,
dt =
dx,
(Z
0
T /2
x=
✓
T
2
◆
or bk =
f (t) dt +
T
2
Z T /2
0
2 Z to +T
f (t) sin k!0 t dt.
T to
)
f (t) dt ;
when t =
T
2
and x = 0 when t = 0.
Therefore
1Z0
1Z0
f (t) dt =
f ( x)( dx) =
T T /2
T T /2
Therefore av =
ak =
1 Z T /2
f (x) dx.
T 0
1 Z T /2
1 Z T /2
f (t) dt +
f (t) dt = 0;
T 0
T 0
2Z0
2 Z T /2
f (t) cos k!0 t dt +
f (t) cos k!0 t dt.
T T /2
T 0
Again, let t =
x in the first integral and we get
2Z0
f (t) cos k!0 t dt =
T T /2
Therefore ak = 0
2 Z T /2
f (x) cos k!0 x dx.
T 0
for all k.
2Z0
2 Z T /2
bk =
f (t) sin k!0 t dt +
f (t) sin k!0 t dt.
T T /2
T 0
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Problems
Using the substitution t =
16–13
x, the first integral becomes
2 Z T /2
f (x) sin k!0 x dx;
T 0
4 Z T /2
Therefore we have bk =
f (t) sin k!0 t dt.
T 0
P 16.8
2Z0
2 Z T /2
bk =
f (t) sin k!0 t dt +
f (t) sin k!0 t dt.
T T /2
T 0
Now let t = x T /2 in the first integral, then dt = dx, x = 0 when t =
and x = T /2 when t = 0, also
sin k!0 (x T /2) = sin(k!0 x k⇡) = sin k!0 x cos k⇡. Therefore
2Z0
f (t) sin k!0 t dt =
T T /2
2
bk = (1
T
cos k⇡)
Z T /2
0
2 Z T /2
f (x) sin k!0 x cos k⇡ dx and
T 0
f (x) sin k!0 t dt.
Now note that 1 cos k⇡ = 0 when k is even, and 1
odd. Therefore bk = 0 when k is even, and
bk =
P 16.9
T /2
cos k⇡ = 2 when k is
4 Z T /2
f (t) sin k!0 t dt when k is odd.
T 0
Because the function is even and has half-wave symmetry, we have av = 0,
ak = 0 for k even, bk = 0 for all k and
4 Z T /2
f (t) cos k!0 t dt,
ak =
T 0
k odd.
The function also has quarter-wave symmetry;
therefore f (t) = f (T /2 t) in the interval T /4  t  T /2;
thus we write
ak =
4 Z T /2
4 Z T /4
f (t) cos k!0 t dt +
f (t) cos k!0 t dt.
T 0
T T /4
Now let t = (T /2 x) in the second integral, then dt =
t = T /4 and x = 0 when t = T /2. Therefore we get
4 Z T /2
f (t) cos k!0 t dt =
T T /4
dx, x = T /4 when
4 Z T /4
f (x) cos k⇡ cos k!0 x dx.
T 0
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16–14
CHAPTER 16. Fourier Series
Therefore we have
4
ak = (1
T
cos k⇡)
Z T /4
0
f (t) cos k!0 t dt.
But k is odd, hence
8 Z T /4
f (t) cos k!0 t dt,
ak =
T 0
k odd.
P 16.10 Because the function is odd and has half-wave symmetry, av = 0, ak = 0 for all
k, and bk = 0 for k even. For k odd we have
bk =
4 Z T /2
f (t) sin k!0 t dt.
T 0
The function also has quarter-wave symmetry, therefore f (t) = f (T /2
the interval T /4  t  T /2. Thus we have
bk =
t) in
4 Z T /2
4 Z T /4
f (t) sin k!0 t dt +
f (t) sin k!0 t dt.
T 0
T T /4
Now let t = (T /2 x) in the second integral and note that dt =
when t = T /4 and x = 0 when t = T /2, thus
4 Z T /2
f (t) sin k!0 t dt =
T T /4
dx, x = T /4
Z T /4
4
cos k⇡
f (x)(sin k!0 x) dx.
T
0
But k is odd, therefore the expression for bk becomes
bk =
8 Z T /4
f (t) sin k!0 t dt.
T 0
P 16.11 [a] !o =
2⇡
= ⇡ rad/s.
T
[b] Yes.
[c] No.
[d] No.
P 16.12 [a] f =
[b] No.
1
1
=
= 62.5 Hz.
T
16 ⇥ 10 3
[c] Yes.
[d] Yes.
[e] Yes.
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Problems
[f ] av = 0,
function is odd;
ak = 0,
for all k; the function is odd;
bk = 0,
for k even, the function has half-wave symmetry;
bk =
8 Z T /4
f (t) sin k!o t,
T 0
8
=
T
=
16–15
(Z
T /8
0
k odd
5t sin k!o t dt +
Z T /4
T /8
)
0.01 sin k!o t dt
8
{Int1 + Int2.}
T
Z T /8
Int1 = 5
0
t sin k!o t dt
"
T /8
t
cos k!o t
k!o
0
1
= 5 2 2 sin k!o t
k !o
=
5
k 2 !o2
sin
Z T /4
Int2 = 0.01
T /8
k⇡
4
#
0.625T
k⇡
;
cos
k!o
4
sin k!o t dt =
T /4
0.01
k⇡
0.01
;
cos k!o t
cos
=
k!o
k!o
4
T /8
✓
k⇡
5
0.01
+
Int1 + Int2 = 2 2 sin
k !o
4
k!o
0.625T
k!o
◆
cos
k⇡
;
4
0.625T = 0.625(16 ⇥ 10 3 ) = 0.01;
.·.
Int1 + Int2 =
bk =

5
k 2 !o2
sin
k⇡
.
4
5
0.16
k⇡
k⇡
8
· 2 2 · T 2 sin
= 2 2 sin
,
T 4⇡ k
4
⇡ k
4
k odd.
P 16.13 [a] v(t) is even and has both half- and quarter-wave symmetry, therefore
av = 0, bk = 0 for all k, ak = 0 for k-even; for odd k we have
!
8 Z T /4
4Vm
k⇡
sin
.
Vm cos k!0 t dt =
ak =
T 0
⇡k
2

1
4Vm X
n⇡
1
v(t) =
sin
cos n!0 t V.
⇡ n=1,3,5,... n
2
[b] v(t) is even and has both half- and quarter-wave symmetry, therefore
av = 0, bk = 0 for k-even, ak = 0 for all k; for k-odd we have
✓
8 Z T /4 4Vp
t
ak =
T 0
T
◆
Vp cos k!0 t dt =
8Vp
.
2
⇡ k2
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16–16
CHAPTER 16. Fourier Series
Therefore v(t) =
1
8Vp X
1
cos n!0 t V.
2
⇡ n=1,3,5,... n2
P 16.14 [a]
[b] av = 0;
bk =
for all k;
bk = 0,
8 Z T /4
f (t) sin k!0 t dt,
T 0
for k odd
=
ak = 0,
for k even.
◆
✓
8 Z T /4
40
8 Z T /8 120t
sin k!0 t dt +
10 + t sin k!0 t dt
T 0
T
T T /8
T
960 Z T /8
80 Z T /4
320 Z T /4
= 2
t sin k!0 t dt +
sin k!0 t dt + 2
t sin k!0 t dt
T 0
T T /8
T T /8
"
#T /8
"
320 sin k!0 t
+ 2
T
k 2 !02
k!0
k⇡
T
=
;
4
2
k!0
80 cos k!0 t T /4
T k!0
T /8
t cos k!0 t
k!0
0
960 sin k!0 t
= 2
T
k 2 !02
#T /4
t cos k!0 t
k!0
T /8
k⇡
T
=
;
8
4
"
#
960 sin(k⇡/4)
bk = 2
T
k 2 !02
T
cos(k⇡/4)
8k!0
80
[cos(k⇡/2)
k!0 T
"
T cos(k⇡/2)
4
k!0
sin(k⇡/4) T cos(k⇡/4)
+
k 2 !02
8k!0
320 sin(k⇡/2)
+ 2
T
k 2 !02
=
320
960
sin(k⇡/4) +
sin(k⇡/2)
2
(k!0 T )
(k!0 T )2
=
640
320
sin(k⇡/4) +
sin(k⇡/2).
2
(k!0 T )
(k!0 T )2
k!0 T = 2k⇡;
cos(k⇡/4)]
#
320
sin(k⇡/4)
(k!o T )2
(k!0 T )2 = 4k 2 ⇡ 2 ;
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Problems
bk =
[c] bk =
160
80
sin(k⇡/4) + 2 2 sin(k⇡/2).
2
2
⇡ k
⇡ k
80
[2 sin(k⇡/4) + sin(k⇡/2)];
⇡2k2
b1 =
80
[2 sin(⇡/4) + sin(⇡/2)] ⇠
= 19.57;
⇡2
b3 =
80
[2 sin(3⇡/4) + sin(3⇡/2)] ⇠
= 0.37;
9⇡ 2
b5 =
80
[2 sin(5⇡/4) + sin(5⇡/2)] ⇠
=
25⇡ 2
f (t) = 19.57 sin(!0 t) + 0.37 sin(3!0 t)
[d] t =
16–17
T
;
4
!0 t =
0.13;
0.13 sin(5!0 t) + . . . .
⇡
2⇡ T
· = ;
T 4
2
f (T /4) ⇠
= 19.57 sin(⇡/2) + 0.37 sin(3⇡/2)
0.13 sin(5⇡/2) ⇠
= 19.81.
P 16.15 [a]
[b] Even, since f (t) = f ( t).
[c] Yes, since f (t) =
f (T /2
t) in the interval 0 < t < 4.
[d] av = 0,
ak = 0,
for k even (half-wave symmetry);
bk = 0,
for all k
(function is even).
Because of the quarter-wave symmetry, the expression for ak is
8 Z T /4
f (t) cos k!0 t dt,
ak =
T 0
k odd
"
#2
k!02 t2 2
8Z 2 2
2t
sin k!0 t ;
4t cos k!0 t dt = 4 2 2 cos k!0 t +
=
8 0
k !0
k 3 !03
0
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16–18
CHAPTER 16. Fourier Series
✓
◆
k⇡
2⇡
k!0 (2) = k
;
(2) =
8
2
cos(k⇡/2) = 0,
since k is odd;
"
#
.·.
4k 2 !02 2
16k 2 !02 8
sin(k⇡/2)
=
sin(k⇡/2).
ak = 4 0 +
k 3 !03
k 3 !03
!0 =
⇡
2⇡
= ;
8
4
ak =
k2⇡2 8
(64) sin(k⇡/2)
k3⇡3
!02 =
⇡2
;
16
!03 =
⇡3
;
64
!
f (t) = 64
1
X
n=1,3,5,···
[e] cos n!0 (t
"
#
n2 ⇡ 2 8
sin(n⇡/2) cos(n!0 t).
⇡ 3 n3
2) = cos(n!0 t
f (t) = 64
1
X
n=1,3,5,···
"
⇡/2) = sin n!0 t;
#
n2 ⇡ 2 8
sin(n⇡/2) sin(n!0 t).
⇡ 3 n3
P 16.16 [a]
[b] Odd, since f ( t) =
f (t).
[c] f (t) has quarter-wave symmetry, since f (T /2
0 < t < 4.
[d] av = 0,
bk = 0,
bk =
(half-wave symmetry);
for all k
(function is odd);
for k even (half-wave symmetry);
8 Z T /4
f (t) sin k!0 t dt,
T 0
=
ak = 0,
t) = f (t) in the interval
k odd
8Z 2 3
t sin k!0 t dt
8 0
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
"
3t2
= 2 2 sin k!0 t
k !0
✓
6
sin k!0 t
4
k !04
16–19
#2
t3
6t
cos k!0 t + 3 3 cos k!0 t .
k!0
k !0
0
◆
k⇡
2⇡
;
k!0 (2) = k
(2) =
8
2
cos(k⇡/2) = 0,
since k is odd;
"
#
6
sin(k⇡/2) .
4
k !02
24
bk = 2 2 sin(k⇡/2)
k !0
.·.
k!0 = k
✓
2⇡
8
◆
=
k⇡
;
4

384
bk = 2 2 1
⇡ k
.·.
k 2 !02 =
4
sin(k⇡/2),
⇡2k2

✓
f (t) =
2) = sin(n!0 t

k 4 !04 =
k4⇡4
;
256
k odd.
◆
1
384 X
1
f (t) =
1
⇡ n=1,3,5,··· n2
[e] sin n!0 (t
k2⇡2
;
16
4
sin(n⇡/2) sin n!0 t.
2
⇡ n2
⇡/2) =
cos n!0 t;
✓
◆
1
384 X
1
1
⇡ n=1,3,5,··· n2
4
sin(n⇡/2) cos n!0 t.
2
⇡ n2
P 16.17 [a] i(t) is even, therefore bk = 0 for all k.
Im
1
1 T
· · Im · 2 · =
A;
2 4
T
4
av =
✓
4 Z T /4
ak =
Im
T 0
=
◆
4Im
t cos k!o t dt
T
4Im Z T /4
cos k!o t dt
T
0
= Int1
Int1 =
Int2 =
16Im Z T /4
t cos k!o t dt
T2 0
Int2 .
k⇡
4Im Z T /4
2Im
sin
;
cos k!o t dt =
T 0
⇡k
2
16Im Z T /4
t cos k!o t dt
T2 0
16Im
=
T2
(
T /4
t
1
cos
k!
t
+
sin
k!
t
o
o
k 2 !o2
k!o
0
k⇡
4Im
= 2 2 cos
⇡ k
2
!
1 +
)
k⇡
2Im
sin
;
k⇡
2
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16–20
CHAPTER 16. Fourier Series
4Im
.·. ak = 2 2 1
⇡ k
.·. i(t) =
k⇡
cos
2
!
1
Im 4Im X
1
+ 2
4
⇡ n=1
A.
cos(n⇡/2)
cos n!o t A.
n2
[b] Shifting the reference axis to the left is equivalent to shifting the periodic
function to the right:
cos n!o (t
T /2) = cos n⇡ cos n!o t.
Thus
i(t) =
1
Im 4Im X
(1
+ 2
4
⇡ n=1
cos(n⇡/2)) cos n⇡
cos n!o t A.
n2
P 16.18 v2 (t + T /8) is even, so bk = 0 for all k.
av =
Vm
(Vm /2)(T /4)
=
;
T
8
k⇡
4 Z T /8 Vm
Vm
ak =
cos k!0 t dt =
sin
.
T 0 2
k⇡
4
Therefore,
1
n⇡
Vm Vm X
1
v2 (t + T /8) =
+
sin
cos n!0 t
8
⇡ n=1 n
4
so v2 (t) =
1
n⇡
Vm Vm X
1
+
sin
cos n!0 (t
8
⇡ n=1 n
4
T /8);
✓
◆
✓
◆
1
n⇡
n⇡
n⇡
Vm Vm Vm X
1
1
+
+
sin
cos
sin2
cos n!0 t +
sin n!0 t
v(t) =
2
8
⇡ n=1 n
4
4
n
4
.·.
✓
◆
✓
1
5Vm Vm X
n⇡
1
=
+
sin
cos n!0 t + 1
8
2⇡ n=1 n
2
◆
n⇡
cos
sin n!0 t V.
2
Thus, since av = 5Vm /8 = 62.5⇡ V,
ak =
k⇡
50
k⇡
Vm
sin
=
sin
2⇡k
2
k
2
and
"
Vm
bk =
1
2⇡k
#
"
k⇡
50
cos
=
1
2
k
#
k⇡
cos
.
2
These equations match the equations for av , ak , and bk derived in Problem
16.4.
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Problems
16–21
P 16.19 From Problem 16.2,
v(t) =
1
320 X
cos(n!o t)
V.
⇡ n=2,4,6,··· (n2 1)
160
+ 20 sin !o t
⇡
Therefore,
av =
an =
160
V;
⇡
320
⇡(n2
b1 = 20
for n even;
1)
and
bn = 0
for all other n.
Therefore,
A1 = 20
and
✓1 = 90 ;
and
An =
320
⇡(n2
Thus, v(t) =
and ✓n = 0
1)
for all even n.
160
+ 20 cos(!o t + 90 )
⇡
1
320 X
cos(n!o t)
V.
⇡ n=2,4,6,··· (n2 1)
P 16.20 The periodic function in Problem 16.12 is odd, so av = 0 and ak = 0 for all k.
Thus,
Ak /
✓k = a k
jbk = 0
jbk = bk /
90 .
From Problem 16.12,
bk =
0.16
k⇡
,
sin
2
2
⇡ k
4
k odd.
Therefore,
Ak =
0.16
k⇡
,
sin
2
2
⇡ k
4
k odd,
and
✓k =
90 ,
Thus, i(t) =
k odd.
1
160 X
sin(n⇡/4)
cos(125n⇡t
2
⇡ n=1,3,5,···
n2
90 ) mA.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16–22
CHAPTER 16. Fourier Series
P 16.21 The periodic function in Problem 16.15 is even, so bk = 0 for all k. Thus,
Ak /
✓k = a k
jbk = ak = ak /0 .
From Problem 16.15,
av = 0 = A 0 ;
!
k2⇡2 8
(64) sin(k⇡/2).
k3⇡3
ak =
Therefore,
Ak =
!
k2⇡2 8
(64) sin(k⇡/2)
k3⇡3
and
✓k = 0 .
1
X
Thus, f (t) = 64
n=1,3,5,···
"
#
n2 ⇡ 2 8
sin(n⇡/2) cos(n⇡t/4).
⇡ 3 n3
P 16.22 [a] The current has half-wave symmetry. Therefore,
av = 0;
ak = bk = 0,
k even.
For k odd,
✓
4 Z T /2
ak =
Im
T 0
=
◆
2Im
t cos k!o t dt
T
4 Z T /2
Im cos k!0 t dt
T 0
4Im sin k!0 t T /2
=
T
k!0
0
=0
✓
8Im
=
T2
=
"
8Im cos k⇡
T 2 k 2 !02
◆
!
1
(1
2
k !02
4Im
20
= 2,
2
2
⇡ k
k
8Im Z T /2
t cos k!0 t dt
T2 0
"
#T /2
t
8Im cos k!o t
+
sin k!0 t
2
T2
k 2 !0
k!0
0
1
2
k !02
#
cos k⇡)
for k odd.
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Problems
◆
✓
4 Z T /2
bk =
Im
T 0
=
2Im
t sin k!o t dt
T
4Im Z T /2
sin k!0 t dt
T 0
4Im
=
T
16–23
"
cos k!0 t
k!0
0
"
#
8Im Z T /2
t sin k!0 t dt
T2 0
#T /2
4Im 1 cos k⇡
=
T
k!0
"
#T /2
8Im sin k!o t
T2
k 2 !02
8Im
T2
"
T cos k⇡
2k!0
t
cos k!0 t
k!0
0
#

8Im 1
=
k!0 T 2
=
ak
10⇡
2Im
=
,
⇡k
k
20
jbk = 2
k
where
i(t) = 10
✓
10 2
10⇡
=
j
k
k k
tan ✓k =
1
X
n=1,3,5,...
for k odd.
◆
10 p 2 2
⇡ k + 4/
k2
j⇡ =
⇡k
.
2
q
(n⇡)2 + 4
n2
p
[b] A1 = 10 4 + ⇡ 2 ⇠
= 37.24 A
cos(n!0 t
✓n ),
tan ✓1 =
⇡
2
✓n = tan 1
10 p
3⇡
tan ✓3 =
4 + 9⇡ 2 ⇠
= 10.71 A
9
2
10 p
5⇡
tan ✓5 =
A5 =
4 + 25⇡ 2 ⇠
= 6.33 A
25
2
10 p
7⇡
tan ✓7 =
A7 =
4 + 49⇡ 2 ⇠
= 4.51 A
49
2
10 p
9⇡
tan ✓9 =
A9 =
4 + 81⇡ 2 ⇠
= 3.50 A
81
2
i(t) ⇠
= 37.24 cos(!o t 57.52 ) + 10.71 cos(3!o t
+6.33 cos(5!o t
82.74 ) + 4.51 cos(7!o t
+3.50 cos(9!o t
85.95 ) + . . . .
i(T /4) ⇠
= 37.24 cos(90
✓3 ⇠
= 78.02 ;
✓5 ⇠
= 82.74 ;
✓7 ⇠
= 84.80 ;
✓9 ⇠
= 85.95 ;
78.02 )
84.80 )
57.52 ) + 10.71 cos(270
82.74 ) + 4.51 cos(630
n⇡
.
2
✓1 ⇠
= 57.52 ;
A3 =
+6.33 cos(450
✓k
78.02 )
84.80 )
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16–24
CHAPTER 16. Fourier Series
85.95 ) ⇠
= 26.23 A.
+3.50 cos(810
Actual value:
✓ ◆
1
T
i
= (5⇡ 2 ) ⇠
= 24.67 A.
4
2
P 16.23 The function has half-wave symmetry, thus ak = bk = 0 for k-even, av = 0; for
k-odd
4 Z T /2
ak =
Vm cos k!0 t dt
T 0
h
8Vm Z T /2 t/RC
e
cos k!0 t dt
⇢T 0
i
where ⇢ = 1 + e T /2RC .
Upon integrating we get
ak =
4Vm sin k!0 t T /2
T
k!0
0
(
8Vm
e t/RC
·
·
⇢T
(1/RC)2 + (k!0 )2
=
8Vm RC
T [1 + (k!0 RC)2 ].
bk =
4 Z T /2
Vm sin k!0 t dt
T 0
=
4Vm cos k!0 t T /2
T
k!0
0
"
#
cos k!0 t
+ k!0 sin k!0 t
RC
(
P 16.24 [a]
4Vm
⇡k
0
)
8Vm Z T /2 t/RC
e
sin k!0 t dt
⇢T 0
"
#
8Vm
e t/RC
sin k!0 t
·
+ k!0 cos k!0 t
·
2
2
⇢T
(1/RC) + (k!0 )
RC
=
T /2
T /2
0
)
8k!0 Vm R2 C 2
T [1 + (k!0 RC)2 ].
a2k + b2k = a2k +
✓
◆
2
4Vm
+ k!0 RCak
⇡k
= a2k [1 + (k!0 RC)2 ] + 8V⇡km
But ak =
(
Therefore
a2k =
h
2Vm
+ k!0 RCak
⇡k
i
.
8Vm RC
}
T [1 + (k!0 RC)2 ]
(
)
64Vm2 R2 C 2
,
T 2 [1 + (k!0 RC)2 ]2
thus we have
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Problems
a2k + b2k =
16Vm2
64Vm2 R2 C 2
+
T 2 [1 + (k!0 RC)2 ]
⇡2k2
16–25
64Vm2 k!0 R2 C 2
.
⇡kT [1 + (k!0 RC)2 ]
Now let ↵ = k!0 RC and note that T = 2⇡/!0 , thus the expression for
a2k + b2k reduces to a2k + b2k = 16Vm2 /⇡ 2 k 2 (1 + ↵2 ). It follows that
q
a2k + b2k =
4Vm
q
⇡k 1 + (k!0 RC)2
[b] bk = k!0 RCak +
.
4Vm
.
⇡k
4Vm
bk
= k!0 RC +
=↵
ak
⇡kak
ak
= ↵ = k!0 RC.
Therefore
bk
Thus
1 + ↵2
=
↵
1
.
↵
P 16.25 Since av = 0 (half-wave symmetry), Eq. 16.20 gives us
vo (t) =
1
X
4Vm
1,3,5,...
n⇡
1
q
1 + (n!0 RC)2
cos(n!0 t
✓n ) where
tan ✓n =
bn
.
an
But tan k = k!0 RC. It follows from Eq. 16.29 that tan k = ak /bk or
tan ✓n = cot n . Therefore ✓n = 90 + n and
cos(n!0 t ✓n ) = cos(n!0 t
90 ) = sin(n!0 t
n
n ), thus our expression
for vo becomes
vo =
1
4Vm X
sin(n!0 t
n)
q
.
⇡ n=1,3,5,... n 1 + (n!0 RC)2
P 16.26 [a] e x ⇠
=1
e
x for small x;
t/RC ⇠
✓
t
RC
= 1
vo ⇠
= Vm
⇠
=
✓
✓
Vm
RC
◆
therefore
and e
T /2RC ⇠
= 1
✓
◆"
Vm
t
RC
Vm T
4RC
2Vm [1 (t/RC)]
Vm
=
2 (T /2RC)
RC
◆✓
◆
t
T
4
✓
◆
=
8
8
[b] ak =
Vp =
2
2
2
⇡ k
⇡ k2
✓
✓
◆✓
◆
Vm T
4RC
◆
=
2t
2
◆
T
.
2RC
(T /2)
(T /2RC)
#
for 0  t 
T
.
2
4Vm
.
⇡!0 RCk 2
P 16.27 [a] Express vg as a constant plus a symmetrical square wave. The constant is
Vm /2 and the square wave has an amplitude of Vm /2, is odd, and has
half- and quarter-wave symmetry. Therefore the Fourier series for vg is
1
Vm 2Vm X
1
vg =
+
sin n!0 t.
2
⇡ n=1,3,5,... n
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16–26
CHAPTER 16. Fourier Series
The dc component of the current is Vm /2R and the kth harmonic phase
current is
2Vm
2Vm /k⇡
q
/ ✓k
Ik =
=
R + jk!0 L
k⇡ R2 + (k!0 L)2
where ✓k = tan
1
!
k!0 L
.
R
Thus the Fourier series for the steady-state current is
i=
[b]
1
sin(n!0 t ✓n )
Vm 2Vm X
q
+
A.
2R
⇡ n=1,3,5,... n R2 + (n!0 L)2
The steady-state current will alternate between I1 and I2 in exponential
traces as shown. Assuming t = 0 at the instant i increases toward
(Vm /R), we have
✓
Vm
i=
+ I1
R
◆
Vm
e t/⌧
R
for 0  t 
T
2
and i = I2 e [t (T /2)]/⌧ for T /2  t  T, where ⌧ = L/R. Now we solve for
I1 and I2 by noting that
I1 = I2 e
T /2⌧
✓
Vm
+ I1
and I2 =
R
◆
Vm
e T /2⌧ .
R
These two equations are now solved for I1 . Letting x = T /2⌧, we get
(Vm /R)e x
.
1+e x
Therefore the equations for i become
I1 =
Vm
i=
R
"
"
#
Vm
e t/⌧
R(1 + e x )
#
Vm
i=
e [t (T /2)]/⌧
x
R(1 + e )
for 0  t 
for
T
2
and
T
 t  T.
2
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Problems
16–27
A check on the validity of these expressions shows they yield an average
value of (Vm /2R):
1
Iavg =
T
1
=
T
=
(Z
T /2 
0
⇢
✓
Vm T
+ ⌧ (1
2R
Vm
2R
)
◆
Z T
Vm
t/⌧
e
dt +
I2 e [t (T /2)]/⌧ dt
R
T /2
Vm
+ I1
R
x
✓
e ) I1
since I1 + I2 =
Vm
+ I2
R
◆
Vm
.
R
P 16.28 From the result of Problem 16.13(a),
✓
◆
1
n⇡
4A X
1
sin
cos n!0 t;
vi =
⇡ n=1,3,5,··· n
2
!0 =
2⇡
⇥ 103 = 500 rad/s;
4⇡
vi = 60
1
X
n=1,3,5,···
✓
4A
= 60;
⇡
◆
n⇡
1
sin
cos 500nt V.
n
2
From the circuit
Vo =
j!
j!
Vi
· j!L =
Vi =
Vi ;
R + j!L
R/L + j!
1000 + j!
Vi1 = 60/0 V;
Vi3 =
! = 500 rad/s;
20/0 = 20/180 V;
Vi5 = 12/0 V;
3! = 1500 rad/s;
5! = 2500 rad/s;
Vo1 =
j500
(60/0 ) = 26.83/63.43 V;
1000 + j500
Vo3 =
j1500
(20/180 ) = 16.64/
1000 + j1500
Vo5 =
j2500
(12/0 ) = 11.14/21.80 V;
1000 + j2500
.·.
vo = 26.83 cos(500t + 63.43 ) + 16.64 cos(1500t
146.31 V;
146.31 )
+11.14 cos(2500t + 21.80 ) + . . . V.
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16–28
CHAPTER 16. Fourier Series
P 16.29 [a] From the solution to Problem 16.13(a) the Fourier series for the input
voltage is
vg = 42
1
X
n=1,3,5,···

✓
n⇡
1
sin
n
2
◆
cos 2000nt V.
Employing the technique used in solving Assessment Problem 16.6 we
have
Vg1 = 42/0
!0 = 2000 rad/s;
Vg3 = 14/180
3!0 = 6000 rad/s;
Vg5 = 8.4/0
5!0 = 10,000 rad/s;
Vg7 = 6/180
7!0 = 14,000 rad/s.
From the circuit in Fig. P16.29 we have
Vo Vo Vg
+
+ (Vo
R
sL
.·.
Vg )sC = 0;
Vo
s2 + 1/LC
= H(s) = 2
Vg
s + (s/RC) + (1/LC).
Substituting in the numerical values gives
H(s) =
s2 + 108
;
s2 + 500s + 108
H(j2000) =
96
= 0.9999/
96 + j1
0.60 ;
H(j6000) =
64
= 0.9989/
64 + j3
2.68 ;
H(j10,000) = 0;
H(j14,000) =
96
= 0.9974/4.17 ;
96 + j7
Vo1 = (42/0 )(0.9999/
0.60 ) = 41.998/
Vo3 = (14/180 )(0.9989/
2.68 ) = 13.985/
0.60 V;
177.32 V;
Vo5 = 0 V;
Vo7 = (6/180 )(0.9974/4.17 ) = 5.984/184.17 V;
vo = 41.998 cos(2000t
0.60 ) + 13.985 cos(6000t
177.32 )
+5.984 cos(14,000t + 184.17 ) + . . . V.
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Problems
16–29
q
[b] The 5th harmonic at the frequency 1/LC = 10,000 rad/s has been
eliminated from the output voltage by the circuit, which is a band reject
filter with a center frequency of 10,000 rad/s.
P 16.30 [a]
V0 Vg
V0
+ V0 (12.5 ⇥ 10 6 s) +
= 0;
16s
1000
V0

1
Vg
1
+ 12.6 ⇥ 10 6 s +
;
=
16s
1000
16s
V0 (1000 + 0.2s2 + 16s) = 1000Vg ;
V0 =
5000Vg
;
s2 + 80s + 5000
I0 =
5Vg
V0
= 2
;
1000
s + 80s + 5000
H(s) =
I0
5
;
= 2
Vg
s + 80s + 5000
H(nj!0 ) =
!0 =
5
;
(5000
n2 !02 ) + j80n!0
2⇡
= 240⇡;
T
!02 = 57,600⇡ 2 ;
H(jn!0 ) =
(5000
80!0 = 19,200⇡;
5
;
57,600⇡ 2 n2 ) + j19,200⇡n
H(0) = 10 3 ;
H(j!0 ) = 8.82 ⇥ 10 6 /
173.89 ;
H(j2!0 ) = 2.20 ⇥ 10 6 /
176.96 ;
H(j3!0 ) = 9.78 ⇥ 10 7 /
177.97 ;
H(j4!0 ) = 5.5 ⇥ 10 7 /
178.48 ;

1360 1
1
1
1
cos !0 t +
cos 2!0 t +
cos 3!0 t +
cos 4!0 t + . . . ;
⇡
3
15
35
63
vg =
680
⇡
i0 =
680
⇥ 10 3
⇡
1360
(8.82 ⇥ 10 6 ) cos(!0 t
3⇡
1360
(2.20 ⇥ 10 6 ) cos(2!0 t
15⇡
176.96 )
1360
(9.78 ⇥ 10 7 ) cos(3!0 t
35⇡
177.97 )
1360
(5.5 ⇥ 10 7 ) cos(4!0 t
63⇡
178.48 )
173.89 )
...
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16–30
CHAPTER 16. Fourier Series
= 216.45 ⇥ 10 3 + 1.27 ⇥ 10 3 cos(240⇡t + 6.11 )
+ 6.35 ⇥ 10 5 cos(480⇡t + 3.04 )
+ 1.21 ⇥ 10 5 cos(720⇡t + 2.03 )
+ 3.8 ⇥ 10 6 cos(960⇡t + 1.11 )
...;
i0 ⇠
= 216.45 + 1.27 cos(240⇡t + 6.11 ) mA.
Note that the sinusoidal component is very small compared to the dc
component, so
i0 ⇠
= 216.45 mA (a dc current).
[b] The circuit is a low pass filter, so the harmonic terms are greatly reduced
in the output.
P 16.31 The function is odd with half-wave and quarter-wave symmetry. Therefore,
ak = 0,
for all k; the function is odd;
bk = 0,
for k even, the function has half-wave symmetry;
8 Z T /4
f (t) sin k!o t,
bk =
T 0
8
=
T
=
(Z
T /10
0
k odd
500t sin k!o t dt +
Z T /4
T /10
)
sin k!o t dt
8
{Int1 + Int2; }
T
Z T /10
Int1 = 500
0
t sin k!o t dt
"
1
= 500 2 2 sin k!o t
k !o
=
Int2 =
k⇡
500
sin
2
2
k !o
5
Z T /4
T /10
T /10
t
cos k!o t
k!o
0
#
50T
k⇡
;
cos
k!o
5
sin k!o t dt =
T /4
1
k⇡
1
;
cos k!o t
cos
=
k!o
k!o
5
T /10
✓
500
k⇡
1
Int1 + Int2 = 2 2 sin
+
k !o
5
k!o
◆
50T
k⇡
.
cos
k!o
5
50T = 50(20 ⇥ 10 3 ) = 1;
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Problems
.·.
Int1 + Int2 =
bk =

i(t) =
16–31
500
k⇡
.
sin
2
2
k !o
5
500
20
k⇡
k⇡
8
· 2 2 · T 2 sin
= 2 2 sin
,
T 4⇡ k
5
⇡ k
5
k odd;
1
20 X
sin(n⇡/5)
sin n!o t A.
2
⇡ n=1,3,5,···
n2
From the circuit,
H(s) =
Vo
= Zeq ;
Ig
Yeq =
1
1
+ sC;
+
R1 R2 + sL
Zeq =
1/C(s + R2 /L)
.
2
s + s(R1 R2 C + L)/R1 LC + (R1 + R2 )/R1 LC
Therefore,
H(s) =
320 ⇥ 104 (s + 32 ⇥ 104 )
.
s2 + 32.8 ⇥ 104 s + 28.8 ⇥ 108
We want the output for the third harmonic:
!0 =
2⇡
2⇡
=
= 100⇡;
T
20 ⇥ 10 3
Ig3 =
3⇡
20 1
sin
= 0.214/0 ;
2
⇡ 9
5
H(j300⇡) =
3!0 = 300⇡;
320 ⇥ 104 (j300⇡ + 32 ⇥ 104 )
= 353.6/
(j300⇡)2 + 32.8 ⇥ 104 (j300⇡) + 28.8 ⇥ 108
5.96 .
Therefore,
Vo3 = H(j300⇡)Ig3 = (353.6/
vo3 = 75.7 sin(300⇡t
5.96 )(0.214/0 ) = 75.7/
5.96 V;
5.96 ) V.
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16–32
CHAPTER 16. Fourier Series
P 16.32 !o =
2⇡
2⇡
=
⇥ 106 = 200 krad/s;
T
10⇡
5 ⇥ 106
= 25.
0.2 ⇥ 106
.·. n =
3 ⇥ 106
= 15;
0.2 ⇥ 106
H(s) =
(1/RC)s
Vo
= 2
Vg
s + (1/RC)s + (1/LC)
n=
1012
1
=
= 106 ;
3
RC
(250 ⇥ 10 )(4)
H(s) =
(103 )(1012 )
1
=
= 25 ⇥ 1012 ;
LC
(10)(4)
106 s
.
s2 + 106 s + 25 ⇥ 1012
H(j!) =
j! ⇥ 106
.
(25 ⇥ 1012 ! 2 ) + j106 !
15th harmonic input:
vg15 = (150)(1/15) sin(15⇡/2) cos 15!o t =
.·. Vg15 = 10/
H(j3 ⇥ 106 ) =
10 cos 3 ⇥ 106 t V;
180 V.
j3
= 0.1843/79.38 ;
16 + j3
Vo15 = (10)(0.1843)/
vo15 = 1.84 cos(3 ⇥ 106 t
100.62 V;
100.62 ) V.
25th harmonic input:
vg25 = (150)(1/25) sin(25⇡/2) cos 5 ⇥ 106 t = 6 cos 5 ⇥ 106 t V;
.·. Vg25 = 6/0 V.
H(j5 ⇥ 106 ) =
j5
= 1/0 ;
0 + j5
Vo25 = 6/0 V;
vo25 = 6 cos 5 ⇥ 106 t V.
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Problems

✓
16–33
◆
1 1 1
3Vm
T
P 16.33 [a] av =
;
Im + Im =
T 2 T
2
4
2Im
t,
T
i(t) =
0  t  T /2;
T /2  t  T.
i(t) = Im ,
2 Z T /2 2Im
2ZT
ak =
t cos k!o t dt +
Im cos k!o t dt
T 0
T
T T /2
Im
(cos k⇡
⇡2k2
=
1);
2 Z T /2 2Im
2ZT
bk =
t sin k!o t dt +
Im sin k!o t dt
T 0
T
T T /2
Im
.
⇡k
=
a1 =
2Im
,
⇡2
a3 =
2Im
;
9⇡ 2
b1 =
Im
,
⇡
.·.
a2 = 0,
b2 =
Irms = Im
s
av =
3Im
;
3
Im
;
2⇡
2
1
1
9
+ 4 + 2 + 2 = 0.8040Im .
16 ⇡
2⇡
8⇡
Irms = 192.95 mA.
P = (0.19295)2 (1000) = 37.23 W.
[b] Area under i2 :
A=
Z T /2
0
=
2
4Im
2 T
t dt + Im
2
T
2
2 3 T /2
4Im
t
2 T
+Im
2
T 3 0
2
2
= Im
T
Irms =
s

2 2
1 3
+
= T Im
;
6 6
3
1 2 2
· TI =
T 3 m
s
2
Im = 195.96 mA;
3
P = (0.19596)2 1000 = 38.5 W.
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16–34
CHAPTER 16. Fourier Series
✓
◆
37.23
[c] Error =
38.40
P 16.34 [a] av =
2
⇣
1 T
V
2 4 m
T
⌘
1 (100) =
Vm
;
4
=

4 Z T /4
Vm
T 0
ak =
"
4Vm
= 2 2 1
⇡ k
bk = 0,
3.05%.
4Vm
t cos k!o t dt
T
#
k⇡
cos
;
2
all k.
av =
60
= 15 V;
4
a1 =
240
;
⇡2
a2 =
240
(1
4⇡ 2
Vrms
v
"✓
u
◆2 ✓
◆2 #
u
1
120
240
= t(15)2 +
+
= 24.38 V.
P =
cos ⇡) =
⇡2
2
⇡2
(24.38)2
= 59.46 W;
10
[b] Area under v 2 ;
A=2
0  t  T /4
28,800
57,600 2
t+
t.
T
T2
v 2 = 3600
Z T /4 
Vrms =
P =
120
;
⇡2
p
0
s
57,600 2
28,800
t+
t dt = 600T.
T
T2
3600
p
1
600T = 600 = 24.49 V;
T
2
600 /10 = 60 W.
✓
59.46
[c] Error =
60.00
◆
1 100 =
0.9041%.
P 16.35 The voltage waveform is even, so bk = 0 for all k. The average value is
av = 0.5(20)(4⇡)/4⇡ = 10.
Z 2⇡

2⇡
80
2t
4
10
= 2;
ak = ⇡
t cos k!0 t dt = 10 2 cos(k/2)t + sin(k/2)t
k
k
k
0 ⇡
0
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Problems
1
80 X
1
cos n!o t V;
2
⇡ n=1,3,5,... n2
.·.
vg = 10
!o =
2⇡
2⇡
=
⇥ 103 = 500 rad/s;
T
4⇡
80
cos 500t
⇡2
vg = 10
Vo
Vg
sL
16–35
+ sCVo +
80
cos 1500t + . . . .
9⇡ 2
Vo
= 0;
R
Vo (RLCs2 + Ls + R) = RVg ;
H(s) =
1/LC
Vo
;
= 2
Vg
s + s/RC + 1/LC
106
1
=
= 106 ;
LC
(0.1)(10)
p
106
1
p
= 1000 2;
=
RC
(50 2)(10)
H(s) =
106
p
;
s2 + 1000 2s + 106
H(j!) =
106
106
p ;
! 2 + j1000! 2
H(j0) = 1;
H(j500) = 0.9701/
43.31 ;
H(j1500) = 0.4061/
120.51 ;
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16–36
CHAPTER 16. Fourier Series
80
(0.9701) cos(500t
⇡2
vo = 10(1) +
+
80
(0.4061) cos(1500t
9⇡ 2
vo = 10 + 7.86 cos(500t
v
u
u
⇠t
Vrms =
P ⇠
=
7.86
102 + p
2
43.31 )
120.51 ) + . . . ;
43.31 ) + 0.3658 cos(1500t
!2
0.3658
p
+
2
!2
120.51 ) + . . . .
= 11.44 V;
2
Vrms
p = 1.85 W.
50 2
Note – the higher harmonics are severely attenuated and can be ignored. For
example, the 5th harmonic component of vo is
✓
◆
80
cos(2500t
vo5 = (0.1580)
25⇡ 2
146.04 ) = 0.0512 cos(2500t
P 16.36 [a] v = 15 + 400 cos 500t + 100 cos(1500t
i = 2 + 5 cos(500t
146.04 ) V.
90 ) V;
30 ) + 3 cos(1500t
15 ) A;
1
1
P = (15)(2) + (400)(5) cos(30 ) + (100)(3) cos( 75 ) = 934.85 W.
2
2
[b] Vrms =
[c] Irms =
v
u
u
t
v
u
u
t
400
(15)2 + p
2
(2)2 +
5
p
2
!2
!2
3
+ p
2
P 16.37 [a] Area under v 2 = A = 4
Therefore Vrms
100
+ p
2
Z T /6
0
!2
!2
= 291.93 V.
= 4.58 A.
=
V 2T
2Vm2 T
+ m .
9
3
v
u
u1
=t
V 2T
2Vm2 T
+ m
9
3
T
✓
36Vm2 2
T
t dt + 2Vm2
2
T
3
!
= Vm
s
T
6
◆
2 1
+ = 74.5356 V.
9 3
[b] From Assessment Problem 16.3,
vg = 105.30 sin !0 t
Therefore Vrms ⇠
=
s
4.21 sin 5!0 t + 2.15 sin 7!0 t + · · · V.
(105.30)2 + (4.21)2 + (2.15)2
= 74.5306 V.
2
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Problems
16–37
P 16.38 [a] v has half-wave symmetry, quarter-wave symmetry, and is odd
.·. av = 0, ak = 0 all k, bk = 0, for k even.
8 Z T /4
f (t) sin k!o t dt,
bk =
T 0
8
=
T
(Z
T /8
8Vm
=
4T
cos k!o t T /8 8Vm
+
k!o
T
0
"
"
k⇡
k⇡
8Vm
cos
cos
+
4
T k!o
4
"
(
1
k⇡
k⇡
cos
+ cos
4
4
4
#
8Vm
1
=
4k!o T
8Vm
=
k!o T
(
4Vm
=
⇡k
)
Z T /4
Vm
sin k!o t dt +
Vm sin k!o t dt
4
T /8
0
"
k-odd
1
4
k⇡
1
+ 0.75 cos
4
4
)
=
cos k!o t T /4
k!o
T /8
#
0
)
1
(10 + 30 cos(k⇡/4).
k
b1 = 10 + 30 cos(⇡/4) = 31.21;
1
b3 = [10 + 30 cos(3⇡/4)] = 3.74;
3
1
b5 = [10 + 30 cos(5⇡/4)] =
5
2.24;
1
b7 = [10 + 30 cos(7⇡/4)] = 4.46.
7
Vg (rms) ⇡ Vm

s
31.212 + 3.742 + 2.242 + 4.462
= 22.51.
2
[b] Area = 2 2(6.25⇡)
Vg (rms) =
s
✓
2
✓
T
8
◆
+ 100⇡
2
✓
T
4
◆
= 53.125⇡ 2 T ;
p
1
(53.125⇡ 2 )t = 53.125⇡ = 22.90.
T
22.51
[c] % Error =
22.90
◆
1 (100) =
1.7%.
P 16.39 [a] Use the Fourier series constructed in Problem 16.3(a):
v(t) =
480
1
1
1
1
{sin !o t + sin 3!o t + sin 5!o t + sin 7!o t + sin 9!o t + · · · .
⇡
3
5
7
9
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16–38
CHAPTER 16. Fourier Series
v
!2
u
480 u
t p1
+
Vrms =
⇡ s 2
!2
1
1
p
+ p
3 2
5 2
1
1
1
1
480
+
+
= p 1+ +
9 25 49 81
⇡ 2
= 117.55 V.
!2
1
+ p
7 2
!2
1
+ p
9 2
!2
◆
✓
117.55
1 (100) = 2.04%.
[b] % error =
120
[c] Use the Fourier series constructed in Example 16.2:
⇢
960
1
1
v(t) = 2 sin !o t + sin 3!o t +
sin 5!o t
⇡
9
25
1
1
sin 7!o t +
sin 9!o t
49
81
··· .
s
1
1
1
1
960
Vrms ⇠
+
+
+
= 2p 1 +
81 625 2401 6561
⇡ 2
⇠
= 69.2765 V.
120
Vrms = p = 69.2820 V.
3
✓
69.2765
% error =
69.2820
◆
1 (100) =
0.0081%.
P 16.40 [a] Use the Fourier series constructed in Problem 16.3(b):
⇢
340
v(t) ⇡
⇡
680 1
1
cos !o t +
cos 2!o t + · · · .
⇡ 3
15
v
2
u✓
u 340 ◆2 ✓ 680 ◆2
u
4
+
Vrms ⇡ t
⇡
s
1
p
3 2
⇡
✓
!2
1
p
+
15 2
◆
1
340
1
+
=
1+4
= 120.0819 V.
⇡
18 450
!2 3
5
170
[b] Vrms = p = 120.2082;
2
% error =
✓
120.0819
120.2082
◆
1 (100) =
0.11%.
[c] Use the Fourier series constructed in Problem 16.3(c):
v(t) ⇡
170
+ 85 sin !o t
⇡
340
cos 2!o t;
3⇡
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
Vrms ⇡
Vrms =
v
u✓
u 170 ◆2
t
⇡
85
+ p
2
!2
340
+ p
3 2⇡
!2
16–39
⇡ 84.8021 V;
170
= 85 V.
2
% error =
0.23%.
P 16.41 [a] From Problem 16.14,
The area under v 2 :
"Z
◆2 #
Z T /4 ✓
T /8 14,400
40t
t2 dt +
10 +
dt
A=4
T2
T
T /8
0
=
T /4
57,600 t3 T /8
3200 t2 T /4 6400 t3 T /4
+400t
+ 2
+
T2 3 0
T 2 T /8
T 3 T /8
T /8
=
T
3T
7T
575
57,600
T + 400 + 1600
+ 6400
=
T.
1536
8
64
1536
3
Vrms =
s
1
T
✓
575
T
3
◆
=
s
575
= 13.84 V.
3
2
Vrms
= 12.78 W.
15
[c] From Problem 16.14,
[b] P =
80
(2 sin 45 + sin 90 ) = 18.57 V;
⇡2
vg ⇠
= 19.57 sin !0 t V;
p
(19.57/ 2)2
P =
= 12.76 W.
15
b1 =
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16–40
CHAPTER 16. Fourier Series
✓
◆
12.76
[d] % error =
12.78
1 (100) =
0.1024%;
P 16.42 [a] Half-wave symmetry av = 0, ak = bk = 0, even k:
ak =
4 Z T /4 4Im
16Im Z T /4
t cos k!0 t dt =
t cos k!0 t dt
T 0 T
T2 0
16Im
=
T2
(
T /4
t
cos k!0 t
+
sin
k!
t
0
k 2 !02
k!0
0
(
1
;
2
k !02
T
k⇡
16Im
0+
sin
=
2
T
4k!0
2
"
2Im
k⇡
ak =
sin
⇡k
2
!
)
)
#
2
,
⇡k
k—odd.
4 Z T /4 4Im
16Im Z T /4
bk =
t sin k!0 t dt =
t sin k!0 t dt
T 0 T
T2 0
16Im
=
T2
[b] ak
a1
a3
a5
a7
(
2Im
jbk =
⇡k
)
!
T /4
t
4Im
k⇡
cos k!0 t
.
= 2 2 sin
k!0
⇡ k
2
0
sin k!0 t
k 2 !02
("
k⇡
sin
2
⇢✓
!
2
⇡k
#
"
2
k⇡
sin
j
⇡k
2
!#)
;
◆
2Im
2
2
jb1 =
1
j
= 0.47Im / 60.28 ;
⇡
⇡
⇡
⇢✓
◆
✓
◆
2Im
2
2
jb3 =
1
+j
= 0.26Im /170.07 ;
3⇡
3⇡
3⇡
⇢✓
◆
✓
◆
2Im
2
2
jb5 =
1
j
= 0.11Im / 8.30 ;
5⇡
5⇡
5⇡
⇢✓
◆
✓
◆
2Im
2
2
jb7 =
1
+j
= 0.10Im /175.23 .
7⇡
7⇡
7⇡
ig = 0.47Im cos(!0 t
60.28 ) + 0.26Im cos(3!0 t + 170.07 )
+0.11Im cos(5!0 t
v
u
u
[c] Ig = t
1
X
n=1,3,5,...
A2n
2
!
8.30 ) + 0.10Im cos(7!0 t + 175.23 ) + · · · .
s
(0.47)2 + (0.26)2 + (0.11)2 + (0.10)2
⇠
= 0.3927Im .
= Im
2
!
!
◆
Z T /4 ✓
2
2
4Im 2
32Im
t3 T /4 Im
T
[d] Area = 2
t dt =
;
=
2
T
T
3 0
6
0
v
u
u1
Ig = t
T
2T
Im
6
!
Im
= p = 0.41Im .
6
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
!
estimated
exact
[e] % error =
1 100 =
P 16.43 Figure P16.43(b): ta = 0.2s;
v = 25t
3.8%.
tb = 0.6s.
0.6  t  1.0.
25,
Area 1 = A1 =
Z 0.2
2500t2 dt =
Area 2 = A2 =
Z 0.6
100(4
20t + 25t2 ) dt =
Area 3 = A3 =
Z 1.0
625(t2
2t + 1) dt =
0
0.2
A1 + A2 + A3 =
Vrms =
1 100 =
0.2  t  0.6;
50t + 20,
s
!
0  t  0.2;
v = 50t,
v=
0.3927Im
p
(Im / 6)
16–41
✓
0.6
20
;
3
40
;
3
40
3
100
.
3
◆
10
1 100
= p V.
1 3
3
Figure P16.43(c): ta = tb = 0.4s
v(t) = 25t,
v(t) =
50
(t
3
A1 =
Z 0.4
A2 =
Z 1.0
0
1),
2500 2
(t
9
0.4
s
0.4  t  1.
625t2 dt =
A1 + A2 =
Vrms =
0  t  0.4;
40
;
3
2t + 1) dt =
60
;
3
100
.
3
1
(A1 + A2 ) =
T
s
✓
◆
10
1 100
= p V.
1 3
3
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16–42
CHAPTER 16. Fourier Series
Figure P16.43(d): ta = tb = 1.
0  t  1;
v = 10t,
A1 =
Z 1
Vrms =
0
100t2 dt =
s
✓
100
;
3
◆
10
1 100
= p V.
1 3
3
P 16.44 Co = Av =
Vm
Vm T 1
· =
.
2
T
2
1 Z T Vm jn!o t
te
dt
Cn =
T 0 T
"
#T
Vm e jn!0 t
( jn!0 t
= 2
T
n2 !02
"
jn2⇡T /T ✓
Vm e
= 2
T
n2 !02
"
1)
0
2⇡
jn T
T
Vm
1
= 2
(1 + jn2⇡)
T n2 !02
=j
Vm
,
2n⇡
P 16.45 [a] Vrms =
=
s
s
1
1
n2 !02
#
◆
1
( 1)
n2 !02
#
n = ±1, ±2, ±3, . . . .
1ZT 2
v dt =
T 0
s
1ZT
T 0
✓
Vm
T
◆2
t2 dt
Vm2 t3 T
T3 3 0
s
Vm
Vm2
=p ;
3
3
p 2
(120/ 3)
= 480 W.
P =
10
=
[b] From the solution to Problem 16.44
c0 =
15
120
=j ;
8⇡
⇡
120
= 60 V;
2
c4 = j
60
120
=j ;
2⇡
⇡
c5 = j
c1 = j
12
120
=j ;
10⇡
⇡
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Problems
c2 = j
30
120
=j ;
4⇡
⇡
c6 = j
10
120
=j ;
12⇡
⇡
c3 = j
20
120
=j ;
6⇡
⇡
c7 = j
8.57
120
=j
.
14⇡
⇡
Vrms
16–43
v
u
1
X
u
= tc2 + 2
|c |2
n
o
n=1
=
q
602 + (2/⇡ 2 )(602 + 302 + 202 + 152 + 122 + 102 + 8.572 )
= 68.58 V.
[c] P =
(68.58)2
= 470.32 W;
10
✓
◆
470.32
% error =
480
1 (100) =
2.02%.
"
1 Z T /4
Vm e jn!o t T /4
Vm e jn!o t dt =
P 16.46 Cn =
T 0
T
jn!o 0

v(t) =
1
X
✓
n⇡
Vm
Vm
n⇡
sin
+j
1)] =
cos
2⇡n
2
2⇡n
2
Vm
[j(e jn⇡/2
=
T n!o
Vm
n⇡
=
sin
2⇡n
2
#
✓
j 1
n⇡
cos
2
◆
◆
1
;
Cn ejn!o t .
n= 1
Co = Av =
Vm
1 Z T /4
Vm dt =
T 0
4
or
"
Vm
sin(n⇡/2)
Co =
lim
2⇡ n!0
n
"
j
1
Vm
(⇡/2) cos(n⇡/2)
lim
=
n!0
2⇡
1

Vm ⇡
=
2⇡ 2
j0 =
cos(n⇡/2)
n
#
(⇡/2) sin(n⇡/2)
j
1
#
Vm
.
4
Note it is much easier to use Co = Av than to use L’Hopital’s rule to find the
limit of 0/0.
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16–44
CHAPTER 16. Fourier Series
P 16.47 [a] Co = av =
Cn =
Vm
(1/2)(T /2)Vm
=
;
T
4
1 Z T /2 2Vm jn!o t
te
dt
T 0
T
"
2Vm e jn!o t
= 2
( jn!o t
T
n2 !ot
=
#T /2
1)
Vm
[e jn⇡ ( jn⇡ + 1)
2n2 ⇡ 2
0
1].
Since e jn⇡ = cos n⇡ we can write
Vm
Vm
Cn = 2 2 (cos n⇡ 1) + j
cos n⇡.
2⇡ n
2n⇡
54
[b] Co =
= 13.5 V;
4
C 1=
54
27
= 10.19/122.48 V;
+j
2
⇡
⇡
C1 = 10.19/
C 2=
j
122.48 V;
13.5
= 4.30/
⇡
90 V;
C2 = 4.30/90 V;
C 3=
6
⇡2
C3 = 2.93/
C 4=
j
+j
9
= 2.93/101.98 V;
⇡
101.98 V;
6.75
= 2.15/
⇡
90 V;
C4 = 2.15/90 V.
[c]
Vo
Vo Vg
Vo
+
+ Vo sC +
= 0;
250 sL
62.5
.·.
(250LCs2 + 5sL + 250)Vo = 4sLVg .
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
16–45
(4/250C)s
Vo
;
= H(s) = 2
Vg
s + 1/50C + 1/LC
H(s) =
!o =
16,000s
s2 + 2 ⇥ 104 s + 4 ⇥ 1010
;
2⇡
2⇡
=
⇥ 106 = 2 ⇥ 105 rad/s;
T
10⇡
H(j0) = 0;
H(j2 ⇥ 105 k) =
j8k
.
100(1 k 2 ) + j10k
Therefore,
H 1 = 0.8/0 ;
H1 = 0.8/0 ;
H 2=
j16
= 0.0532/86.19 ;
300 j20
H2 = 0.0532/
86.19 ;
H 3=
j24
= 0.0300/87.85 ;
800 j30
H2 = 0.0300/
87.85 ;
H 4=
j32
= 0.0213/88.47 ;
1500 j40
H2 = 0.0213/
88.47 .
The output voltage coefficients:
C0 = 0;
C 1 = (10.19/122.48 )(0.8/0 ) = 8.15/122.48 V;
C1 = 8.15/
C 2 = (4.30/
122.48 V;
90 )(0.05/86.19 ) = 0.2287/
3.81 V;
C2 = 0.2287/3.81 V;
C 3 = (2.93/101.98 )(0.03/87.85 ) = 0.0878/
170.17 V;
C3 = 0.0878/170.17 V;
C 4 = (2.15/
90 )(0.02/88.47 ) = 0.0458/
1.53 V;
C4 = 0.0458/1.53 V.
[d] Vrms
v
v
u
u 4
4
u
u X
X
t
2
2
⇠
⇠
|C | = t2
|C |2
= C +2
n
o
n=1
⇠
=
P =
q
n
n=1
2(8.152 + 0.22872 + 0.08782 + 0.04582 ) ⇠
= 11.53 V;
(11.53)2
= 531.95 mW.
250
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16–46
CHAPTER 16. Fourier Series
P 16.48 [a] Vrms =
s
=
◆
✓
1 Z T /2 2Vm 2
t dt
T 0
T
v
#
u "
u 1 4V 2 t3 T /2
t
m
T2 3
T
0
v
u
u 4Vm2
Vm
=p ;
=t
(3)(8)
6
54
Vrms = p = 22.05 V.
6
[b] From the solution to Problem 16.47
|C3 | = 2.93;
C0 = 13.5;
|C1 | = 10.19;
|C4 | = 2.15;
|C2 | = 4.30.
Vg (rms) ⇠
=
q
13.52 + 2(10.192 + 4.302 + 2.932 + 2.152 ) ⇠
= 21.29 V.
✓
21.29
[c] % Error =
22.05
◆
1 (100) =
3.44%.
P 16.49 [a] From Example 16.3 we have:
40
av =
= 10 V,
4
"
40
bk =
1
⇡k
A1 = 18.01 V
A3 = 6 V,
A6 = 4.24 V,
!
40
k⇡
ak =
sin
;
⇡k
2
k⇡
cos
2
!#
,
✓1 = 45 ,
✓3 = 135 ,
✓6 = 90 ,
Ak /
✓k = ak
A2 = 12.73 V,
A4 = 0,
jbk .
✓2 = 90 ,
A5 = 3.6 V,
A7 = 2.57 V,
✓5 = 45 ,
✓7 = 135 ;
A8 = 0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[b] Cn =
an
jbn
2
,
C n=
an + jbn
= Cn⇤ .
2
C6 = 2.12/90 V;
C0 = av = 10 V;
C3 = 3/135 V;
C1 = 9/45 V;
C 3 = 3/
C 1 = 9/
C4 = C 4 = 0
C7 = 1.29/135 V;
C5 = 1.8/45 V;
C 7 = 1.29/
45 V;
C2 = 6.37/90 V;
C 2 = 6.37/
16–47
135 V; C 6 = 2.12/
90 V; C 5 = 1.8/
90 V;
135 V;
45 V.
P 16.50 [a] From the solution to Problem 16.33 we have
Ak = a k
jbk =
Im
(cos k⇡
⇡2k2
1) + j
Im
.
⇡k
A0 = 0.75Im = 180 mA;
A1 =
240
240
= 90.56/122.48 mA;
( 2) + j
2
⇡
⇡
A2 = j
A3 =
240
240
= 26.03/101.98 mA;
( 2) + j
2
9⇡
3⇡
A4 = j
A5 =
240
= 38.20/90 mA;
2⇡
240
= 19.10/90 mA;
4⇡
240
240
= 15.40/97.26 mA;
( 2) + j
2
25⇡
5⇡
A6 = j
240
= 12.73/90 mA.
6⇡
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16–48
CHAPTER 16. Fourier Series
[b] C0 = A0 = 180 mA;
1
C1 = A1 /✓1 = 45.28/122.48 mA;
2
C 1 = 45.28/
122.48 mA;
1
C2 = A2 /✓2 = 19.1/90 mA;
2
C 2 = 19.1/
90 mA;
1
C3 = A3 /✓3 = 13.02/101.98 mA;
2
C 3 = 13.02/
101.98 mA;
1
C4 = A4 /✓4 = 9.55/90 mA;
2
C 4 = 9.55/
90 mA;
1
C5 = A5 /✓5 = 7.70/97.26 mA;
2
C 5 = 7.70/
97.26 mA;
1
C6 = A6 /✓6 = 6.37/90 mA;
2
C61 = 6.37/
90 mA.
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Problems
P 16.51 [a] v = A1 cos(!o t
90 ) + A3 cos(3!o t + 90 )
+A5 cos(5!o t
v=
v(t);
⇡)
+A5 sin(5!o t
=
A5 sin 5!o t + A7 sin 7!o t;
odd function.
T /2) = A1 sin(!o t
.·. v(t
A5 sin 5!o t + A7 sin 7!o t.
A1 sin !o t + A3 sin 3!o t
.·. v( t) =
[c] v(t
90 ) + A7 cos(7!o t + 90 );
A1 sin !o t + A3 sin 3!o t
[b] v( t) =
16–49
A3 sin(3!o t
5⇡)
A7 sin(7!o t
A1 sin !o t + A3 sin 3!o t
T /2) =
3⇡)
7⇡)
A5 sin 5!o t + A7 sin 7!o t;
v(t), yes, the function has half-wave symmetry.
[d] Since the function is odd, with hws, we test to see if
f (T /2
t) = f (t);
f (T /2
t) = A1 sin(⇡
!o t)
+ A5 sin(5⇡
= A1 sin !o t
.·. f (T /2
A3 sin(3⇡
5!o t)
3!o t)
A7 sin(7⇡
7!o t)
A3 sin 3!o t + A5 sin 5!o t
A7 sin 7!o t;
t) = f (t) and the voltage has quarter-wave symmetry.
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16–50
CHAPTER 16. Fourier Series
P 16.52 [a] i = 11,025 cos 10,000t + 1225 cos(30,000t
180 ) + 441 cos(50,000t
180 )
+ 225 cos 70,000t µA
= 11,025 cos 10,000t
1225 cos 30,000t
441 cos 50,000t
+ 225 cos 70,000t µA.
[b] i(t) = i( t),
Function is even.
[c] Yes,
An = 0 for n even.
A0 = 0,
s
11,0252 + 12252 + 4412 + 2252
= 7.85 mA.
2
[e] A1 = 11,025/0 µA;
C1 = 5512.50/0 µA;
[d] Irms =
A3 = 1225/180 µA;
C3 = 612.5/180 µA;
A5 = 441/180 µA;
C5 = 220.5/180 µA;
A7 = 225/0 µA;
C 1 = 5512.50/0 µA;
C 5 = 220.5/
180 µA;
C7 = 112.50/0 µA.
C 3 = 612.5/
180 µA;
C 7 = 112.50/0 µA.
i = 112.5e j70,000t + 220.5e j180 e j50,000t + 612.5e j180 e j30,000t
+ 5512.5e j10,000t + 5512.5ej10,000t + 612.5ej180 ej30,000t
+ 220.5ej180 ej50,000t + 112.5ej70,000t µA.
[f ]
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Problems
16–51
P 16.53 From Table 15.1 we have
H(s) =
1
.
(s + 1)(s2 + s + 1)
After scaling we get
H 0 (s) =
!o =
106
.
(s + 100)(s2 + 100s + 104 )
2⇡
2⇡
=
⇥ 103 = 400 rad/s;
T
5⇡
.·. H 0 (jn!o ) =
1
16n2 ) + j4n]
(1 + j4n)[(1
.
It follows that
H(j0) = 1/0 ;
H(j!o ) =
1
(1 + j4)( 15 + j4)
H(j2!o ) =
vg (t) =
= 0.0156/
1
(1 + j8)( 63 + j8)
A A
+ sin !o t
⇡
2
241.03 ;
= 0.00195/
255.64 .
1
2A X
cos n!o t
⇡ n=2,4,6, n2 1
= 54 + 27⇡ sin !o t
.·. vo = 54 + 1.33 sin(400t
36 cos 2!o t
241.03 )
· · · V;
0.07 cos(800t
255.64 )
· · · V.
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16–52
CHAPTER 16. Fourier Series
P 16.54 Using the technique outlined in Problem 16.18 we can derive the Fourier series
for vg (t). We get
vg (t) = 100 +
1
800 X
1
cos n!o t.
⇡ 2 n=1,3,5, n2
The transfer function of the prototype second-order low pass Butterworth
filter is
H(s) =
1
p
,
s2 + 2s + 1
where !c = 1 rad/s.
Now frequency scale using kf = 2000 to get !c = 2 krad/s:
H(s) =
4 ⇥ 106
p
.
s2 + 2000 2s + 4 ⇥ 106
H(j0) = 1;
H(j5000) =
4 ⇥ 106
p
= 0.1580/
(j5000)2 + 2000 2(j5000)2 + 4 ⇥ 106
H(j15,000) =
146.04 ;
4 ⇥ 106
p
= 0.0178/
(j15,000)2 + 2000 2(j15,000)2 + 4 ⇥ 106
169.13 ;
Vdc = 100 V;
Vg1 =
800
/0 V;
n2
Vg3 =
800
/0 V;
9⇡ 2
Vodc = 100(1) = 100 V;
Vo1 =
800
(0.1580/
⇡2
146.04 ) = 12.81/
146.04 V;
Vo3 =
800
(0.0178/
9⇡ 2
169.13 ) = 0.16/
169.13 V;
vo (t) = 100 + 12.81 cos(5000t
+ 0.16 cos(15,000t
146.04 )
169.13 ) + · · · V.
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Problems
P 16.55 vg =
4(2.5⇡) (cos 5000t)
=5
⇡
4 1
2(2.5⇡)
⇡
(10/3) cos 5000t
16–53
· · · V.
H(j0) = 0;
H(j5000) =
25 ⇥ 106
p
= 0.999/16.42 ;
25 ⇥ 106 ) + j5 2 ⇥ 106
(106
.·. vo (t) = 0
3.33 cos(5000t + 16.42 )
· · · V.
P 16.56 [a] Let Va represent the node voltage across R2 , then the node-voltage
equations are
Va
Vg
R1
(0
+
Va
+ Va sC2 + (Va
R2
Va )sC2 +
0
Vo
Vo )sC1 = 0;
= 0.
R3
Solving for Vo in terms of Vg yields
1
s
Vo
R1 C1
⇣
⌘
= H(s) =
Vg
s2 + 1 1 + 1 s +
R3
C1
C2
.
R1 +R2
R1 R2 R3 C1 C2
⌘
⇣
⌘
It follows that
R 1 + R2
!o2 =
R1 R2 R3 C1 C2
=
1
R3
✓
R3
Ko =
R1
◆
1
1
+
;
C1 C2
✓
◆
C2
.
C1 + C2
Note that
⇣
C2
1
1
R3
+ C12 s
R1 C1 +C2 R3 C1
⇣
⌘
⇣
⌘.
H(s) =
2
s2 + R13 C11 + C12 s + R1 RR21R+R
C
C
3 1 2
[b] For the given values of R1 , R2 , R3 , C1 , and C2 we have
R3
R1
1
R3
✓
✓
C2
C1 + C2
1
1
+
C1 C2
◆
◆
=
R3
=
2R1
400
;
313
= 2000;
R1 + R2
= 0.16 ⇥ 1010 = 16 ⇥ 108 ;
R1 R2 R3 C1 C2
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16–54
CHAPTER 16. Fourier Series
H(s) =
!o =
(400/313)(2000)s
s2 + 2000s + 16 ⇥ 108
.
2⇡
2⇡
=
⇥ 106 = 4 ⇥ 104 rad/s;
T
50⇡
H(jn!o ) =
=
(400/313)(2000)jn!o
16 ⇥ 108 n2 !o2 + j2000n!o
j(20/313)n
;
(1 n2 ) + j0.05n
400
=
313
H(j!o ) =
j(20/313)
=
j(0.050)
H(j3!o ) =
j(20/313)(3)
= 0.0240/91.07 ;
8 + j0.15
H(j5!o ) =
j(100/313)
= 0.0133/90.60 ;
24 + j0.25
vg (t) =
1.28;
1
4A X
1
sin(n⇡/2) cos n!o t;
⇡ n=1,3,5,··· n
A = 15.65⇡ V;
vg (t) = 62.60 cos !o t
20.87 cos 3!o t + 12.52 cos 5!o t
vo (t) =
0.50 cos(3!o t + 91.07 )
80 cos !o t
+ 0.17 cos(5!o t + 90.60 )
···;
· · · V.
[c] The fundamental frequency component dominates the output, so we
expect the quality factor Q to be quite high.
[d] !o = 4 ⇥ 104 rad/s and = 2000 rad/s. Therefore, Q = 40,000/2000 = 20.
We expect the output voltage to be dominated by the fundamental
frequency component since the bandpass filter is tuned to this frequency!
P 16.57 [a] Using the equations derived in Problem 16.56(a),
Ko =
R3
R1
1
=
R3
!o2 =
✓
✓
C2
C1 + C2
1
1
+
C1 C2
◆
◆
=
400
;
313
= 2000 rad/s;
R 1 + R2
= 16 ⇥ 108 .
R1 R2 R3 C1 C2
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Problems
[b] H(jn!o ) =
=
(400/313)(2000)jn!o
16 ⇥ 108 n2 !o2 + j2000n!o
j(20/313)n
;
(1 n2 ) + j0.05n
400
=
313
H(j!o ) =
j(20/313)
=
j(0.050)
H(j3!o ) =
j(20/313)(3)
= 0.0240/91.07 ;
8 + j0.15
H(j5!o ) =
j(100/313)
= 0.0133/90.60 ;
24 + j0.25
vg (t) =
16–55
1.28;
1
4A X
1
sin(n⇡/2) cos n!o t;
⇡ n=1,3,5,... n
A = 15.65⇡ V;
vg (t) = 62.60 cos !o t
20.87 cos 3!o t + 12.52 cos 5!o t
vo (t) =
0.50 cos(3!o t + 91.07 )
80 cos !o t
+ 0.17 cos(5!o t + 90.60 )
···;
· · · V.
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The Fourier Transform
17
Assessment Problems
AP 17.1 [a] F (!) =
Z 0
( Ae
⌧ /2
j!t
) dt +
Z ⌧ /2
0
Ae j!t dt
A
[2 ej!⌧ /2 e j!⌧ /2 ]
j! "
#
ej!⌧ /2 + e j!⌧ /2
2A
1
=
j!
2
j2A
[1 cos(!⌧ /2)].
=
!
=
[b] F (!) =
AP 17.2
1
f (t) =
2⇡
=
Z 1
0
⇢Z
at
te
2
3
e
jt!
4e
1
{4e j2t
j2⇡t
j!t
dt =
Z 1
te (a+j!)t dt =
Z 2
jt!
Z 3
d! +
2
0
e
4e j3t + ej2t
d! +
"
1
(4 sin 3t
⇡t
4ejt! d!
e j2t + 4ej3t
1 3e j2t 3ej2t 4ej3t 4e j3t
=
+
⇡t
j2
j2
=
2
1
.
(a + j!)2
4ej2t }
#
3 sin 2t).
AP 17.3 [a] F (!) = F (s) |s=j! = L{e at sin!0 t}s=j!
=
!0
!0
=
.
2
2
(s + a) + !0 s=j! (a + j!)2 + !02
#
"
1
1
[b] F (!) = L{f (t)}s= j! =
=
.
2
(s + a) s= j! (a j!)2
17–1
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17–2
CHAPTER 17. The Fourier Transform
[c] f + (t) = te at ,
1
,
(s + a)2
L{f + (t)} =
Therefore F (!) =
2A
,
⌧
0<t<
⌧
;
2
f 0 (t) =
2A
[u(t + ⌧ /2)
⌧
u(t)]
2A
[u(t)
⌧
u(t
⌧ /2)]
2A
u(t + ⌧ /2)
⌧
f 00 (t) =

2A
⌧
✓
t+
2A j!⌧ /2
e
⌧
⌧
2
4A
2A
u(t) +
u(t
⌧
⌧
◆
Thus we have F (!) =
#
⇢
◆
;
◆
"
#
"
#
⇢ ✓
⌧
F u t+
2
F u t
1
!2
⌧
2
⌧
2
◆
◆
✓
u t
◆
⌧
.
2
t

⌧
2
✓
4A
!⌧
1 =
cos
⌧
2
! 2 F (!);
 ✓
v(t) = Vm u t +
⌧ /2);
4A 2A j!⌧ /2
+
e
⌧
⌧
"
[c] F{f 00 (t)} = (j!)2 F (!) =
✓
4A 2A
+
⌧
⌧
4A ej!⌧ /2 + e j!⌧ /2
=
⌧
2
⇢ ✓
1
j4a!
= 2
.
2
(a j!)
(a + ! 2 )2
1
(a + j!)2
f 0 (t) =
[b] F{f 00 (t)} =
AP 17.5
1
.
(s + a)2
⌧
< t < 0;
2
=
.·.
L{f (t)} =
2A
,
⌧
AP 17.4 [a] f 0 (t) =
.·.
te at ;
f (t) =
◆
1 .
therefore F (!) =

✓
4A
!⌧
cos
⌧
2
◆
1
1
F{f 00 (t)}.
2
!
.
1
ej!⌧ /2 ;
= ⇡ (!) +
j!
1
e j!⌧ /2 .
= ⇡ (!) +
j!
"
#
1 h j!⌧ /2
e
Therefore V (!) = Vm ⇡ (!) +
j!
= j2Vm ⇡ (!) sin
=
✓
!⌧
2
◆
+
e j!⌧ /2
✓
i
2Vm
!⌧
sin
!
2
◆
(Vm ⌧ ) sin(!⌧ /2)
.
!⌧ /2
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Problems
AP 17.6 [a] Ig (!) = F{10sgn t} =
17–3
20
.
j!
Vo
.
Ig
Using current division and Ohm’s law,

4s
4
Ig ;
( Ig )s =
Vo = I 2 s =
4+1+s
5+s
[b] H(s) =
H(s) =
4s
,
s+5
H(j!) =
[c] Vo (!) = H(j!) · Ig (!) =
j4!
.
5 + j!
j4!
5 + j!
!
20
j!
!
=
80
.
5 + j!
[d] vo (t) = 80e 5t u(t) V.
[e] Using current division,
1
1
i1 (0 ) = ig = ( 10) = 2 A.
5
5
+
[f ] i1 (0 ) = ig + i2 (0+ ) = 10 + i2 (0 ) = 10 + 8 = 18 A.
[g] Using current division,
4
i2 (0 ) = (10) = 8 A.
5
[h] Since the current in an inductor must be continuous,
i2 (0+ ) = i2 (0 ) = 8 A.
[i] Since the inductor behaves as a short circuit for t < 0,
vo (0 ) = 0 V.
[j] vo (0+ ) = 1i2 (0+ ) + 4i1 (0+ ) = 80 V.
AP 17.7 [a] Vg (!) =
H(s) =
1
1
j!
+ ⇡ (!) +
1
;
j!
1
0.5k(1/s)
Va
=
,
=
Vg
1 + 0.5k(1/s)
s+3
H(j!) =
1
;
3 + j!
Va (!) = H(j!)Vg (j!)
1
1
⇡ (!)
=
+
+
(1 j!)(3 + j!) j!(3 + j!) 3 + j!
1/4
1/3
1/3
⇡ (!)
1/4
+
+
+
=
1 j! 3 + j!
j!
3 + j! 3 + j!
1/3
1/12
⇡ (!)
1/4
+
+
.
=
1 j!
j!
3 + j! 3 + j!
Therefore va (t) =

1
1 t
e u( t) + sgn t
4
6
1 3t
1
e u(t) +
V.
12
6
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17–4
CHAPTER 17. The Fourier Transform
[b] va (0 ) =
1
1
1
+ 0 + = V;
6
6
4
1
4
1
1
1
+ = V;
12 6
4
va (0+ ) = 0 +
1
6
va (1) = 0 +
1
1
1
+ 0 + = V.
6
6
3
AP 17.8
v(t) = 4te t u(t);
V (!) =
Therefore |V (!)| =
4
.
1 + !2
"
p
1Z 3
4
W1⌦ =
⇡ 0
(1 + ! 2 )
(
#2
4
.
(1 + j!)2
d!
p )

3
1
!
1 !
+
tan
2 !2 + 1
1 0
#
"p
3 1
+
= 16
= 3.769 J;
8⇡
6
16
=
⇡
W1⌦ (total) =

3.769
(100) = 94.23%.
4
Therefore % =
AP 17.9
|V (!)| = 6

1
8
8
⇡
!
1 !
+
tan
=
0+
= 4 J.
2
⇡ ! +1
1 0
⇡
2
✓
|V (!)|2 = 36
◆
6
!,
2000⇡
0  !  2000⇡;
✓
✓
◆
"
1 Z 2000⇡
36
W1⌦ =
⇡ 0
"
1
=
36!
⇡
"
◆
72
36
!+
!2;
2
2000⇡
4⇡ ⇥ 106
#
36 ⇥ 10 6 2
72!
+
! d!
2000⇡
4⇡ 2
#2000⇡
36 ⇥ 10 6 ! 3
72! 2
+
4000⇡
12⇡ 2
0
1
=
36(2000⇡)
⇡
36 ⇥ 10 6 (2000⇡)3
72
(2000⇡)2 +
4000⇡
12⇡ 2
#
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Problems
= 36(2000)
17–5
72(2000)2 36 ⇥ 10 6 (2000)3
+
4000
12
= 24 kJ.
W6k⌦ =
24 ⇥ 103
= 4 J.
6 ⇥ 103
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17–6
CHAPTER 17. The Fourier Transform
Problems
P 17.1
[a] F (!) =
Z ⌧ /2
2A j!t
te
dt
⌧ /2 ⌧
"
2A e j!t
=
( j!t
⌧
!2

✓
#⌧ /2
1)
⌧ /2
◆
◆
✓
2A
j!⌧
j!⌧
= 2 e j!⌧ /2
+1
+1 ;
ej!⌧ /2
! ⌧
2
2

⌘
!⌧ ⇣ j!⌧ /2
2A
e
+ ej!⌧ /2 ;
F (!) = 2 e j!⌧ /2 ej!⌧ /2 + j
! ⌧
2
"
#
2A !⌧ cos(!⌧ /2) 2 sin(!⌧ /2)
.
F (!) = j
⌧
!2
[b] Using L’Hopital’s rule,
"
!⌧ (⌧ /2)[ sin(!⌧ /2)] + ⌧ cos(!⌧ /2)
F (0) = lim 2A
!!0
2!⌧
= lim 2A
"
!⌧ (⌧ /2) sin(!⌧ /2)
2!⌧
= lim 2A
"
⌧ sin(!⌧ /2)
= 0;
4
!!0
!!0
.·.
2(⌧ /2) cos(!⌧ /2)
#
#
#
F (0) = 0.
[c] When A = 10 and ⌧ = 0.1
"
#
0.1! cos(!/20) 2 sin(!/20)
;
F (!) = j200
!2
|F (!)| =
20! cos(!/20) 400 sin(!/20)
.
!2
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Problems
P 17.2
[a] F (!) = A +
2A
!,
!o
!o /2  !  0;
F (!) = A
2A
!,
!o
0  !  !o /2;
F (!) = 0
17–7
elsewhere;
✓
◆
✓
2A
! ejt! d!;
!o
1 Z0
2A
f (t) =
! ejt! d!
A+
2⇡ !o /2
!o
+
◆
1 Z !o /2
A
2⇡ 0
Z 0
1
f (t) =
2⇡
Ae
d! +
Z 0
Aejt! d!
Z !o /2
2A jt!
!e d!
!o
jt!
!o /2
2A jt!
!e d!
!o /2 !o
+
Z !o /2
=
1
[ Int1 + Int2 + Int3
2⇡
0
0
Int4 ].
Int1 =
Z 0
Int2 =
Z 0
Int3 =
Z !o /2
Aejt! d! =
Int4 =
Z !o /2
2A
t!o jt!o /2
2A jt!
e
!e d! =
( j
+ ejt!o /2
2
!o
!o t
2
!o /2
Aejt! d! =
A
(1
jt
2A
2A jt!
!e d! =
(1
!o t2
!o /2 !o
0
0
A jt!o /2
(e
jt
Int1 + Int3 =
2A
sin(!o t/2);
t
Int2
4A
[1
!o t2
.·.
e jt!o /2 );
Int4 =

i
2A h
2
2
sin
(!
t/4)
o
⇡!o t2
=
4!o A
sin2 (!o t/4)
⇡!o2 t2
#2
e jt!o /2 );
1);
2A
sin(!o t/2);
t
cos(!o t/2))
=
"
t!o jt!o /2
e
2
1);
cos(!o t/2)]
1 4A
(1
f (t) =
2⇡ !o t2
!o A sin(!o t/4)
=
4⇡
(!o t/4)
j
.
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17–8
CHAPTER 17. The Fourier Transform
!o A 2
(1) = 79.58 ⇥ 10 3 !o A.
4⇡
[c] A = 5⇡;
!o = 100 rad/s;
[b] f (0) =
"
sin(t/25)
f (t) = 125
(t/25)
P 17.3
Z 2 
.
✓ ◆
j4⇡A
⇡
sin 2!.
t e j!t dt = 2
2
⇡
4! 2
2
✓
◆
◆
Z 0
Z ⌧ /2 ✓
2A
2A
j!t
[b] F (!) =
t+A e
t + A e j!t dt
dt +
⌧
⌧
⌧ /2
0
[a] F (!) =
A sin

4A
= 2 1
! ⌧
P 17.4
#2
F{sin !0 t} = F
=
(
✓
!⌧
cos
2
ej!0 t
2j
)
F
1
[2⇡ (!
2j
(
(!
1
.
(s + a)2
F (!) = F (s)
+F (s)
s=j!
)
2⇡ (! + !0 )]
[a] F (s) = L{te at } =
"
.
e j!0 t
2j
!0 )
= j⇡[ (! + !0 )
P 17.5
◆
!0 )].
;
s= j!
#
"
1
1
+
F (!) =
2
(a + j!)
(a j!)2
=
#
2(a2 ! 2 )
2(a2 ! 2 )
=
.
(a2 ! 2 )2 + 4a2 ! 2
(a2 + ! 2 )2
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Problems
[b] F (s) = L{t3 e at } =
6
.
(s + a)4
F (!) = F (s)
F (s)
s=j!
F (!) =
[c] F (s) = L{e at cos !0 t} =
+F (s)
0.5
(a + j!)
+
=
j!0
!0 )2
[d] F (s) = L{e at sin !0 t} =
F (!) = F (s)
[e] F (!) =
;
+
0.5
(a + j!) + j!0
+
0.5
j!) + j!0
(a
a
a2 + (! + !0 )2
.
!0
j0.5
j0.5
=
+
.
2
(s + a)2 + !0
(s + a) j!0 (s + a) + j!0
F (s)
s=j!
F (!) =
+
j!0
a
a2 + (!
a2 ! 2
.
(a2 + ! 2 )4
s= j!
0.5
j!)
(a
j48a!
s+a
0.5
0.5
=
+
.
2
2
(s + a) + !0
(s + a) j!0 (s + a) + j!0
s=j!
F (!) =
;
s= j!
6
6
+
=
4
(a + j!)
(a j!)4
F (!) = F (s)
17–9
;
s= j!
ja
ja
+
.
a2 + (! !0 )2 a2 + (! + !0 )2
Z 1
1
(t
to )e j!t dt = e j!to .
(Use the sifting property of the Dirac delta function.)
P 17.6
1 Z1
[A(!) + jB(!)][cos t! + j sin t!] d!
f (t) =
2⇡ 1
1 Z1
=
[A(!) cos t!
2⇡ 1
B(!) sin t!] d!
j Z1
+
[A(!) sin t! + B(!) cos t!] d!.
2⇡ 1
But f (t) is real, therefore the second integral in the sum is zero.
P 17.7
By hypothesis, f (t) =
f ( t). From Problem 17.6, we have
1 Z1
f ( t) =
[A(!) cos t! + B(!) sin t!] d!.
2⇡ 1
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17–10
CHAPTER 17. The Fourier Transform
For f (t) = f ( t), the integral
f (t) is real and odd, we have
R1
1 A(!) cos t! d! must be zero. Therefore, if
1Z 1
f (t) =
B(!) sin t! d!.
2⇡ 1
P 17.8
F (!) =
j2
;
!
f (t) =
1 Z1
2⇡ 1
But
sin t!
!
therefore B(!) =
✓
2
;
!
thus we have
◆
1 Z 1 sin t!
2
d!.
sin t! d! =
!
⇡ 1 !
is even;
therefore f (t) =
2 Z 1 sin t!
d!.
⇡ 0
!
Therefore,
2 ⇡
f (t) = · = 1,
⇡ 2
✓
◆
2
⇡
f (t) = ·
=
⇡
2
9
>
>
t > 0; >
=
>
>
;
1, t < 0. >
from a table of definite integrals
Therefore f (t) = sgn t.
P 17.9
From Problem 17.5[c] we have
F (!) =
✏
✏2 + (!
✏
+
.
!0 )2 ✏2 + (! + !0 )2
Note that as ✏ ! 0, F (!) ! 0 everywhere except at ! = ±!0 . At ! = ±!0 ,
F (!) = 1/✏, therefore F (!) ! 1 at ! = ±!0 as ✏ ! 0. The area under each
bell-shaped curve is independent of ✏, that is
Z 1
✏d!
✏d!
=
= ⇡.
2
2
2
!0 )
1 ✏ + (! + !0 )2
1 ✏ + (!
Z 1
Therefore as ✏ ! 0,
P 17.10 A(!) =
Z 0
1
F (!) ! ⇡ (!
f (t) cos !t dt +
Z 1
0
!0 ) + ⇡ (! + !0 ).
f (t) cos !t dt = 0
since f (t) cos !t is an odd function.
B(!) =
2
Z 1
0
f (t) sin !t dt,
since f (t) sin !t is an even function.
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Problems
P 17.11 A(!) =
=
Z 1
1
Z 0
Z 1
=2
0
f (t) cos !t dt
f (t) cos !t dt +
Z 1
f (t) cos !t dt,
since f (t) cos !t is also even.
1
17–11
0
f (t) cos !t dt
B(!) = 0, since f (t) sin !t is an odd function and
Z 0
1
Z 1
f (t) sin !t dt =
0
f (t) sin !t dt.
)
(
Z 1
df (t) j!t
df (t)
e
dt.
P 17.12 [a] F
=
dt
1 dt
Let u = e j!t , then du = j!e j!t dt; let dv = [df (t)/dt] dt, then
v = f (t).
Therefore F
(
df (t)
dt
)
= f (t)e j!t
1
1
Z 1
1
f (t)[ j!e j!t dt]
= 0 + j!F (!).
[b] Fourier transform of f (t) exists, i.e., f (1) = f ( 1) = 0.
[c] To find F
Then F
(
(
)
let g(t) =
)
(
d2 f (t)
,
dt2
d2 f (t)
dt2
But G(!) = F
(
=F
)
= j!G(!).
)
= j!F (!).
(
)
df (t)
dt
Therefore we have F
dg(t)
dt
df (t)
.
dt
d2 f (t)
dt2
= (j!)2 F (!).
Repeated application of this thought process gives
F
P 17.13 [a] F
(
dn f (t)
dtn
⇢Z t
1
)
= (j!)n F (!).
f (x) dx =
Now let u =
Let dv = e
Z t
j!t
1
Z 1 Z t
1
1
f (x) dx,
dt,
f (x) dx e j!t dt.
then du = f (t)dt.
e j!t
then v =
.
j!
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17–12
CHAPTER 17. The Fourier Transform
Therefore,
F
⇢Z t
1
e j!t Z t
f (x) dx =
f (x) dx
j!
1
1
1
= 0+
Z 1
[b] We require
1
Z 1
[c] No, because
P 17.14 [a] F{f (at)} =
Z 1
Let u = at,
1
#
e j!t
f (t) dt
j!
F (!)
.
j!
f (x) dx = 0.
1
1
Z 1 "
e ax u(x) dx =
1
6= 0.
a
f (at)e j!t dt.
du = adt,
u = ±1 when t = ±1.
Therefore,
F{f (at)} =
[b] F{e |t| } =
Z 1
1
f (u)e
du
a
j!u/a
!
1
= F
a
✓
◆
!
,
a
a > 0.
1
2
1
+
=
.
1 + j! 1 j!
1 + !2
Therefore F{e a|t| } =
(1/a)2
.
(!/a)2 + 1
Therefore F{e 0.5|t| } =
4
4! 2 + 1
,
F{e |t| } =
2
!2 + 1
;
F{e 2|t| } = 1/[0.25! 2 + 1], yes as “a” increases, the sketches show that
f (t) approaches zero faster and F (!) flattens out over the frequency
spectrum.
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Problems
P 17.15 [a] F{f (t
a)} =
Z 1
a)e j!t dt.
f (t
1
17–13
Let u = t a, then du = dt, t = u + a, and u = ±1 when t = ±1.
Therefore,
F{f (t
a)} =
Z 1
1
j!a
=e
[b] F{ej!0 t f (t)} =
f (u)e j!(u+a) du
Z 1
1
Z 1
f (u)e j!u du = e j!a F (!).
f (t)e j(! !0 )t dt = F (!
1
(
"
ej!0 t + e j!0 t
[c] F{f (t) cos !0 t} = F f (t)
2
P 17.16 Y (!) =
=
Z 1 Z 1
1
Z 1
Let u = t
1
1
x( )
#)
1
= F (!
2
1
!0 ) + F (! + !0 )
2
x( )h(t
)d
Z 1
1
!0 .)
.
e j!t dt
)e j!t dt d .
h(t
, du = dt, and u = ±1, when t = ±1.
Therefore Y (!) =
=
=
P 17.17 F{f1 (t)f2 (t)} =
=
=
Z 1
1
Z 1
1
Z 1
1
Z 1 
1
x( )
Z 1

1
x( ) e j!
h(u)e j!(u+ ) du d
Z 1
h(u)e j!u du d
x( )e j! H(!) d = H(!)X(!).
1Z1
F1 (u)ejtu du f2 (t)e j!t dt
2⇡
1
1Z1
2⇡
1
Z 1

1
F1 (u)f2 (t)e j!t ejtu du dt
Z 1
1Z1
F1 (u)
f2 (t)e j(! u)t dt du
2⇡
1
1
1Z1
=
F1 (u)F2 (!
2⇡
1
P 17.18 (i) F{e at u(t)} =
"
1
1
= F (!);
a + j!
u) du.
dF (!)
j
=
.
d!
(a + j!)2
#
1
dF (!)
Therefore j
.
=
d!
(a + j!)2
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17–14
CHAPTER 17. The Fourier Transform
Therefore F{te at u(t)} =
1
.
(a + j!)2
(ii) F{|t|e a|t| } = F{te at u(t)}
1
=
(a + j!)2
F{teat u( t)}
d
j
d!
1
a
j!
!
1
1
+
.
2
(a + j!)
(a j!)2
=
(iii) F{te a|t| } = F{te at u(t)} + F{teat u( t)}
1
d
=
+j
2
(a + j!)
d!
=
1
(a + j!)2
P 17.19 [a] f1 (t) = cos !0 t,
f2 (t) = 1,
a
j!
!
1
.
(a j!)2
F1 (u) = ⇡[ (u + !0 ) + (u
⌧ /2 < t < ⌧ /2,
Thus F2 (u) =
!0 )];
and f2 (t) = 0 elsewhere;
⌧ sin(u⌧ /2)
.
u⌧ /2
Using convolution,
1Z1
F (!) =
F1 (u)F2 (!
2⇡
1
=
1
u) du
1Z1
⇡[ (u + !0 ) + (u
2⇡
1
!0 )]⌧
sin[(! u)⌧ /2]
du
(! u)(⌧ /2)
sin[(! u)⌧ /2]
⌧Z 1
du
(u + !0 )
=
2
(! u)(⌧ /2)
1
+
=
⌧Z 1
(u
2
1
!0 )
sin[(! u)⌧ /2]
du
(! u)(⌧ /2)
⌧ sin[(! + !0 )⌧ /2] ⌧ sin[(! !0 )⌧ /2]
·
+ ·
.
2 (! + !0 )(⌧ /2)
2
(! !0 )⌧ /2
[b] As ⌧ increases, the amplitude of F (!) increases at ! = ±!0 and at the
same time the width of the frequency band of F (!) approaches zero as !
deviates from ±!0 .
The area under the [sin x]/x function is independent of ⌧, that is
Z 1
⌧ Z 1 sin[(! !0 )(⌧ /2)]
sin[(! !0 )(⌧ /2)]
d! =
[(⌧ /2) d!] = ⇡.
2 1 (! !0 )(⌧ /2)
(! !0 )(⌧ /2)
1
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Problems
17–15
Therefore as t ! 1,
f1 (t)f2 (t) ! cos !0 t and F (!) ! ⇡[ (!
!0 ) + (! + !0 )].
P 17.20 [a] Find the Thévenin equivalent with respect to the terminals of the
capacitor:
5
vTh = vg ;
6
Io =
RTh = 60k12 = 10 k⌦;
2sVTh
VTh
=
;
6
10,000 + 10 /2s
20,000s + 106
H(s) =
Io
10 4 s
;
=
VTh
s + 50
H(j!) =
5
vTh = vg = 30sgn(t);
6
Io = H(j!)VTh (j!) =
VTh =
60
j!
!
j! ⇥ 10 4
;
j! + 50
60
;
j!
j! ⇥ 10 4
j! + 50
!
=
6 ⇥ 10 3
;
j! + 50
io (t) = 6e 50t u(t) mA.
[b] At t = 0 the circuit is
At t = 0+ the circuit is
ig (0+ ) =
30 + 36
= 5.5 mA;
12
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17–16
CHAPTER 17. The Fourier Transform
i60k (0+ ) =
30
= 0.5 mA;
60
io (0+ ) = 5.5 + 0.5 = 6 mA,
which agrees with our solution.
We also know io (1) = 0, which agrees with our solution.
The time constant with respect to the terminals of the capacitor is RTh C
Thus,
1
= 50,
.·.
⌧
⌧ = (10,000)(2 ⇥ 10 6 ) = 20 ms;
which also agrees with our solution.
Thus our solution makes sense in terms of known circuit behavior.
P 17.21 [a] From the solution of Problem 17.20 we have
Vo =
106
VTh
·
;
104 + (106 /2s) 2s
H(s) =
Vo
50
;
=
VTh
s + 50
H(j!) =
50
;
j! + 50
VTh (!) =
60
;
j!
Vo (!) = H(j!)VTh (!) =
=
!
50
j! + 50
3000
60
=
(j!)(j! + 50)
j!
vo (t) = 30sgn(t)
[b] vo (0 ) =
60
j!
60
;
j! + 50
60e 50t u(t) V.
30 V;
vo (0+ ) = 30
60 =
30 V.
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Problems
17–17
This makes sense because there cannot be an instantaneous change in the
voltage across a capacitor.
vo (1) = 30 V.
This agrees with vTh (1) = 30 V.
As in Problem 17.20 we know the time constant is 20 ms.
P 17.22 [a] vg = 100u(t);
#
"
1
;
Vg (!) = 100 ⇡ (!) +
j!
H(s) =
10
2
=
.
5s + 10
s+2
H(!) =
2
;
j! + 2
Vo (!) = H(!)Vg (!) =
200
200⇡ (!)
+
j! + 2
j!(j! + 2)
= V1 (!) + V2 (!);
1
1 Z 1 200⇡ejt!
(!) d! =
v1 (t) =
2⇡ 1 j! + 2
2⇡
V2 (!) =
K2
100
K1
+
=
j!
j! + 2
j!
v2 (t) = 50sgn(t)
✓
200⇡
2
◆
= 50 (sifting property);
100
;
j! + 2
100e 2t u(t);
vo (t) = v1 (t) + v2 (t) = 50 + 50sgn(t)
= 100u(t)
100e 2t u(t);
vo (t) = 100(1
e 2t )u(t) V.
100e 2t u(t)
[b]
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17–18
CHAPTER 17. The Fourier Transform
P 17.23 [a] From the solution to Problem 17.22
H(!) =
2
.
j! + 2
Now,
Vg (!) =
200
.
j!
Then,
Vo (!) = H(!)Vg (!) =
.·.
K1
K2
200
400
=
+
=
j!(j! + 2)
j!
j! + 2
j!
200
;
j! + 2
200e 2t u(t) V.
vo (t) = 100sgn(t)
[b]
P 17.24 [a] Io =
RCsIg
Ig R
=
;
R + 1/sC
RCs + 1
106
1
=
= 40;
RC
25 ⇥ 103
ig = 200sgn(t) µA;
H(s) =
H(j!) =
Io
s
;
=
Ig
s + 1/RC
j!
;
j! + 40
2
Ig = (200 ⇥ 10 )
j!
6
!
=
400 ⇥ 10 6
;
j!
j!
400 ⇥ 10 6
400 ⇥ 10 6
Io = Ig [H(j!)] =
·
=
;
j!
j! + 40
j! + 40
io (t) = 400e 40t u(t) µA.
[b] Yes, at the time the source current jumps from 200 µA to +200 µA the
capacitor is charged to (200)(50) ⇥ 10 3 = 10 V, positive at the lower
terminal. The circuit at t = 0 is
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Problems
17–19
At t = 0+ the circuit is
The time constant is (50 ⇥ 103 )(0.5 ⇥ 10 6 ) = 25 ms.
1
.·.
= 40
⌧
P 17.25 [a] Vo =
.·.
for t > 0,
io = 400e 40t µA.
Ig R
Ig R(1/sC)
=
;
R + (1/sC)
RCs + 1
H(s) =
Vo
2 ⇥ 106
1/C
=
;
=
Ig
s + (1/RC)
s + 40
H(j!) =
2 ⇥ 106
;
40 + j!
Ig (!) =
Vo (!) = H(j!)Ig (!) =
=
400 ⇥ 10 6
j!
20
800
=
j!(40 + j!)
j!
vo (t) = 10sgn(t)
400 ⇥ 10 6
;
j!
!
2 ⇥ 106
40 + j!
!
20
;
40 + j!
20e 40t u(t) V.
[b] Yes, at the time the current source jumps from 200 to +200 µA the
capacitor is charged to 10 V. That is, at t = 0 ,
vo (0 ) = (50 ⇥ 103 )( 200 ⇥ 10 6 ) = 10 V.
At t = 1 the capacitor will be charged to +10 V. That is,
vo (1) = (50 ⇥ 103 )(200 ⇥ 10 6 ) = 10 V.
The time constant of the circuit is (50 ⇥ 103 )(0.5 ⇥ 10 6 ) = 25 ms, so
1/⌧ = 40. The function vo (t) is plotted below:
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17–20
CHAPTER 17. The Fourier Transform
P 17.26 [a]
Vo
4/s
;
= H(s) =
Vg
0.5 + 0.01s + 4/s
H(s) =
400
s2 + 50s + 400
H(j!) =
Vg (!) =
6
;
j!
2400
= 6;
400
K3 =
2400
= 2;
( 40)( 30)
[b] vo (0 ) =
[c] vo (0+ ) = 3
K2 =
2400
=
( 10)(30)
8;
8
2
+
;
j! + 10 j! + 40
6
j!
vo (t) = 3sgn(t)
[d] For t
2400
;
j!(j! + 10)(j! + 40)
K2
K3
K1
+
+
;
j!
j! + 10 j! + 40
K1 =
Vo (!) =
400
;
(s + 10)(s + 40)
400
;
(j! + 10)(j! + 40)
Vo (!) = Vg (!)H(j!) =
Vo (!) =
=
8e 10t u(t) + 2e 40t u(t) V.
3 V.
8+2=
3 V.
0+ :
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Problems
17–21
(Vo + 3/s)s
Vo 3/s
+
= 0;
0.5 + 0.01s
4
Vo

s
300
100
+
=
s + 50 4
s(s + 50)
0.75;
Vo =
K1
K2
K3
1200 3s2 150s
=
+
+
;
s(s + 10)(s + 40)
s
s + 10 s + 40
K1 =
1200
= 3;
400
K3 =
1200 4800 + 6000
= 2;
( 40)( 30)
K2 =
1200 300 + 1500
=
( 10)(30)
8;
8e 10t + 2e 40t )u(t) V.
vo (t) = (3
[e] Yes.
P 17.27 [a] Io =
Vg
;
0.5 + 0.01s + 4/s
H(s) =
Io
100s
100s
=
;
= 2
Vg
s + 50s + 400
(s + 10)(s + 40)
H(j!) =
100(j!)
;
(j! + 10)(j! + 40)
Vg (!) =
6
;
j!
Io (!) = H(j!)Vg (!) =
=
20
j! + 10
io (t) = (20e 10t
600
(j! + 10)(j! + 40)
20
;
j! + 40
20e 40t )u(t) A.
[b] io (0 ) = 0.
[c] io (0+ ) = 0.
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17–22
CHAPTER 17. The Fourier Transform
[d]
Io =
=
600
6/s
= 2
0.5 + 0.01s + 4/s
s + 50s + 400
600
20
=
(s + 10)(s + 40)
s + 10
io (t) = (20e 10t
20
;
s + 40
20e 40t )u(t) A.
[e] Yes.
P 17.28 [a] ig = 3e 5|t| ;
.·.
Ig (!) =
3
+
j! + 5
3
30
=
.
j! + 5
(j! + 5)( j! + 5)
Vo Vo s
+
= Ig ;
10
10
Vo
10
;
= H(s) =
Ig
s+1
.·.
Vo (!) = Ig (!)H(!) =
=
300
= 12.5;
(4)(6)
K2 =
300
=
( 4)(10)
K3 =
300
= 5;
(6)(10)
Vo (!) =
12.5
j! + 1
vo (t) = [12.5e t
10
.
j! + 1
300
(j! + 1)(j! + 5)( j! + 5)
K1
K2
+
+
j! + 1 j! + 5
K1 =
H(!) =
K3
.
j! + 5
7.5;
7.5
+
j! + 5
5
;
j! + 5
7.5e 5t ]u(t) + 5e5t u( t) V.
[b] vo (0 ) = 5 V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
[c] vo (0+ ) = 12.5
5t
[d] ig = 3e
7.5 = 5 V.
u(t),
3
;
s+5
Ig =
17–23
0+ ;
t
H(s) =
vo (0+ ) = 5 V;
10
;
s+1
C = 0.5.
Vo Vo s
+
= Ig + 0.5;
10
10
Vo (s + 1) =
Vo =
5
30
+
(s + 5)(s + 1) s + 1
=
.·.
30
+ 5;
s+5
7.5
7.5
5
12.5
+
+
=
s+5 s+1 s+1
s+1
vo (t) = (12.5e t
7.5e 5t )u(t) V.
0+ the solution in part (a) is also
[e] Yes, for t
vo (t) = (12.5e t
P 17.29 [a] Io =
7.5
;
s+5
7.5e 5t )u(t) V.
10s
s
10k(10/s)
Ig =
Ig =
Ig ;
10/s
10s + 10
s+1
H(s) =
Io
s
;
=
Ig
s+1
H(j!) =
Ig (!) =
j!
;
j! + 1
30
=
(j! + 5)( j! + 5)
j!
Io (!) = H(j!)Ig (j!) =
j! + 1
=
3
3
+
;
j! + 5 j! + 5
"
3
3
+
j! + 5 j! + 5
#
3j!
3j!
+
(j! + 1)( j! + 5) (j! + 1)(j! + 5)
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17–24
CHAPTER 17. The Fourier Transform
=
0.5
+
j! + 1
.·. Io (!) =
2.5
0.75
3.75
+
+
.
j! + 5 j! + 1 j! + 5
1.25
+
j! + 1
2.5
3.75
+
.
j! + 5 j! + 5
io (t) = 2.5e5t u( t) + [ 1.25e t + 3.75e 5t ]u(t) A.
[b] io (0 ) = 2.5 A.
[c] io (0+ ) = 2.5 A.
[d] From Problem 17.28, vo (0+ ) = 5 V.
sVg 5
Vg (5/s)
=
;
10 + (10/s)
10s + 10
Io =
.·. Io =
Vg =
30
.
s+5
1.25
3.75
2.5(s 1)
=
+
.
(s + 1)(s + 5)
s+1
s+5
io (t) = ( 1.25e t + 3.75e 5t )u(t) A.
[e] Yes, for t
0+ the solution in part (a) is also
io (t) = ( 1.25e t + 3.75e 5t )u(t) A.
P 17.30
Vo
Vg
2s
+
.·. Vo =
Io =
100Vo
Vo s
+
= 0;
s
100s + 125 ⇥ 104
s(100s + 125 ⇥ 104 )Vg
.
125(s2 + 12,000s + 25 ⇥ 106 )
sVo
;
100s + 125 ⇥ 104
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Problems
H(s) =
17–25
Io
s2
.
=
Vg
125(s2 + 12,000s + 25 ⇥ 106 )
8 ⇥ 10 3 ! 2
;
(25 ⇥ 106 ! 2 ) + j12,000!
H(j!) =
Vg (!) = 300⇡[ (! + 5000) + (!
Io (!) = H(j!)Vg (!) =
5000)];
2.4⇡! 2 [ (! + 5000) + (! 5000)]
;
(25 ⇥ 106 ! 2 ) + j12,000!
2.4⇡ Z 1 ! 2 [ (! + 5000) + (! 5000)] jt!
e d!
2⇡
! 2 ) + j12,000!
1 (25 ⇥ 106
io (t) =
=
1.2
(
(
e j5000t ej5000t
+
j
j
6
=
12
25 ⇥ 106 ej5000t
25 ⇥ 106 e j5000t
+
j(12,000)(5000) j(12,000)(5000)
)
)
= 0.5[e j(5000t+90 ) + ej(5000t+90 ) ];
io (t) = 1 cos(5000t + 90 ) A.
P 17.31 [a]
Vo
Vo
Vo Vg
= 0;
+
+
sL1
sL2
R
.·. Vo =
Io =
.·.
RVg
✓
◆ .
1
1
+
L1 s + R
L1 L2

Vo
;
sL2
Io
R/L1 L2
.
= H(s) =
Vg
s(s + R[(1/L1 ) + (1/L2 )])
R
= 12 ⇥ 105 ;
L1 L2
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17–26
CHAPTER 17. The Fourier Transform
✓
1
1
R
+
L1 L2
◆
.·. H(s) =
12 ⇥ 105
.
s(s + 3 ⇥ 104 )
H(j!) =
= 3 ⇥ 104 ;
12 ⇥ 105
;
j!(j! + 3 ⇥ 104 )
Vg (!) = 125⇡[ (! + 4 ⇥ 104 ) + (!
4 ⇥ 104 )];
1500⇡ ⇥ 105 [ (! + 4 ⇥ 104 ) + (!
Io (!) = H(j!)Vg (!) =
j!(j! + 3 ⇥ 104 )
io (t) =
4 ⇥ 104 )]
;
1500⇡ ⇥ 105 Z 1 [ (! + 4 ⇥ 104 ) + (! 4 ⇥ 104 )]ejt!
d!;
2⇡
j!(j! + 3 ⇥ 104 )
1
5
io (t) = 750 ⇥ 10
(
e j40,000t
j40,000(30,000 j40,000)
ej40,000t
+
j40,000(30,000 + j40,000)
75 ⇥ 106
=
4 ⇥ 108
75
=
400
(
(
)
ej40,000t
e j40,000t
+
j(3 + j4) j(3 + j4)
ej40,000t
e j40,000t
+
5/ 143.13
5/143.13
= 0.075 cos(40,000t
io (t) = 75 cos(40,000t
)
)
143.13 ) A;
143.13 ) mA.
[b] In the phasor domain:
Vo
Vo
Vo 125
+
+
= 0;
j200
j800 120
1500 + 3Vo + j20Vo = 0;
12Vo
Vo =
Io =
1500
= 60/
15 + j20
Vo
= 75 ⇥ 10 3 /
j800
53.13 V;
143.13 A;
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Problems
io (t) = 75 cos(40,000t
17–27
143.13 ) mA.
P 17.32 [a]
(Vo
Vg )s
106
.·. Vo =
Vo
Vo
+
= 0;
4s 800
+
s2 Vg
.
s2 + 1250s + 25 ⇥ 104
s2
Vo
;
= H(s) =
Vg
(s + 250)(s + 1000)
H(j!) =
(j!)2
;
(j! + 250)(j! + 1000)
vg = 45e 500|t| ;
Vg (!) =
.·. Vo (!) = H(j!)Vg (!) =
=
45,000
;
(j! + 500)( j! + 500)
45,000(j!)2
(j! + 250)(j! + 500)(j! + 1000)( j! + 500)
K1
K2
K3
+
+
+
j! + 250 j! + 500 j! + 1000
K4
.
j! + 500
45,000( 250)2
= 20;
K1 =
(250)(750)(750)
K2 =
45,000( 500)2
=
( 250)(500)(1000)
K3 =
45,000( 1000)2
= 80;
( 750)( 500)(1500)
K4 =
45,000(500)2
= 10;
(750)(1000)(1500)
.·. vo (t) = [20e 250t
[b] vo (0 ) = 10 V;
90;
90e 500t + 80e 1000t ]u(t) + 10e500t u( t) V.
Vo (0+ ) = 20
90 + 80 = 10 V;
vo (1) = 0 V.
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17–28
CHAPTER 17. The Fourier Transform
[c] IL =
0.25sVg
Vo
=
;
4s
(s + 250)(s + 1000)
H(s) =
0.25s
IL
;
=
Vo
(s + 250)(s + 1000)
H(j!) =
IL (!) =
=
K4 =
0.25(j!)
;
(j! + 250)(j! + 1000)
0.25(j!)(45,000)
(j! + 250)(j! + 500)(j! + 1000)( j! + 500)
K2
K3
K1
+
+
+
j! + 250 j! + 500 j! + 1000
K4
;
j! + 500
(0.25)(500)(45,000)
= 5 mA.
(750)(1000)(1500)
iL (t) = 5e500t u( t);
.·. iL (0 ) = 5 mA;
K1 =
(0.25)( 250)(45,000)
=
(250)(750)(750)
K2 =
(0.25)( 500)(45,000)
= 45 mA;
( 250)(500)(1000)
K3 =
(0.25)( 1000)(45,000)
=
( 750)( 500)(1500)
20 mA;
.·. iL (0+ ) = K1 + K2 + K3 =
20 + 45
20 mA;
Checks, i.e.,
At t = 0 :
iL (0+ ) = iL (0 ) = 5 mA.
vC (0 ) = 45
10 = 35 V.
20 = 5 mA.
At t = 0+ :
vC (0+ ) = 45
10 = 35 V.
[d] We can check the correctness of out solution for t
Laplace transform. Our circuit becomes
0+ by using the
Vo (Vo Vg )s
5 ⇥ 10 3
Vo
6
+
+
= 0;
+
35
⇥
10
+
800 4s
106
s
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Problems
.·. (s2 + 1250s + 24 ⇥ 104 )Vo = s2 Vg
vg (t) = 45e 500t u(t) V;
Vg =
45s
.·. (s + 250)(s + 1000)Vo =
.·. Vo =
=
17–29
(35s + 5000).
45
;
s + 500
2
(35s + 5000)(s + 500)
.
(s + 500)
10s2 22,500s 250 ⇥ 104
(s + 250)(s + 500)(s + 1000)
20
s + 250
.·. vo (t) = [20e 250t
90
80
+
;
s + 500 s + 1000
90e 500t + 80e 1000t ]u(t) V.
This agrees with our solution for vo (t) for t
0+ .
P 17.33 [a]
From the plot of vg note that vg is
before t = 0. Therefore
vo (0 ) =
10 V for an infinitely long time
10 V.
There cannot be an instantaneous change in the voltage across a
capacitor, so
vo (0+ ) =
10 V.
[b] io (0 ) = 0 A.
At t = 0+ the circuit is
io (0+ ) =
30
40
( 10)
=
= 8 A.
5
5
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17–30
CHAPTER 17. The Fourier Transform
[c] The s-domain circuit is
"
Vg
Vo =
5 + (10/s)
#✓
◆
2Vg
10
;
=
s
s+2
2
Vo
.
= H(s) =
Vg
s+2
H(j!) =
2
;
j! + 2
2
Vg (!) = 5
j!
!
5[2⇡ (!)] +
10
30
=
j! + 5
j!
"
10⇡ (!) +
30
10⇡ (!) +
j! + 5
2
10
Vo (!) = H(!)Vg (!) =
j! + 2 j!
20
j!(j! + 2)
=
K0
K1
K2
K3
+
+
+
j!
j! + 2 j! + 2 j! + 5
20⇡ (!)
.
j! + 2
20
= 10;
2
60
= 20;
3
Vo (!) =
#
20⇡ (!)
60
+
j! + 2
(j! + 2)(j! + 5)
=
K0 =
30
j! + 5;
K1 =
20
=
2
10
10
+
j! j! + 2
vo (t) = 5sgn(t) + [10e 2t
20
j! + 5
10;
K2 =
K3 =
60
=
3
20;
20⇡ (!)
.
j! + 2
20e 5t ]u(t)
5 V;
P 17.34 [a]
36
=
4 + j!
(4
Vg (!) =
36
4 j!
Vo (s) =
(16/s)
Vg (s);
10 + s + (16/s)
72j!
;
j!)(4 + j!)
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Problems
H(s) =
17–31
16
16
Vo (s)
= 2
=
.
Vg (s)
s + 10s + 16
(s + 2)(s + 8)
H(j!) =
16
;
(j! + 2)(j! + 8)
Vo (j!) = H(j!) · Vg (!) =
=
1152j!
j!)(4 + j!)(2 + j!)(8 + j!)
(4
K1
K2
K3
K4
+
+
+
;
4 j! 4 + j! 2 + j! 8 + j!
K1 =
1152(4)
= 8;
(8)(6)(12)
K2 =
1152( 4)
= 72;
(8)( 2)(4)
K3 =
1152( 2)
=
(6)(2)(6)
K4 =
1152( 8)
=
(12)( 4)( 6)
32;
32;
72
4 + j!
32
2 + j!
32
.
8 + j!
.·. vo (t) = 8e4t u( t) + [72e 4t
32e 2t
32e 8t ]u(t)V.
.·. Vo (j!) =
8
4
j!
+
[b] vo (0 ) = 8V.
[c] vo (0+ ) = 72 32 32 = 8V.
The voltages at 0 and 0+ must be the same since the voltage cannot
change instantaneously across a capacitor.
P 17.35 [a]
Vo =
V g s2
Vg s
= 2
;
25 + (100/s) + s
s + 25s + 100
H(s) =
Vo
s2
;
=
Vg
(s + 5)(s + 20)
H(j!) =
(j!)2
.
(j! + 5)(j! + 20)
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17–32
CHAPTER 17. The Fourier Transform
vg = 25ig = 450e10t u( t)
450
j! + 10
Vg =
450e 10t u(t) V;
450
;
j! + 10
Vo (!) = H(j!)Vg =
450(j!)2
( j! + 10)(j! + 5)(j! + 20)
450(j!)2
+
(j! + 10)(j! + 5)(j! + 20)
=
K1
K2
K3
K4
K5
K6
+
+
+
+
+
.
j! + 10 j! + 5 j! + 20 j! + 5 j! + 10 j! + 20
K1 =
450(100)
= 100
(15)(30)
K2 =
450(25)
= 50
(15)(15)
K3 =
450(400)
=
(30)( 15)
Vo (!) =
450(25)
=
(5)(15)
K4 =
K5 =
400
150;
450(100)
= 900;
( 5)(10)
K6 =
450(400)
=
( 15)( 10)
1200;
100
1600
900
100
+
+
+
j! + 10 j! + 5 j! + 20 j! + 10
vo = 100e10t u( t) + [900e 10t
100e 5t
1600e 20t ]u(t) V.
[b] vo (0 ) = 100 V.
[c] vo (0+ ) = 900
100
1600 =
800 V.
[d] At t = 0 the circuit is
Therefore, the solution predicts v1 (0 ) will be 350 V.
Now v1 (0+ ) = v1 (0 ) because the inductor will not let the current in the
25 ⌦ resistor change instantaneously, and the capacitor will not let the
voltage across the 0.01 F capacitor change instantaneously.
At t = 0+ the circuit is
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Problems
17–33
From the circuit at t = 0+ we see that vo must be 800 V, which is
consistent with the solution for vo obtained in part (a).
It is informative to solve for either the current in the circuit or the
voltage across the capacitor and note the solutions for io and vC are
consistent with the solution for vo
The solutions are
io = 10e10t u( t) + [20e 5t + 80e 20t
vC = 100e10t u( t) + [900e 10t
P 17.36 Vo (s) =
30
10
+
s
s + 20
90e 10t ]u(t) A;
400e 5t
400e 20t ]u(t) V.
40
600(s + 10)
=
;
s + 30
s(s + 20)(s + 30)
Vo (s) = H(s) ·
15
;
s
.·.
H(s) =
40(s + 10)
;
(s + 20)(s + 30)
.·.
H(!) =
40(j! + 10)
;
(j! + 20)(j! + 30)
.·.
Vo (!) =
40(j! + 10)
1200(j! + 10)
30
·
=
.
j! (j! + 20)(j! + 30)
j!(j! + 20)(j! + 30)
vo (!) =
60
20
+
j! j! + 20
80
.
j! + 30
vo (t) = 10sgn(t) + [60e 20t
1
P 17.37 [a] f (t) =
2⇡
[b] W = 2
[c] W =
1
⇡
⇢Z 0
Z 1
0
Z 1
0
1
! jt!
e e
2
80e 30t ]u(t) V.
d! +
Z 1
2
(1/⇡)
dt = 2
2
2
(1 + t )
⇡
e 2! d! =
1 e 2!
⇡ 2
0
e ! ejt! d! =
Z 1
0
1
0
1/⇡
.
1 + t2
1
dt
J.
=
2
2
(1 + t )
2⇡
=
1
J.
2⇡
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17–34
CHAPTER 17. The Fourier Transform
1 Z !1 2!
0.9
[d]
,
1 e 2!1 = 0.9,
e
d! =
⇡ 0
2⇡
!1 = (1/2) ln 10 ⇠
= 1.15 rad/s.
P 17.38 Io =
e2!1 = 10;
sIg
0.5sIg
=
;
0.5s + 25
s + 50
H(s) =
Io
s
.
=
Ig
s + 50
H(j!) =
I(!) =
j!
;
j! + 50
12
;
j! + 10
Io (!) = H(j!)I(!) =
|Io (!)| = q
|Io (!)|2 =
=
12(j!)
;
(j! + 10)(j! + 50)
12!
(! 2 + 100)(! 2 + 2500)
;
144! 2
(! 2 + 100)(! 2 + 2500)
6
+
! 2 + 100
150
! 2 + 2500
.
Wo (total) =
1 Z 1 150d!
⇡ 0 ! 2 + 2500
1 Z 1 6d!
⇡ 0 ! 2 + 100
=
3
! 1
tan 1
⇡
50 0
0.6
! 1
tan 1
⇡
10 0
= 1.5
✓
◆
✓
◆
0.3 = 1.2 J.
Wo (0 — 100 rad/s) =
3
tan 1 (2)
⇡
= 1.06
0.6
tan 1 (10)
⇡
0.28 = 0.78 J.
Therefore, the percent between 0 and 100 rad/s is
0.78
(100) = 64.69%.
1.2
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Problems
17–35
P 17.39
Io =
RCsIg
Ig R
=
;
R + (1/sC)
RCs + 1
H(s) =
Io
s
;
=
Ig
s + (1/RC)
RC = (100 ⇥ 103 )(1.25 ⇥ 10 6 ) = 125 ⇥ 10 3 ;
H(s) =
s
;
s+8
Ig (!) =
30 ⇥ 10 6
;
j! + 2
H(j!) =
Io (!) = H(j!)Ig (!) =
1
1
=
= 8;
RC
0.125
j!
;
j! + 8
30 ⇥ 10 6 j!
;
(j! + 2)(j! + 8)
!(30 ⇥ 10 6 )
p
|Io (!)| = p 2
;
( ! + 4)( ! 2 + 64)
K1
K2
900 ⇥ 10 12 ! 2
|Io (!)| = 2
= 2
+ 2
.
2
(! + 4)(! + 64)
! + 4 ! + 64
2
K1 =
(900 ⇥ 10 12 )( 4)
=
(60)
K2 =
(900 ⇥ 10 12 )( 64)
= 960 ⇥ 10 12 .
( 60)
|Io (!)|2 =
960 ⇥ 10 12
! 2 + 64
60 ⇥ 10 12 ;
60 ⇥ 10 12
.
!2 + 4
1Z1
960 ⇥ 10 12 Z 1 d!
2
|Io (!)| d! =
W1⌦ =
⇡ 0
⇡
! 2 + 64
0
60 ⇥ 10 12 Z 1 d!
⇡
!2 + 4
0
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17–36
CHAPTER 17. The Fourier Transform
=
120 ⇥ 10 12
! 1
tan 1
⇡
8 0
=
✓
120 ⇡
·
⇡ 2
30 ⇥ 10 12
! 1
tan 1
⇡
2 0
◆
30 ⇡
·
⇥ 10 12 = (60
⇡ 2
15) ⇥ 10 12 = 45 pJ.
Between 0 and 4 rad/s
W1⌦ =
%=

30
tan 1 2 ⇥ 10 12 = 7.14 pJ;
⇡
1
120
tan 1
⇡
2
7.14
(100) = 15.86%.
45
P 17.40 [a] Vg (!) =
60
;
(j! + 1)( j! + 1)
H(s) =
Vo
0.4
;
=
Vg
s + 0.5
Vo (!) =
24
;
(j! + 1)(j! + 0.5)( j! + 1)
Vo (!) =
32
24
+
+
j! + 1 j! + 0.5
H(!) =
0.4
.
(j! + 0.5)
8
;
j! + 1
vo (t) = [ 24e t + 32e t/2 ]u(t) + 8et u( t) V.
[b] |Vg (!)| =
60
.
(! 2 + 1)
[c] |Vo (!)| =
(! 2 + 1)
24
p
.
! 2 + 0.25
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Problems
[d] Wi = 2
[e] Wo =
Z 1
0
Z 0
900e 2t dt = 1800
2t
1
64e dt +
Z 1
0
R
= 32 + 01 [576e 2t
= 32 + 288
[f ] |Vg (!)| =
60
!2 + 1
1
e 2t
= 900 J.
2 0
( 24e t + 32e t/2 )2 dt
1536e 3t/2 + 1024e t ] dt
1024 + 1024 = 320 J.
|Vg2 (!)| =
,
3600
;
(! 2 + 1)2
3600 Z 2
d!
Wg =
2
⇡
0 (! + 1)2
3600
=
⇡
=
(
✓
17–37
✓
1
!
+ tan 1 !
2
2 ! +1
◆
◆ 2)
0
1800 2
+ tan 1 2 = 863.53 J;
⇡
5
✓
◆
863.53
.·. % =
⇥ 100 = 95.95%.
900
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17–38
CHAPTER 17. The Fourier Transform
576
[g] |Vo (!)|2 =
(! 2 + 1)2 (! 2 + 0.25)
=
1024
768
! 2 + 0.25
(! 2 + 1)2
(
2
1
1024 · 2 · tan 1 2!
Wo =
⇡
0
1024 tan
1
2
!
0
2048
tan 1 4
=
⇡
1024
;
(! 2 + 1)
✓ ◆✓
1
768
2
)
◆
✓
384 2
+ tan 1 2
⇡ 5
◆
2
!
1
+
tan
!
!2 + 1
0
1024
tan 1 2
⇡
= 319.2 J.
%=
319.2
⇥ 100 = 99.75%.
320
P 17.41 [a] |Vi (!)|2 =
[b] Vo =
4 ⇥ 104
;
!2
|Vi (100)|2 =
4 ⇥ 104
= 4;
1002
|Vi (200)|2 =
4 ⇥ 104
= 1.
2002
sRCVi
Vi R
=
;
R + (1/sC)
RCs + 1
H(s) =
s
Vo
;
=
Vi
s + (1/RC)
1
106 10 3
1000
=
=
= 100.
RC
(0.5)(20)
10
H(j!) =
j!
;
j! + 100
|Vo (!)| =
|!|
200
200
·p 2
=p 2
;
4
|!|
! + 10
! + 104
|Vo (!)|2 =
4 ⇥ 104
,
! 2 + 104
100  !  200 rad/s;
|Vo (!)|2 = 0,
elsewhere;
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Problems
|Vo (100)|2 =
4 ⇥ 104
= 2;
104 + 104
|Vo (200)|2 =
1 Z 200 4 ⇥ 104
4 ⇥ 104
[c] W1⌦ =
d!
=
⇡ 100
!2
⇡

4 ⇥ 104 1
=
⇡
100

17–39
4 ⇥ 104
= 0.8.
5 ⇥ 104
1 200
! 100
1
200 ⇠
=
= 63.66 J.
200
⇡
200
4 ⇥ 104
1 Z 200 4 ⇥ 104
1 !
[d] W1⌦ =
·
tan
d!
=
⇡ 100 ! 2 + 104
⇡
100 100
400
[tan 1 2 tan 1 1] ⇠
=
= 40.97 J.
⇡
P 17.42 [a] Vi (!) =
A
;
a + j!
A
|Vi (!)| = p 2
;
a + !2
H(s) =
s
;
s+↵
H(j!) =
j!
;
↵ + j!
!
|H(!)| = p 2
.
↵ + !2
!A
Therefore |Vo (!)| = q
(a2 + ! 2 )(↵2 + ! 2 )
Therefore |Vo (!)|2 =
WIN =
Z 1
0
.
! 2 A2
.
(a2 + ! 2 )(↵2 + ! 2 )
A2 e 2at dt =
A2
;
2a
when ↵ = a we have
(
A2 R a ! 2 d!
A2 Z a d!
=
WOUT =
⇡ 0 (! 2 + a2 )2
⇡
0 a2 + ! 2
A2
=
4a⇡
✓
⇡
2
Z a
a2 d!
0 (a2 + ! 2 )2
)
◆
1 .
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17–40
CHAPTER 17. The Fourier Transform
#
"
A2
A2 Z 1
!2
WOUT (total) =
.
d!
=
⇡ 0
(a2 + ! 2 )2
4a
WOUT (a)
= 0.5
WOUT (total)
Therefore
1
= 0.1817 or 18.17%.
⇡
[b] When ↵ 6= a we have
WOUT (↵) =
1Z ↵
! 2 A2 d!
⇡ 0 (a2 + ! 2 )(↵2 + ! 2 )
A2
=
⇡
where K1 =
⇢Z ↵ 
0
K2
K1
+ 2
2
2
a +!
↵ + !2
a2
a2
and K2 =
↵2
d! ,
↵2
.
a2 ↵ 2
Therefore
WOUT (↵) =
A2
⇡(a2
WOUT (total) =
Therefore
↵2 )

a tan 1
✓ ◆
↵
a
↵⇡
;
4

↵
⇡
A2
.
=
2
2(a + ↵)
A2
⇡(a2
↵2 )
a
⇡
2

✓ ◆
2
WOUT (↵)
↵
=
· a tan 1
WOUT (total)
⇡(a ↵)
a
↵⇡
.
4
p
For ↵ = a 3, this ratio is 0.2723, orp
27.23% of the output energy lies in
the frequency band between 0 and a 3.
p
[c] For ↵ = a/ 3, the ratio is 0.1057, or 10.57%
of the output energy lies in
p
the frequency band between 0 and a/ 3.
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18
Two-Port Circuits
Assessment Problems
AP 18.1 With port 2 short-circuited, we have
V1 V1
+ ;
I1 =
20
5
I1
= y11 = 0.25 S;
V1
When V2 = 0, we have I1 = y11 V1
Therefore I2 =
Thus y21 =
0.8(y11 V1 ) =
0.8y11 =
I2 =
✓
◆
20
I1 =
25
0.8I1 .
and I2 = y21 V1 .
0.8y11 V1 .
0.2 S.
With port 1 short-circuited, we have
I2 =
V2 V2
+ ;
15
5
✓
4
I2
= y22 =
V2
15
◆
S;
18–1
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18–2
CHAPTER 18. Two-Port Circuits
I1 =
✓
◆
15
I2 =
20
0.75I2 =
Therefore y12 = ( 0.75)
AP 18.2
✓
◆
I1
g11 =
V1
✓
V2
V1
◆
✓
I1
I2
◆
V2
g22 =
I2
◆
g21 =
g12 =
✓
◆
✓
V1
h12 =
V2
◆
✓
◆
AP 18.3
=
( 15/20)I2
=
I2
V1 =0
I2
I1
I2
h22 =
V2
(15/20)V1
= 0.75;
V1
= 15k5 =
✓
h21 =
=
V1 =0
◆
0.2 S.
1
1
+
= 0.1 S;
20 20
I2 =0
V1
I1
4
=
15
=
I2 =0
✓
h11 =
0.75y22 V2 .
0.75;
75
= 3.75 ⌦.
20
= 20k5 = 4 ⌦;
V2 =0
=
( 20/25)I1
=
I1
=
(20/25)V2
= 0.8;
V2
=
1
8
1
+
=
S.
15 25
75
V2 =0
I1 =0
I1 =0
0.8;
g11 =
I1
5 ⇥ 10 6
= 0.1 mS;
=
V1 I2 =0 50 ⇥ 10 3
g21 =
V2
200 ⇥ 10 3
= 4;
=
V1 I2 =0 50 ⇥ 10 3
g12 =
I1
2 ⇥ 10 6
= 4;
=
I2 V1 =0 0.5 ⇥ 10 6
g22 =
V2
10 ⇥ 10 3
= 20 k⌦.
=
I2 V1 =0 0.5 ⇥ 10 6
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Problems
18–3
AP 18.4 First calculate the b-parameters:
b11 =
V2
15
= 1.5 ⌦;
=
V1 I1 =0 10
b12 =
V2
=
I1 V1 =0
b21 =
10
= 2 ⌦;
5
I2
30
= 3 S;
=
V1 I1 =0 10
b22 =
I2
=
I1 V1 =0
4
= 0.8.
5
Now the z-parameters are calculated:
z11 =
4
b22
0.8
=
⌦;
=
b21
3
15
1
1
= ⌦;
b21
3
(1.5)(0.8)
3
6
z11 = z22 ,
z12 = z21 ,
95 = z11 (5) + z12 (0).
Therefore,
z11 = z22 = 95/5 = 19 ⌦.
z21 =
b
z12 =
b21
=
=
1.6 ⌦;
z22 =
1
b11
1.5
= ⌦.
=
b21
3
2
AP 18.5
11.52 = 19I1
0 = z12 I1
z12 (2.72);
19(2.72).
Solving these simultaneous equations for z12 yields the quadratic equation
2
z12
+
✓
◆
72
z12
17
6137
= 0.
17
For a purely resistive network, it follows that z12 = z21 = 17 ⌦.
AP 18.6 [a] I2 =
=
Vg
a11 ZL + a12 + a21 Zg ZL + a22 Zg
50 ⇥ 10 3
(5 ⇥ 10 4 )(5 ⇥ 103 ) + 10 + (10 6 )(100)(5 ⇥ 103 ) + ( 3 ⇥ 10 2 )(100)
=
50 ⇥ 10 3
=
10
5 mA.
1
PL = (5 ⇥ 10 3 )2 (5 ⇥ 103 ) = 62.5 mW.
2
[b] ZTh =
=
a12 + a22 Zg
10 + ( 3 ⇥ 10 2 )(100)
=
a11 + a21 Zg
5 ⇥ 10 4 + (10 6 )(100)
7
70
k⌦.
=
4
6 ⇥ 10
6
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18–4
CHAPTER 18. Two-Port Circuits
[c] VTh =
Vg
50 ⇥ 10 3
500
V;
=
=
4
a11 + a21 Zg
6 ⇥ 10
6
Therefore V2 =
250
V;
6
Pmax =
(1/2)(250/6)2
= 74.4 mW.
(70/6) ⇥ 103
AP 18.7 [a] For the given bridged-tee circuit, we have
1
S,
a012 = 11.25 ⌦.
20
The a-parameters of the cascaded networks are
a011 = a022 = 1.25,
a021 =
a11 = (1.25)2 + (11.25)(0.05) = 2.125;
a12 = (1.25)(11.25) + (11.25)(1.25) = 28.125 ⌦;
a21 = (0.05)(1.25) + (1.25)(0.05) = 0.125 S;
a22 = a11 = 2.125,
RTh = (45.125/3.125) = 14.44 ⌦.
100
= 32 V;
therefore V2 = 16 V.
3.125
162
= 17.73 W.
[c] P =
14.44
[b] Vt =
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Problems
18–5
Problems
P 18.1
✓
◆
✓
◆
✓
◆
✓
◆
✓
◆
V1
h11 =
I1
I2
h21 =
I1
V1
h12 =
V2
I2
h22 =
V2
I1
g11 =
V1
V2
V1
◆
✓
I1
I2
◆
V2
g22 =
I2
◆
g12 =
✓
=
( 20/25)I1
=
I1
=
(20/25)V2
= 0.8;
V2
=
1
8
1
+
=
S.
15 25
75
=
1
1
+
= 0.1 S;
20 20
=
(15/20)V1
= 0.75;
V1
=
( 15/20)I2
=
I2
V2 =0
✓
g21 =
= 20k5 = 4 ⌦;
V2 =0
I1 =0
I1 =0
I2 =0
I2 =0
V1 =0
= 15k5 =
V1 =0
0.8;
0.75;
75
= 3.75 ⌦.
20
P 18.2
z11 =
V1
= 1 + 12 = 13 ⌦;
I1 I2 =0
z21 =
V2
= 12 ⌦;
I1 I2 =0
z22 =
V2
= 4 + 12 = 16 ⌦;
I2 I1 =0
z12 =
V1
= 12 ⌦.
I2 I1 =0
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18–6
P 18.3
CHAPTER 18. Two-Port Circuits
z = (13)(16)
(12)(12) = 64.
y11 =
16
z22
=
= 0.25 S;
z
64
y12 =
12
z12
=
=
z
64
0.1875 S;
y21 =
12
z21
=
=
z
64
0.1875 S;
y22 =
13
z11
=
= 0.203125 S.
z
64
P 18.4
✓
◆
540
7
I2 ;
V2 = 20 I2 + 50I2 =
8
8
◆
500
1
I2 ;
V1 = 100 I2 + 50I2 =
8
8
.·.
b11 =
V2
540
= 1.08;
=
V1 I1 =0 500
.·.
b21 =
I2
8
= 0.016 S.
=
V1 I1 =0 500
R1 = 20 +
✓
8000
160
5000
=
=
⌦;
150
150
3
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Problems
Ia =
160/3
4
I2 = I2 ;
(160/3) + 40
7
Ic =
1 3
1
50
Ib =
I2 = I2 ;
150
3 7
7
✓
◆
I1 + Ia + Ic = 0;
.·.
I2
=
I1 V1 =0
b22 =
V2 = Re I2 ,
P 18.5
I1 =
V2
=
I1 V1 =0
40
3
I2 = I2 ;
(160/3) + 40
7
5
I2 ;
7
( 7/5) = 1.4;
Re = 40kR1 =
160
160
V2 =
I2 =
7
7
b12 =
Ib =
18–7
✓
160
40(160/3)
=
⌦;
40 + (160/3)
7
◆
7
I1 =
5
32 V;
( 160/5) = 32 ⌦.
For V2 = 0:
15k20 =
60
⌦;
7
5k10 =
10
⌦;
3
I1 =
21
V1
=
V1 ;
(60/7) + (10/3)
250
.·.
y11 =
Ia =
2
10
I1 = I1 ;
15
3
I2 = Ib
I1
21
= 84 mS.
=
V1 V2 =0 250
Ia =
4
7
Ib =
2
=
3
20
4
I1 = I1 ;
35
7
2
I1 ;
21
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18–8
CHAPTER 18. Two-Port Circuits
I2 =
✓
.·.
y21 =
2
21
◆✓
◆
21
V1 ;
250
2
I2
=
=
V1 V2 =0 250
8 mS.
For V1 = 0:
10k20 =
P 18.6
40
200
=
⌦;
30
6
5k15 =
I2 =
24
V2
=
V2 ;
(40/6) + (15/4)
250
.·.
y22 =
Ia =
3
15
I2 = I2 ;
20
4
75
15
=
⌦;
20
4
I2
24
= 96 mS.
=
V2 V1 =0 250
Ib =
3
=
4
I1 = Ib
Ia =
2
3
I1 =
1 24
V2 ;
12 250
✓
◆
.·.
y12 =
I1
=
V2 V1 =0
20
2
I2 = I2 ;
30
3
1
I2 ;
12
8 mS.
For I2 = 0:
Calculate g11 :
20k(60 + 20) = 16 ⌦;
16 + 4 = 20 ⌦;
20k80 = 16 ⌦;
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
g11 =
18–9
I1
1
= 62.5 mS.
=
V1 I2 =0 16
Calculate g21 :
Va
V1
4
+
Va Va V2
+
= 0;
20
20
V2 =
3
Va
(60) = Va ;
80
4
.·.
4
Va = V2 ;
3
.·.
28
V2
3
g21 =
V2
15
= 0.60.
=
V1 I2 =0 25
V2 = 5V1
so
25
V2 = 5V1 ;
3
For V1 = 0:
Va Va Va V2
+
+
= 0;
4
20
20
7Va = V2
I1 =
so Va =
V2
;
7
V2
Va
=
;
4
28

V2
24
V2
2000
1
I2 =
+
= V2
+
V2 ;
=
60 20 + (80/24)
60 560
33,600
.·.
V2 =
336
I2 ;
20
.·.
I1 =
336
I2 =
(28)(20)
0.6I2 ;
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18–10
CHAPTER 18. Two-Port Circuits
g12 =
I1
=
I2 V1 =0
g22 =
336
V2
= 16.8 ⌦.
=
I2 V1 =0 20
0.60;
Summary:
g11 = 62.5 mS;
P 18.7
g12 =
0.60;
g21 = 0.60;
h11 =
V1
= R1 kR2 = 4
I1 V2 =0
.·.
h21 =
I2
R2
=
=
I1 V2 =0 R1 + R2
0.8;
.·.
R2 = 0.8R1 + 0.8R2
so R1 =
g22 = 16.8 ⌦.
R 1 R2
= 4;
R 1 + R2
R2
.
4
Substituting,
(R2 /4)R2
= 4 so R2 = 20 ⌦ and R1 = 5 ⌦;
(R2 /4) + R2
h22 =
1
I2
1
=
= 0.14;
=
V2 I1 =0 R3 k(R1 + R2 )
R3 k25
.·.
R3 = 10.
Summary:
R1 = 5 ⌦;
P 18.8
R2 = 20 ⌦;
R3 = 10 ⌦.
For V2 = 0:
V1 = (400 + 1200)I1 ;
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Problems
18–11
V1
1600
= 1600 ⌦;
=
I1 V2 =0
1
h11 =
Vp = 1200(1 A) = 1200 V = Vn .
At Vn ,
1200 1200 Vo
+
= 0 so Vo = 3600 V;
500
1000
3600
=
200
.·.
I2 =
h21 =
I2
=
I1 I1 =0
18 A;
18
=
1
18.
For I1 = 0:
V1 = 0;
h12 =
V1
0
= = 0.
V2 I1 =0 1
At Vn ,
Vn Vo
Vn
+
= 0.
500
100
But Vn = Vp = 0 so Vo = 0; therefore,
I2 =
1V
= 5 mS;
200 ⌦
h22 =
I2
5m
= 5 mS.
=
V2 I1 =0
1
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18–12
P 18.9
CHAPTER 18. Two-Port Circuits
For I2 = 0:
50I1
40I2 = 1;
40I1 + 48I2
3(40)(I1
I2 ) = 0 so
160I1 + 168I2 = 0.
Solving,
I1 = 84 mA;
V2 = 3I2
I2 = 80 mA;
3(40)(I1
I2 ) =
0.24 V;
g11 =
I1
84 m
= 84 mS;
=
V1 I2 =0
1
g21 =
V2
=
V1 I2 =0
0.24
=
1
0.24.
For V1 = 0:
50I1
40I2 = 0;
40I1 + 48I2 + 3
3(40)(I1
I2 ) = 0 so
160I1 + 168I2 =
3.
Solving,
I1 =
60 mA;
I2 =
75 mA;
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Problems
V2 = 3(I2 + 1)
3(40)(I1
I2 ) = 0.975 V;
g12 =
I1
=
I2 V1 =0
60 m
=
1
0.06;
g22 =
V2
0.975
= 0.975 ⌦.
=
I2 V1 =0
1
P 18.10 V1 = a11 V2
a12 I2 ;
I1 = a21 V2
a22 I2 ;
a11 =
V1
;
V2 I2 =0
a21 =
18–13
I1
.
V2 I2 =0
V1 = 103 I1 + 10 4 V2 = 103 ( 0.5 ⇥ 10 6 )V2 + 10 4 V2 ;
.·. a11 =
5 ⇥ 10 4 + 10 4 =
4 ⇥ 10 4 .
V2 =
(50I1 )(40 ⇥ 103 );
.·. a21 =
a12 =
V1
;
I2 V2 =0
I1
.
I2 V2 =0
I2 = 50I1 ;
a22 =
.·. a22 =
I1
=
I2
1
=
2 ⇥ 106
0.5 µS;
1
;
50
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18–14
CHAPTER 18. Two-Port Circuits
V1
=
I2
.·. a12 =
V1 = 1000I1 ;
V1 I1
=
I1 I2
(1000)(1/50) =
20 ⌦.
Summary
a11 =
P 18.11 g11 =
g12 =
4 ⇥ 10 4 ;
a11
20 ⌦;
a21 =
0.5 µS;
a22 =
0.02.
0.5 ⇥ 10 6
= 1.25 mS;
4 ⇥ 10 4
a21
=
a11
a
a12 =
=
( 4 ⇥ 10 4 )( 1/50) ( 0.5 ⇥ 10 6 )( 20)
=
4 ⇥ 10 4
g21 =
1
=
a11
1
=
4 ⇥ 10 4
g22 =
a12
=
a11
( 20)
= 5 ⇥ 104 ⌦.
6
400 ⇥ 10
0.005;
2500;
P 18.12 For V2 = 0:
Ia =
1
50I2
= I2 =
200
4
h21 =
I2 ;
.·.
I2 = 0;
.·.
h11 =
V1
= 10 + j20 ⌦.
I1 V2 =0
I2
= 0;
I1 V2 =0
V1 = (10 + j20)I1
For I1 = 0:
V1 = 50I2 ;
I2 =
V2 50I2
V2
+
;
j100
200
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Problems
200I2 = j2V2 + V2
18–15
50i2 ;
250I2 = V2 (1 + j2);
50I2 = V2
✓
◆
1 + j2
= (0.2 + j0.4)V2 ;
5
.·.
V1 = (0.2 + j0.4)V2 .
h12 =
V1
= 0.2 + j0.4;
V2 I1 =0
h22 =
I2
1 + j2
= 4 + j8 mS.
=
V2 I1 =0
250
Summary:
h11 = 10 + j20 ⌦;
P 18.13 I1 = g11 V1 + g12 I2 ;
h12 = 0.2 + j0.4;
h21 = 0;
h22 = 4 + j8 mS.
V2 = g21 V1 + g22 I2 ;
g11 =
I1
0.25 ⇥ 10 6
= 12.5 ⇥ 10 6 = 12.5 µS;
=
V1 I2 =0
20 ⇥ 10 3
g21 =
V2
5
⇥ 103 =
=
V1 I2 =0 20
0=
250(10) + g22 (50 ⇥ 10 6 );
g22 =
2500
= 50 M⌦;
50 ⇥ 10 6
250;
200 ⇥ 10 6 = 12.5 ⇥ 10 6 (10) + g12 (50 ⇥ 10 6 )
(200
125)10 6 = g12 (50 ⇥ 10 6 ; )
g12 =
75
= 1.5.
50
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18–16
CHAPTER 18. Two-Port Circuits
P 18.14 [a] I1 = y11 V1 + y12 V2 ;
y21 =
I2 = y21 V1 + y22 V2 ;
I2
50 ⇥ 10 6
= 5 µS;
=
V1 V2 =0
10
0 = y21 (20 ⇥ 10 3 ) + y22 ( 5);
1
y22 = y21 (20 ⇥ 10 3 ) = 20 nS
5
.·.
200 ⇥ 10 6 = y11 (10) so y11 = 20 µS.
0.25 ⇥ 10 6 = 20 ⇥ 10 6 (20 ⇥ 10 3 ) + y12 ( 5);
y12 =
0.25 ⇥ 10 6
5
0.4 ⇥ 10 6
= 30 nS.
Summary:
y11 = 20 µS;
g
[b] y11 =
g22
;
y12 = 30 nS;
y21 = 5 µS;
y22 = 20 nS.
g12
;
g22
g21
;
g22
1
;
g22
y12 =
y21 =
y22 =
g12 g21 = (12.5 ⇥ 10 6 )(50 ⇥ 106 )
g = g11 g22
1.5( 250)
= 625 + 375 = 1000;
y11 =
y12 =
1000
= 20 µS;
50 ⇥ 106
y21 =
1.5
= 30 nS;
50 ⇥ 106
y22 =
250
= 5 µS;
5 ⇥ 106
1
= 20 nS.
5 ⇥ 106
These values are the same as those in part (a).
P 18.15 I1 = g11 V1 + g12 I2 ;
V2 = g21 V1 + g22 I2 ;
V1 =
I1
g11
g12
I2
g11
and I2 =
V2
g22
g21
V1 .
g22
Substituting,
"
I1
V1 =
g11
g12 V2
g11 g22
V1 = 1
g12 g21
g11 g22
!
=
#
g21
V1 ;
g22
I1
g11
g12
V2 ;
g11 g22
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Problems
g22
V1 =
g11 g22
g12 g21
g12
I1
g11 g22
g12 g21
18–17
V2 ;
V1 = h11 I1 + h12 V2 .
Therefore,
h11 =
g22
;
g
h12 =
g12
g
"
g12
I2 ;
g11
V2
I2 =
g22
g21 I1
g22 g11
I2 = 1
g12 g21
g11 g22
I2 =
g11
V2
g
!
=
where
g = g11 g22
g12 g21 .
#
V2
g22
g21
I1 ;
g11 g22
g21
I1 ;
g
I2 = h21 I1 + h22 V2 .
Therefore,
h21 =
g21
;
g
h22 =
P 18.16 V1 = h11 I1 + h12 V2 ;
g11
.
g
I2 = h21 I1 + h22 V2 .
Rearranging the first equation,
1
V1
h12
h11
I1 ;
h12
V2 = b11 V1
b12 I1 .
V2 =
Therefore,
b11 =
1
;
h11
b12 =
h11
.
h12
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18–18
CHAPTER 18. Two-Port Circuits
Solving the second h-parameter equation for I2 :
I2 = h21 I1 + h22
1
V1
h12
= I1 h21
h22 h11
h12
=
h
h12
I2 = b21 V1
I1 +
!
h11
I1
h12
+
!
h22
V1
h12
h22
V1 ;
h12
b22 I1 .
Therefore,
b21 =
h22
;
h12
b22 =
h
h12
where
h = h11 h22
P 18.17 I1 = g11 V1 + g12 I2 ;
V2 = g21 V1 + g22 I2 ;
V1 = z11 I1 + z12 I2 ;
V2 = z21 I1 + z22 I2 ;
I1 =
V1
z11
z12
I2 ;
z11
.·. g11 =
1
;
z11
✓
V1
z11
.·. g21 =
z21
;
z11
V2 = z21
h12 h21 .
g12 =
z12
.
z11
◆
✓
◆
z12
z21
z11 z22 z12 z21
I2 + z22 I2 =
V1 +
I2 ;
z11
z11
z11
g22 =
z
z11
.
P 18.18 For I2 = 0:
V2 =
4sV1
V1
(4) =
;
4 + (1/s)
4s + 1
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Problems
a11 =
s + 0.25
V1
4s + 1
=
;
=
V2 I2 =0
4s
s
sV1
V1
=
4 + (1/s)
4s + 1
I1 =
a21 =
18–19
so V2 = 4I1 =
4sV1
;
4s + 1
I1
1
= = 0.25.
V2 I2 =0 4
For V2 = 0:
I1 (4)
;
s+4
I2 =
I1
s+4
;
=
I2 V2 =0
s
a22 =
✓
◆
✓
4s
4s
1
4s
1
1
I1 =
+
+
V1 = I1 +
I1 =
s
s+4
s s+4
s s+4
✓
=
!
4s2 + s + 4
I2 ;
4s
s+1
V1
(1 + 1/s)(1)
+s=
+s
=
I1 I2 =0
2 + 1/s
2s + 1
=
2s2 + 2s + 1
s + 1 + 2s2 + 2
=
;
2s + 1
2s + 1
z22 = z11
z21 =
(s + 4)
I2
4
V1
s2 + 0.25s + 1
.
=
I2 V2 =0
s
a12 =
P 18.19 z11 =
◆
s+4
+ s I2 =
4s
◆
(the circuit is reciprocal and symmetrical).
V2
;
I1 I2 =0
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18–20
CHAPTER 18. Two-Port Circuits
V2 = I1
z21 =
s + 2s2 + s
V2
s
+s=
;
=
I1
2s + 1
2s + 1
1
(1) + sI1 ;
2 + 1/s
2s(s + 1)
2s2 + 2s
=
;
2s + 1
2s + 1
z12 = z21
(the circuit is reciprocal and symmetrical).
V1
;
I1 V2 =0
h21 =
V1 = (R + sL)I1
sM I2 l
P 18.20 [a] h11 =
0=
sM I1 + (R + sL)I2 l
=
(R + sL)
sM
sM
(R + sL)
V1
N1 =
I1 =
I2
.
I1 V2 =0
sM
0 (R + sL)
N1
=
= (R + sL)2
s2 M 2 l
= (R + sL)V1 l
(R + sL)V1
;
(R + sL)2 s2 M 2
0=
sM I1 + (R + sL)I2 ;
h12 =
V1
;
V2 I1 =0
h22 =
V1 =
sM I2 ;
I2 =
h11 =
.·. h21 =
V1
(R + sL)2 s2 M 2
.
=
I1
R + sL
I2
sM
;
=
I1
R + sL
I2
.
V2 I1 =0
V2
;
R + sL
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
V1 =
sM V2
;
R + sL
h22 =
I2
1
.
=
V2
R + sL
[b] h12 =
h21
h11 h22
V1
sM
;
=
V2
R + sL
h12 =
(reciprocal);
h12 h21 = 1 (symmetrical, reciprocal);
sM
;
R + sL
h12 =
18–21
h11 h22
=
h21 =
h12 h21 =
sM
R + sL
(checks).
(R + sL)2 s2 M 2
1
·
R + sL
R + sL
(sM )( sM )
(R + sL)2
(R + sL)2 s2 M 2 + s2 M 2
= 1 (checks).
(R + sL)2
P 18.21 For I2 = 0:
V2
V1
20
+ 0.025V1 +
V2
= 0;
40
2V2
2V1 + V1 + V2 = 0 so 3V2 = V1 ;
.·.
a11 =
I1 =
V1 5V2 V1 V2
V1
+
+
j50
100
20
V1
= 3.
V2 I2 =0
= V1

1
1
j
+
+
50 100 20
= V1

6 + j2
100
V2

V2

1
5
+
100 20
1
.
10
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18–22
CHAPTER 18. Two-Port Circuits
But V1 = 3V2 so
I1 =

a21 =
18 + j6 10
V2 = (0.08 + j0.06)V2 ;
100
I1
= 0.08 + j0.06 S = 80 + j60 mS.
V2 I2 =0
For V2 = 0:
I1 =
V1
V1
V1
(6 + j2)
+
+
= V1
;
j50 100 20
100
I2 = 0.025V1
V1
=
20
0.025V1 ;
a12 =
V1
1
= 40 ⌦;
=
I2 V2 =0 0.025
a22 =
I1
2V1 (3 + j1)
= 2.4 + j0.8.
=
I2 V2 =0 100( 0.025)V1
Summary:
a11 = 3;
P 18.22 h11 =
h12 =
a12 = 40 ⌦;
40
a12
= 15
=
a22
(0.8)(3 + j1)
a
a22
a22 = 2.4 + j0.8.
j5 ⌦;
;
a = 3(2.4 + j0.8)
h12 =
a21 = 80 + j60 mS;
40(0.08 + j0.06) = 7.2 + j2.4
4
= 1.5
(0.8)(3 + j1)
3.2
j2.4 = 4;
j0.50;
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
h21 =
h22 =
1
1
=
=
a22
(0.8)(3 + j1)
18–23
0.375 + j0.125;
a21
0.08 + j0.06
= 0.0375 + j0.0125 = 37.5 + j12.5 mS.
=
a22
(0.8)(3 + j1)
P 18.23 First we note that
z11 =
(Zb + Zc )(Za + Zb )
Za + 2Zb + Zc
Therefore
z12 =
Ix =
z21 =
Use the circuit below:
Zc (I2
so
V1
(Zb + Zc )2
=
I2
Za + 2Zb + Zc
V2
;
I1 I2 =0
V2 = Zb Ix
Ix =
Zc Iy = Zb Ix
Zb + Zc
I2
Za + 2Zb + Zc
.·. Z12 =
(Za + Zb )(Zb + Zc )
Za + 2Zb + Zc
z11 = z22 .
V1
;
I2 I1 =0
V1 = Zb Ix
and z22 =
Zc Iy = Zb Ix
Zb + Zc
I1
Za + 2Zb + Zc
Ix ) = (Zb + Zc )Ix
V1 =
(Zb + Zc )2
I2
Za + 2Zb + Zc
Zc =
Zb2 Za Zc
.
Za + 2Zb + Zc
Zc I2 ;
Zc I2 ;
Use the circuit below:
Zc (I1
so
Ix ) = (Zb + Zc )Ix
V2 =
(Zb + Zc )2
I1
Za + 2Zb + Zc
Zc I1 ;
Zc I1 ;
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18–24
CHAPTER 18. Two-Port Circuits
.·. z21 =
V2
(Zb + Zc )2
=
I1
Za + 2Zb + Zc
Zc =
Zb2 Za Zc
= z12 .
Za + 2Zb + Zc
Thus the network is symmetrical and reciprocal.
P 18.24
V2 = b11 V1
b12 I1 ;
V1 = Vg
I2 = b21 V1
b22 I1 ;
V2 =
V2
=
ZL
I2 =
ZL I2 ;
b11 V1 + b12 I1
;
ZL
b11 V1 + b12 I1
= b21 V1
ZL
.·. V1
I1 Zg ;
b22 I1 ;
!
!
b12
b11
+ b21 = b22 +
I1 ;
ZL
ZL
b22 ZL + b12
V1
=
= Zin .
I1
b21 ZL + b11
P 18.25 I1 = g11 V1 + g12 I2 ;
V1 = Vg
V2 = g21 V1 + g22 I2 ;
V2 =
ZL I2 = g21 V1 + g22 I2 ;
Zg I1 ;
ZL I2 ;
V1 =
I1
g12 I2
g11
.·.
ZL I2 =
g21
(I1
g11
.·.
ZL I2 +
g12 g21
I2
g11
g22 I2 =
g21
I1 ;
g11
.·. (ZL g11 +
g)I2 =
g21 I1 ;
.·.
;
g12 I2 ) + g22 I2 ;
g21
I2
=
I1
g11 ZL +
g
.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
P 18.26 I1 = y11 V1 + y12 V2 ;
V1 = Vg
I2 = y21 V1 + y22 V2 ;
V2 =
18–25
Zg I1 ;
ZL I2 ;
V2
= y21 V1 + y22 V2 ;
ZL
✓
◆
.·.
1
+ y22 V2 ;
y21 V1 =
ZL
.·.
y21 ZL
V2
=
.
V1
1 + y22 ZL
P 18.27 V1 = z11 I1 + z12 I2 ;
V1 = Vg
V2 = z21 I1 + z22 I2 ;
VTh = V2
.·. I1 =
ZTh =
;
V2 =
Vg
;
z11 + Zg
V2
;
I2 Vg =0
.·.
V2
= z22
I2
ZL I2 ;
I1 =
.·. V2 =
Vg I1 Zg
V1
=
;
z11
z11
z21 Vg
= Vt .
z11 + Zg
V2 = z21 I1 + z22 I2 ;
I1 Zg = z11 I1 + z12 I2 ;
"
Zg I1 ;
V2 = z21 I1 ;
I2 =0
.·. V2 = z21
y21 ZL V1 = (1 + y22 ZL )V2 ;
I1 =
z12 I2
;
z11 + Zg
#
z12 I2
+ z22 I2 ;
z11 + Zg
z12 z21
= ZTh .
z11 + Zg
P 18.28 V1 = h11 I1 + h12 V2 ;
I2 = h21 I1 + h22 V2 ;
V1 = Vg
V2 =
.·. Vg
Zg I1 = h11 I1 + h12 V2 ;
.·. I1 =
Vg h12 V2
;
h11 + Zg
Zg I1 ;
ZL I2 ;
Vg = (h11 + Zg )I1 + h12 V2 ;
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18–26
CHAPTER 18. Two-Port Circuits
"
#
Vg h12 V2
V2
= h21
+ h22 V2 .
ZL
h11 + Zg
.·.
V2 (h11 + Zg )
= h21 Vg
ZL
h12 h21 V2 + h22 (h11 + Zg )V2 ;
V2 (h11 + Zg ) = h21 ZL Vg
h12 h21 ZL V2 + h22 ZL (h11 + Zg )V2 ;
h21 ZL Vg = (h11 + Zg ) [V2 + h22 ZL V2 ]
h21 ZL
V2
=
.·.
Vg
(h11 + Zg )(1 + h22 ZL )
h12 h21 ZL V2 ;
h12 h21 ZL
.
P 18.29 V1 = h11 I1 + h12 V2 ;
I2 = h21 I1 + h22 V2 .
From the first measurement:
h11 =
h21 =
.·.
4
V1
= ⇥ 103 = 800 ⌦;
I1
5
200
=
5
40;
V1 = 800I1 + h12 V2 ;
I2 =
40I1 + h22 V2 .
From the second measurement:
h22 V2 = 40I1 ;
h22 =
40(20 ⇥ 10 6 )
= 20 µS;
40
20 ⇥ 10 3 = 800(20 ⇥ 10 6 ) + 40h12 ;
.·.
h12 =
4 ⇥ 10 3
= 10 4 .
40
Summary:
h11 = 800 ⌦;
h12 = 10 4 ;
h21 =
40;
h22 = 20 µS.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
18–27
From the circuit,
Zg = 250 ⌦;
Vg = 5.25 mV;
h11 + Zg
;
h22 Zg + h
ZTh =
h = 800(20 ⇥ 10 6 ) + 40 ⇥ 10 4 = 20 ⇥ 10 3 ;
800 + 250
ZTh =
20 ⇥ 10
6 (250) + 20 ⇥ 10 3
= 42 k⌦;
40(5.25 ⇥ 10 3 )
h21 Vg
=
= 8.4 V.
VTh =
h22 Zg + h
25 ⇥ 10 3
i=
8.4
= 0.10 mA;
84,000
P = (0.10 ⇥ 10 3 )2 (42,000) = 420 µW.
P 18.30 [a] ZTh =
b11 Zg + b12
;
b21 Zg + b22
b11 Zg = 6 + j2;
.·.
ZTh =
b21 Zg = 2;
6 + j2 1 + j4
5 + j6
=
= 2.1 + j1.3 ⌦.
2 + 1 + j1
3 + j1
⇤
ZL = ZTh
= 2.1
j1.3 ⌦;
bZL
V2
=
;
Vg
b12 + b11 Zg + b22 ZL + b21 Zg ZL
b=
3 + j1
(1 + j1)
3
( 1 + j4)(1/3) = 1;
b11 Zg = 6 + j2;
b22 ZL = (1 + j1)(2.1
j1.3) = 3.4 + j0.8;
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18–28
CHAPTER 18. Two-Port Circuits
b21 Zg ZL = 4.2
V2
=
Vg
.·.
j2.6;
2.1 j1.3
1 + j4 + 6 + j2 + 3.4 + j0.8 + 4.2
V2 =
2.1 j1.3
(90/0 ) = 10.71
12.6 + j4.2
j2.6
=
2.1 j13
;
12.6 + j4.2
j12.86 = 16.74/
50.19 V.
16.74
V2 (rms) = p = 11.84 V.
2
[b] I2 =
(10.71 j12.86)
= 6.78/161.55 A;
2.1 j1.3
6.78
I2 (rms) = p = 4.79 A;
2
P = (4.79)2 (2.1) = 48.18 W.
[c]
b
I2
=
;
I1
b11 + b21 ZL
1
(3 + j1)/3 + (2.1
.·.
j1.3)/3
5.1 j0.3
I1
=
=
I2
3
=
3
;
5.1 j0.3
1.7 + j0.1.
I1 = ( 1.7 + j0.1)(6.78/161.55 ) = 11.55/
21.82 A;
1
Pg (dev) = (11.55)(90) cos(21.82 ) = 482.51 W.
2
48.18
(100) ⇠
%=
= 10%.
482.51
P 18.31 [a]
V2
h21 ZL
=
Vg
(h11 + Zg )(1 + h22 ZL )
h12 h21 ZL
;
h21 ZL = 50 ⇥ 104 ;
h11 + Zg = 2000;
1 + h22 ZL = 1 + 50 ⇥ 10 6 ⇥ 104 = 1.5;
h12 h21 ZL = 10 3 (50)(104 ) = 500;
50 ⇥ 104
V2
=
=
Vg
2000(1.5) 500
V2 =
200;
200(250 ⇥ 10 3 /0 ) =
50/0 V;
50
V2 (rms) = p = 35.36 V.
2
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Problems
18–29
p
(50/ 2)2
= 125 mW.
[b] P =
104
I2
h21
50
[c]
;
=
=
I1
1 + h22 ZL
1.5
.·.
I1
1.5
= 0.03.
=
I2
50
I2 =
V2
50/0
=
= 5/0 mA;
ZL
10 ⇥ 103
.·.
I1 = (0.03)(5/0 ) = 0.15/0 mA.
1
Pg (dev) = (0.15)(250) ⇥ 10 6 = 18.75 µW.
2
P 18.32 [a] ZTh =
Zg + h11
;
h22 Zg + h
h = (500)(50 ⇥ 10 6 )
50 ⇥ 10 3 =
25 ⇥ 10 3 ;
h22 Zg = 50 ⇥ 10 6 (1500) = 75 ⇥ 10 3 ;
ZTh =
2000
= 40,000 + j0 ⌦;
75 ⇥ 10 3 25 ⇥ 10 3
⇤
= 40 + j0 k⌦.
ZL = ZTh
[b] VTh =
P =
[c] I2 =
h21 Vg
=
50 ⇥ 10 3
50(250) ⇥ 10 3
=
50 ⇥ 10 3
250 V.
1 (125)2
= 196.3125 mW.
2 40,000
125
= 3.125 mA;
40,000
50
h21
50
I2
=
;
=
=
I1
1 + h22 ZL
1 + (50 ⇥ 10 6 )(40,000)
3
3
I1
= 0.06;
=
I2
50
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18–30
CHAPTER 18. Two-Port Circuits
I1 = 0.06I2 = 187.5 µA;
1
Pg (dev) = (250)(187.5) ⇥ 10 9 = 23.4375 µW.
2
P 18.33
V2
bZL
=
;
Vg
b12 + b11 Zg + b22 ZL + b21 Zg ZL
b = b11 b22
b12 b21 = (25)( 40)
(1000)( 1.25) = 250;
.·.
250(100)
V2
=
Vg
1000 + 25(20) 40(100)
V2 =
5(120/0 ) = 600/180 V(rms);
I2 =
V2
=
100
1.25(2000)
=
5.
600/180
= 6 A(rms);
100
b
I2
=
=
I1
b11 + b21 ZL
25
250
= 2.5;
1.25(100)
6
I2
=
= 2.4 A(rms);
2.5
2.5
.·.
I1 =
.·.
Pg = (120)(2.4) = 288 W;
.·.
3600
Po
= 12.5.
=
Pg
288
Po = 36(100) = 3600 W;
P 18.34 [a] For I2 = 0:
j150
j3V1
V1
=
;
50 + j50
1 + j1
V2 =
j150I1 =
a11 =
V1
1 + j1
=
=
V2 I2 =0
j3
a21 =
I1
=
V2 I2 =0
1 + j1
;
3
j
1
=
S.
j150
150
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
18–31
For V2 = 0:
V1 = (50 + j50)I1
0=
j150I2 ;
j150I1 + (400 + j800)I2 ;
50 + j50
=
j150
= 2500(1 + j24);
j150 400 + j800
50 + j50 V1
N2 =
I2 =
j150 0
N2
=
j150V1
;
2500(1 + j24)
V1
=
I2 V2 =0
a12 =
= j150V1 ;
50
(24
3
j1) ⌦;
j150I1 = (400 + j800)I2 ;
I1
=
I2 V2 =0
a22 =
[b] VTh =
8
(2
3
j1).
Vg
260/0
=
=
a11 + a21 Zg
( 1 + j1)/3 + j25/150
= 120( 2
ZTh =
=
j3) = 432.47/
(260/0 )6
1560/0
=
2 + j2 + j1
2 + j3
123.69 V.
a12 + a22 Zg
[ (50/3)(24 j1)] + [( 8/3)(2 j1)(25)]
=
a11 + a21 Zg
[( 1 + j1)/3] + [(j/150)(25)]
100(24
j1) 16(2 j1)(25)
=
2 + j2 + j1
3200 + j500
2 + j3
= 607.69 + j661.54 ⌦.
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18–32
CHAPTER 18. Two-Port Circuits
[c] V2 =
1000
(432.67/
1607.69 + j661.54
v2 (t) = 248.88 cos(4000t
123.69 ) = 248.88/
146.06 ;
146.06 ) V.
P 18.35 When V2 = 0
V1 = 20 V,
I1 = 1 A,
I2 =
1 A.
When I1 = 0
V2 = 80 V,
V1 = 400 V,
I2 = 3 A;
h11 =
V1
20
= 20 ⌦;
=
I1 V2 =0 1
h12 =
V1
400
= 5;
=
V2 I1 =0 80
h21 =
I2
1
=
=
I1 V2 =0
1
h22 =
I2
3
= 37.5 mS.
=
V2 I1 =0 80
ZTh =
Zg + h11
= 10 ⌦.
h22 Zg + h
1;
Source-transform the current source and parallel resistance to get Vg = 240 V.
Then,
I2 =
h21 Vg
(1 + h22 ZL )(h11 + Zg )
h12 h21 ZL
=
1.5 A;
P = ( 1.5)2 (10) = 22.5 W.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
18–33
P 18.36 [a] V1 = z11 I1 + z12 I2 ;
V2 = z21 I1 + z22 I2 ;
z11 =
s2 + 1
V1
1
;
=s+ =
I1 I2 =0
s
s
z21 =
V2
1
= ;
I1 I2 =0 s
z12 =
V1
1
= ;
I2 I1 =0 s
s2 + 1
V2
1
.
=s+ =
z22 =
I2 I1 =0
s
s
[b]
V2
z21 ZL
=
Vg
(z11 + Zg )(z22 + ZL )
=
z21
(z11 + 1)(z22 + 1)
z12 z21
1/s
= ⇣ s2 +1
z12 z21
⌘⇣ 2
⌘
+ 1 s s+1 + 1
s
s
= 2
(s + s + 1)2 1
=
=
=
1
s2
s
s4 + 2s3 + 3s2 + 2s + 1
1
1
s3 + 2s2 + 3s + 2
1
;
(s + 1)(s2 + s + 2)
50
.
s(s + 1)(s2 + s + 2)
p
1
7
±j
;
2
2
.·. V2 =
s1,2 =
V2 =
K2
K3
K1
K3⇤
p +
p ;
+
+
s
s+1 s+ 1 j 7
s+ 1 +j 7
2
K1 = 25;
K2 =
.·. v2 (t) = [25
25;
2
2
2
K3 = 9.45/90 ;
25e t + 18.90e 0.5t cos(1.32t + 90 )]u(t) V.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18–34
CHAPTER 18. Two-Port Circuits
CHECK
v2 (0) = 25
25 + 18.90 cos 90 = 0;
v2 (1) = 25 + 0 + 0 = 25 V.
P 18.37 [a] h11 =
V1
;
I1 V2 =0
h21 =
I2
.
I1 V2 =0
h11 =
(1/C)s
(1/sC)(sL)
= 2
;
(1/sC) + sL
s + (1/LC)
I2 =
Ia ;
I2 =
I1
;
2
s LC + 1
Ia =
I1 (1/sC)
;
sL + (1/sC)
h21 =
I2
(1/LC)
;
= 2
I1
s + (1/LC)
h12 =
V1
;
V2 I1 =0
V1 =
V2
V2 (1/sC)
= 2
;
sL + (1/sC)
s LC + 1
h22 =
I2
.
V2 I1 =0
1/LC
V1
;
= h12 = 2
V2
s + (1/LC)
s2 + (1/LC)
(1/sC)[sL + (1/LC)]
V2
=
;
=
I2
sL + (2/LC)
sC[s2 + (2/LC)]
Cs[s2 + (2/LC)]
I2
.
= h22 =
V2
s2 + (1/LC)
[b]
(103 )(106 )
1
=
= 25 ⇥ 106 ;
LC
(0.2)(200)
h11 =
h21 =
1
= 5 ⇥ 106 ;
C
5 ⇥ 106 s
;
s2 + 25 ⇥ 106
25 ⇥ 106
;
s2 + 25 ⇥ 106
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problems
18–35
h21 ZL
.
hZL + h11
V2
=
V1
h=1
(the circuit is reciprocal and symmetrical).
25 ⇥ 106 (400)
V2
= 2
V1
[s + 25 ⇥ 106 ][400 + (5 ⇥ 106 s)/(s2 + 25 ⇥ 106 )]
25 ⇥ 106
1010
= 2
=
400s2 + 1010 + 5 ⇥ 106 s
s + 12,500s + 25 ⇥ 106
=
25 ⇥ 106
.
(s + 2500)(s + 10,000)
v1 = 30u(t);
V1 =
30
;
s
V2 =
K1
K2
K3
(25 ⇥ 106 )(30)
=
+
+
;
s(s + 2500)(s + 10,000)
s
s + 2500 s + 10,000
K1 =
(25 ⇥ 106 )(30)
= 30;
25 ⇥ 106
K3 =
K2 =
(25 ⇥ 106 )(30)
=
( 2500)(7500)
40;
(25 ⇥ 106 )(30)
= 10;
( 10,000)( 7500)
v2 (t) = [30
40e 2500t + 10e 10,000t ]u(t) V.
P 18.38 The Thevenin equivalent seen looking into the g-network from the right is
VTh =
g21 Vg
(800/7)(30)
= 1846.154 V;
=
1 + g11 Zg
1 + (3/35)(10)
ZTh = g22
=
g12 g21 Zg
1 + g11 Zg
50,000
7
(20/7)(800/7)10)
= 5384.615 ⌦.
1 + (3/35)(10)
The simplified circuit is shown here:
Vo =
h11 ZL Vg
(h11 + Zg )(1 + h22 ZL )
h12 h21 ZL
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18–36
CHAPTER 18. Two-Port Circuits
=
4(15,000)(1846.154)
(5000 + 5384.6)[1 + (0.0002)(15,000)]
(0.8)(15,000)
= 3750 V.
P 18.39 The a parameters of the first two port are
a011 =
h
h21
=
5 ⇥ 10 3
=
40
1000
=
40
125 ⇥ 10 6 ;
a012 =
h11
=
h21
a021 =
h22
25
⇥ 10 6 =
=
h21
40
a022 =
1
h21
=
1
=
40
25 ⌦;
625 ⇥ 10 9 S;
25 ⇥ 10 3 .
The a parameters of the second two port are
5
a0011 = ;
4
a0012 =
or a0011 = 1.25;
3R
;
4
a0021 =
a0012 = 54 k⌦;
3
;
4R
a0021 =
5
a0022 = ,
4
1
mS;
96
a0022 = 1.25.
The a parameters of the cascade connection are
a11 =
125 ⇥ 10 6 (1.25) + ( 25)(10 3 /96) =
10 2
;
24
a12 =
125 ⇥ 10 6 (54 ⇥ 103 ) + ( 25)(1.25) =
38 ⌦;
a21 =
625 ⇥ 10 9 (1.25) + ( 25 ⇥ 10 3 )(10 3 /96) =
10 4
S;
96
a22 =
625 ⇥ 10 9 (54 ⇥ 103 ) + ( 25 ⇥ 10 3 )(1.25) =
65 ⇥ 10 3 .
ZL
Vo
=
.
Vg
(a11 + a21 Zg )ZL + a12 + a22 Zg
a21 Zg =
10 4
(800) =
96
a11 + a21 Zg =
10 2
+
24
10 2
;
12
10 2
=
12
10 2
;
8
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Problems
10 2
(72,000) =
8
(a11 + a21 Zg )ZL =
a22 Zg =
Vo
=
Vg
65 ⇥ 10 3 (800) =
72,000
=
90 38 52
v o = Vo =
400Vg =
90;
52;
400;
3.6 V.
P 18.40 [a] From reciprocity and symmetry
a 0 = a0 ,
a0 = 1;
.·. 4
11
18–37
22
2a021 = 1,
a021 = 1.5 S.
For network B
a0011 =
V1
;
V2 I2 =0
V1 = (1 + j6
V2 = ( j5
a0011 =
a0021 =
j6)I1 = I1 ;
j6)I1 =
j11I1 ;
j
1
= ;
j11
11
I1
=
V2 I2 =0
j
1
= ;
j11
11
j
;
11
a00 = 1 = (j/11)(j/11)
a0022 = a0011 =
(j/11)a0012 ;
j
.
11
[b] a11 = a011 a0011 + a012 a0021 = 2(j/11) + 5(j/11) = j7/11;
.·. a0012 =
a12 = a011 a0012 + a012 a0022 = 2(j/11) + 5(j/11) = j7/11;
a21 = a021 a0011 + a022 a0021 = 1.5(j/11) + 2(j/11) = j3.5/11;
a22 = a021 a0012 + a022 a0022 = j3.5/11.
I2 =
V2 =
Vg
= j5.357;
a11 ZL + a12 + a21 Zg ZL
14I2 =
j75 V.
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18–38
CHAPTER 18. Two-Port Circuits
P 18.41 [a] At the input port: V1 = h11 I1 + h12 V2 ;
At the output port: I2 = h21 I1 + h22 V2 .
[b]
V2
+ (100 ⇥ 10 6 V2 ) + 100I1 = 0,
104
therefore I1 =
2 ⇥ 10 6 V2 .
V20 = 1000I1 + 15 ⇥ 10 4 V2 =
5 ⇥ 10 4 V2 ;
100I10 + 10 4 V20 + ( 2 ⇥ 10 6 )V2 = 0,
therefore I10 = 205 ⇥ 10 10 V2 .
Vg = 1500I10 + 15 ⇥ 10 4 V20 = 3000 ⇥ 10 8 V2 ;
105
V2
= 33,333.
=
Vg
3
P 18.42 [a] V1 = I2 (z12
z21 ) + I1 (z11
z21 ) + z21 (I1 + I2 )
= I2 z12
I2 z21 + I1 z11
I1 z21 + z21 I1 + z21 I2 = z11 I1 + z12 I2 .
V2 = I2 (z22
z21 ) + z21 (I1 + I2 ) = z21 I1 + z22 I2 .
[b] Short circuit Vg and apply a test current source to port 2 as shown. Note
that IT = I2 . We have
V
z21
IT +
V + IT (z12 z21 )
= 0.
Zg + z11 z21
Therefore
"
#
z21 (Zg + z11 z12 )
V =
IT
Zg + z11
Thus
VT
= ZTh = z22
IT
and VT = V + IT (z22
z21 ).
!
z12 z21
.
Zg + z11
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Problems
For VTh note that Voc =
P 18.43 [a] V1 = (z11
V2 = (z21
18–39
z21
Vg since I2 = 0.
Zg + z11
z12 )I1 + z12 (I1 + I2 ) = z11 I1 + z12 I2 ;
z12 )I1 + (z22
z12 )I2 + z12 (I2 + I1 ) = z21 I1 + z22 I2 .
[b] With port 2 terminated in an impedance ZL , the two mesh equations are
V1 = (z11
z12 )I1 + z12 (I1 + I2 );
0 = ZL I2 + (z21
z12 )I1 + (z22
z12 )I2 + z12 (I1 + I2 ).
Solving for I1 :
I1 =
V1 (z22 + ZL )
.
z11 (ZL + z22 ) z12 z21
Therefore
V1
Zin =
= z11
I1
z12 z21
.
z22 + ZL
P 18.44 [a] I1 = y11 V1 + y21 V2 + (y12
I1 = y11 V1 + y12 V2 ;
y21 )V2 ;
I2 = y21 V1 + y22 V2 .
I2 = y12 V1 + y22 V2 + (y21
y12 )V1 .
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18–40
CHAPTER 18. Two-Port Circuits
[b] Using the second circuit derived in part [a], we have
where ya = (y11 + y12 ) and yb = (y22 + y12 ).
At the input port we have
I1 = ya V1
y12 (V1
V2 ) = y11 V1 + y12 V2 .
At the output port we have
V2
+ (y21
ZL
y12 )V1 + yb V2
y12 (V2
V1 ) = 0.
Solving for V1 gives
V1 =
!
1 + y22 ZL
V2 .
y21 ZL
Substituting Eq. 18.2 into Eq. 18.1 and at the same time using
V2 = ZL I2 , we get
y21
I2
=
.
I1
y11 + yZL
P 18.45 [a] The g-parameter equations are I1 = g11 V1 + g12 I2 and V2 = g21 V1 + g22 I2 .
These equations are satisfied by the following circuit:
[b] Replace the two-port network described by g parameters with its
equivalent circuit from part (a) and replace the two-port network
described by h parameters with its equivalent circuit from Fig. P18.41.
Attach the source and the load to get the circuit shown here:
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Problems
18–41
Note from the mesh in the middle of the circuit that I10 = Ix and that
I2 = Ix .
Write a KCL equation at the node labeled V1 :
V1
30
10
+
V1
20
+ ( Ix ) = 0.
35/3
7
Write a KVL equation for the mesh whose current is Ix :
✓
◆
50,000
+ 5000 Ix
7
0.2V20
800
V1 = 0.
7
Write a KCL equation at the node labeled V20 :
4Ix + 200 ⇥ 10 6 V20 +
V20
= 0.
15,000
Solving these three equations, we get
V1 = 20 V;
Ix = 0.25 A;
V20 = 3750 V.
Thus, the voltage across the load is 3750 V, which matches the solution
to Problem 18.38.
P 18.46 [a] To determine b11 and b21 create an open circuit at port 1. Apply a voltage
at port 2 and measure the voltage at port 1 and the current at port 2. To
determine b12 and b22 create a short circuit at port 1. Apply a voltage at
port 2 and measure the currents at ports 1 and 2.
[b] The equivalent b-parameters for the black-box amplifier can be calculated
as follows:
1
1
b11 =
=
= 1000;
h12
10 3
b12 =
h11
500
=
= 500 k⌦;
h12
10 3
b21 =
h22
0.05
=
= 50 S;
h12
10 3
b22 =
h
h12
=
23.5
= 23,500.
10 3
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18–42
CHAPTER 18. Two-Port Circuits
Create an open circuit a port 1. Apply 1 V at port 2. Then,
b11 =
V2
1
= 1000 so V1 = 1 mV measured;
=
V1 I1 =0 V1
b21 =
I2
I2
= 50 S so I2 = 50 mA measured.
=
V1 I1 =0 10 3
Create a short circuit a port 1. Apply 1 V at port 2. Then,
b12 =
V2
1
= 500 k⌦ so I1 =
=
I1 V1 =0 I1
b22 =
I2
=
I1 V1 =0
2 µA measured;
I2
= 23,500 so I2 = 47 mA measured.
2 ⇥ 10 6
P 18.47 [a] To determine y11 and y21 create a short circuit at port 2. Apply a voltage
at port 1 and measure the currents at ports 1 and 2. To determine y12
and y22 create a short circuit at port 1. Apply a voltage at port 2 and
measure the currents at ports 1 and 2.
[b] The equivalent y-parameters for the black-box amplifier can be calculated
as follows:
1
1
y11 =
= 2 mS;
=
h11
500
y12 =
h12
10 3
=
=
h11
500
y21 =
h21
1500
= 3 S;
=
h11
500
y22 =
h
h11
=
2 µS;
23.5
= 47 mS.
500
Create a short circuit at port 2. Apply 1 V at port 1. Then,
y11 =
I1
I1
= 2 mS so I1 = 2 mA measured;
=
V1 V2 =0 1
y21 =
I2
I2
= 3 S so I2 = 3 A measured.
=
V1 V2 =0 1
Create a short circuit at port 1. Apply 1 V at port 2. Then,
y12 =
I1
I1
=
=
V2 V1 =0 1
y22 =
I2
I2
= 47 mS so I2 = 47 mA measured.
=
V2 V1 =0 1
2 µS so I1 =
2 µA measured;
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