Solutions Manual — Electric Circuits, 12th Edition — James W. Nilsson & Susan A. Riedel — ISBN 13: 9780137648375

Electric Circuits – 12th Edition
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SOLUTIONS
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MANUAL
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James W. Nilsson
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Susan A. Riedel
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Comprehensive Solutions Manual for
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Instructors and Students
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© James W. Nilsson & Susan A. Riedel
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All rights reserved. Reproduction or distribution without permission is prohibited




©STUDYSTREAM

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Circuit Variables
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Assessment Problems
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AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
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second to miles per second:
3 × 108 m 100 cm 1 in 1 ft 1 mile 177,090.79 miles
(0.95) · · · · = .
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
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950 miles:
177,090.79 miles 950 miles
= .
1s xs
Therefore,
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950
x= = 0.00536 = 5.36 × 10−3 s = 5.36 ms.
177,090.79
AP 1.2 We begin by expressing $1 trillion in scientific notation:
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$1 trillion = $1 × 1012 .

Divide by 100 = 102 to find the number of $100 bills:
1012
$1 trillion = = 1010 $100 bills.
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102
Calculate the height of a stack of 1010 $100 bills:
0.11 mm 1m
1010 bills · · = 1.1 × 106 m.
bill 1000 mm
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Now we can convert from meters to miles, again with a product of ratios:
100 cm 1 in 1 ft 1 mi
1.1 × 106 m · · · · = 683.51 miles.
1m 2.54 cm 12 in 5280 ft

1–1

, 1–2 CHAPTER 1. Circuit Variables


AP 1.3 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 0.25te−2000t .
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dt
Therefore, dq = 0.25te−2000t dt.

To find the charge, we can integrate both sides of the last equation. Note
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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) Z t
dx = 0.25 ye−2000y dy.
q(0) 0

We solve the integral and make the substitutions for the limits of the
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integral:
t
e−2000y
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
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= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9

= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).

But q(0) = 0 by hypothesis, so
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q(t) = 62.5(1 − 2000te−2000t − e−2000t ) nC.
[b] q(0.001) = (62.5)[1 − 2000(0.001)e−2000(0.001) − e−2000(0.001) ] = 37.12 nC.
75 × 10−6 C/s
= 4.681 × 1014 elec/s.
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AP 1.4 n =
1.6022 × 10−19 C/elec
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
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Also sketch the four figures from Fig. 1.6:
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, Problems 1–3


[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 250 mA of current
entering Terminal 2 is the same as 250 mA of current leaving Terminal 1.
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We get
(a) v = 50 V, i = −0.25 A; (b) v = 50 V, i = 0.25 A;
(c) v = −50 V, i = −0.25 A; (d) v = −50 V, i = 0.25 A.
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[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (50)(−0.25) = −12.5 W.
[c] Since the power is less than 0, the box is delivering power.
Z t
AP 1.6 p = vi; w= p dx.
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Since the energy is the area under the power vs. time plot, let us plot p vs. t.
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Note that in constructing the plot above, we used the fact that 60 hr
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= 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;
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w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16,200 J.
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AP 1.7 [a] p = vi = (15e−250t )(0.04e−250t ) = 0.6e−500t W;
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p(0.01) = 0.6e−500(0.01) = 0.6e−5 = 0.00404 = 4.04 mW.
Z ∞ Z ∞ ∞
0.6 −500x
[b] wtotal = p(x) dx = 0.6e−500x dx = e
0 0 −500 0
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= −0.0012(e∞ − e0 ) = 0.0012 = 1.2 mJ.