Solutions Manual for Electric Circuits 12 Edition by James Nilsson, Susan Riedel

Solutions Manual for Electric Circuits 12 Edition by James Nilsson, Susan Riedel




Solutions Manual for Electric Circuits 12 Edition by James Nilsson, Susan Riedel

, 1
Circuit Variables



Assessment Problems

AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per
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
950 miles:
177,090.79 miles 950 miles
= .
1s xs
Therefore,
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:

$1 trillion = $1 × 1012.

Divide by 100 = 102 to find the number of $100 bills:
1012 10
$1 trillion = 2 = 10 $100 bills.
10
Calculate the height of a stack of 1010 $100 bills:
0.11 mm 1m
1010 bills · · = 1.1 × 106 m.
bill 1000 mm
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.
dt
Therefore, dq = 0.25te−2000t 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:
∫ q(t) ∫ t
dx = 0.25 ye−2000y dy.
q(0) 0

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

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

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

But q(0) = 0 by hypothesis, so
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 14
AP 1.4 n = = 4.681 × 10 elec/s.
1.6022 × 10−19 C/elec
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:




Solutions Manual for Electric Circuits 12 Edition by James Nilsson, Susan Riedel

, 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.
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.
[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.
∫ t
AP 1.6 p = vi; w= p dx.
0
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 − 60 × 10−3 )(216 × 103) = 16,200 J.
2

AP 1.7 [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.
∫∞ ∫∞ ∞
0.6 −500x
[b] wtotal = p(x) dx = 0.6e −500x dx = e
0 0 −500 0

= −0.0012(e∞ − e0) = 0.0012 = 1.2 mJ.