SOLUTION MANUAL
Electric Circuits, 12th edition
By James Nilsson, Susan Reidel
TE
ST
SO
LU
TI
O
N
, TABLE OF CONTENT
Circuit Variables
Circuit Elements
Simple Resistive Circuits
Techniques of Circuit Analysis
The Operational Amplifier
TE
Inductance, Capacitance, and Mutual Inductance
Response of First-Order RL and RC Circuits
Natural and Step Responses of RLC Circuits
ST
Sinusoidal Steady-State Analysis
Sinusoidal Steady-State Power Calculations
Balanced Three-Phase Circuits
Introduction to the Laplace Transform
SO
The Laplace Transform in Circuit Analysis
Introduction to Frequency Selective Circuits
Active Filter Circuits
LU
Fourier Series
The Fourier Transform
Two-Port Circuits
TI
O
N
, Circuit Variables
TE
Assessment Problems
ST
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
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(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
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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:
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$1 trillion = $1 × 1012 .
Divide by 100 = 102 to find the number of $100 bills:
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1012
$1 trillion = = 1010 $100 bills.
102
Calculate the height of a stack of 1010 $100 bills:
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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
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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
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
SO
= 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.
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75 × 10−6 C/s
AP 1.4 n = = 4.681 × 1014 elec/s.
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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Electric Circuits, 12th edition
By James Nilsson, Susan Reidel
TE
ST
SO
LU
TI
O
N
, TABLE OF CONTENT
Circuit Variables
Circuit Elements
Simple Resistive Circuits
Techniques of Circuit Analysis
The Operational Amplifier
TE
Inductance, Capacitance, and Mutual Inductance
Response of First-Order RL and RC Circuits
Natural and Step Responses of RLC Circuits
ST
Sinusoidal Steady-State Analysis
Sinusoidal Steady-State Power Calculations
Balanced Three-Phase Circuits
Introduction to the Laplace Transform
SO
The Laplace Transform in Circuit Analysis
Introduction to Frequency Selective Circuits
Active Filter Circuits
LU
Fourier Series
The Fourier Transform
Two-Port Circuits
TI
O
N
, Circuit Variables
TE
Assessment Problems
ST
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
SO
(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
LU
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:
TI
$1 trillion = $1 × 1012 .
Divide by 100 = 102 to find the number of $100 bills:
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1012
$1 trillion = = 1010 $100 bills.
102
Calculate the height of a stack of 1010 $100 bills:
N
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
TE
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
ST
integral:
t
e−2000y
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
SO
= 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.
LU
75 × 10−6 C/s
AP 1.4 n = = 4.681 × 1014 elec/s.
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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