SOLUTION MANUAL
Electric Circuits, 12th edition
by James Nilsson, Susan Reidel
LU
XE
LI
BR
AR
Y
, TABLE OF CONTENT
1. Circuit Variables
2. Circuit Elements
3. Simple Resistive Circuits
4. Techniques of Circuit Analysis
5. The Operational Amplifier
6. Inductance, Capacitance, and Mutual Inductance
LU
7. Response of First-Order RL and RC Circuits
8. Natural and Step Responses of RLC Circuits
9. Sinusoidal Steady-State Analysis
10. Sinusoidal Steady-State Power Calculations
XE
11. Balanced Three-Phase Circuits
12. Introduction to the Laplace Transform
13. The Laplace Transform in Circuit Analysis
LI
14. Introduction to Frequency Selective Circuits
15. Active Filter Circuits
BR
16. Fourier Series
17. The Fourier Transform
18. Two-Port Circuits
AR
Y
, Circuit Variables
LU
Assessment Problems
XE
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
LI
Now set up a proportion to determine how long it takes this signal to travel
950 miles:
177,090.79 miles 950 miles
= .
BR
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:
AR
$1 trillion = $1 × 1012 .
Divide by 100 = 102 to find the number of $100 bills:
1012
$1 trillion = = 1010 $100 bills.
102
Y
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:
LU
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
integral:
t
e−2000y
XE
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).
LI
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.
BR
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:
AR
Also sketch the four figures from Fig. 1.6:
Y
Electric Circuits, 12th edition
by James Nilsson, Susan Reidel
LU
XE
LI
BR
AR
Y
, TABLE OF CONTENT
1. Circuit Variables
2. Circuit Elements
3. Simple Resistive Circuits
4. Techniques of Circuit Analysis
5. The Operational Amplifier
6. Inductance, Capacitance, and Mutual Inductance
LU
7. Response of First-Order RL and RC Circuits
8. Natural and Step Responses of RLC Circuits
9. Sinusoidal Steady-State Analysis
10. Sinusoidal Steady-State Power Calculations
XE
11. Balanced Three-Phase Circuits
12. Introduction to the Laplace Transform
13. The Laplace Transform in Circuit Analysis
LI
14. Introduction to Frequency Selective Circuits
15. Active Filter Circuits
BR
16. Fourier Series
17. The Fourier Transform
18. Two-Port Circuits
AR
Y
, Circuit Variables
LU
Assessment Problems
XE
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
LI
Now set up a proportion to determine how long it takes this signal to travel
950 miles:
177,090.79 miles 950 miles
= .
BR
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:
AR
$1 trillion = $1 × 1012 .
Divide by 100 = 102 to find the number of $100 bills:
1012
$1 trillion = = 1010 $100 bills.
102
Y
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:
LU
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
integral:
t
e−2000y
XE
q(t) − q(0) = 0.25 (−2000y − 1)
(−2000)2 0
= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9
= 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ).
LI
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.
BR
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:
AR
Also sketch the four figures from Fig. 1.6:
Y
