Electric Circuits – 11th Edition
ST
SOLUTIONS
UV
MANUAL
IA
_A
James W. Nilsson
PP
Susan A. Riedel
RO
Comprehensive Solutions Manual for
VE
Instructors and Students
D?
© James W. Nilsson & Susan A. Riedel
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All rights reserved. Reproduction or distribution without permission is prohibited
©STUDYSTREAM
,ST
Circuit Variables
UV
Assessment Problems
IA
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
_A
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
PP
1100 miles:
124,274.24 miles 1100 miles
= .
1s xs
Therefore,
RO
1100
x= = 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
VE
scientific notation:
$100 billion = $100 × 109 .
Now we determine the number of milliseconds in one year, again using a
D?
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 × 109 1 year 100
· 9
= = $3.17/ms.
1 year 31.5576 × 10 ms 31.5576
1–1
, 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 = dqdt
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
ST
of current:
Z t
q(t) = i(x) dx.
0
UV
We are given the expression for current, i, which can be substituted into the
above expression. To find the total charge, we let t → ∞ in the integral. Thus
we have
Z ∞ ∞
20 −5000x 20
qtotal = 20e−5000x dx = e = (e−∞ − e0 )
0 −5000 0 −5000
IA
20 20
= (0 − 1) = = 0.004 C = 4000 µC.
−5000 5000
AP 1.4 Recall from Eq. 1.2 that current is the time rate of change of charge, or
_A
i = dq
dt
. In this problem we are given an expression for the charge, and asked to
find the maximum current. First we will find an expression for the current
using Eq. 1.2:
dq d 1 t 1
PP
i= = 2
− + 2 e−αt
dt dt α α α
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
RO
1 −αt t 1
= 0− e − α e−αt − −α 2 e−αt
α α α
1 1 −αt
= − +t+ e
α α
VE
= te−αt .
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:
D?
di d
= (te−αt ) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0.
dt dt
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only
??
when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For
this value of t, the current is
1 −α/α 1
i= e = e−1 .
α α
, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e−1 ∼
= 10 A.
ST
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
UV
Also sketch the four figures from Fig. 1.6:
IA
_A
PP
[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
RO
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4 A;
(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (−20)(−4) = 80 W.
VE
[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
D?
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 − 120 × 107 t2 e−1000t = 0.
dt
Therefore,
240 × 104 − 120 × 107 t = 0.
ST
SOLUTIONS
UV
MANUAL
IA
_A
James W. Nilsson
PP
Susan A. Riedel
RO
Comprehensive Solutions Manual for
VE
Instructors and Students
D?
© James W. Nilsson & Susan A. Riedel
??
All rights reserved. Reproduction or distribution without permission is prohibited
©STUDYSTREAM
,ST
Circuit Variables
UV
Assessment Problems
IA
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
_A
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
PP
1100 miles:
124,274.24 miles 1100 miles
= .
1s xs
Therefore,
RO
1100
x= = 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
VE
scientific notation:
$100 billion = $100 × 109 .
Now we determine the number of milliseconds in one year, again using a
D?
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 × 109 1 year 100
· 9
= = $3.17/ms.
1 year 31.5576 × 10 ms 31.5576
1–1
, 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 = dqdt
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
ST
of current:
Z t
q(t) = i(x) dx.
0
UV
We are given the expression for current, i, which can be substituted into the
above expression. To find the total charge, we let t → ∞ in the integral. Thus
we have
Z ∞ ∞
20 −5000x 20
qtotal = 20e−5000x dx = e = (e−∞ − e0 )
0 −5000 0 −5000
IA
20 20
= (0 − 1) = = 0.004 C = 4000 µC.
−5000 5000
AP 1.4 Recall from Eq. 1.2 that current is the time rate of change of charge, or
_A
i = dq
dt
. In this problem we are given an expression for the charge, and asked to
find the maximum current. First we will find an expression for the current
using Eq. 1.2:
dq d 1 t 1
PP
i= = 2
− + 2 e−αt
dt dt α α α
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
RO
1 −αt t 1
= 0− e − α e−αt − −α 2 e−αt
α α α
1 1 −αt
= − +t+ e
α α
VE
= te−αt .
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:
D?
di d
= (te−αt ) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0.
dt dt
Since e−αt never equals 0 for a finite value of t, the expression equals 0 only
??
when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For
this value of t, the current is
1 −α/α 1
i= e = e−1 .
α α
, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e−1 ∼
= 10 A.
ST
0.03679
AP 1.5 Start by drawing a picture of the circuit described in the problem statement:
UV
Also sketch the four figures from Fig. 1.6:
IA
_A
PP
[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
RO
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4 A;
(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4 A.
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention,
p = vi = (−20)(−4) = 80 W.
VE
[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
D?
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 − 120 × 107 t2 e−1000t = 0.
dt
Therefore,
240 × 104 − 120 × 107 t = 0.
