Solution Manual For Electric Circuits, 10th Edition

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1Circuit Variables
Assessment Problems
AP 1.1 Use a product of ratios to convert two-thirds the speed of light from meters
per second to miles per second:
( 2
3
) 3 × 108 m
1 s · 100 cm
1 m · 1 in
2.54 cm · 1 ft
12 in · 1 mile
5280 feet = 124,274.24 miles
1 s
Now set up a proportion to determine how long it takes this signal to travel
1100 miles:
124,274.24 miles
1 s = 1100 miles
x s
Therefore,
x = 1100
124,274.24 = 0.00885 = 8.85 × 10−3 s = 8.85 ms
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
scientific notation:
$100 billion = $100 × 109
Now we determine the number of milliseconds in one year, again using a
product of ratios:
1 year
365.25 days · 1 day
24 hours · 1 hour
60 mins · 1 min
60 secs · 1 sec
1000 ms = 1 year
31.5576 × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a
product of ratios:
$100 × 109
1 year · 1 year
31.5576 × 109 ms = 100
31.5576 = $3.17/ms
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 = dq
dt 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 of current:
q(t) =
∫ t
0
i(x) dx
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
qtotal =
∫ ∞
0
20e−5000x dx = 20
−5000 e−5000x
∣
∣
∣
∣
∞
0
= 20
−5000 (e−∞ − e0)
= 20
−5000 (0 − 1) = 20
5000 = 0.004 C = 4000 μC
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or
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):
i = dq
dt = d
dt
[ 1
α2 −
( t
α + 1
α2
)
e−αt
]
= d
dt
( 1
α2
)
− d
dt
( t
αe−αt
)
− d
dt
( 1
α2 e−αt
)
= 0 −
( 1
αe−αt − α t
αe−αt
)
−
(
−α 1
α2 e−αt
)
=
(
− 1
α + t + 1
α
)
e−αt
= 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:
di
dt = d
dt (te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
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
i = 1
αe−α/α = 1
αe−1

Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
i = 1
0.03679 e−1 ∼= 10 A
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:
[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
(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. Since the power is greater than 0, the box is
absorbing power.
[c] From the calculation in part (b), the box is absorbing 80 W.
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
with respect to time, set the resulting expression equal to zero, and solve
for time:
p = (80,000te−500t)(15te−500t) = 120 × 104t2e−1000t
dp
dt = 240 × 104te−1000t − 120 × 107t2e−1000t = 0

1–4 CHAPTER 1. Circuit Variables
Therefore,
240 × 104 − 120 × 107t = 0
Solving,
t = 240 × 104
120 × 107 = 2 × 10−3 = 2 ms
[b] The maximum power occurs at 2 ms, so find the value of the power at 2
ms:
p(0.002) = 120 × 104(0.002)2e−2 = 649.6 mW
[c] From Eq. (1.3), we know that power is the time rate of change of energy,
or p = dw/dt. If we know the power, we can find the energy by
integrating Eq. (1.3). To find the total energy, the upper limit of the
integral is infinity:
wtotal =
∫ ∞
0
120 × 104x2e−1000x dx
= 120 × 104
(−1000)3 e−1000x[(−1000)2x2 − 2(−1000)x + 2)
∣
∣
∣
∣
∣
∞
0
= 0 − 120 × 104
(−1000)3 e0(0 − 0 + 2) = 2.4 mJ
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and
thus entering the lower terminal where the polarity marking of the voltage is
negative. Thus, using the passive sign convention, p = −vi. Substituting the
values of voltage and current given in the figure,
p = −(800 × 103)(1.8 × 103) = −1440 × 106 = −1440 MW
Thus, because the power associated with the Oregon end of the line is
negative, power is being generated at the Oregon end of the line and
transmitted by the line to be delivered to the California end of the line.

Problems 1–5
Chapter Problems
P 1.1 (260 × 106)(540)
109 = 104.4 gigawatt-hours
P 1.2 (480)(320) pixels
1 frame · 2 bytes
1 pixel · 30 frames
1 sec = 9.216 × 106 bytes/sec
(9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes
x = 32 × 230
9.216 × 106 = 3728 sec = 62 min ≈ 1 hour of video
P 1.3 [a] 20,000 photos
(11)(15)(1) mm3 = x photos
1 mm3
x = (20,000)(1)
(11)(15)(1) = 121 photos
[b] 16 × 230 bytes
(11)(15)(1) mm3 = x bytes
(0.2)3 mm3
x = (16 × 230)(0.008)
(11)(15)(1) = 832,963 bytes
P 1.4 (4 cond.) · (845 mi) · 5280 ft
1 mi · 2526 lb
1000 ft · 1 kg
2.2 lb = 20.5 × 106 kg
P 1.5 Volume = area × thickness
Convert values to millimeters, noting that 10 m2 = 106 mm2
106 = (10 × 106)(thickness)
⇒ thickness = 106
10 × 106 = 0.10 mm
P 1.6 [a] We can set up a ratio to determine how long it takes the bamboo to grow
10 μm First, recall that 1 mm = 103μm. Let’s also express the rate of
growth of bamboo using the units mm/s instead of mm/day. Use a
product of ratios to perform this conversion:
250 mm
1 day · 1 day
24 hours · 1 hour
60 min · 1 min
60 sec = 250
(24)(60)(60) = 10
3456 mm/s
Use a ratio to determine the time it takes for the bamboo to grow 10 μm:
10/3456 × 10−3 m
1 s = 10 × 10−6 m
x s so x = 10 × 10−6
10/3456 × 10−3 = 3.456 s

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1–6 CHAPTER 1. Circuit Variables
[b] 1 cell length
3.456 s · 3600 s
1 hr · (24)(7) hr
1 week = 175,000 cell lengths/week
P 1.7 [a] First we use Eq. (1.2) to relate current and charge:
i = dq
dt = 0.125e−2500t
Therefore, dq = 0.125e−2500t 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)
q(0)
dx = 0.125
∫ t
0
e−2500y dy
We solve the integral and make the substitutions for the limits of the
integral:
q(t) − q(0) = 0.125e−2500y
−2500
∣
∣
∣
∣
∣
t
0
= 50 × 10−6(1 − e−2500t)
But q(0) = 0 by hypothesis, so
q(t) = 50(1 − e−2500t) μC
[b] As t → ∞, qT = 50 μC.
[c] q(0.5 × 10−3) = (50 × 10−6)(1 − e(−2500)(0.0005)) = 35.675 μC.
P 1.8 First we use Eq. (1.2) to relate current and charge:
i = dq
dt = 20 cos 5000t
Therefore, dq = 20 cos 5000t 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)
q(0)
dx = 20
∫ t
0
cos 5000y dy
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
q(t) − q(0) = 20sin 5000y
5000
∣
∣
∣
∣
t
0
= 20
5000 sin 5000t − 20
5000 sin 5000(0) = 20
5000 sin 5000t
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum
value at t = 0, so q(t) = 4 × 10−3 sin 5000t C = 4 sin 5000t mC

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Problems 1–7
P 1.9 [a] First we use Eq. (1.2) to relate current and charge:
i = dq
dt = 40te−500t
Therefore, dq = 40te−500t 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)
q(0)
dx = 40
∫ t
0
ye−500y dy
We solve the integral and make the substitutions for the limits of the
integral:
q(t) − q(0) = 40 e−500y
(−500)2 (−500y − 1)
∣
∣
∣
∣
∣
t
0
= 160 × 10−6e−500t(−500t − 1) + 160 × 10−6
= 160 × 10−6(1 − 500te−500t − e−500t)
But q(0) = 0 by hypothesis, so
q(t) = 160(1 − 500te−500t − e−500t) μC
[b] q(0.001) = (160)[1 − 500(0.001)e−500(0.001) − e−500(0.001) = 14.4 μC.
P 1.10 n = 35 × 10−6 C/s
1.6022 × 10−19 C/elec = 2.18 × 1014 elec/s
P 1.11 w = qV = (1.6022 × 10−19)(6) = 9.61 × 10−19 = 0.961 aJ
P 1.12 [a]
p = vi = (40)(−10) = −400 W
Power is being delivered by the box.
[b] Entering
[c] Gaining
P 1.13 [a] p = vi = (−60)(−10) = 600 W, so power is being absorbed by the box.
[b] Entering

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1–8 CHAPTER 1. Circuit Variables
[c] Losing
P 1.14 Assume we are standing at box A looking toward box B. Use the passive sign
convention to get p = vi, since the current i is flowing into the + terminal of
the voltage v. Now we just substitute the values for v and i into the equation
for power. Remember that if the power is positive, B is absorbing power, so
the power must be flowing from A to B. If the power is negative, B is
generating power so the power must be flowing from B to A.
[a] p = (30)(6) = 180 W 180 W from A to B
[b] p = (−20)(−8) = 160 W 160 W from A to B
[c] p = (−60)(4) = −240 W 240 W from B to A
[d] p = (40)(−9) = −360 W 360 W from B to A
P 1.15 [a] In Car A, the current i is in the direction of the voltage drop across the 12
V battery(the current i flows into the + terminal of the battery of Car
A). Therefore using the passive sign convention,
p = vi = (30)(12) = 360 W.
Since the power is positive, the battery in Car A is absorbing power, so
Car A must have the ”dead” battery.
[b] w(t) =
∫ t
0
p dx; 1 min = 60 s
w(60) =
∫ 60
0
360 dx
w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ
P 1.16 p = vi; w =
∫ t
0
p dx
Since the energy is the area under the power vs. time plot, let us plot p vs. t.

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Problems 1–9
Note that in constructing the plot above, we used the fact that 40 hr
= 144,000 s = 144 ks
p(0) = (1.5)(9 × 10−3) = 13.5 × 10−3 W
p(144 ks) = (1)(9 × 10−3) = 9 × 10−3 W
w = (9 × 10−3)(144 × 103) + 1
2 (13.5 × 10−3 − 9 × 10−3)(144 × 103) = 1620 J
P 1.17 p = (12)(100 × 10−3) = 1.2 W; 4 hr · 3600 s
1 hr = 14,400 s
w(t) =
∫ t
0
p dt w(14,400) =
∫ 14,400
0
1.2 dt = 1.2(14,400) = 17.28 kJ
P 1.18 [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
[b] wtotal =
∫ ∞
0
p(x) dx =
∫ ∞
0
0.6e−500x dx = 0.6
−500 e−500x
∣
∣
∣
∣
∞
0
= −0.0012(e−∞ − e0) = 0.0012 = 1.2 mJ
P 1.19 [a] p = vi = (0.05e−1000t)(75 − 75e−1000t) = (3.75e−1000t − 3.75e−2000t) W
dp
dt = −3750e−1000t + 7500e−2000t = 0 so 2e−2000t = e−1000t
2 = e1000t so ln 2 = 1000t thus p is maximum at t = 693.15 μs
pmax = p(693.15 μs) = 937.5 mW
[b] w =
∫ ∞
0
[3.75e−1000t − 3.75e−2000t] dt =
[ 3.75
−1000 e−1000t − 3.75
−2000 e−2000t
∣
∣
∣
∣
∞
0
]
= 3.75
1000 − 3.75
2000 = 1.875 mJ
P 1.20 [a] p = vi = 0.25e−3200t − 0.5e−2000t + 0.25e−800t
p(625 μs) = 42.2 mW
[b] w(t) =
∫ t
0
(0.25e−3200t − 0.5e−2000t + 0.25e−800t)
= 140.625 − 78.125e−3200t + 250e−2000t − 312.5e−800tμJ
w(625 μs) = 12.14 μJ
[c] wtotal = 140.625 μJ

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1–10 CHAPTER 1. Circuit Variables
P 1.21 [a] p = vi
= [(1500t + 1)e−750t](0.04e−750t)
= (60t + 0.04)e−1500t
dp
dt = 60e−1500t − 1500e−1500t(60t + 0.04)
= −90,000te−1500t
Therefore, dp
dt = 0 when t = 0
so pmax occurs at t = 0.
[b] pmax = [(60)(0) + 0.04]e0 = 0.04
= 40 mW
[c] w =
∫ t
0
pdx
w =
∫ t
0
60xe−1500x dx +
∫ t
0
0.04e−1500x dx
= 60e−1500x
(−1500)2 (−1500x − 1)
∣
∣
∣
∣
∣
t
0
+ 0.04e−1500x
−1500
∣
∣
∣
∣
∣
t
0
When t = ∞ all the upper limits evaluate to zero, hence
w = 60
225 × 104 + 0.04
1500 = 53.33 μJ.
P 1.22 [a] p = vi
= [(3200t + 3.2)e−1000t][(160t + 0.16)e−1000t]
= e−2000t[512,000t2 + 1024t + 0.512]
dp
dt = e−2000t[1,024,000t + 1024] − 2000e−2000t[512,000t2 + 1024t + 0.512]
= −e−2000t[1024 × 106t2 + 1,024,000t]
Therefore, dp
dt = 0 when t = 0
so pmax occurs at t = 0.
[b] pmax = e−0[0 + 0 + 0.512]
= 512 mW
[c] w =
∫ t
0
pdx
w =
∫ t
0
512,000x2e−2000x dx +
∫ t
0
1024xe−2000x dx +
∫ t
0
0.512e−2000x dx
= 512,000e−2000x
−8 × 109 [4 × 106x2 + 4000x + 2]
∣
∣
∣
∣
∣
t
0
+
1024e−2000x
4 × 106 (−2000x − 1)
∣
∣
∣
∣
∣
t
0
+ 0.512e−2000x
−2000
∣
∣
∣
∣
∣
t
0

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Problems 1–11
When t → ∞ all the upper limits evaluate to zero, hence
w = (512,000)(2)
8 × 109 + 1024
4 × 106 + 0.512
2000
w = 128 × 10−6 + 256 × 10−6 + 256 × 10−6 = 640 μJ.
P 1.23 [a] We can find the time at which the power is a maximum by writing an
expression for p(t) = v(t)i(t), taking the first derivative of p(t)
and setting it to zero, then solving for t. The calculations are shown below:
p = 0 t < 0, p = 0 t > 40 s
p = vi = t(1 − 0.025t)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W 0 ≤ t ≤ 40 s
dp
dt = 4 − 0.6t + 0.015t2 = 0.015(t2 − 40t + 266.67)
dp
dt = 0 when t2 − 40t + 266.67 = 0
t1 = 8.453 s; t2 = 31.547 s
(using the polynomial solver on your calculator)
p(t1) = 4(8.453) − 0.3(8.453)2 + 0.005(8.453)3 = 15.396 W
p(t2) = 4(31.547) − 0.3(31.547)2 + 0.005(31.547)3 = −15.396 W
Therefore, maximum power is being delivered at t = 8.453 s.
[b] The maximum power was calculated in part (a) to determine the time at
which the power is maximum: pmax = 15.396 W (delivered)
[c] As we saw in part (a), the other “maximum” power is actually a
minimum, or the maximum negative power. As we calculated in part (a),
maximum power is being extracted at t = 31.547 s.
[d] This maximum extracted power was calculated in part (a) to determine
the time at which power is maximum: pmax = 15.396 W (extracted)
[e] w =
∫ t
0
pdx =
∫ t
0
(4x − 0.3x2 + 0.005x3)dx = 2t2 − 0.1t3 + 0.00125t4
w(0) = 0 J w(30) = 112.5 J
w(10) = 112.5 J w(40) = 0 J
w(20) = 200 J
To give you a feel for the quantities of voltage, current, power, and energy
and their relationships among one another, they are plotted below:

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Loading page 13...

Problems 1–13
[b] p = vi = 2000e−200t sin2 200t
= 2000e−200t
[ 1
2 − 1
2 cos 400t
]
= 1000e−200t − 1000e−200t cos 400t
w =
∫ ∞
0
1000e−200t dt −
∫ ∞
0
1000e−200t cos 400t dt
= 1000 e−200t
−200
∣
∣
∣
∣
∣
∞
0
−1000
{ e−200t
(200)2 + (400)2 [−200 cos 400t + 400 sin 400t]
}∣
∣
∣
∣
∣
∞
0
= 5 − 1000
[ 200
4 × 104 + 16 × 104
]
= 5 − 1
w = 4 J
P 1.25 [a] p = vi = 2000 cos(800πt) sin(800πt) = 1000 sin(1600πt) W
Therefore, pmax = 1000 W
[b] pmax(extracting) = 1000 W
[c] pavg = 1
2.5 × 10−3
∫ 2.5×10−3
0
1000 sin(1600πt) dt
= 4 × 105
[ − cos 1600πt
1600π
]2.5×10−3
0
= 250
π [1 − cos 4π] = 0
[d]
pavg = 1
15.625 × 10−3
∫ 15.625×10−3
0
1000 sin(1600πt) dt
= 64 × 103
[ − cos 1600πt
1600π
]15.625×10−3
0
= 40
π [1 − cos 25π] = 25.46 W
P 1.26 [a] q = area under i vs. t plot
= 1
2 (8)(12,000) + (16)(12,000) + 1
2 (16)(4000)
= 48,000 + 192,000 + 32,000 = 272,000 C
[b] w =
∫
p dt =
∫
vi dt
v = 250 × 10−6t + 8 0 ≤ t ≤ 16 ks
0 ≤ t ≤ 12,000s:
i = 24 − 666.67 × 10−6t
p = 192 + 666.67 × 10−6t − 166.67 × 10−9t2
w1 =
∫ 12,000
0
(192 + 666.67 × 10−6t − 166.67 × 10−9t2) dt
= (2304 + 48 − 96)103 = 2256 kJ

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1–14 CHAPTER 1. Circuit Variables
12,000 s ≤ t ≤ 16,000 s:
i = 64 − 4 × 10−3t
p = 512 − 16 × 10−3t − 10−6t2
w2 =
∫ 16,000
12,000
(512 − 16 × 10−3t − 10−6t2) dt
= (2048 − 896 − 789.33)103 = 362.667 kJ
wT = w1 + w2 = 2256 + 362.667 = 2618.667 kJ
P 1.27 [a] 0 s ≤ t < 10 ms:
v = 8 V; i = 25t A; p = 200t W
10 ms < t ≤ 30 ms:
v = −8 V; i = 0.5 − 25t A; p = 200t − 4 W
30 ms ≤ t < 40 ms:
v = 0 V; i = −250 mA; p = 0 W
40 ms < t ≤ 60 ms:
v = 8 V; i = 25t − 1.25 A; p = 200t − 10 W
t > 60 ms:
v = 0 V; i = 250 mA; p = 0 W
[b] Calculate the area under the curve from zero up to the desired time:
w(0.01) = 1
2 (2)(0.01) = 10 mJ
w(0.03) = w(0.01) − 1
2 (2)(0.01) + 1
2 (2)(0.01) = 10 mJ
w(0.08) = w(0.03) − 1
2 (2)(0.01) + 1
2 (2)(0.01) = 10 mJ

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Solution Manual For Electric Circuits, 10th Edition

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