,1
Circuit 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 × 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 1100
miles:
124,274.24 miles 1100 miles
=
1s xs
Therefore,
1100
= 0.00885 = 8.85 × 10−3 s = 8.85 ms
x=
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 scientific notation:
$100 billion = $100 ×109
Now we determine the number of milliseconds in one year, again using a product
of ratios:
1 year1 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
1 year · = = $3.17/ms
31.5576 × 109 ms 31.5576
1–1
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, 1–2 CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or
dt In this problem, we are given the current and asked to find the total charge.
i = dq
To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of
current:
Z t
q ( t) = i(x) dx
0
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 20
5000x −5000x
qtotal = 20e dx = = (e−∞ − e0)
0 −5000 e
0
−5000
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
dt . In this problem we are given an expression for the charge, and asked to find
i = dq
the maximum current. First we will find an expression for the current using Eq. (1.2):
dq d 1 t 1
i= = — e− αt
dt dt α2 +
d 1 d α α2
= d 1
dt α2 − dt t e− αt
e−αt − dt α2
α
1 t 1
= 0− e−αt − α e−αt − −α e−αt
α α α2
1 1
= − +t+ e−αt
α α
= 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 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 1
i= e−α/α = e−1
α α
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, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e−1 ∼
= 10 A
0.03679
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 = 4A
(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4A
[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
= 240 × 104te−1000t − 120 × 107 t2e−1000t = 0
dt
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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, 1–4 CHAPTER 1. Circuit Variables
Therefore,
240 × 104 − 120 × 107t = 0
Solving,
240 × 10 4 −3
t=
= 2 × 10 = 2 ms
120 × 107
[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:
Z ∞
wtotal = 120 ×104x2e−1000x dx
0
∞
120 ×104
= e−1000x[(−1000)2 x2 − 2(−1000)x + 2)
(−1000) 3
0
4
120 ×10
=0− e0(0 − 0 + 2) = 2.4 mJ
(−1000)3
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.
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, Problems 1–5
Chapter Problems
(260 ×106)(540)
P 1.1 = 104.4 gigawatt-hours
109
(480)(320) pixels 2 bytes 30 frames
P 1.2 · · = 9.216 × 106 bytes/sec
1 frame 1 pixel 1 sec
(9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes
32 ×230
x= = 3728 sec = 62 min ≈ 1 hour of video
9.216 × 106
20,000 photos x photos
P 1.3 [a] 3 =
(11)(15)(1) mm 1 mm3
(20,000)(1)
x= = 121 photos
(11)(15)(1)
16 × 230 bytes x bytes
[b]
(11)(15)(1) mm3 =
(0.2)3 mm3
(16 × 230)(0.008)
= 832,963 bytes
x=
(11)(15)(1)
5280 ft 2526 lb 1 kg
P 1.4 (4 cond.) · (845 mi) · · · = 20.5 × 106 kg
1 mi 1000 ft 2.2 lb
P 1.5 Volume = area × thickness
Convert values to millimeters, noting that 10 m2 = 106 mm2 106 =
(10 × 106)(thickness)
⇒ thickness = 106 = 0.10 mm
10 × 106
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 hour 1 min 250 10
· · · = = mm/s
1 day 24 hours 60 min 60 sec (24)(60)(60) 3456
Use a ratio to determine the time it takes for the bamboo to grow 10 µm:
10/3456 × 10−3 m 10 × 10−6 m 10 × 10−6
= xs so x = 10/3456 ×10−3 = 3.456 s
1s
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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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, 1–6 CHAPTER 1. Circuit Variables
[b] 1 cell length · 3600 s (24)(7) hr
· = 175,000 cell lengths/week
3.456 s 1 hr 1 week
P 1.7 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 0.125e−2500t
dt
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:
Z q(t) Z t
dx = 0.125 e−2500y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral:
−2500y t
q(t) − q(0) = 0.125 e = 50 × 10−6 (1 − e−2500t)
−2500 0
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:
dq
i = = 20 cos 5000t
dt
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:
Z q(t) Zt
dx = 20 cos 5000y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
t
sin 5000y 20 20 20
q(t) −q(0) = 20 sin 5000t − sin 5000(0) = sin 5000t
5000 = 5000 5000
5000
0
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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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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, Problems 1–7
P 1.9 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 40te−500t
dt
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:
Z q(t) Z t
dx = 40 ye−500y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral:
t
q(t) −q(0) = 40 e−500y (−500y − 1) = 160 × 10−6 e−500t(−500t − 1) + 160 × 10−6
(−500) 2
0
−500t
−6
= 160 × 10 (1 − 500te − 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.
35 × 10−6 C/s
P 1.10 n = 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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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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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.
t
Z p dx; 1 min = 60 s
[b] w(t) =
0
Z 60
w(60) = 360 dx
0
w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ
Z t
P 1.16 p = vi; w= p dx
0
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
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Circuit 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 × 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 1100
miles:
124,274.24 miles 1100 miles
=
1s xs
Therefore,
1100
= 0.00885 = 8.85 × 10−3 s = 8.85 ms
x=
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 scientific notation:
$100 billion = $100 ×109
Now we determine the number of milliseconds in one year, again using a product
of ratios:
1 year1 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
1 year · = = $3.17/ms
31.5576 × 109 ms 31.5576
1–1
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, 1–2 CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or
dt In this problem, we are given the current and asked to find the total charge.
i = dq
To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of
current:
Z t
q ( t) = i(x) dx
0
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 20
5000x −5000x
qtotal = 20e dx = = (e−∞ − e0)
0 −5000 e
0
−5000
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
dt . In this problem we are given an expression for the charge, and asked to find
i = dq
the maximum current. First we will find an expression for the current using Eq. (1.2):
dq d 1 t 1
i= = — e− αt
dt dt α2 +
d 1 d α α2
= d 1
dt α2 − dt t e− αt
e−αt − dt α2
α
1 t 1
= 0− e−αt − α e−αt − −α e−αt
α α α2
1 1
= − +t+ e−αt
α α
= 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 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 1
i= e−α/α = e−1
α α
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= e−1 ∼
= 10 A
0.03679
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 = 4A
(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4A
[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
= 240 × 104te−1000t − 120 × 107 t2e−1000t = 0
dt
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, 1–4 CHAPTER 1. Circuit Variables
Therefore,
240 × 104 − 120 × 107t = 0
Solving,
240 × 10 4 −3
t=
= 2 × 10 = 2 ms
120 × 107
[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:
Z ∞
wtotal = 120 ×104x2e−1000x dx
0
∞
120 ×104
= e−1000x[(−1000)2 x2 − 2(−1000)x + 2)
(−1000) 3
0
4
120 ×10
=0− e0(0 − 0 + 2) = 2.4 mJ
(−1000)3
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.
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, Problems 1–5
Chapter Problems
(260 ×106)(540)
P 1.1 = 104.4 gigawatt-hours
109
(480)(320) pixels 2 bytes 30 frames
P 1.2 · · = 9.216 × 106 bytes/sec
1 frame 1 pixel 1 sec
(9.216 × 106 bytes/sec)(x secs) = 32 × 230 bytes
32 ×230
x= = 3728 sec = 62 min ≈ 1 hour of video
9.216 × 106
20,000 photos x photos
P 1.3 [a] 3 =
(11)(15)(1) mm 1 mm3
(20,000)(1)
x= = 121 photos
(11)(15)(1)
16 × 230 bytes x bytes
[b]
(11)(15)(1) mm3 =
(0.2)3 mm3
(16 × 230)(0.008)
= 832,963 bytes
x=
(11)(15)(1)
5280 ft 2526 lb 1 kg
P 1.4 (4 cond.) · (845 mi) · · · = 20.5 × 106 kg
1 mi 1000 ft 2.2 lb
P 1.5 Volume = area × thickness
Convert values to millimeters, noting that 10 m2 = 106 mm2 106 =
(10 × 106)(thickness)
⇒ thickness = 106 = 0.10 mm
10 × 106
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 hour 1 min 250 10
· · · = = mm/s
1 day 24 hours 60 min 60 sec (24)(60)(60) 3456
Use a ratio to determine the time it takes for the bamboo to grow 10 µm:
10/3456 × 10−3 m 10 × 10−6 m 10 × 10−6
= xs so x = 10/3456 ×10−3 = 3.456 s
1s
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
, 1–6 CHAPTER 1. Circuit Variables
[b] 1 cell length · 3600 s (24)(7) hr
· = 175,000 cell lengths/week
3.456 s 1 hr 1 week
P 1.7 [a] First we use Eq. (1.2) to relate current and charge:
dq
i= = 0.125e−2500t
dt
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:
Z q(t) Z t
dx = 0.125 e−2500y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral:
−2500y t
q(t) − q(0) = 0.125 e = 50 × 10−6 (1 − e−2500t)
−2500 0
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:
dq
i = = 20 cos 5000t
dt
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:
Z q(t) Zt
dx = 20 cos 5000y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral,
remembering that sin 0 = 0:
t
sin 5000y 20 20 20
q(t) −q(0) = 20 sin 5000t − sin 5000(0) = sin 5000t
5000 = 5000 5000
5000
0
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:
dq
i= = 40te−500t
dt
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:
Z q(t) Z t
dx = 40 ye−500y dy
q (0) 0
We solve the integral and make the substitutions for the limits of the integral:
t
q(t) −q(0) = 40 e−500y (−500y − 1) = 160 × 10−6 e−500t(−500t − 1) + 160 × 10−6
(−500) 2
0
−500t
−6
= 160 × 10 (1 − 500te − 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.
35 × 10−6 C/s
P 1.10 n = 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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from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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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.
t
Z p dx; 1 min = 60 s
[b] w(t) =
0
Z 60
w(60) = 360 dx
0
w = 360(60 − 0) = 360(60) = 21,600 J = 21.6 kJ
Z t
P 1.16 p = vi; w= p dx
0
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
© 2015 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained
from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.
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