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 × 10 8 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
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
scientific notation:
$100 billion = $100 × 109
Now we determine the number of milliseconds in one year, again using a
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
· = = $3.17/ms
1 year 31.5576 × 10 9 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,
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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
i = dtdq 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:
∫ 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
∫ ∞ 20 ∞ 20
qtotal = 20e −5000x dx = e −5000x = (e−∞ — e0)
0 −5000 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
i = dqdt. 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
i= = − + e−αt
dt dt α2 α α2
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
1 −αt t 1
=0− e — 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 −α/ α 1 −1
i= e = e
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 Saddl e River, NJ 07458.
, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= ∼ 10 A
e−1 =
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 × 10 4t2e−1000t
dp
= 240 × 10 4te−1000t − 120 × 107t2e−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 × 10 4 − 120 × 10 7t = 0
Solving,
240 × 104 −3
t= = 2 × 10 = 2 ms
120 × 10 7
[b] The maximum power occurs at 2 ms, so find the value of the power at 2
ms:
p(0.002) = 120 × 10 4(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 = 120 × 104 x2e−1000x dx
0
∞
120 × 104
= e−1000x[(−1000)2x2 − 2(−1000)x + 2)
(−1000)3 0
120 × 104 0
=0− e (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 × 10 3)(1.8 × 10 3) = −1440 × 10 6 = −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.
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 × 10 8 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
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
scientific notation:
$100 billion = $100 × 109
Now we determine the number of milliseconds in one year, again using a
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
· = = $3.17/ms
1 year 31.5576 × 10 9 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
i = dtdq 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:
∫ 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
∫ ∞ 20 ∞ 20
qtotal = 20e −5000x dx = e −5000x = (e−∞ — e0)
0 −5000 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
i = dqdt. 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
i= = − + e−αt
dt dt α2 α α2
d 1 d t −αt d 1 −αt
= − e − e
dt α2 dt α dt α2
1 −αt t 1
=0− e — 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 −α/ α 1 −1
i= e = e
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 Saddl e River, NJ 07458.
, Problems 1–3
Remember in the problem statement, α = 0.03679. Using this value for α,
1
i= ∼ 10 A
e−1 =
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 × 10 4t2e−1000t
dp
= 240 × 10 4te−1000t − 120 × 107t2e−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 × 10 4 − 120 × 10 7t = 0
Solving,
240 × 104 −3
t= = 2 × 10 = 2 ms
120 × 10 7
[b] The maximum power occurs at 2 ms, so find the value of the power at 2
ms:
p(0.002) = 120 × 10 4(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 = 120 × 104 x2e−1000x dx
0
∞
120 × 104
= e−1000x[(−1000)2x2 − 2(−1000)x + 2)
(−1000)3 0
120 × 104 0
=0− e (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 × 10 3)(1.8 × 10 3) = −1440 × 10 6 = −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.
