S
, Circuit Variables 1
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
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
1 year = $3.17/ms.
· = 31.5576
31.5576 ⇥ 10 ms
9
1–1
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
,
, 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:
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 ! 1 in the integral. Thus we have
Z∞ ∞
− 20 20
5000 x −5000 x
qtotal = 20e dx = = (e−∞ e0)
0 5000 e 5000
0
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 1
i= = dt α2
✓
t e−αt
α2
dt α+
✓ ◆ ✓ ◆ ✓ ◆
d 1 d t −αt
d 1 −αt
=
e — e
dt α2 dt α dt α2
✓ ◆ ✓ ◆
1 −αt t −α t 1 −αt
=0 e α e α e
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
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 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 = 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.
[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 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 ⇥ 107t2e−1000t = 0.
dt
Therefore,
240 ⇥ 104 120 ⇥ 107t = 0.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 1–4 CHAPTER 1. Circuit Variables
Solving,
240 ⇥ 104
t= = 2 ⇥ 10−3 = 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:
∞
wtotal = Z
120 ⇥ 104x2e−1000x dx
0
∞
120 ⇥ 104
= e−1000x[( 1000)2x2 2( 1000)x + 2)
( 1000)3
0
4
120 ⇥ 10 e (0 0 + 2) = 2.4 mJ.
=0 ( 1000)03
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.
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Problems 1–5
Chapter Problems
5280 ft 2526 lb 1 kg = 20.5 ⇥ 106 kg.
P 1.1 (4 cond.) · (845 mi) ·
· · 2.2 lb
1 mi 1000 ft
P 1.2 [a] To begin, we calculate the number of pixels that make up the display:
npixels = (3840)(2160) = 8,294,400 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one byte, each
pixel requires 3 bytes of information. We can calculate the number of bytes of
information required for the display by multiplying the number of pixels in the
display by 3 bytes per pixel:
8,294,400 pixels 3 bytes
nbytes = · = 24,883,200 bytes/display.
1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB:
24,883,200 bytes 1 MB
· 106 bytes = 24.88 MB/display.
1 display
24,883,200 bytes 30 images 60 s 60 min 2 hr
[b]
· · · · 1 video
1 image 1s 1 min 1 hr
= 5.375 ⇥ 1012 bytes/video = 5.375 TB/video.
24,883,200 bytes 8 bits 30 images
[c] = 5,971,968,000 bits/s
· · 1 sec
1 image 1 byte
= 5.972 Gb/s.
P 1.3 [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
· · · = = 3456 mm/s.
1 day 24 hours 60 min 60 sec (24)(60)(60)
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
= so x = 10/3456 ⇥ 10−3 = 3.456 s.
1s
xs
1 cell length 3600 s (24)(7) hr
[b] · · 1 week = 175,000 cell lengths/week.
3.456 s 1 hr
(480)(320) pixels 2 bytes 30 frames
P 1.4 · · = 9.216 ⇥ 106 bytes/sec;
1 sec
1 frame 1 pixel
(9.216 ⇥ 106 bytes/sec)(x secs) = 32 ⇥ 230 bytes;
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 1–6 CHAPTER 1. Circuit Variables
32 ⇥230
x= = 3728 sec = 62 min ⇡ 1 hour of video.
9.216 ⇥ 106
20,000 photos x photos
P 1.5 [a] =
(11)(15)(1) mm 3 ;
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)
x= = 832,963 bytes.
(11)(15)(1)
(260 ⇥ 106)(540)
P 1.6 109 = 104.4 gigawatt-hours.
P 1.7 First we use Eq. 1.2 to relate current and charge:
dq
i = = 24 cos 4000t.
dt
Therefore, dq = 24 cos 4000t 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 = 24 cos 4000y 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 4000y 24 24 24
q(t) q(0) = 24 = sin 4000t sin 4000(0) = sin 4000t.
4000 4000 4000
4000
0
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t
= 0, so q(t) = 6 ⇥ 10−3 sin 4000t C = 6 sin 4000t mC.
P 1.8 w = qV = (1.6022 ⇥ 10−19)(6) = 9.61 ⇥ 10−19 = 0.961 aJ.
35 ⇥ 10−6 C/s
P 1.9 n= = 2.18 ⇥ 1014 elec/s.
1.6022 ⇥10−19 C/elec
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Problems 1–7
P 1.10 [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 ) Zt
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
e
q(t) q(0) = 0.125 = 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 !1, qT = 50 µC.
[c] q(0.5 ⇥ 10−3) = (50 ⇥ 10−6)(1 e(−2500)(0.0005)) = 35.675 µC.
P 1.11 [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) Zt
dx = 40 ye−500y dy.
q(0) 0
We solve the integral and make the substitutions for the limits of the integral:
−500y t
e = 160 ⇥ 10−6e−500t( 500t 1) + 160 ⇥ 10−6
q (t) q (0) = 40 ( 500y 1)
( 500)2
0
= 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.
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 1–8 CHAPTER 1. Circuit Variables
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, Circuit Variables 1
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
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
1 year = $3.17/ms.
· = 31.5576
31.5576 ⇥ 10 ms
9
1–1
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
,
, 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:
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 ! 1 in the integral. Thus we have
Z∞ ∞
− 20 20
5000 x −5000 x
qtotal = 20e dx = = (e−∞ e0)
0 5000 e 5000
0
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 1
i= = dt α2
✓
t e−αt
α2
dt α+
✓ ◆ ✓ ◆ ✓ ◆
d 1 d t −αt
d 1 −αt
=
e — e
dt α2 dt α dt α2
✓ ◆ ✓ ◆
1 −αt t −α t 1 −αt
=0 e α e α e
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
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 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 = 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.
[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 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 ⇥ 107t2e−1000t = 0.
dt
Therefore,
240 ⇥ 104 120 ⇥ 107t = 0.
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 1–4 CHAPTER 1. Circuit Variables
Solving,
240 ⇥ 104
t= = 2 ⇥ 10−3 = 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:
∞
wtotal = Z
120 ⇥ 104x2e−1000x dx
0
∞
120 ⇥ 104
= e−1000x[( 1000)2x2 2( 1000)x + 2)
( 1000)3
0
4
120 ⇥ 10 e (0 0 + 2) = 2.4 mJ.
=0 ( 1000)03
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.
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Problems 1–5
Chapter Problems
5280 ft 2526 lb 1 kg = 20.5 ⇥ 106 kg.
P 1.1 (4 cond.) · (845 mi) ·
· · 2.2 lb
1 mi 1000 ft
P 1.2 [a] To begin, we calculate the number of pixels that make up the display:
npixels = (3840)(2160) = 8,294,400 pixels.
Each pixel requires 24 bits of information. Since 8 bits equal one byte, each
pixel requires 3 bytes of information. We can calculate the number of bytes of
information required for the display by multiplying the number of pixels in the
display by 3 bytes per pixel:
8,294,400 pixels 3 bytes
nbytes = · = 24,883,200 bytes/display.
1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB:
24,883,200 bytes 1 MB
· 106 bytes = 24.88 MB/display.
1 display
24,883,200 bytes 30 images 60 s 60 min 2 hr
[b]
· · · · 1 video
1 image 1s 1 min 1 hr
= 5.375 ⇥ 1012 bytes/video = 5.375 TB/video.
24,883,200 bytes 8 bits 30 images
[c] = 5,971,968,000 bits/s
· · 1 sec
1 image 1 byte
= 5.972 Gb/s.
P 1.3 [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
· · · = = 3456 mm/s.
1 day 24 hours 60 min 60 sec (24)(60)(60)
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
= so x = 10/3456 ⇥ 10−3 = 3.456 s.
1s
xs
1 cell length 3600 s (24)(7) hr
[b] · · 1 week = 175,000 cell lengths/week.
3.456 s 1 hr
(480)(320) pixels 2 bytes 30 frames
P 1.4 · · = 9.216 ⇥ 106 bytes/sec;
1 sec
1 frame 1 pixel
(9.216 ⇥ 106 bytes/sec)(x secs) = 32 ⇥ 230 bytes;
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, 1–6 CHAPTER 1. Circuit Variables
32 ⇥230
x= = 3728 sec = 62 min ⇡ 1 hour of video.
9.216 ⇥ 106
20,000 photos x photos
P 1.5 [a] =
(11)(15)(1) mm 3 ;
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)
x= = 832,963 bytes.
(11)(15)(1)
(260 ⇥ 106)(540)
P 1.6 109 = 104.4 gigawatt-hours.
P 1.7 First we use Eq. 1.2 to relate current and charge:
dq
i = = 24 cos 4000t.
dt
Therefore, dq = 24 cos 4000t 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 = 24 cos 4000y 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 4000y 24 24 24
q(t) q(0) = 24 = sin 4000t sin 4000(0) = sin 4000t.
4000 4000 4000
4000
0
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t
= 0, so q(t) = 6 ⇥ 10−3 sin 4000t C = 6 sin 4000t mC.
P 1.8 w = qV = (1.6022 ⇥ 10−19)(6) = 9.61 ⇥ 10−19 = 0.961 aJ.
35 ⇥ 10−6 C/s
P 1.9 n= = 2.18 ⇥ 1014 elec/s.
1.6022 ⇥10−19 C/elec
© 2019 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
, Problems 1–7
P 1.10 [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 ) Zt
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
e
q(t) q(0) = 0.125 = 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 !1, qT = 50 µC.
[c] q(0.5 ⇥ 10−3) = (50 ⇥ 10−6)(1 e(−2500)(0.0005)) = 35.675 µC.
P 1.11 [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) Zt
dx = 40 ye−500y dy.
q(0) 0
We solve the integral and make the substitutions for the limits of the integral:
−500y t
e = 160 ⇥ 10−6e−500t( 500t 1) + 160 ⇥ 10−6
q (t) q (0) = 40 ( 500y 1)
( 500)2
0
= 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.
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, 1–8 CHAPTER 1. Circuit Variables
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