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: 1100 miles 124,274.24 miles = . 1s xs Therefore, x=
1100 = 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 $100 ⇥ 109 1 year · = = $3.17/ms. 9 1 year 31.5576 ⇥ 10 ms 31.5576 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: q(t) =
Z 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 qtotal =
Z 1 0
20 (0 5000
=
1 20 e 5000x = 5000 0
20e 5000x dx = 1) =
20 (e 1 5000
e0 )
20 = 0.004 C = 4000 µC. 5000
AP 1.4 Recall from Eq. 1.2 that current is the time rate of change of charge, or i = dq . In this problem we are given an expression for the charge, and asked to dt find the maximum current. First we will find an expression for the current using Eq. 1.2: dq d 1 i= = dt dt ↵2
✓
1 t + 2 e ↵t ↵ ↵
✓
✓
◆
=
=0 =
◆
d t ↵t e dt ↵
1 ↵t e ↵
t ↵ e ↵t ↵
d 1 dt ↵2
✓
✓
◆
◆
◆
✓
d 1 ↵t e dt ↵2 ✓
◆
1 ↵ 2 e ↵t ↵
◆
1 1 +t+ e ↵t ↵ ↵
= te ↵t . 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 ↵t ) = e ↵t + t( ↵)e↵t = (1 dt dt
↵t)e ↵t = 0.
Since e ↵t never equals 0 for a finite value of t, the expression equals 0 only when (1 ↵t) = 0. Thus, t = 1/↵ will cause the current to be maximum. For this value of t, the current is i=
1 ↵/↵ 1 e = e 1. ↵ ↵
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Problems
1–3
Remember in the problem statement, ↵ = 0.03679. Using this value for ↵, i=
1 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,
(c) v = 20 V,
i= i=
4 A; 4 A;
(b) v =
20 V,
(d) v = 20 V,
i = 4 A; 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 ⇥ 104 t2 e 1000t ; dp = 240 ⇥ 104 te 1000t dt
120 ⇥ 107 t2 e 1000t = 0.
Therefore, 240 ⇥ 104
120 ⇥ 107 t = 0.
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1–4
CHAPTER 1. Circuit Variables Solving, t=
240 ⇥ 104 = 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)2 e 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 1 0
120 ⇥ 104 x2 e 1000x dx
120 ⇥ 104 1000x = e [( 1000)2 x2 ( 1000)3 =0
120 ⇥ 104 0 e (0 ( 1000)3
2( 1000)x + 2)
1 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.
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Problems
1–5
Chapter Problems 5280 ft 2526 lb 1 kg · · = 20.5 ⇥ 106 kg. 1 mi 1000 ft 2.2 lb
P 1.1
(4 cond.) · (845 mi) ·
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: nbytes =
8,294,400 pixels 3 bytes · = 24,883,200 bytes/display. 1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB: 1 MB 24,883,200 bytes · 6 = 24.88 MB/display. 1 display 10 bytes [b]
2 hr 24,883,200 bytes 30 images 60 s 60 min · · · · 1 image 1s 1 min 1 hr 1 video = 5.375 ⇥ 1012 bytes/video = 5.375 TB/video.
[c]
24,883,200 bytes 8 bits 30 images · · = 5,971,968,000 bits/s 1 image 1 byte 1 sec = 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 = 103 µm. 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 ⇥ 10 6 m 10/3456 ⇥ 10 3 m = 1s xs [b]
P 1.4
so
10 ⇥ 10 6 x= = 3.456 s. 10/3456 ⇥ 10 3
1 cell length 3600 s (24)(7) hr · · = 175,000 cell lengths/week. 3.456 s 1 hr 1 week
(480)(320) pixels 2 bytes 30 frames · · = 9.216 ⇥ 106 bytes/sec; 1 frame 1 pixel 1 sec (9.216 ⇥ 106 bytes/sec)(x secs) = 32 ⇥ 230 bytes;
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1–6
P 1.5
CHAPTER 1. Circuit Variables
x=
32 ⇥ 230 = 3728 sec = 62 min ⇡ 1 hour of video. 9.216 ⇥ 106
[a]
x photos 20,000 photos = ; 3 (11)(15)(1) mm 1 mm3 x=
[b]
(20,000)(1) = 121 photos. (11)(15)(1)
16 ⇥ 230 bytes x bytes = ; 3 (11)(15)(1) mm (0.2)3 mm3 x=
(16 ⇥ 230 )(0.008) = 832,963 bytes. (11)(15)(1)
P 1.6
(260 ⇥ 106 )(540) = 104.4 gigawatt-hours. 109
P 1.7
First we use Eq. 1.2 to relate current and charge: i=
dq = 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) q(0)
dx = 24
Z t 0
cos 4000y dy.
We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: q(t)
q(0) = 24
sin 4000y 4000
t
= 0
24 sin 4000t 4000
24 24 sin 4000(0) = sin 4000t. 4000 4000
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.
P 1.9
n=
35 ⇥ 10 6 C/s = 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.