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 1 s 1 m 2.54 cm 12 in 5280 feet 1 s (cid:18) (cid:19) Now set up a proportion to determine how long it takes this signal to travel 1100 miles: 124,274.24 miles 1100 miles = 1 s x s 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 ×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 i = dq In this problem, we are given the current and asked to find the total dt 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 Z0 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 q = 20e−5000xdx = e−5000x = (e−∞ −e0) total −5000 −5000 Z 0 (cid:12)0 (cid:12) (cid:12) 20 20 (cid:12) = (0−1) = = 0.004 C = 4000µC −5000 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 t 1 i = = − + e−αt dt dt α2 α α2 (cid:20) (cid:18) (cid:19) (cid:21) d 1 d t d 1 = − e−αt − e−αt dt α2 dt α dt α2 (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) 1 t 1 = 0− e−αt −α e−αt − −α e−αt α α α2 (cid:18) (cid:19) (cid:18) (cid:19) 1 1 = − +t+ e−αt α α (cid:18) (cid:19) = 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 = −20V, i = −4A; (b) v = −20V, i = 4A (c) v = 20V, i = −4A; (d) v = 20V, i = 4A [b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p = vi = (−20)(−4) = 80W. 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−120×107t2e−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, 240×104 −120×107t = 0 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: ∞ w = 120×104x2e−1000xdx total Z 0 ∞ 120×104 = e−1000x[(−1000)2x2−2(−1000)x+2) (−1000)3 (cid:12) (cid:12)0 (cid:12) 120 ×104 (cid:12) = 0− e0(0−0+2) = 2.4 mJ (cid:12) (−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] = (11)(15)(1) mm3 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) 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) 106 ⇒ thickness = = 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 10µm 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 10µm: 10/3456 ×10−3 m 10×10−6 m 10×10−6 = so x = = 3.456 s 1 s x s 10/3456 ×10−3 © 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 1 cell length 3600 s (24)(7) hr [b] · · = 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−2500tdt 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) t dx = 0.125 e−2500ydy Zq(0) Z0 We solve the integral and make the substitutions for the limits of the integral: e−2500y t q(t)−q(0) = 0.125 = 50×10−6(1−e−2500t) −2500 (cid:12) (cid:12)0 (cid:12) But q(0) = 0 by hypothesis(cid:12), so (cid:12) q(t) = 50(1−e−2500t)µC [b] As t → ∞, q = 50µC. T [c] q(0.5×10−3) = (50×10−6)(1−e(−2500)(0.0005)) = 35.675µC. P 1.8 First we use Eq. (1.2) to relate current and charge: dq i = = 20cos5000t dt Therefore, dq = 20cos5000tdt 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) t dx = 20 cos5000ydy Zq(0) Z0 We solve the integral and make the substitutions for the limits of the integral, remembering that sin0 = 0: sin5000y t 20 20 20 q(t)−q(0) = 20 = sin5000t− sin5000(0) = sin5000t 5000 5000 5000 5000 (cid:12)0 (cid:12) (cid:12) But q(0) = 0 by hypothesi(cid:12)s, i.e., the current passes through its maximum value at t = 0, so q(t) = 4×10−3sin5000tC = 4sin5000tmC © 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 1–7 P 1.9 [a] First we use Eq. (1.2) to relate current and charge: dq i = = 40te−500t dt Therefore, dq = 40te−500tdt 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) t dx = 40 ye−500ydy Zq(0) Z0 We solve the integral and make the substitutions for the limits of the integral: e−500y t q(t)−q(0) = 40 (−500y −1) = 160×10−6e−500t(−500t−1)+160 ×10−6 (−500)2 (cid:12) (cid:12)0 (cid:12) = 160×10−6(1−500te−500t −(cid:12)(cid:12) 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)= 14.4µC. 35×10−6 C/s P 1.10 n = = 2.18×1014 elec/s 1.6022×10−19 C/elec 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 © 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. 1–8 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 [b] w(t) = pdx; 1 min = 60 s Z0 60 w(60) = 360dx Z0 w = 360(60 −0) = 360(60) = 21,600 J = 21.6 kJ t P 1.16 p = vi; w = pdx Z0 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. 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 1 w = (9×10−3)(144×103)+ (13.5×10−3 −9×10−3)(144 ×103) = 1620 J 2 3600 s P 1.17 p = (12)(100×10−3) = 1.2 W; 4 hr· = 14,400 s 1 hr t 14,400 w(t) = pdt w(14,400) = 1.2dt = 1.2(14,400) = 17.28 kJ Z0 Z0 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 ∞ ∞ 0.6 ∞ [b] w = p(x)dx = 0.6e−500xdx = e−500x total −500 Z0 Z0 (cid:12)0 (cid:12) = −0.0012(e−∞ −e0) = 0.0012 = 1.2 mJ (cid:12)(cid:12) P 1.19 [a] p = vi = (0.05e−1000t)(75−75e−1000t) = (3.75e−1000t−3.75e−2000t) W dp = −3750e−1000t+7500e−2000t = 0 so 2e−2000t = e−1000t dt 2 = e1000t so ln2 = 1000t thus p is maximum at t = 693.15µs p = p(693.15µs) = 937.5 mW max ∞ 3.75 3.75 ∞ [b] w = [3.75e−1000t−3.75e−2000t]dt = e−1000t− e−2000t −1000 −2000 Z0 (cid:20) (cid:12)0 (cid:21) (cid:12) 3.75 3.75 (cid:12) (cid:12) = − = 1.875 mJ 1000 2000 P 1.20 [a] p = vi = 0.25e−3200t −0.5e−2000t+0.25e−800t p(625µs) = 42.2 mW t [b] w(t) = (0.25e−3200t−0.5e−2000t+0.25e−800t) Z0 = 140.625 −78.125e−3200t +250e−2000t −312.5e−800tµJ w(625µs) = 12.14µJ [c] w = 140.625µJ total © 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.