Solutions Manual For Electric Circuits 10th Edition By Nilsson & Riedel.All Chapters Included

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: 23 3 ×1 s 108 m · 101 m 0 cm · 2.54 cm 1 in · 12 in 1 ft · 5280 feet 1 mile = 124,2741 s .24 miles Now set up a proportion to determine how long it takes this signal to travel 1100 miles: 124,274.24 miles 1 s = 1100 miles x s Therefore, x = 1100 124,274.24 = 0.00885 = 8.85 × 10−3 s = 8.85 ms 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 365.25 days · 1 day 24 hours · 1 hour 60 mins · 1 min 60 secs · 1 sec 1000 ms = 1 year 31.5576 × 109 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: $100 × 109 1 year · 1 year 31.5576 × 109 ms = 100 31.5576 = $3.17/ms 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 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) = Z0t i(x) dx 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 qtotal = Z ∞ 0 20e−5000x dx = −5000 20 e−5000x ∞ 0 = −5000 20 (e−∞ − e0) = 20 −5000 (0 − 1) = 20 5000 = 0.004 C = 4000 µC AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq dt . 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): i = dq dt = d dt α12 − αt + α12  e−αt = d dt α12  − dt d αt e−αt − dt d α12e−αt = 0 − α1 e−αt − ααt e−αt − −αα12e−αt = −α1 + t + α1  e−αt = 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 dt = d dt(te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0 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 i = 1 α e−α/α = 1 α e−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.Problems 1–3 Remember in the problem statement, α = 0.03679. Using this value for α, i = 1 0.03679 e−1 ∼ = 10 A 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. 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 dt = 240 × 104te−1000t − 120 × 107t2e−1000t = 0 © 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, t = 240 × 104 120 × 107 = 2 × 10−3 = 2 ms [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 ∞ 0 120 × 104x2e−1000x dx = 120 × 104 (−1000)3 e−1000x[(−1000)2x2 − 2(−1000)x + 2) ∞ 0 = 0 − 120 × 104 (−1000)3 e0(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. © 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 P 1.1 (260 × 106)(540) 109 = 104.4 gigawatt-hours P 1.2 (480)(320) pixels 1 frame · 2 bytes 1 pixel · 30 frames 1 sec = 9.216 × 106 bytes/sec (9.216 × 106 bytes/sec)(x secs)