Exam (elaborations) TEST BANK FOR Electric Circuits 8th Edition By Nilsson, J.W. and Riedel, S (Solution Manual) Circuit Variables Assessment Problems AP
... [Show More] 1.1 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 AP 1.2 First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal will travel in 10−9 s if it is traveling at 80% of the speed of light. Remember that the speed of light c = 3× 108 m/s. Therefore, 80% of c is (0.8)(3 × 108) = 2.4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 2.4 × 108 m 1s · 1 s 109 ns · 100 cm 1 m · 1 in 2.54 cm = (2.4 × 108)(100) (109)(2.54) = 9.45 in 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond. 1–1 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) = t 0 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 = ∞ 0 20e−5000x dx = 20 −5000e−5000x ∞ 0 = 20 −5000 (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 1 α2 − t α + 1 α2 e−αt = d dt 1 α2 − d dt t α e−αt − d dt 1 α2 e−αt = 0− 1 α e−αt − α t α e−αt − −α 1 α2 e−αt = −1 α + t + 1 α 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 Remember in the problem statement, α = 0.03679. Using this value for α, i = 1 0.03679e−1 ∼= 10 A Problems 1–3 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= −4 A; (b) v = −20 V, i = 4A (c) v = 20V, i = −4 A; (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 Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig. 1.5, p = vi. 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 begin, find the expression for power: p = vi = (10,000e−5000t)(20e−5000t) = 200,000e−10,000t = 2× 105e−10,000t W Now find the expression for energy by integrating Eq. (1.3): w(t) = t 0 p(x) dx 1–4 CHAPTER 1. Circuit Variables Substitute the expression for power, p, above. Note that to find the total energy, we let t→∞in the integral. Thus we have w = ∞ 0 2 × 105e−10,000x dx = 2 × 105 −10,000e−10,000x ∞ 0 = 2 × 105 −10,000 (e−∞ − e0) = 2 × 105 −10,000 (0 − 1) = 2 × 105 10,000 = 20 J 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. Chapter Problems P 1.1 To begin, we calculate the number of pixels that make up the display: npixels = (1280)(1024) = 1,310,720 pixels Each pixel requires 24 bits of information. Since 8 bits comprise a 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 = 1,310,720 pixels 1 display · 3 bytes 1 pixel = 3,932,160 bytes/display Finally, we use the fact that there are 106 bytes per MB: 3,932,160 bytes 1 display · 1 MB 106bytes = 3.93 MB/display Problems 1–5 P 1.2 c = 3× 108 m/s so 1 2c = 1.5 × 108 m/s 1.5 × 108 m 1 s = 5 × 106 m x s so x = 5 × 106 1.5 × 108 = 33.3 ms P 1.3 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 day 24 hours · 1 hour 60 min · 1 min 60 sec = 250 (24)(60)(60) = 10 3456 mm/s Use a ratio to determine the time it takes for the bamboo to grow 10 μm: 10/3456 × 10−3 m 1 s = 10 × 10−6 m x s so x = 10 × 10−6 10/3456 × 10−3 = 3.456 s P 1.4 Volume = area × thickness 106 = (10 × 106)(thickness) ⇒ thickness = 106 10 × 106 = 0.10 mm P 1.5 300 × 109 dollars 1 year · 100 pennies 1 dollar · 1 year 365.25 days · 1 day 24 hr · 1 hr 3600 s · 1.5 mm 1 penny · 1 m 1000 mm = 1426 m/s P 1.6 Our approach is as follows: To determine the area of a bit on a track, we need to know the height and width of the space needed to store the bit. The height of the space used to store the bit can be determined from the width of each track on the disk. The width of the space used to store the bit can be determined by calculating the number of bits per track, calculating the circumference of the inner track, and dividing the number of bits per track by the circumference of the track. The calculations are shown below. Width of track = 1 in 77 tracks 25,400μm in = 329.87μm/track Bits on a track = 1.4 MB 2 sides 8 bits byte 1 side 77 tracks = 72,727.273 bits/track Circumference of inner track = 2π(1/2)(25,400μm/in) = 79,796.453μm Width of bit on inner track = 79,796.453μm 72,727.273 bits = 1.0972μm/bit Area of bit on inner track = (1.0972)(329.87) = 361.934μm2 1–6 CHAPTER 1. Circuit Variables P 1.7 C/m3 = 1.6022 × 10−19 × 1029 = 1.6022 × 1010 C/m3 C/m = (1.6022 × 1010)(5.4 × 10−4) = 8.652 × 106 C/m Therefore, (8.652 × 106) C m × ave vel m s = i Thus, average velocity = 1400 8.652 × 10−6 = 161.81 μm/s P 1.8 [a] n = 20 × 10−6 C/s 1.6022 × 10−19 C/elec = 1.25 × 1014 elec/s [b] m = 3303 mi · 5280 ft 1 mi · 12 in 1 ft · 2.54 cm 1 in · 104 μm 1 cm = 5.32 × 1012 μm Therefore, n m = 23.5 The number of electrons/second is approximately 23.5 times the number of micrometers between Sydney and San Francisco. P 1.9 First we use Eq. (1.2) to relate current and charge: i = dq dt = 20 cos 5000t Therefore, dq = 20 cos 5000t 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: q(t) q(0) dx = 20 t 0 cos 5000y dy We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: q(t) − q(0) = 20 sin 5000y 5000 t 0 = 20 5000 sin 5000t − 20 5000 sin 5000(0) = 20 5000 sin 5000t But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 4 × 10−3 sin 5000tC = 4 sin 5000tmC P 1.10 w = qV = (1.6022 × 10−19)(9) = 14.42 × 10−19 = 1.442 aJ Problems 1–7 P 1.11 p = (6)(100 × 10−3) = 0.6 W; 3 hr · 3600 s 1 hr = 10,800 s w(t) = t 0 p dt w(10,800) = 10,800 0 0.6 dt = 0.6(10,800) = 6480 J P 1.12 Assume we are standing at box A looking toward box B. Then, using the passive sign convention 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 = (20)(15) = 300 W 300 W from A to B [b] p = (100)(−5) = −500 W 500 W from B to A [c] p = (−50)(4) = −200 W 200 W from B to A [d] p = (−25)(−16) = 400 W 400 W from A to B P 1.13 [a] p = vi = (−20)(5) = −100 W Power is being delivered by the box. [b] Leaving [c] Gaining P 1.14 [a] p = vi = (−20)(−5) = 100 W, so power is being absorbed by the box. [b] Entering [c] Losing 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 = (−40)(12) = −480 W. Since the power is negative, the battery in Car A is generating power, so Car B must have the ”dead” battery. 1–8 CHAPTER 1. Circuit Variables [b] w(t) = t 0 p dx; 1.5 min = 1.5 · 60 s 1 min = 90 s w(90) = 90 0 480 dx w = 480(90 − 0) = 480(90) = 43,200 J = 43.2 kJ P 1.16 p = vi; w = t 0 p dx Since the energy is the area under the power vs. time plot, let us plot p vs. t. Note that in constructing the plot above, we used the fact that 60 hr = 216,000 s = 216 ks p(0) = (6)(15 × 10−3) = 90 × 10−3 W p(216 ks) = (4)(15 × 10−3) = 60 × 10−3 W w = (60 × 10−3)(216 × 103) + 1 2 (90 × 10−3 − 60 × 10−3)(216 × 103) = 16.2 kJ P 1.17 [a] To find the power at 625 μs, we substitute this value of time into both the equations for v(t) and i(t) and multiply the resulting numbers to get p(625 μs): v(625 μs) = 50e−1600(625×10−6) − 50e−400(625×10−6) = 18.394 − 38.94 = −20.546 V i(625 μs) = 5 × 10−3e−1600(625×10−6) − 5 × 10−3e−400(625×10−6) = 0. − 0. = −0. A p(625 μs) = (−20.546)(−0.) = 42.2 mW [b] To find the energy at 625 μs, we need to integrate the equation for p(t) from 0 to 625 μs. To start, we need an expression for p(t): p(t) = v(t)i(t) = (50)(5 × 10−3)(e−1600t − e−400t)(e−1600t − e−400t) Problems 1–9 = 1 4 (e−3200t − 2e−2000t + e−800t) Now we integrate this expression for p(t) to get an expression for w(t). Note we substitute x for t on the right side of the integral. w(t) = 1 4 t 0 (e−3200x − 2e−2000x + e−800x)dx = 1 4 e−3200x −3200 + e−2000x 1000 − e−800x 800 t 0 = 1 4 e−3200t −3200 + e−2000t 1000 − e−800t 800 − 1 −3200 + 1 1000 − 1 800 = 1 4 e−3200t −3200 + e−2000t 1000 − e−800t 800 + 5.625 × 10−4 Finally, substitute t = 625 μs into the equation for w(t): w(625 μs) = 1 4 [−4.2292 × 10−5 + 2.865 × 10−4 − 7.5816 × 10−4 + 5.625 × 10−4] = 12.137 μJ [c] To find the total energy, we let t→∞in the above equation for w(t). Note that this will cause all expressions of the form e−nt to go to zero, leaving only the constant term 5.625 × 10−4. Thus, wtotal = 1 4 [5.625 × 10−4] = 140.625 μJ P 1.18 [a] v(20 ms) = 100e−1 sin 3 = 5.19 V i(20 ms) = 20e−1 sin 3 = 1.04 A p(20 ms) = vi = 5.39 W [b] p = vi = 2000e−100t sin.2 150t = 2000e−100t 1 2 − 1 2 cos 300t = 1000e−100t − 1000e−100t cos 300t w = ∞ 0 1000e−100t dt − ∞ 0 1000e−100t cos 300t dt = 1000 e−100t −100 ∞ 0 −1000 e−100t (100)2 + (300)2 [−100 cos 300t + 300 sin 300t] ∞ 0 = 10 − 1000 100 1 × 104 + 9 × 104 = 10 − 1 w = 9 J 1–10 CHAPTER 1. Circuit Variables P 1.19 [a] 0 s ≤ t < 1 s: v = 5 V; i = 20t A; p = 100t W 1 s < t ≤ 2 s: v = 0 V; i = 20 A; p = 0 W 2 s ≤ t < 3 s: v = 0 V; i = 20 A; p = 0 W 3 s < t ≤ 4 s: v = −5 V; i = 80 − 20t A; p = −400 + 100t W 4 s ≤ t < 5 s: v = −5 V; i = 80 − 20t A; p = −400 + 100t W 5 s < t ≤ 6 s: v = 5 V; i = −120 + 20t A; p = −600 + 100t W 6 s ≤ t < 7 s: v = 5 V; i = −120 + 20t A; p = −600 + 100t W t > 7 s: v = 0 V; i = 20 A; p = 0 W [b] Calculate the area under the curve from zero up to the desired time: w(1) = 12 (1)(100) = 50 J w(6) = 12 (1)(100) − 12 (1)(100) + 1 2(1)(100) − 12 (1)(100) = 0 J w(10) = 12 (1)(100) − 12 (1)(100) + 1 2(1)(100) − 12 (1)(100) + 12 (1)(100) = 50 J P 1.20 [a] p = vi = (100e−500t)(0.02 − 0.02e−500t) = (2e−500t − 2e−1000t) W dp dt = −1000e−500t + 2000e−1000t = 0 so 2e−1000t = e−500t 2 = e500t so ln 2 = 500t thus p is maximum at t = 1.4 ms Problems 1–11 pmax = p(1.4 ms) = 0.5 W [b] w = ∞ 0 [2e−500t − 2e−1000t] dt = 2 −500e−500t − 2 −1000e−1000t ∞ 0 = 4 1000 − 2 1000 = 2 mJ P 1.21 [a] p = vi = 200 cos(500πt)4.5 sin(500πt) = 450 sin(1000πt) W Therefore, pmax = 450 W [b] pmax(extracting) = 450 W [c] pavg = 1 4 × 10−3 4×10−3 0 450 sin(1000πx) dx = 450 4 × 10−3 −cos 1000πt 1000π 4×10−3 0 = −450 4π [cos 4π − cos 0] = −450 4π [1 − 1] = 0 W [d] pavg = −450 4π [cos 15π − cos 0] = −450 4π [−1 − 1] = 900 4π = 71.62 W P 1.22 [a] q = area under i vs. t plot = 6(5000) 2 + 6(5000) + 6(10,000) 2 + 8(15,000) + 8(5000) 2 = 15,000 + 30,000 + 30,000 + 120,000 + 20,000 = 215,000 C [b] w = pdt = vi dt v = 0.2 × 10−3t + 8 0 ≤ t ≤ 20 ks 0 ≤ t ≤ 5000s i = 20 − 1.2 × 10−3t p = (8 + 0.2 × 10−3t)(20 − 1.2 × 10−3t) = 160 − 5.6 × 10−3t − 2.4 × 10−7t2 w1 = 5000 0 (160 − 5.6 × 10−3t − 2.4 × 10−7t2) dt = 160t − 5.6 × 10−3 2 t2 − 2.4 × 10−7 3 t3 5000 0 = 720 kJ 5000 ≤ t ≤ 15,000s i = 17 − 0.6 × 10−3t p = (8 + 0.2 × 10−3t)(17 − 0.6 × 10−3t) = 136 − 1.4 × 10−3t − 1.2 × 10−7t2 w2 = 15,000 5000 (136 − 1.4 × 10−3t − 1.2 × 10−7t2) dt = 136t − 1.4 × 10−3 2 t2 − 1.2 × 10−7 3 t3 15,000 5000 = 1090 kJ 1–12 CHAPTER 1. Circuit Variables 15,000 ≤ t ≤ 20,000s i = 32 − 1.6 × 10−3t p = (8 + 0.2 × 10−3t)(32 − 1.6 × 10−3t) = 256 − 6.4 × 10−3t − 3.2 × 10−7t2 w3 = 20,000 15,000 (256 − 6.4 × 10−3t − 3.2 × 10−7t2) dt = 256t − 6.4 × 10−3 2 t2 − 3.2 × 10−7 3 t3 20,000 15,000 = 226,666.67 J wT = w1 + w2 + w3 = 720,000 + 1,090,000 + 226,666.67 = 2036.67 kJ P 1.23 [a] p = vi = [104t + 5)e−400t][(40t + 0.05)e−400t] = 400 × 103t2e−800t + 700te−800t + 0.25e−800t = e−800t[400,000t2 + 700t + 0.25] dp dt = {e−800t[800 × 103t + 700] − 800e−800t[400,000t2 + 700t + 0.25]} = [−3,200,000t2 + 2400t + 5]100e−800t Therefore, dp dt = 0 when 3,200,000t2 − 2400t − 5 = 0 so pmax occurs at t = 1.68 ms. [b] pmax = [400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168) = 666.34 mW [c] w = t 0 pdx w = t 0 400,000x2e−800x dx + t 0 700xe−800x dx + t 0 0.25e−800x dx = 400,000e−800x −512 × 106 [64 × 104x2 + 1600x + 2] t 0 + 700e−800x 64 × 104 (−800x − 1) t 0 + 0.25e−800x −800 t 0 When t→∞all the upper limits evaluate to zero, hence w = (400,000)(2) 512 × 106 + 700 64 × 104 + 0.25 800 = 2.97 mJ. P 1.24 [a] We can find the time at which the power is a maximum by writing an expression for p(t) = v(t)i(t), taking the first derivative of p(t) and setting it to zero, then solving for t. The calculations are shownbelow: p = 0 t < 0, p= 0 t > 40 s p = vi = (t − 0.025t2)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W 0 < t < 40 s dp dt = 4 − 0.6t + 0.015t2 = 0 Problems 1–13 Use a calculator to find the two solutions to this quadratic equation: t1 = 8.453 s; t2 = 31.547 s Now we must find which of these two times gives the minimum power by substituting each of these values for t into the equation for p(t): p(t1) = (8.453 − 0.025(8.453)2)(4 − 0.2 · 8.453) = 15.396 W p(t2) = (31.547 − 0.025(31.547)2)(4 − 0.2 · 31.547) = −15.396 W Therefore, maximum power is being delivered at t = 8.453 s. [b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 15.396 W (delivered) [c] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t = 31.547 s. [d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmaxext = 15.396 W (extracted) [e] w = t 0 pdx = t 0 (4x − 0.3x2 + 0.005x3)dx = 2t2 − 0.1t3 + 0.00125t4 w(0) = 0J w(30) = 112.50 J w(10) = 112.50 J w(40) = 0J w(20) = 200 J To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below: 1–14 CHAPTER 1. Circuit Variables Problems 1–15 P 1.25 [a] p = vi = (8 × 104te−500t)(15te−500t) = 12 × 105t2e−1000t W dp dt = 12 × 105[t2(−1000)e−1000t + e−1000t(2t)] = 12 × 105e−1000t[t(2 − 1000t)] dp dt = 0att = 0, t= 2 ms We know p is a minimum at t = 0 since v and i are zero at t = 0. [b] pmax = 12 × 105(2 × 10−3)2e−2 = 649.61 mW [c] w = 12 × 105 ∞ 0 t2e−1000t dt = 12 × 105 e−1000t (−1000)3 [106t2 + 2,000t + 2] ∞ 0 = 2400 μJ P 1.26 We use the passive sign convention to determine whether the power equation is p = vi or p = −vi and substitute into the power equation the values for v and i, as shown below: pa = −vaia = −(−18)(−51) = −918 W pb = vbib = (−18)(45) = −810 W pc = vcic = (2)(−6) = −12 W pd = −vdid = −(20)(−20) = 400 W pe = −veie = −(16)(−14) = 224 W pf = vf if = (36)(31) = 1116 W Remember that if the power is positive, the circuit element is absorbing power, whereas is the power is negative, the circuit element is developing power. We can add the positive powers together and the negative powers together — if the power balances, these power sums should be equal: Pdev = 918 + 810 + 12 = 1740 W; Pabs = 400 + 224 + 1116 = 1740 W Thus, the power balances and the total power developed in the circuit is 1740 W. 1–16 CHAPTER 1. Circuit Variables P 1.27 [a] From the diagram and the table we have pa = −vaia = −(900)(−22.5) = 20,250 W pb = −vbib = −(105)(−52.5) = 5512.5 W pc = −vcic = −(−600)(−30) = −18,000 W pd = vdid = (585)(−52.5) = −30,712.5 W pe = −veie = −(−120)(30) = 3600 W pf = vf if = (300)(60) = 18,000 W pg = −vgig = −(585)(82.5) = −48,262.5 W ph = −vhih = −(−165)(82.5) = 13,612.5 W Pdel = 18,000 + 30,712.5 + 48,262.5 = 96,975 W Pabs = 20,250 + 5512.5 + 3600 + 18,000 + 13,612.5 = 60,975 W Therefore, Pdel = Pabs and the subordinate engineer is correct. [b] The difference between the power delivered to the circuit and the power absorbed by the circuit is 96,975 − 60,975 = 36,000 One-half of this difference is 18,000W, so it is likely that pc or pf is in error. Either the voltage or the current probably has the wrong sign. (In Chapter 2, we will discover that using KCL at the top node, the current ic should be 30 A, not −30 A!) If the sign of pc is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: Pdel = 30,712.5 + 48,262.5 = 78,975 W Pabs = 20,250 + 5512.5 + 18,000 + 3600 + 18,000 + 13,612.5 = 78,975 W Now the power delivered equals the power absorbed and the power balances for the circuit. P 1.28 pa = vaia = (9)(1.8) = 16.2 W pb = −vbib = −(−15)(1.5) = 22.5 W pc = −vcic = −(45)(−0.3) = 13.5 W pd = −vdid = −(54)(−2.7) = 145.8 W pe = veie = (−30)(−1) = 30 W pf = −vf if = −(−240)(4) = 960 W pg = −vgig = −(294)(4.5) = −1323 W ph = vhih = (−270)(−0.5) = 135 W Problems 1–17 Therefore, Pabs = 16.2 + 22.5 + 13.5 + 145.8 + 3 − +960 + 135 = 1323 W Pdel = 1323 W Pabs = Pdel Thus, the interconnection satisfies the power check P 1.29 pa = vaia = (−160)(−10) = 1600 W pb = vbib = (−100)(−20) = 2000 W pc = −vcic = −(−60)(6) = 360 W pd = vdid = (800)(−50) = −40,000 W pe = −veie = −(800)(−20) = 16,000 W pf = −vf if = −(−700)(14) = 9800 W pg = −vgig = −(640)(−16) = 10,240 W Pdel = 40,000 W Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,000 = 40,000 W Therefore, Pdel = Pabs = 40,000 W P 1.30 [a] From an examination of reference polarities, the following elements employ the passive convention: a, c, e, and f. [b] pa = −56 W pb = −14 W pc = 150 W pd = −50 W pe = −18 W pf = −12 W Pabs = 150 W; Pdel = 56 + 14 + 50 + 18 + 12 = 150 W. 2 Circuit Elements Assessment Problems AP 2.1 [a] To find vg write a KVL equation clockwise around the left loop, starting below the dependent source: −ib 4 + vg = 0 so vg = ib 4 To find ib write a KCL equation at the upper right node. Sum the currents leaving the node: ib + 8A = 0 so ib = −8A Thus, vg = −8 4 = −2V [b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi. To do this, write a KVL equation clockwise around the left loop, starting below the voltage source: −vg + vi = 0 so vi = vg = −2V Using the passive sign convention, ps = (8A)(vi) = (8A)(−2V) = −16W Thus the current source generated 16 W of power. 2–1 2–2 CHAPTER 2. Circuit Elements AP 2.2 [a] Note from the circuit that vx = −25 V. To find α write a KCL equation at the top left node, summing the currents leaving: 15A + αvx = 0 Substituting for vx, 15A + α(−25V) = 0 so α(25V) = 15A Thus α = 15A 25V = 0.6 A/V [b] To find the power associated with the voltage source we need to know the current, iv. To find this current, write a KCL equation at the top left node, summing the currents leaving the node: −αvx + iv = 0 so iv = αvx = (0.6)(−25) = −15A Using the passive sign convention, ps = −(iv)(25V) = −(−15A)(25V) = 375W. Thus the voltage source dissipates 375 W. AP 2.3 [a] A KVL equation gives −vg + vR = 0 so vR = vg = 1kV Note from the circuit that the current through the resistor is ig = 5 mA. Use Ohm’s law to calculate the value of the resistor: R = vR ig = 1 kV 5mA = 200 kΩ Using the passive sign convention to calculate the power in the resistor, pR = (vR)(ig) = (1kV)(5mA) = 5W The resistor is dissipating 5 W of power. Problems 2–3 [b] Note from part (a) the vR = vg and iR = ig. The power delivered by the source is thus psource = −vgig so vg = −psource ig = −(−3W) 75mA = 40V Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value: R = vg ig = 40V 75mA = 533.33Ω The power absorbed by the resistor must equal the power generated by the source. Thus, pR = −psource = −(−3W) = 3W [c] Again, note the iR = ig. The power dissipated by the resistor can be determined from the resistor’s current: pR = R(iR)2 = R(ig)2 Solving for ig, i2g = pr R = 480mW 300Ω = 0.0016 so ig = √ 0.0016 = 0.04A = 40mA Then, since vR = vg vR = RiR = Rig = (300 Ω)(40mA) = 12V so vg = 12V AP 2.4 [a] Note from the circuit that the current throught the conductance G is ig, flowing from top to bottom (from KCL), and the voltage drop across the current source is vg, positive at the top (from KVL). From a version of Ohm’s law, vg = ig G = 0.5A 50mS = 10V Now that we know the voltage drop across the current source, we can find the power delivered by this source: psource = −vgig = −(10)(0.5) = −5W Thus the current source delivers 5 W to the circuit. 2–4 CHAPTER 2. Circuit Elements [b] We can find the value of the conductance using the power, and the value of the current using Ohm’s law and the conductance value: pg = Gv2 g so G = pg v2 g = 9 152 = 0.04 S = 40mS ig = Gvg = (40mS)(15V) = 0.6A [c] We can find the voltage from the power and the conductance, and then use the voltage value in Ohm’s law to find the current: pg = Gv2 g so v2 g = pg G = 8W 200 μS = 40,000 Thus vg = 40,000 = 200V ig = Gvg = (200 μS)(200V) = 0.04A = 40mA AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every circuit element. Write a KVL equation clockwise around the circuit, starting below the voltage source: −24V + v2 + v5 − v1 = 0 Next, use Ohm’s law to calculate the three unknown voltages from the three currents: v2 = 3i2; v5 = 7i5; v1 = 2i1 A KCL equation at the upper right node gives i2 = i5; a KCL equation at the bottom right node gives i5 = −i1; a KCL equation at the upper left node gives is = −i2. Now replace the currents i1 and i2 in the Ohm’s law equations with i5: v2 = 3i2 = 3i5; v5 = 7i5; v1 = 2i1 = −2i5 Now substitute these expressions for the three voltages into the first equation: 24 = v2 + v5 − v1 = 3i5 + 7i5 − (−2i5) = 12i5 Therefore i5 = 24/12 = 2 A Problems 2–5 [b] v1 = −2i5 = −2(2) = −4V [c] v2 = 3i5 = 3(2) = 6V [d] v5 = 7i5 = 7(2) = 14V [e] A KCL equation at the lower left node gives is = i1. Since i1 = −i5, is = −2 A. We can now compute the power associated with the voltage source: p24 = (24)is = (24)(−2) = −48W Therefore 24 V source is delivering 48 W. AP 2.6 Redraw the circuit labeling all voltages and currents: We can find the value of the unknown resistor if we can find the value of its voltage and its current. To start, write a KVL equation clockwise around the right loop, starting below the 24Ω resistor: −120V + v3 = 0 Use Ohm’s law to calculate the voltage across the 8Ω resistor in terms of its current: v3 = 8i3 Substitute the expression for v3 into the first equation: −120V + 8i3 = 0 so i3 = 120 8 = 15A Also use Ohm’s law to calculate the value of the current through the 24Ω resistor: i2 = 120V 24Ω = 5A Now write a KCL equation at the top middle node, summing the currents leaving: −i1 + i2 + i3 = 0 so i1 = i2 + i3 = 5+15 = 20A Write a KVL equation clockwise around the left loop, starting below the voltage source: −200V + v1 + 120V = 0 so v1 = 200 − 120 = 80V 2–6 CHAPTER 2. Circuit Elements Now that we know the values of both the voltage and the current for the unknown resistor, we can use Ohm’s law to calculate the resistance: R = v1 i1 = 80 20 = 4Ω AP 2.7 [a] Plotting a graph of vt versus it gives Note that when it = 0, vt = 25 V; therefore the voltage source must be 25 V. Since the plot is a straight line, its slope can be used to calculate the value of resistance: R = Δv Δi = 25 − 0 0.25 − 0 = 25 0.25 = 100Ω A circuit model having the same v − i characteristic is a 25 V source in series with a 100Ω resistor, as shown below: [b] Draw the circuit model from part (a) and attach a 25Ω resistor: To find the power delivered to the 25Ω resistor we must calculate the current through the 25Ω resistor. Do this by first using KCL to recognize that the current in each of the components is it, flowing in a clockwise direction. Write a KVL equation in the clockwise direction, starting below the voltage source, and using Ohm’s law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors: −25V + 100it + 25it = 0 so 125it = 25 so it = 25 125 = 0.2A Thus, the power delivered to the 25Ω resistor is p25 = (25)i2t = (25)(0.2)2 = 1W. Problems 2–7 AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0, it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot is a straight line, its slope can be used to calculate the value of resistance: R = Δv Δi = 25 − 0 0.25 − 0 = 25 0.25 = 100Ω A circuit model having the same v − i characteristic is a 0.25 A current source in parallel with a 100Ω resistor, as shown below: [b] Draw the circuit model from part (a) and attach a 25Ω resistor: Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt. Write a KCL equation at the top center node, summing the currents leaving the node. Use Ohm’s law to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors. −0.25 + vt 100 + vt 25 = 0, so 5vt = 25, thus vt = 5V p25 = v2 t 25 = 1W. AP 2.9 First note that we know the current through all elements in the circuit except the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1; the current in the three elements to the right of the 6 kΩ resistor is 30i1). To find the current in the 6 kΩ resistor, write a KCL equation at the top node: i1 + 30i1 = i6k = 31i1 We can then use Ohm’s law to find the voltages across each resistor in terms of i1. 2–8 CHAPTER 2. Circuit Elements The results are shown in the figure below: [a] To find i1, write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5V source: −5V + 54,000i1 − 1V + 186,000i1 = 0 Solving for i1 54,000i1 + 189,000i1 = 6V so 240,000i1 = 6V Thus, i1 = 6 240,000 = 25μA [b] Now that we have the value of i1, we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage v of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: +v − 54,000i1 + 8V − 186,000i1 = 0 Thus, v = 240,000i1 − 8V = 240,000(25 × 10−6) − 8V = 6V − 8V = −2V We now know the values of voltage [Show Less]