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Solution Manual For Electric Circuits, 9th Edition

Solution Manual For Electric Circuits, 9th Edition offers textbook solutions that are easy to follow, helping you ace your assignments.

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    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 ) 3 × 10 8 m 1 s · 100 cm 1 m · 1 in 2 . 54 cm · 1 ft 12 in · 1 mile 5280 feet = 124 , 274 . 24 miles 1 s 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 × 10 9 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 × 10 9 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: $100 × 10 9 1 year · 1 year 31 . 5576 × 10 9 ms = 100 31 . 5576 = $3 . 17 / ms 1–1

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    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 q total = ∫ ∞ 0 20 e − 5000 x dx = 20 − 5000 e − 5000 x ∣ ∣ ∣ ∣ ∞ 0 = 20 − 5000 ( e −∞ − e 0 ) = 20 − 5000 (0 − 1) = 20 5000 = 0 . 004 C = 4000 μ C 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 α 2 − ( t α + 1 α 2 ) e − αt ] = d dt ( 1 α 2 ) − d dt ( t α e − αt ) − d dt ( 1 α 2 e − αt ) = 0 − ( 1 α e − αt − α t α e − αt ) − ( − α 1 α 2 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 ) e αt = (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

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    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 , 000 te − 500 t )(15 te − 500 t ) = 120 × 10 4 t 2 e − 1000 t dp dt = 240 × 10 4 te − 1000 t − 120 × 10 7 t 2 e − 1000 t = 0

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    1–4 CHAPTER 1. Circuit Variables Therefore, 240 × 10 4 − 120 × 10 7 t = 0 Solving, t = 240 × 10 4 120 × 10 7 = 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 × 10 4 (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: w total = ∫ ∞ 0 120 × 10 4 x 2 e − 1000 x dx = 120 × 10 4 ( − 1000) 3 e − 1000 x [( − 1000) 2 x 2 − 2( − 1000) x + 2) ∣ ∣ ∣ ∣ ∣ ∞ 0 = 0 − 120 × 10 4 ( − 1000) 3 e 0 (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 × 10 3 )(1 . 8 × 10 3 ) = − 1440 × 10 6 = − 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 P 1.1 [a] We can set up a ratio to determine how long it takes the bamboo to grow 10 μ m First, recall that 1 mm = 10 3 μ m. 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 μ m: 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 [b] 1 cell length 3 . 456 s · 3600 s 1 hr · (24)(7) hr 1 week = 175 , 000 cell lengths/week P 1.2 Volume = area × thickness Convert values to millimeters, noting that 10 m 2 = 10 6 mm 2 10 6 = (10 × 10 6 )(thickness) ⇒ thickness = 10 6 10 × 10 6 = 0 . 10 mm P 1.3 (260 × 10 6 )(540) 10 9 = 104 . 4 gigawatt-hours P 1.4 [a] 20 , 000 photos (11)(15)(1) mm 3 = x photos 1 mm 3 x = (20 , 000)(1) (11)(15)(1) = 121 photos [b] 16 × 2 30 bytes (11)(15)(1) mm 3 = x bytes (0 . 2) 3 mm 3 x = (16 × 2 30 )(0 . 008) (11)(15)(1) = 832 , 963 bytes P 1.5 (480)(320) pixels 1 frame · 2 bytes 1 pixel · 30 frames 1 sec = 9 . 216 × 10 6 bytes/sec (9 . 216 × 10 6 bytes/sec)( x secs) = 32 × 2 30 bytes x = 32 × 2 30 9 . 216 × 10 6 = 3728 sec = 62 min ≈ 1 hour of video

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    1–6 CHAPTER 1. Circuit Variables P 1.6 (4 cond.) · (845 mi) · 5280 ft 1 mi · 2526 lb 1000 ft · 1 kg 2 . 2 lb = 20 . 5 × 10 6 kg P 1.7 w = qV = (1 . 6022 × 10 − 19 )(6) = 9 . 61 × 10 − 19 = 0 . 961 aJ P 1.8 n = 35 × 10 − 6 C/s 1 . 6022 × 10 − 19 C/elec = 2 . 18 × 10 14 elec/s P 1.9 C/m 3 = 1 . 6022 × 10 − 19 C 1 electron × 10 29 electrons 1 m 3 = 1 . 6022 × 10 10 C/m 3 Cross-sectional area of wire = (0 . 4 × 10 − 2 m)(16 × 10 − 2 m) = 6 . 4 × 10 − 4 m 2 C/m = (1 . 6022 × 10 10 C/m 3 )(6 . 4 × 10 − 4 m 2 ) = 10 . 254 × 10 6 C/m Therefore, i ( C sec ) = (10 . 254 × 10 6 ) ( C m ) × avg vel ( m s ) Thus, average velocity = i 10 . 254 × 10 6 = 1600 10 . 254 × 10 6 = 156 . 04 μ m/s P 1.10 First we use Eq. (1.2) to relate current and charge: i = dq dt = 20 cos 5000 t Therefore, dq = 20 cos 5000 t 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 5000 y 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 5000 y 5000 ∣ ∣ ∣ ∣ t 0 = 20 5000 sin 5000 t − 20 5000 sin 5000(0) = 20 5000 sin 5000 t 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 5000 t C = 4 sin 5000 t mC

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    Problems 1–7 P 1.11 [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. [b] w ( t ) = ∫ t 0 p dx ; 1 min = 60 s w (60) = ∫ 60 0 360 dx w = 360(60 − 0) = 360(60) = 21 , 600 J = 21 . 6 kJ P 1.12 p = (12)(100 × 10 − 3 ) = 1 . 2 W; 4 hr · 3600 s 1 hr = 14 , 400 s w ( t ) = ∫ t 0 p dt w (14 , 400) = ∫ 14 , 400 0 1 . 2 dt = 1 . 2(14 , 400) = 17 . 28 kJ P 1.13 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 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 w = (9 × 10 − 3 )(144 × 10 3 ) + 1 2 (13 . 5 × 10 − 3 − 9 × 10 − 3 )(144 × 10 3 ) = 1620 J P 1.14 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.

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    1–8 CHAPTER 1. Circuit Variables [a] p = − (125)(10) = − 1250 W 1250 W from B to A [b] p = − ( − 240)(5) = 1200 W 1200 W from A to B [c] p = − (480)( − 12) = 5760 W 5760 W from A to B [d] p = − ( − 660)( − 25) = − 16 , 500 W 16 , 500 W from B to A P 1.15 [a] p = vi = (40)( − 10) = − 400 W Power is being delivered by the box. [b] Entering [c] Gaining P 1.16 [a] p = vi = ( − 60)( − 10) = 600 W, so power is being absorbed by the box. [b] Entering [c] Losing P 1.17 [a] p = vi = (0 . 05 e − 1000 t )(75 − 75 e − 1000 t ) = (3 . 75 e − 1000 t − 3 . 75 e − 2000 t ) W dp dt = − 3750 e − 1000 t + 7500 e − 2000 t = 0 so 2 e − 2000 t = e − 1000 t 2 = e 1000 t so ln 2 = 1000 t thus p is maximum at t = 693 . 15 μ s p max = p (693 . 15 μ s) = 937 . 5 mW [b] w = ∫ ∞ 0 [3 . 75 e − 1000 t − 3 . 75 e − 2000 t ] dt = [ 3 . 75 − 1000 e − 1000 t − 3 . 75 − 2000 e − 2000 t ∣ ∣ ∣ ∣ ∞ 0 ] = 3 . 75 1000 − 3 . 75 2000 = 1 . 875 mJ P 1.18 [a] p = vi = 0 . 25 e − 3200 t − 0 . 5 e − 2000 t + 0 . 25 e − 800 t p (625 μ s) = 42 . 2 mW [b] w ( t ) = ∫ t 0 (0 . 25 e − 3200 t − 0 . 5 e − 2000 t + 0 . 25 e − 800 t ) = 140 . 625 − 78 . 125 e − 3200 t + 250 e − 2000 t − 312 . 5 e − 800 t μ J w (625 μ s) = 12 . 14 μ J

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    Problems 1–9 [c] w total = 140 . 625 μ J P 1.19 [a] 0 s ≤ t < 1 s: v = 5 V; i = 20 t A; p = 100 t W 1 s < t ≤ 3 s: v = 0 V; i = 20 A; p = 0 W 3 s ≤ t < 5 s: v = − 5 V; i = 80 − 20 t A; p = 100 t − 400 W 5 s < t ≤ 7 s: v = 5 V; i = 20 t − 120 A; p = 100 t − 600 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) = 1 2 (1)(100) = 50 J w (6) = 1 2 (1)(100) − 1 2 (1)(100) + 1 2 (1)(100) − 1 2 (1)(100) = 0 J w (10) = w (6) + 1 2 (1)(100) = 50 J P 1.20 [a] v (10 ms) = 400 e − 1 sin 2 = 133 . 8 V i (10 ms) = 5 e − 1 sin 2 = 1 . 67 A p (10 ms) = vi = 223 . 80 W

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    1–10 CHAPTER 1. Circuit Variables [b] p = vi = 2000 e − 200 t sin 2 200 t = 2000 e − 200 t [ 1 2 − 1 2 cos 400 t ] = 1000 e − 200 t − 1000 e − 200 t cos 400 t w = ∫ ∞ 0 1000 e − 200 t dt − ∫ ∞ 0 1000 e − 200 t cos 400 t dt = 1000 e − 200 t − 200 ∣ ∣ ∣ ∣ ∣ ∞ 0 − 1000 { e − 200 t (200) 2 + (400) 2 [ − 200 cos 400 t + 400 sin 400 t ] }∣ ∣ ∣ ∣ ∣ ∞ 0 = 5 − 1000 [ 200 4 × 10 4 + 16 × 10 4 ] = 5 − 1 w = 4 J P 1.21 [a] p = vi = [16 , 000 t + 20) e − 800 t ][(128 t + 0 . 16) e − 800 t ] = 2048 × 10 3 t 2 e − 1600 t + 5120 te − 1600 t + 3 . 2 e − 1600 t = 3 . 2 e − 1600 t [640 , 000 t 2 + 1600 t + 1] dp dt = 3 . 2 { e − 1600 t [1280 × 10 3 t + 1600] − 1600 e − 1600 t [640 , 000 t 2 + 1600 t + 1] } = − 3 . 2 e − 1600 t [128 × 10 4 (800 t 2 + t )] = − 409 . 6 × 10 4 e − 1600 t t (800 t + 1) Therefore, dp dt = 0 when t = 0 so p max occurs at t = 0. [b] p max = 3 . 2 e − 0 [0 + 0 + 1] = 3 . 2 W [c] w = ∫ t 0 pdx w 3 . 2 = ∫ t 0 640 , 000 x 2 e − 1600 x dx + ∫ t 0 1600 xe − 1600 x dx + ∫ t 0 e − 1600 x dx = 640 , 000 e − 1600 x − 4096 × 10 6 [256 × 10 4 x 2 + 3200 x + 2] ∣ ∣ ∣ ∣ ∣ t 0 + 1600 e − 1600 x 256 × 10 4 ( − 1600 x − 1) ∣ ∣ ∣ ∣ ∣ t 0 + e − 1600 x − 1600 ∣ ∣ ∣ ∣ ∣ t 0 When t → ∞ all the upper limits evaluate to zero, hence w 3 . 2 = (640 , 000)(2) 4096 × 10 6 + 1600 256 × 10 4 + 1 1600 w = 10 − 3 + 2 × 10 − 3 + 2 × 10 − 3 = 5 mJ.

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    Problems 1–11 P 1.22 [a] p = vi = 400 × 10 3 t 2 e − 800 t + 700 te − 800 t + 0 . 25 e − 800 t = e − 800 t [400 , 000 t 2 + 700 t + 0 . 25] dp dt = { e − 800 t [800 × 10 3 t + 700] − 800 e − 800 t [400 , 000 t 2 + 700 t + 0 . 25] } = [ − 3 , 200 , 000 t 2 + 2400 t + 5]100 e − 800 t Therefore, dp dt = 0 when 3 , 200 , 000 t 2 − 2400 t − 5 = 0 so p max occurs at t = 1 . 68 ms. [b] p max = [400 , 000( . 00168) 2 + 700( . 00168) + 0 . 25] e − 800( . 00168) = 666 mW [c] w = ∫ t 0 pdx w = ∫ t 0 400 , 000 x 2 e − 800 x dx + ∫ t 0 700 xe − 800 x dx + ∫ t 0 0 . 25 e − 800 x dx = 400 , 000 e − 800 x − 512 × 10 6 [64 × 10 4 x 2 + 1600 x + 2] ∣ ∣ ∣ ∣ ∣ t 0 + 700 e − 800 x 64 × 10 4 ( − 800 x − 1) ∣ ∣ ∣ ∣ ∣ t 0 + 0 . 25 e − 800 x − 800 ∣ ∣ ∣ ∣ ∣ t 0 When t = ∞ all the upper limits evaluate to zero, hence w = (400 , 000)(2) 512 × 10 6 + 700 64 × 10 4 + 0 . 25 800 = 2 . 97 mJ. P 1.23 [a] p = vi = 2000 cos(800 πt ) sin(800 πt ) = 1000 sin(1600 πt ) W Therefore, p max = 1000 W [b] p max (extracting) = 1000 W [c] p avg = 1 2 . 5 × 10 − 3 ∫ 2 . 5 × 10 − 3 0 1000 sin(1600 πt ) dt = 4 × 10 5 [ − cos 1600 πt 1600 π ] 2 . 5 × 10 − 3 0 = 250 π [1 − cos 4 π ] = 0 [d] p avg = 1 15 . 625 × 10 − 3 ∫ 15 . 625 × 10 − 3 0 1000 sin(1600 πt ) dt = 64 × 10 3 [ − cos 1600 πt 1600 π ] 15 . 625 × 10 − 3 0 = 40 π [1 − cos 25 π ] = 25 . 46 W P 1.24 [a] q = area under i vs. t plot = [ 1 2 (5)(4) + (10)(4) + 1 2 (8)(4) + (8)(6) + 1 2 (3)(6) ] × 10 3 = [10 + 40 + 16 + 48 + 9]10 3 = 123 , 000 C

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    1–12 CHAPTER 1. Circuit Variables [b] w = ∫ p dt = ∫ vi dt v = 0 . 2 × 10 − 3 t + 9 0 ≤ t ≤ 15 ks 0 ≤ t ≤ 4000 s i = 15 − 1 . 25 × 10 − 3 t p = 135 − 8 . 25 × 10 − 3 t − 0 . 25 × 10 − 6 t 2 w 1 = ∫ 4000 0 (135 − 8 . 25 × 10 − 3 t − 0 . 25 × 10 − 6 t 2 ) dt = (540 − 66 − 5 . 3333)10 3 = 468 . 667 kJ 4000 ≤ t ≤ 12 , 000 i = 12 − 0 . 5 × 10 − 3 t p = 108 − 2 . 1 × 10 − 3 t − 0 . 1 × 10 − 6 t 2 w 2 = ∫ 12 , 000 4000 (108 − 2 . 1 × 10 − 3 t − 0 . 1 × 10 − 6 t 2 ) dt = (864 − 134 . 4 − 55 . 467)10 3 = 674 . 133 kJ 12 , 000 ≤ t ≤ 15 , 000 i = 30 − 2 × 10 − 3 t p = 270 − 12 × 10 − 3 t − 0 . 4 × 10 − 6 t 2 w 3 = ∫ 15 , 000 12 , 000 (270 − 12 × 10 − 3 t − 0 . 4 × 10 − 6 t 2 ) dt = (810 − 486 − 219 . 6)10 3 = 104 . 4 kJ w T = w 1 + w 2 + w 3 = 468 . 667 + 674 . 133 + 104 . 4 = 1247 . 2 kJ P 1.25 [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 shown below: p = 0 t < 0 , p = 0 t > 40 s p = vi = t (1 − 0 . 025 t )(4 − 0 . 2 t ) = 4 t − 0 . 3 t 2 + 0 . 005 t 3 W 0 ≤ t ≤ 40 s dp dt = 4 − 0 . 6 t + 0 . 015 t 2 = 0 . 015( t 2 − 40 t + 266 . 67) dp dt = 0 when t 2 − 40 t + 266 . 67 = 0 t 1 = 8 . 453 s; t 2 = 31 . 547 s (using the polynomial solver on your calculator) p ( t 1 ) = 4(8 . 453) − 0 . 3(8 . 453) 2 + 0 . 005(8 . 453) 3 = 15 . 396 W p ( t 2 ) = 4(31 . 547) − 0 . 3(31 . 547) 2 + 0 . 005(31 . 547) 3 = − 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: p max = 15 . 396 W (delivered)

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    Problems 1–13 [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: p max = 15 . 396 W (extracted) [e] w = ∫ t 0 pdx = ∫ t 0 (4 x − 0 . 3 x 2 + 0 . 005 x 3 ) dx = 2 t 2 − 0 . 1 t 3 + 0 . 00125 t 4 w (0) = 0 J w (30) = 112.5 J w (10) = 112.5 J w (40) = 0 J 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:

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    1–14 CHAPTER 1. Circuit Variables 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: p a = v a i a = (150 × 10 3 )(0 . 6 × 10 − 3 ) = 90 W p b = v b i b = (150 × 10 3 )( − 1 . 4 × 10 − 3 ) = − 210 W p c = − v c i c = − (100 × 10 3 )( − 0 . 8 × 10 − 3 ) = 80 W p d = v d i d = (250 × 10 3 )( − 0 . 8 × 10 − 3 ) = − 200 W p e = − v e i e = − (300 × 10 3 )( − 2 × 10 − 3 ) = 600 W p f = v f i f = ( − 300 × 10 3 )(1 . 2 × 10 − 3 ) = − 360 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: ∑ P dev = 210 + 200 + 360 = 770 W; ∑ P abs = 90 + 80 + 600 = 770 W Thus, the power balances and the total power developed in the circuit is 770 W. P 1.27 p a = − v a i a = − (990)( − 0 . 0225) = 22 . 275 W p b = − v b i b = − (600)( − 0 . 03) = 18 W p c = v c i c = (300)(0 . 06) = 18 W p d = v d i d = (105)(0 . 0525) = 5 . 5125 W p e = − v e i e = − ( − 120)(0 . 03) = 3 . 6 W p f = v f i f = (165)(0 . 0825) = 13 . 6125 W p g = − v g i g = − (585)(0 . 0525) = − 30 . 7125 W p h = v h i h = ( − 585)(0 . 0825) = − 48 . 2625 W Therefore, ∑ P abs = 22 . 275 + 18 + 18 + 5 . 5125 + 3 . 6 + 13 . 6125 = 81 W

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    Problems 1–15 ∑ P del = 30 . 7125 + 48 . 2625 = 78 . 975 W ∑ P abs 6 = ∑ P del Thus, the interconnection does not satisfy the power check. P 1.28 [a] From the diagram and the table we have p a = − v a i a = − (46 . 16)(6) = − 276 . 96 W p b = v b i b = (14 . 16)(4 . 72) = 66 . 8352 W p c = v c i c = ( − 32)( − 6 . 4) = 204 . 8 W p d = − v d i d = − (22)(1 . 28) = − 28 . 16 W p e = − v e i e = − (33 . 6)(1 . 68) = − 56 . 448 W p f = v f i f = (66)( − 0 . 4) = − 26 . 4 W p g = v g i g = (2 . 56)(1 . 28) = 3 . 2768 W p h = − v h i h = − ( − 0 . 4)(0 . 4) = 0 . 16 W ∑ P del = 276 . 96 + 28 . 16 + 56 . 448 + 26 . 4 = 387 . 968 W ∑ P abs = 66 . 8352 + 204 . 8 + 3 . 2768 + 0 . 16 = 275 . 072 W Therefore, ∑ P del 6 = ∑ P abs and the subordinate engineer is correct. [b] The difference between the power delivered to the circuit and the power absorbed by the circuit is − 387 . 986 + 275 . 072 = − 112 . 896 W One-half of this difference is − 56 . 448 W, so it is likely that p e 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 node connecting components b, c, and e, the current i e should be − 1 . 68 A, not 1 . 68 A!) If the sign of p e is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: ∑ P del = 276 . 96 + 28 . 16 + 26 . 4 = 331 . 52 W ∑ P abs = 66 . 8352 + 204 . 8 + 56 . 448 + 3 . 2768 + 0 . 16 = 331 . 52 W Now the power delivered equals the power absorbed and the power balances for the circuit. P 1.29 [a] From an examination of reference polarities, elements a , e , f , and h use a + sign in the power equation, so would be expected to absorb power. Elements b , c , d , and g use a − sign in the power equation, so would be expected to supply power.

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