Solution Manual For Electric Circuits, 10th Edition

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1Circuit VariablesAssessment ProblemsAP 1.1Use a product of ratios to convert two-thirds the speed of light from metersper second to miles per second:(23)3×108m1 s·100 cm1 m·1 in2.54 cm·1 ft12 in·1 mile5280 feet = 124,274.24 miles1 sNow set up a proportion to determine how long it takes this signal to travel1100 miles:124,274.24 miles1 s= 1100 milesxsTherefore,x=1100124,274.24 = 0.00885 = 8.85×10−3s = 8.85 msAP 1.2To solve this problem we use a product of ratios to change units fromdollars/year to dollars/millisecond. We begin by expressing $10 billion inscientific notation:$100 billion = $100×109Now we determine the number of milliseconds in one year, again using aproduct of ratios:1 year365.25 days·1 day24 hours·1 hour60 mins·1 min60 secs·1 sec1000 ms =1 year31.5576×109msNow we can convert from dollars/year to dollars/millisecond, again with aproduct of ratios:$100×1091 year·1 year31.5576×109ms =10031.5576 = $3.17/ms1–1

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1–2CHAPTER 1. Circuit VariablesAP 1.3Remember from Eq. (1.2), current is the time rate of change of charge, ori=dqdtIn this problem, we are given the current and asked to find the totalcharge. To do this, we must integrate Eq. (1.2) to find an expression forcharge in terms of current:q(t) =∫t0i(x)dxWe are given the expression for current,i, which can be substituted into theabove expression. To find the total charge, we lett→ ∞in the integral. Thuswe haveqtotal =∫∞020e−5000xdx=20−5000e−5000x∣∣∣∣∞0=20−5000 (e−∞−e0)=20−5000 (0−1) =205000 = 0.004 C = 4000μCAP 1.4Recall from Eq. (1.2) that current is the time rate of change of charge, ori=dqdt. In this problem we are given an expression for the charge, and asked tofind the maximum current. First we will find an expression for the currentusing Eq. (1.2):i=dqdt=ddt[1α2−(tα+1α2)e−αt]=ddt(1α2)−ddt(tαe−αt)−ddt(1α2e−αt)= 0−(1αe−αt−α tαe−αt)−(−α1α2e−αt)=(−1α+t+ 1α)e−αt=te−αtNow that we have an expression for the current, we can find the maximumvalue of the current by setting the first derivative of the current to zero andsolving fort:didt=ddt(te−αt) =e−αt+t(−α)eαt= (1−αt)e−αt= 0Sincee−αtnever equals 0 for a finite value oft, the expression equals 0 onlywhen (1−αt) = 0. Thus,t= 1/αwill cause the current to be maximum. Forthis value oft, the current isi= 1αe−α/α= 1αe−1

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Problems1–3Remember in the problem statement,α= 0.03679. Using this value forα,i=10.03679e−1∼= 10 AAP 1.5Start 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 figurewith the polarities shown in Fig. 1.6. Remember that 4A of currententering Terminal 2 is the same as 4A of current leaving Terminal 1. Weget(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 isabsorbing 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 thevoltage and current polarities shown in Fig. 1.5,p=vi. To find the timeat which the power is maximum, find the first derivative of the powerwith respect to time, set the resulting expression equal to zero, and solvefor time:p= (80,000te−500t)(15te−500t) = 120×104t2e−1000tdpdt= 240×104te−1000t−120×107t2e−1000t= 0

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1–4CHAPTER 1. Circuit VariablesTherefore,240×104−120×107t= 0Solving,t= 240×104120×107= 2×10−3= 2 ms[b]The maximum power occurs at 2 ms, so find the value of the power at 2ms: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,orp=dw/dt. If we know the power, we can find the energy byintegrating Eq. (1.3). To find the total energy, the upper limit of theintegral is infinity:wtotal=∫∞0120×104x2e−1000xdx= 120×104(−1000)3e−1000x[(−1000)2x2−2(−1000)x+ 2)∣∣∣∣∣∞0= 0−120×104(−1000)3e0(0−0 + 2) = 2.4 mJAP 1.7At the Oregon end of the line the current is leaving the upper terminal, andthus entering the lower terminal where the polarity marking of the voltage isnegative. Thus, using the passive sign convention,p=−vi. Substituting thevalues of voltage and current given in the figure,p=−(800×103)(1.8×103) =−1440×106=−1440 MWThus, because the power associated with the Oregon end of the line isnegative, power is being generated at the Oregon end of the line andtransmitted by the line to be delivered to the California end of the line.

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Problems1–5Chapter ProblemsP 1.1(260×106)(540)109= 104.4 gigawatt-hoursP 1.2(480)(320) pixels1 frame·2 bytes1 pixel·30 frames1 sec= 9.216×106bytes/sec(9.216×106bytes/sec)(xsecs) = 32×230bytesx=32×2309.216×106= 3728 sec = 62 min≈1 hour of videoP 1.3[a]20,000 photos(11)(15)(1) mm3=xphotos1 mm3x= (20,000)(1)(11)(15)(1) = 121 photos[b]16×230bytes(11)(15)(1) mm3=xbytes(0.2)3mm3x= (16×230)(0.008)(11)(15)(1)= 832,963 bytesP 1.4(4 cond.)·(845 mi)·5280 ft1 mi·2526 lb1000 ft·1 kg2.2 lb = 20.5×106kgP 1.5Volume =area×thicknessConvert values to millimeters, noting that 10 m2= 106mm2106= (10×106)(thickness)⇒thickness =10610×106= 0.10 mmP 1.6[a]We can set up a ratio to determine how long it takes the bamboo to grow10μm First, recall that 1 mm = 103μm. Let’s also express the rate ofgrowth of bamboo using the units mm/s instead of mm/day. Use aproduct of ratios to perform this conversion:250 mm1 day·1 day24 hours·1 hour60 min·1 min60 sec =250(24)(60)(60) =103456 mm/sUse a ratio to determine the time it takes for the bamboo to grow 10μm:10/3456×10−3m1 s= 10×10−6mxssox=10×10−610/3456×10−3= 3.456 s

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1–6CHAPTER 1. Circuit Variables[b]1 cell length3.456 s·3600 s1 hr·(24)(7) hr1 week= 175,000 cell lengths/weekP 1.7[a]First we use Eq. (1.2) to relate current and charge:i=dqdt= 0.125e−2500tTherefore,dq= 0.125e−2500tdtTo find the charge, we can integrate both sides of the last equation. Notethat we substitutexforqon the left side of the integral, andyfortonthe right side of the integral:∫q(t)q(0)dx= 0.125∫t0e−2500ydyWe solve the integral and make the substitutions for the limits of theintegral:q(t)−q(0) = 0.125e−2500y−2500∣∣∣∣∣t0= 50×10−6(1−e−2500t)Butq(0) = 0 by hypothesis, soq(t) = 50(1−e−2500t)μC[b]Ast→ ∞,qT= 50μC.[c]q(0.5×10−3) = (50×10−6)(1−e(−2500)(0.0005)) = 35.675μC.P 1.8First we use Eq. (1.2) to relate current and charge:i=dqdt= 20 cos 5000tTherefore,dq= 20 cos 5000t dtTo find the charge, we can integrate both sides of the last equation. Note thatwe substitutexforqon the left side of the integral, andyforton the rightside of the integral:∫q(t)q(0)dx= 20∫t0cos 5000y dyWe solve the integral and make the substitutions for the limits of the integral,remembering that sin 0 = 0:q(t)−q(0) = 20sin 5000y5000∣∣∣∣t0=205000 sin 5000t−205000 sin 5000(0) =205000 sin 5000tButq(0) = 0 by hypothesis, i.e., the current passes through its maximumvalue att= 0, soq(t) = 4×10−3sin 5000tC = 4 sin 5000tmC

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Problems1–7P 1.9[a]First we use Eq. (1.2) to relate current and charge:i=dqdt= 40te−500tTherefore,dq= 40te−500tdtTo find the charge, we can integrate both sides of the last equation. Notethat we substitutexforqon the left side of the integral, andyfortonthe right side of the integral:∫q(t)q(0)dx= 40∫t0ye−500ydyWe solve the integral and make the substitutions for the limits of theintegral:q(t)−q(0) = 40e−500y(−500)2(−500y−1)∣∣∣∣∣t0= 160×10−6e−500t(−500t−1) + 160×10−6= 160×10−6(1−500te−500t−e−500t)Butq(0) = 0 by hypothesis, soq(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.P 1.10n=35×10−6C/s1.6022×10−19C/elec = 2.18×1014elec/sP 1.11w=qV= (1.6022×10−19)(6) = 9.61×10−19= 0.961 aJP 1.12[a]p=vi= (40)(−10) =−400 WPower is being delivered by the box.[b]Entering[c]GainingP 1.13[a]p=vi= (−60)(−10) = 600 W, so power is being absorbed by the box.[b]Entering

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1–8CHAPTER 1. Circuit Variables[c]LosingP 1.14Assume we are standing at box A looking toward box B. Use the passive signconvention to getp=vi, since the currentiis flowing into the + terminal ofthe voltagev. Now we just substitute the values forvandiinto the equationfor power. Remember that if the power is positive, B is absorbing power, sothe power must be flowing from A to B. If the power is negative, B isgenerating power so the power must be flowing from B to A.[a]p= (30)(6) = 180 W180 W from A to B[b]p= (−20)(−8) = 160 W160 W from A to B[c]p= (−60)(4) =−240 W240 W from B to A[d]p= (40)(−9) =−360 W360 W from B to AP 1.15[a]In Car A, the currentiis in the direction of the voltage drop across the 12V battery(the currentiflows into the + terminal of the battery of CarA). 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, soCar A must have the ”dead” battery.[b]w(t) =∫t0p dx;1 min = 60 sw(60) =∫600360dxw= 360(60−0) = 360(60) = 21,600 J = 21.6 kJP 1.16p=vi;w=∫t0p dxSince the energy is the area under the power vs. time plot, let us plotpvs.t.

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Problems1–9Note that in constructing the plot above, we used the fact that 40 hr= 144,000 s = 144 ksp(0) = (1.5)(9×10−3) = 13.5×10−3Wp(144 ks) = (1)(9×10−3) = 9×10−3Ww= (9×10−3)(144×103) + 12 (13.5×10−3−9×10−3)(144×103) = 1620 JP 1.17p= (12)(100×10−3) = 1.2 W;4 hr·3600 s1 hr= 14,400 sw(t) =∫t0p dtw(14,400) =∫14,40001.2dt= 1.2(14,400) = 17.28 kJP 1.18[a]p=vi= (15e−250t)(0.04e−250t) = 0.6e−500tWp(0.01) = 0.6e−500(0.01)= 0.6e−5= 0.00404 = 4.04 mW[b]wtotal=∫∞0p(x)dx=∫∞00.6e−500xdx=0.6−500e−500x∣∣∣∣∞0=−0.0012(e−∞−e0) = 0.0012 = 1.2 mJP 1.19[a]p=vi= (0.05e−1000t)(75−75e−1000t) = (3.75e−1000t−3.75e−2000t) Wdpdt=−3750e−1000t+ 7500e−2000t= 0so2e−2000t=e−1000t2 =e1000tsoln 2 = 1000tthuspis maximum att= 693.15μspmax=p(693.15μs) = 937.5 mW[b]w=∫∞0[3.75e−1000t−3.75e−2000t]dt=[3.75−1000e−1000t−3.75−2000e−2000t∣∣∣∣∞0]= 3.751000−3.752000 = 1.875 mJP 1.20[a]p=vi= 0.25e−3200t−0.5e−2000t+ 0.25e−800tp(625μs) = 42.2 mW[b]w(t)=∫t0(0.25e−3200t−0.5e−2000t+ 0.25e−800t)=140.625−78.125e−3200t+ 250e−2000t−312.5e−800tμJw(625μs)=12.14μJ[c]wtotal= 140.625μJ

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1–10CHAPTER 1. Circuit VariablesP 1.21[a]p=vi=[(1500t+ 1)e−750t](0.04e−750t)=(60t+ 0.04)e−1500tdpdt=60e−1500t−1500e−1500t(60t+ 0.04)=−90,000te−1500tTherefore,dpdt= 0 whent= 0sopmaxoccurs att= 0.[b]pmax=[(60)(0) + 0.04]e0= 0.04=40 mW[c]w=∫t0pdxw=∫t060xe−1500xdx+∫t00.04e−1500xdx=60e−1500x(−1500)2(−1500x−1)∣∣∣∣∣t0+ 0.04e−1500x−1500∣∣∣∣∣t0Whent=∞all the upper limits evaluate to zero, hencew=60225×104+ 0.041500 = 53.33μJ.P 1.22[a]p=vi=[(3200t+ 3.2)e−1000t][(160t+ 0.16)e−1000t]=e−2000t[512,000t2+ 1024t+ 0.512]dpdt=e−2000t[1,024,000t+ 1024]−2000e−2000t[512,000t2+ 1024t+ 0.512]=−e−2000t[1024×106t2+ 1,024,000t]Therefore,dpdt= 0 whent= 0sopmaxoccurs att= 0.[b]pmax=e−0[0 + 0 + 0.512]=512 mW[c]w=∫t0pdxw=∫t0512,000x2e−2000xdx+∫t01024xe−2000xdx+∫t00.512e−2000xdx=512,000e−2000x−8×109[4×106x2+ 4000x+ 2]∣∣∣∣∣t0+1024e−2000x4×106(−2000x−1)∣∣∣∣∣t0+ 0.512e−2000x−2000∣∣∣∣∣t0

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Problems1–11Whent→ ∞all the upper limits evaluate to zero, hencew= (512,000)(2)8×109+10244×106+ 0.5122000w= 128×10−6+ 256×10−6+ 256×10−6= 640μJ.P 1.23[a]We can find the time at which the power is a maximum by writing anexpression forp(t) =v(t)i(t), taking the first derivative ofp(t)and setting it to zero, then solving fort. The calculations are shown below:p=0t <0,p= 0t >40 sp=vi=t(1−0.025t)(4−0.2t) = 4t−0.3t2+ 0.005t3W0≤t≤40 sdpdt=4−0.6t+ 0.015t2= 0.015(t2−40t+ 266.67)dpdt=0whent2−40t+ 266.67 = 0t1=8.453 s;t2= 31.547 s(using the polynomial solver on your calculator)p(t1)=4(8.453)−0.3(8.453)2+ 0.005(8.453)3= 15.396 Wp(t2)=4(31.547)−0.3(31.547)2+ 0.005(31.547)3=−15.396 WTherefore, maximum power is being delivered att= 8.453 s.[b]The maximum power was calculated in part (a) to determine the time atwhich the power is maximum:pmax= 15.396 W (delivered)[c]As we saw in part (a), the other “maximum” power is actually aminimum, or the maximum negative power. As we calculated in part (a),maximum power is being extracted att= 31.547 s.[d]This maximum extracted power was calculated in part (a) to determinethe time at which power is maximum:pmax= 15.396 W (extracted)[e]w=∫t0pdx=∫t0(4x−0.3x2+ 0.005x3)dx= 2t2−0.1t3+ 0.00125t4w(0)=0 Jw(30)=112.5 Jw(10)=112.5 Jw(40)=0 Jw(20)=200 JTo give you a feel for the quantities of voltage, current, power, and energyand their relationships among one another, they are plotted below:

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1–12CHAPTER 1. Circuit VariablesP 1.24[a]v(10 ms) = 400e−1sin 2 = 133.8 Vi(10 ms) = 5e−1sin 2 = 1.67 Ap(10 ms) =vi= 223.80 W

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Problems1–13[b]p=vi= 2000e−200tsin2200t=2000e−200t[12−12 cos 400t]=1000e−200t−1000e−200tcos 400tw=∫∞01000e−200tdt−∫∞01000e−200tcos 400t dt=1000e−200t−200∣∣∣∣∣∞0−1000{e−200t(200)2+ (400)2[−200 cos 400t+ 400 sin 400t]}∣∣∣∣∣∞0=5−1000[2004×104+ 16×104]= 5−1w=4JP 1.25[a]p=vi= 2000 cos(800πt) sin(800πt) = 1000 sin(1600πt) WTherefore,pmax= 1000 W[b]pmax(extracting) = 1000 W[c]pavg=12.5×10−3∫2.5×10−301000 sin(1600πt)dt=4×105[−cos 1600πt1600π]2.5×10−30= 250π[1−cos 4π] = 0[d]pavg=115.625×10−3∫15.625×10−301000 sin(1600πt)dt=64×103[−cos 1600πt1600π]15.625×10−30= 40π[1−cos 25π] = 25.46 WP 1.26[a]q=area underivs.tplot=12(8)(12,000) + (16)(12,000) +12(16)(4000)=48,000 + 192,000 + 32,000 = 272,000 C[b]w=∫p dt=∫vi dtv=250×10−6t+ 80≤t≤16 ks0≤t≤12,000s:i=24−666.67×10−6tp=192 + 666.67×10−6t−166.67×10−9t2w1=∫12,0000(192 + 666.67×10−6t−166.67×10−9t2)dt=(2304 + 48−96)103= 2256 kJ

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1–14CHAPTER 1. Circuit Variables12,000 s≤t≤16,000 s:i=64−4×10−3tp=512−16×10−3t−10−6t2w2=∫16,00012,000(512−16×10−3t−10−6t2)dt=(2048−896−789.33)103= 362.667 kJwT=w1+w2= 2256 + 362.667 = 2618.667 kJP 1.27[a]0 s≤t <10 ms:v= 8 V;i= 25tA;p= 200tW10 ms< t≤30 ms:v=−8 V;i= 0.5−25tA;p= 200t−4 W30 ms≤t <40 ms:v= 0 V;i=−250 mA;p= 0 W40 ms< t≤60 ms:v= 8 V;i= 25t−1.25 A;p= 200t−10 Wt >60 ms:v= 0 V;i= 250 mA;p= 0 W[b]Calculate the area under the curve from zero up to the desired time:w(0.01)=12(2)(0.01) = 10 mJw(0.03)=w(0.01)−12(2)(0.01) +12(2)(0.01) = 10 mJw(0.08)=w(0.03)−12(2)(0.01) +12(2)(0.01) = 10 mJ

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Problems1–15P 1.28[a][b]i(t)=10 + 0.5×10−3tmA,0≤t≤10 ksi(t)=15 mA,10 ks≤t≤20 ksi(t)=25−0.5×10−3tmA,20 ks≤t≤30 ksi(t)=0,t >30 ksp=vi= 120isop(t)=1200 + 0.06tmW,0≤t≤10 ksp(t)=1800 mW,10 ks≤t≤20 ksp(t)=3000−0.06tmW,20 ks≤t≤30 ksp(t)=0,t >30 ks

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