电路分析基础(英文版)课后答案第一章.pdf

1Circuit Variables and CircuitElementsDrill ExercisesDE 1.1q=Z1020e5000tdt=4000CDE 1.2i=dqdt=tet;didt=(1 t)et;didt=0when t=1;Therefore imax=1e=10:03679e=10 ADE 1.3aTherefore(a)v=20V;i=4A;(b)v=20V,i=4A(c)v=20V,i=4A;(d)v=20V,i=4Ab Using the reference system in Fig.1.3(a),p=vi=(20)(4)=80W,sothe box is absorbing power.c The box is absorbing 80 W.12CHAPTER 1.Circuit Variables and Circuit ElementsDE 1.4p=vi=20 104e10;000tW;w=Z1020 104e10;000tdt=20 JDE 1.5p=800 103 1:8 103=1440 106=1440 MWfrom Oregon to CaliforniaDE 1.6The interconnection is valid:is=10+15=25 Ap100V=100is=2500 W(absorbing)p10A=100(10)=1000 W(generating)100+vs 40=0so vs=140 Vp15A=15(140)=2100 W(generating)p40V=15(40)=600 W(absorbing)Xpdev=p10A+p15A=3100 WXpabs=p100V+p40V=3100 WXpdev=Xpabs=3100 WDE 1.7a vl vc+v1 vs=0;ilRl icRc+i1R1 vs=0isRl+isRc+isR1 vs=0b is=vs=(Rl+Rc+R1)DE 1.8a 24=v2+v5 v1=3i5+7i5(2i5)=12i5Therefore i5=24=12=2 Ab v1=2i5=4 Vc v2=3i5=6 Vd v5=7i5=14 VProblems3e p24=(24)(2)=48 W;therefore 24 V source is delivering 48 W.DE 1.9i2=120=24=5 Ai3=120=8=15 Ai1=i2+i3=20 A200+20R+120=0R=80=20=4DE 1.10 a Plotting a graph of vtversus itgivesNote that when it=0,vt=25 V;therefore the voltage source must be 25V.When vtis zero,it=0:25 A,hence the resistor must be 25=0:25 or100.A circuit model having the same v i characteristic is a 25 V source inseries with a 100 resistor.4CHAPTER 1.Circuit Variables and Circuit Elementsbit=25125=0:2 A;p=(0:2)2(25)=1 W:DE 1.11 a Since we are constructing the model from two elements,we have twochoices on interconnecting them|series or parallel.From the v icharacteristic we require vt=25 V when it=0.The only way we cansatisfy this requirement is with a parallel connection.The constraint thatvt=0 when it=0:25 A tells us the ideal current source must produce0:25 A.Therefore the parallel resistor must be 25=0:25 or 100.b0:25+vt100+vt25=0;5vt=25;vt=5 Vp=v2t25=1 W:ProblemsP 1.1i=dqdt=24cos4000tTherefore,dq=24cos4000tdtProblems5Zq(t)q(0)dx=24Zt0cos4000y dyq(t)q(0)=24sin4000y4000t0But q(0)=0 by hypothesis,i.e.,the current passes through its maximumvalue at t=0,so q(t)=6 103sin4000tC=6sin4000tmCP 1.2p=(6)(100)103=0:6 W;w=(0:6)(3)(60)(60)=6480 JP 1.3Assume we are standing at box A looking toward box B,then p=vi.a p=(120)(5)=600 Wfrom A to Bb p=(250)(8)=2000 Wfrom B to Ac p=(150)(16)=2400 Wfrom B to Ad p=(480)(10)=4800 Wfrom A to BP 1.4ap=vi=(40)(10)=400 WPower is being delivered by the box.b Enteringc GainP 1.5a p=vi=(60)(10)=600 W,so power is being absorbed by the box.b Enteringc LoseP 1.6a Looking from A to B the current i is in the direction of the voltage riseacross the 12 V battery,thereforep=vi=12(30)=360 W.Thus the power ow is from B to A,and Car A has the dead battery.b w=Zt0pdx=Zt0360dxw=360t=360(1 60)=21:6 kJ6CHAPTER 1.Circuit Variables and Circuit ElementsP 1.7p=vi;w=Zt0pdxSince the energy is the area under the power vs.time plot,let us plot p vs.t.p(0)=(6)(15 103)=90 103Wp(216 ks)=(4)(15 103)=60 103Ww=(60 103)(216 103)+12(216)(30)=16:2 kJNote:60 hr 216;000 s=216 ksP 1.8a p=vi=30e500t 30e1500t 40e1000t+50e2000t 10e3000tp(1 ms)=3:1 mWbw(t)=Zt0(30e500 x 30e1500 x 40e1000 x+50e2000 x 10e3000 x)dx=21:67 60e500t+20e1500t+40e1000t25e2000t+3:33e3000tJw(1 ms)=1:24Jc wtotal=21:67JP 1.9a v(20 ms)=100e1sin3=5:19 Vi(20 ms)=20e1sin3=1:04 Ap(20 ms)=vi=5:39 WProblems7bp=vi=2000e100tsin:2150t=2000e100t1212cos300t=1000e100t 1000e100tcos300tw=Z101000e100tdt Z101000e100tcos300tdt=1000e100t10010 1000(e100t(100)2+(300)2100cos300t+300sin300t)10=10 10001001 104+9 104=10 1w=9 JP 1.10a0 t 10 ms:v=1000t V;i=0:6 mA;p=0:6t mW10 t 25 ms:v=10 V;i=0:6 mA;p=6 mW25 t 35 ms:v=75 2500t V;i=0 mA;p=0 mW35 t 60 ms:v=50+1000t V;i=0:4 mA;p=20 400t mW60 t 70 ms:v=50+1000t V;i=0 mA;p=0 mW70 t 80 ms:v=20 V;i=0:5 mA;p=10 mW80 t 90 ms:v=180 2000t V;i=0 mA;p=0 mW90 t 95 ms:v=180 2000t V;i=0:9 mA;p=162 1800t mW95 t 100 ms:v=200+2000t V;i=0:9 mA;p=180+1800t mW8CHAPTER 1.Circuit Variables and Circuit Elementsbw(25)=12(6)(10)+(6)(15)=120Jw(60)=120+12(15)(6)12(10)(4)=145Jw(90)=145 (10)(10)=45Jw(100)=45 12(10)(9)=0JP 1.11a p=vi=(2e500t 2e1000t)Wdpdt=1000e500t+2000e1000t=0 at t=1.4 mspmax=p(1:4 ms)=0:5 Wb w=Z102e500t 2e1000tdt=2500e500t21000e1000t10=2 mJP 1.12a p=vi=900sin(200t)cos(200t)=450sin(400t)WTherefore,pmax=450 Wb pmax(extracting)=450 Wcpavg=200Z51030450sin(400t)dt=9 104cos400t4002:51030=2251 cos2=0d pavg=1801 cos2:5=180=57:3 WP 1.13aq=area under i vs.t plot=h12(5)(4)+(10)(4)+12(8)(4)+(8)(6)+12(3)(6)i 103=10+40+16+48+9103=123;000 CProblems9bw=Zpdt=Zvidtv=0:2 103t+90 t 15 ks0 t 4000si=15 1:25 103tp=135 8:25 103t 0:25 106t2w1=Z40000(135 8:25 103t 0:25 106t2)dt=(540 66 5:3333)103=468:667 kJ4000 t 12;000i=12 0:5 103tp=108 2:1 103t 0:1 106t2w2=Z12;0004000(108 2:1 103t 0:1 106t2)dt=(864 134:4 55:467)103=674:133 kJ12;000 t 15;000i=30 2 103tp=270 12 103t 0:4 106t2w3=Z15;00012;000(270 12 103t 0:4 106t2)dt=(810 486 219:6)103=104:4 kJwT=w1+w2+w3=468:667+674:133+104:4=1247:2 kJP 1.14ap=vi=400 103t2e800t+700te800t+0:25e800t=e800t400;000t2+700t+0:25dpdt=fe800t800 103t+700 800e800t400;000t2+700t+0:25g=3;200;000t2+2400t+5100e800tTherefore,dpdt=0 when 3;200;000t2 2400t 5=0so pmaxoccurs at t=1:68 ms.bpmax=400;000(:00168)2+700(:00168)+0:25e800(:00168)=666 mW10CHAPTER 1.Circuit Variables and Circuit Elementscw=Zt0pdxw=Zt0400;000 x2e800 xdx+Zt0700 xe800 xdx+Zt00:25e800 xdx=400;000e800 x512 10664 104x2+1600 x+2t0+700e800 x64 104(800 x 1)t0+0:25e800 x800t0When t=1 all the upper limits evaluate to zero,hencew=(400;000)(2)512 106+70064 104+0:25800=2:97 mJ.P 1.15ap=0t 3 sp=vi=t(3 t)(6 4t)=18t 18t2+4t3mW0 t 3 sdpdt=18 36t+12t2=12(t2 3t+1:5)dpdt=0when t2 3t+1:5=0t=3 p9 62=3 p32t1=3=2 p3=2=0:634 s;t2=3=2+p3=2=2:366 sp(t1)=18(0:634)18(0:634)2+4(0:634)3=5:196 mWp(t2)=18(2:366)18(2:366)2+4(2:366)3=5:196 mWTherefore,maximum power is being delivered at t=0:634 s.b pmax=5:196 mW(delivered)c Maximum power is being extracted at t=2:366 s.d pmax=5:196 mW(extracted)e w=Zt0pdx=Zt0(18x 18x2+4x3)dx=9t2 6t3+t4w(0)=0 mJw(2)=4 mJw(1)=4 mJw(3)=0 mJProblems11P 1.16ap=vi=12 105t2e1000tWdpdt=12 105t2(1000)e1000t+e1000t(2t)=12 105te1000tt(2 1000t)dpdt=0 at t=0;t=2 msWe know p is a minimum at t=0 since v and i are zero at t=0.b pmax=12 105(2 103)2e2=649:61 mW12CHAPTER 1.Circuit Variables and Circuit Elementscw=12 105Z10t2e1000tdt=12 105(e1000t(1000)3106t2+2;000t+210)=2:4 mJP 1.17a From the diagram and the table we havepa=vaia=(46:16)(6)=276:96 W(del)pb=vbib=(14:16)(4:72)=66:8352 W(abs)pc=vcic=(32)(6:4)=204:80 W(abs)pd=vdid=(22)(1:28)=28:16 W(del)pe=veie=(33:60)(1:68)=56:448 W(del)pf=vfif=(66)(0:4)=26:40 W(del)pg=vgig=(2:56)(1:28)=3:2768 W(abs)ph=vhih=(0:4)(0:4)=0:16 W(abs)XPdel=276:96+28:16+56:448+26:40=387:9680 WXPabs=66:8352+204:80+3:2768+0:16=275:072 WTherefore,XPdel6=XPabsand the subordinate engineer is correct.b We can also check the data using Kirchhos laws.From Fig.P1.17 the following equations should be satised:ia ib id=0(ok)ib+ic ie=0(no)if ia ic=0(ok)id=ig(ok)ig+ie+ih=0(no)ih=if(ok)Using Kirchhos current law,it appears ieis in error.From Kirchhos voltage law we havevb va vc=0(ok)vd vb+ve+vg=0(ok)ve+vc+vf+vh=0(ok)Therefore all the voltages are consistent with Kirchhos voltage law.Assume ieis in error.Therefore,ie=ib+ic=ig ih=4:72 6:40=1:28 0:4=1:68 ASo the error is in the sign of ie;ieequals minus 1:68 A.Correcting ieleads toXPdel=XPabs=331:52 WProblems13P 1.18pa=vaia=(48)(12)=576 W(abs)pb=vbib=(18)(4)=72 W(del)pc=vcic=(30)(10)=300 W(abs)pd=vdid=(36)(16)=576 W(abs)pe=veie=(36)(8)=288 W(del)pf=vfif=(54)(14)=756 W(abs)pg=vgig=(84)(22)=1848 W(del)XPdel=72+288+1848=2208 WXPabs=576+300+576+756=2208 WTherefore,XPdel=XPabs=2208 WP 1.19a From an examination of reference polarities,the following elements employthe passive convention:a;c;e;and f.bpa=56 W(del)pb=14 W(del)pc=150 W(abs)pd=50 W(del)pe=18 W(del)pf=12 W(del)XPabs=150 W;XPdel=56+14+50+18+12=150 W.P 1.20(a)9(b)7(c)4(d)vaRa,vbRb,vcRc(e)6(f)(1)va Ra Rd Rb vb(2)Rd Rf Re(3)vb Rb Rd Rf Rc vc(4)vc Rc Rf Ra va(5)va Ra Rf Re Rb vb(6)va Ra Rd Re Rc vc(7)vb Rb Re Rc vc14CHAPTER 1.Circuit Variables and Circuit ElementsP 1.21The interconnect is valid since it does not violate Kirchhos laws.60+20+40=0(KVL)8+4 12=0(KCL)XPdev=4(60)+8(60)=720 WXPabs=12(20)+12(40)=720 WXPdev=XPabs=720 WP 1.22a Yes,Kirchhos laws are not violated.b No,because the voltages across the independent and dependent currentsources are indeterminate.For example,dene v1,v2,and v3as shown:Kirchhos voltage law requiresv1+20=v3v2+100=v3Conservation of energy requires8(20)+8v1+16v2+1600 24v3=0orv1+2v2 3v3=220Problems15Now arbitrarily select a value of v3and show the conservation of energywill be satised.Examples:If v3=200 V then v1=180 V and v2=100 V.Then180+200 600=220(CHECKS)If v3=100 V,then v1=120 V and v2=200 V.Then120 400+300=220(CHECKS)P 1.23a Yes,independent voltage sources can carry whatever current is required bythe connection;independent current source can support any voltagerequired by the connection.b30 V source:absorbing10 V source:delivering8 A source:deliveringcP30V=(30)(8)=240 W(abs)P10V=10(8)=80 W(del)P8A=20(8)=160 W(del)XPabs=XPdel=240 Wd Yes,30 V source is delivering,the 10 V source is delivering,and the 8 Asource is absorbingP30V=30(8)=240 W(del)P10V=10(8)=80 W(del)P8A=+40(8)=320 W(abs)P 1.24The interconnection is valid because it does not violate Kirchhos laws.i=25 A;6i=150 V200+50 (150)=0But the power developed in the circuit cannot be determined,as the currentsin the 200 V,50 V,and 6isources are unspecied.P 1.25The interconnection is not valid because it violates Kirchhos current law:3 A+(5 A)6=8 A:16CHAPTER 1.Circuit Variables and Circuit ElementsP 1.26i=4 A so ig=12 Avo=100 V60+v1=100;so v1=160 Vv2 80=100;so v2=180 VXPdev=180(4)+100(8)+60(12)=2240 WCHECK:XPdiss=160(12)+80(4)=1920+320=2240 W|CHECKSP 1.27The interconnection is valid because it does not violate Kirchhos laws:pVsources=(100 60)(5)=200 W:P 1.28First there is no violation of Kirchhos laws,hence the interconnection isvalid.Kirchhos voltage law requiresv1+v2=150 50=100 VThe conservation of energy law requires20v1 10v1+10v2+500 1500=0orv1+v2=100Hence any combination of v1and v2that adds to 100 is a valid solution.Forexample if v1=80 V and v2=20 VPabs=80(20)+10(20)+50(10)=2300 WProblems17Pdev=1500+80(10)=2300 WIf v1=60 V and v2=40 VPabs=60(20)+10(40)+500=2100 WPdev=60(10)+1500=2100 WIf v1=100 V and v2=200 VPabs=10(100)+10(200)+10(50)=3500 WPdev=20(100)+10(150)=3500 WP 1.29a1:6=ig ia80ia=1:6(30+90)=192therefore,ia=2:4 Aig=ia+1:6=2:4+1:6=4 Ab vg=90(1:6)=144 VcXPdis=2:42(80)+1:62(120)=768 WXPdev=(4)(192)=768 WTherefore,XPdis=XPdev=768 WP 1.30avo=8ia+14ia+18ia=40(20)=800 V800=10ioio=800=10=80 Ab ig=ia+io=20+80=100 Ac pg(delivered)=(100)(800)=80;000 W=80 kWP 1.31a20ia=80ibig=ia+ib=5ibia=4ib50=4ig+80ib=20ib+80ib=100ibib=0:5 A,therefore,ia=2 Aandig=2:5 A18CHAPTER 1.Circuit Variables and Circuit Elementsb ib=0:5 Ac vo=80ib=40 Vdp4=i2g(4)=6:25(4)=25 Wp20=i2a(20)=(4)(20)=80 Wp80=i2b(80)=0:25(80)=20 We p5V(delivered)=5ig=125 WCheck:XPdis=25+80+20=125 WXPdel=125 WP 1.32avo=20(8)+16(15)=400 Vio=400=80=5 Aia=25 AP230(supplied)=(230)(25)=5750 Wib=5+15=20 AP260(supplied)=(260)(20)=5200 WbXPdis=(25)2(2)+(20)2(8)+(5)2(4)+(15)216+(20)22+(5)2(80)=1250+3200+100+3600+800+2000=10;950 WXPsup=5750+5200=10;950 WTherefore,XPdis=XPsup=10;950 WProblems19P 1.33av2=80+4(12)=128 Vv1=128 24(2)=80 Vi1=v116=8016=5 Ai3=i1 2=5 2=3 Avg=v1+24i3=80+72=152 Vvg 4i4=v24i4=vg v2=152 128=24 Vi4=24=4=6 Aig=(i3+i4)=(3+6)=9 Abp8=(2)2(8)=32 Wp4=(6)2(4)=144 Wp12=(2)2(12)=48 Wp6=(5)2(6)=150 Wp4=(2)2(4)=16 Wp10=(5)2(10)=250 Wp24=(3)2(24)=216 Wp12=(4)2(12)=192 Wc vg=152 VdXPdis=32+48+16+216+144+150+250+192+80(4)=1368 WXPdel=(152)(9)=1368 WTherefore,XPdis=XPdel20CHAPTER 1.Circuit Variables and Circuit ElementsP 1.34av2=180 100=80 Vi2=v28=10 Ai3+4=i2;i3=10 4=6 Av1=v2+v3=80+6(10)=140 Vi1=v170=14070=2 Abp5=82(5)=320 Wp25=(4)2(25)=400 Wp70=22(70)=280 Wp10=62(10)=360 Wp8=102(8)=800 WcXPdis=320+400+280+360+800=2160 WPdev=180ig=180(12)=2160 WP 1.35aProblems21b v=20 V;i=10 mA;R=vi=2 kc 2i1=3is;i1=1:5is40=i1+is=2:5is;is=16 mAd vs(open circuit)=(40 103)(2 103)=80 Ve vs(open circuit)=55 Vf Linear model cannot predict the nonlinear behavior of the practicalcurrent source.P 1.36a Plot the v i characteristicFrom the plot:R=vi=(125 50)(15 0)=5When it=0,vt=50 V;therefore the ideal current source has a currentof 10 A22CHAPTER 1.Circuit Variables and Circuit Elementsb10+it=i1and5i1=20itTherefore,10+it=4itso it=2 AP 1.37ab R=24 1824 0=624=0:25c i=241:25=19:2 A;v=24 19:2(0:25)=19:2 Vd isc=240:25=96 Ae isc=48 A(from graph)f Linear model cannot predict nonlinear behavior of voltage source.Problems23P 1.38a Plot the v|i characteristic:From the plot:R=vi=(420 100)(16 0)=20When it=0,vt=100 V;therefore the ideal voltage source has a voltageof 100 Vbit=100=(20+20)=2:5 A;Therefore,p20=(2:5)2(20)=125 WP 1.39avb=5(20+12)=160 Vvb+va=250 V,so va=90 Vib=90=(20+10)=3 A24CHAPTER 1.Circuit Variables and Circuit Elementsid=5 ib=2 Avc=vb+10(id)=180 Vvd=250 vc=70 V=14(ia);therefore,ia=5 Aic=ia id=5 2=3 AR=vc=ic=180=3=60b ig=5+3=8 Apg(supplied)=(250)(8)=2000 WP 1.40vab=240 180=60 V;therefore,ie=60=15=4 Aic=ie 1=4 1=3 A;therefore,vbc=10ic=30 Vvcd=180 vbc=180 30=150 V;therefore,id=vcd=(12+18)=150=30=5 Aib=id ic=5 3=2 Avac=vab+vbc=60+30=90 VR=vac=ib=90=2=45CHECK:ig=ib+ie=2+4=6 Apdev=(240)(6)=1440 WXPdis=1(180)+4(45)+9(10)+25(12)+25(18)+16(15)=1440 W(CHECKS)Problems25P 1.41a 15:2=10;000i 0:80+(200)30i16=(16;000)ii=1 mA200(30i)+vy+500(29i)25=0vy=25 6000i 14;500iTherefore,vy=4:5 VbXPgen=15:2i+25(29)i+0:8i=741i=741 mWXPdis=104(i)2+200(30i)2+29i(4:5)+500(29i)2=741 mW.P 1.42a i2=0 because no current can exist in a single conductor connecting twoparts of a circuit.b60=6000igig=10 mAv=5000ig=50 V6 103v=300 mA2000i1=500io;so i1+4i1=300 mA;therefore,i1=60 mAc 300 60+i2=0,so io=240 mA.P 1.43a26CHAPTER 1.Circuit Variables and Circuit Elementsb