电路分析基础(英文版)课后答案第三章.pdf

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1、3Techniques of Circuit AnalysisDrill ExercisesDE 3.1 a 11,8 resistors,2 independent sources,1 dependent sourceb 9c 9,R4 R5forms an essential branch as does R8 10 V.The remainingseven branches contain a single element.d 7e 6f 4g 6DE 3.2 Solution given in text.DE 3.3 Solution given in text.DE 3.4 Solu

2、tion given in text.DE 3.5 a The two node voltage equations are15+v160+v115+v1 v25=05+v22+v2 v15=0Solving,v1=60 V and v2=10 V;Therefore,i1=(v1 v2)=5=10 Ab p15A(del)=(15)(60)=900 Wc p5A=5(10)=50 WDE 3.6 Use the lower node as the reference node.Let v1=node voltage across 1 resistor and v2=node voltage

3、across 12 resistor.Thenv11+v1 v28=4:55354CHAPTER 3.Techniques of Circuit Analysisv212+v2 v18+v2 304=0Solving,v1=6 Vv2=18 V Thus,i=(v1 v2)=8=1:5 Av=v2+2i=15 VDE 3.7 Use the lower node as the reference node.Let v1=node voltage across the 8 resistor,let v2=node voltage across the 4 resistor.Thenv1 506+

4、v18+v1 v22 3i1=05+v24+v2 v12+3i1=0i1=50 v16Solving,v1=32 V;v2=16 V;i1=3 A p50V=50i1=150 W(delivering)p5A=5(v2)=80 W(delivering)p3i1=3i1(v2 v1)=144 W(delivering)DE 3.8 Use the lower node as the reference node.Let v1=node voltage across the 7.5 resistor and v2=node voltage across the 2.5 resistor.Plac

5、e thedependent voltage source inside a supernode between the node voltages v andv2.The node voltage equations arenode 1:v17:5+v1 v2:5=4:8supernode:v v12:5+v10+v22:5+v2 121=0We also have:v+ix=v2and ix=v1=7:5.Solving this set of equations for vgives v=8 VDE 3.9v1 602+v124+v1(60+6i)3=0;i=60+6i v13There

6、forev1=48 VDE 3.10vo40+vo 1010+vo+20i20=0;i=10 vo10+10+20i30Thereforevo=24 VProblems55DE 3.11 Dene three clockwise mesh currents i1,i2,and i3in the lower left,upper,andlower right windows.The three mesh-current equations are80=31i1 5i2 26i30=5i1+125i2 90i30=26i1 90i2+124i3a Solving,i1=5 A;therefore

7、the 80 V source is delivering 400 W to thecircuit.b Solving,i3=2:5 A;therefore p8=(6:25)(8)=50 WDE 3.12 a b=8,n=6,b n+1=3b Dene three clockwise mesh currents i1,i2,and i3in the upper,lower left,and lower right windows.The three mesh-current equations are(3v)+19i1 2i2 3i3=025 10=2i1+7i2 5i310=3i1 5i2

8、+9i3We also havev=3(i3 i1)Solving for i1and i3givesi1=1 A,i3=3 A Thereforev=12Vandp3v=(3v)i1=36 WDE 3.13 Let ia=lower left mesh current cw,let ib=upper mesh current cw,and ic=lower right mesh current cw.Then25=14ia 6ib 8ic0=6ia+16ib 8ic0=8ia 8ib+16ic+5ii=ia;ia=4 A;ic=2Avo=8(ia ic)=16 VDE 3.14Mesh 1:

9、30=11i1 2i2Mesh 2:0=2i1+19i2 5i356CHAPTER 3.Techniques of Circuit AnalysisCurrent source:i3=16 ASolution givesi1=2 A;i2=4 A;i3=16 AThe current in the 2 resistor isi1 i2=6 A:P2=(6)2(2)=72 WDE 3.15Mesh a:7ia 2ib 5ic=75Current sources:ib=10 A;ic=2v5Dependent variable:v=5(ia ic)Solution:ia=15 A;ib=10 A;

10、ic=10 A;v=25 VDE 3.16Supermesh a,b:2ia+4ib 4ic=10Mesh c:2ia 2ib+5ic=6Current source:ia ib=2 ASolution:ia=7 A;ib=5 A;ic=6 A:p1=i2c(1)=(6)2(1)=36 WProblems57DE 3.17 Let v1denote the voltage across the 2 A source.Let v1be a voltage rise in thedirection of the 2 A current.v1 2015 2+v1 2510=0;v1=35 Vp2A=

11、35(2)=70 Wp2A(del)=70 WDE 3.18Mesh 1:12i1 6i2 4i3=128Mesh 2:6i1+14i2 3i3+30ix=0Current source:i3=4 ADependent variable:ix=i1 i3Solution:i1=9 A;i2=6 A;i3=4 A;ix=5 A:v4A=3(i3 i2)4ix=10 VThe power delivered by the 4A source isp4A=(10)(4)=40 WDE 3.19 To nd the Th evenin resistance,deactivate the indepen

12、dent voltage sourceand note that RTh=5k20+8k12=6.With the terminals a,b open,thecurrent delivered by the 72 V source is 72/24 or 3 A.The current(left-to-right)in the 5 resistor is(20=25)(3)=2:4 A,and the current(left-to-right)in the 12 resistor is(5/25)3 or 0:6 A.The Th evenin voltagevTh=vabis the d

13、rop across the 8 resistor plus the drop across the 20 resistor.Thus vTh=(8)(0:6)+(20)(3)=64:8 V.58CHAPTER 3.Techniques of Circuit AnalysisDE 3.20 After one source transformation,the circuit becomesThereforeIN=6 A,RN=20k12=7:5DE 3.21 Find the Th evenin equivalent with respect to A,B.VTh+3612;000+VTh6

14、0;000 0:018=0;VTh=150 VRTh=15;000+(60;000)(12;000)72;000=25 k;Therefore,vmeas=150100;000125;000!=120 VDE 3.22 Summing the currents away from node a,where vTh=vabWe havevTh8+4+3ix+vTh 242=0;ix=vTh8Solving for vThyields vTh=8 ViT=4ix+vT=2;ix=vT=8ThereforeiT=vTandRTh=vT=iT=1Problems59DE 3.23 Use the bo

15、ttom node as the reference.Let v1be the node voltage across the60 resistor.Thenv160+v1(vTh+160i)20 4=0;vTh40+vTh80+vTh+160i v120=0i=vTh40;thereforevTh=30 VLet iTbe the test current into terminal a:iT=vT80+vT40+vT+160(vT=40)80;iTvT=110Therefore,RTh=10DE 3.24 First nd the Th evenin equivalent circuit.

16、To nd vTh,use the bottom nodeas the reference.Let vTh=vaband v1=node voltage across the 20 V 4 branch.The two node Voltage equations arevTh 100 v4+vTh v14=0;(v=v1 20)v1 1004+v1 204+v1 vTh4=0Solving for vThgives vTh=120 V.To nd RTh,deactivate the twoindependent sources and apply a test voltage source

17、 across a,b.Let vTbepositive at a and iTdirected into a.Then the two node Voltage equations arevT v4+vTh v4=iT;v4+v4+v vT4=0Thereforev=vT=3and12iT=4vTSoRTh=vT=iT=3a For maximum power transfer,RL=RTh=3b pmax=(120=6)2(3)=1200 W60CHAPTER 3.Techniques of Circuit AnalysisDE 3.25 When RL=3,the voltage acr

18、oss RLis 60 V.As before,let v1be the nodevoltage across the 20 V|4 branch,then v=v1 20 and603+60 v14+60 100 v4=0Therefore v1=60 V and v=40 V.The current out of the plus terminal ofthe 100 V source isi1=100 604+100+40 604=10+20=30 Aa Therefore 100 V is delivering 3000 W to the circuit.b The current o

19、ut of the plus terminal of the dependent source is 20 A.Therefore the dependent source is delivering 800 W to the circuit.c The load power is(1200=3800)100 or 31:58%of this generated power.ProblemsP 3.1 a Fiveb ThreecSum the currents at any three of the four essential nodes a,b,c,and d.Using nodes a

20、,b,and c we getig+i1+i2=0i1+i4+i3=0i5 i2 i3=0Problems61d Two.e Sum the voltages around two independent closed paths,avoiding a paththat contains the independent current source since the voltage across thecurrent source is not known.Using the upper and lower meshes formedby the ve resistors givesR1i1

21、+R3i3 R2i2=0R3i3+R5i5 R4i4=0P 3.22+vo4+vo 555=0vo=20 Vp2A=(20)(2)=40 W(absorbing)P 3.3 Let v2be the node voltage across the 80 resistor,positive at the upperterminal.Then 4+v120+v280+v240=0(Note we have created a super node in writing this expression.)v1+60=v2:v1=20 V:v2=80 Vpdel=60igwhere igis the

22、current out of the positive terminal4=ig+v120;ig=3 A:pdel=60(3)=180 W62CHAPTER 3.Techniques of Circuit AnalysisP 3.4 av148+v1 1288+v1 v218=0v220+v2 v118+v2 7010=0Solving,v1=96 V;v2=60 Via=128 968=4 Aib=9648=2 Aic=96 6018=2 Aid=6020=3 Aie=60 7010=1 Ab pdev=128(4)+70(1)=582 WP 3.5 Use the lower termin

23、al of the 5 resistor as the reference node.vo 6010+vo5+3=0Solving,vo=10 VP 3.6 aProblems63v1 1102+v1 v28+v1 v316=0v2 v18+v23+v2 v324=0v3+1102+v3 v224+v3 v116=0Solving,v1=74:64 V;v2=11:79 V;v3=82:5 VThus,i1=110 v12=17:68 Ai4=v1 v28=7:86 Ai2=v23=3:93 Ai5=v2 v324=3:93 Ai3=v3+1102=13:75 Ai6=v1 v316=9:82

24、 AbXPdev=110i1+110i3=3457:14 WXPdis=i21(2)+i22(3)+i23(2)+i24(8)+i25(24)+i26(16)=3457:14 WP 3.7 2:4+v1125+v1 v225=0v2 v125+v2250+v2375 3:2=0Solving,v1=25 V;v2=90 VCHECK:p125=(25)2125=5 Wp25=(90 25)225=169 Wp250=(90)2250=32:4 Wp375=(90)2375=21:6 Wp2:4A=(25)(2:4)=60 WXpabs=5+169+32:4+21:6+60=288 WXpdev

25、=(90)(3:2)=288 W(CHECKS)64CHAPTER 3.Techniques of Circuit AnalysisP 3.8 av150+v1 6405+v1 v22:5=0v2 v12:5+v2 v35+12:8=0v32:5+v3 v25 12:8=0Solving,v1=380 V;v2=269 V;v3=111 V;b ig=640 3805=52 Apg(del)=(640)(52)=33;280 WP 3.9v1 1444+v110+v1 v280=0so29v1 v2=28803+v2 v180+v25=0sov1+17v2=240Solving,v1=100

26、V;v2=20 VP 3.10v1 7520+v150+v1 v240=0Problems65v2 v140+v2 75800 6+v2200=0Solving,v1=115 V;v2=287 V:vo=115 75=40 VP 3.11va=4 V7+vb3+vb vc1=02vx+vc vb1+vc va2=0vx=vc va=vc 4Solving,vo=vb=1:5 VP 3.12v1(v2+30)15+v1 v231:25+v125 4=0v1(v2+30)15#+v2 v350+v2 v131:25=066CHAPTER 3.Techniques of Circuit Analys

27、isv3 v250+v350+1=0Solving,v1=76 V;v2=46 V;v3=2 V;i30V=0 Ap4A=4v1=4(76)=304 W(del)p1A=(1)(2)=2 W(del)p30V=(30)(0)=0 Wp15=(0)2(15)=0 Wp25=v2125=76225=231:04 Wp31:25=(v1 v2)231:25=30231:25=28:8 Wp50(lower)=(v2 v3)250=48250=46:08 Wp50(right)=v2350=450=0:08 WXpdiss=0+231:04+28:8+46:8+0:08=306 WXpdev=304+

28、2=306 W(CHECKS)P 3.13v170+v1 v210+v1 805=0v212+v2 v110+v2 805=0Solving,v1=70 V;v2=60 VThus,io=v1 v210=1 AProblems67P 3.14 av0 6010+vo5+3=0;vo=10 Vb Let vx=voltage drop across 3 A sourcevx=vo(100)(3)=290 Vp3A(developed)=(3)(290)=870 Wc Let ig=current into positive terminal of 60 V sourceig=(10 60)=10

29、=5 Ap60V(developed)=(5)(60)=300 WdXpdis=(5)2(10)+(3)2(100)+(10)2=5=1170 WXpdis=300+870=1170 We vois independent of any nite resistance connected in series with the 3 Acurrent sourceP 3.15 a From the solution to Problem 3.5 we know vo=10 V,thereforep3A=3vo=30 W:p3A(developed)=30 Wb The current into t

30、he negative terminal of the 60 V source isig=60 1010=5 Ap60V=60(5)=300 W:p60V(developed)=300 Wc p10=(5)2(10)=250 Wp5=(10)2=5=20 WXpdev=300 WXpdis=250+20+30=300 WP 3.16 av1 2025 103+v10:25 103+11 103+v1+100:5 103=068CHAPTER 3.Techniques of Circuit Analysisv1=5 Vi1=20+525;000=1 mAi2=v1250=5250=20 mAi5

31、=10+5500=10 mAi4=101000=10 mAi4+i3 11+i5=0:i3=11 i4 i5=11+10+10=31 mAb p20V=20i1=20(1 103)=20 mWp10V=10i3=10(31 103)=310 mWv11mA+v1=10;v11mA=10+5=5 Vp11mA=11v11mA=55 mW(del)Xpdev=20+310=330 mWp25k=25 103i21=25 mWp0:25k=0:25 103i22=100 mWp0:5k=0:5 103i25=50 mWp1k=1 103i24=100 mWXpdiss=25+100+50+100+5

32、5=330 mWXpdiss=Xpdev=330 mWP 3.17 aProblems69v2 5004+v2 v42+v2 v33=0v3 v23+v36+v3 v52=0v4 v22+v4 50011+v4 v54=0v5 v32+v54+v5 v44=0Solving,v2=300 V;v3=180 V;v4=280 V;v5=160 Vi5=500 v411=500 28011=20 Ap5=(20)2(5)=2000 Wbi500V=v1 v24+v1 v411=500 3004+500 28011=50+20=70 Ap500V=35;000 WCheck:XPdis=(50)2(

33、4)+(40)2(3)+(30)2(6)+(20)2(11)+(10)2(2)+(30)2(4)+(10)2(2)+(40)2(4)=35;000 WP 3.18 avo v1R+vo v2R+vo v3R+vo vnR=0:nvo=v1+v2+v3+vn:vo=1nv1+v2+v3+vn=1nXnk=1vkb vo=13(150+200 50)=100 V70CHAPTER 3.Techniques of Circuit AnalysisP 3.19 Place v=5 inside a supernode and use the lower node as a reference.Then

34、v1 5010+v130+v1 v=539+v1 v=578=0134v1 6v=3900;v=50 v1Solving,v1=30 V;v=20 V;vo=30 v=5=30 4=26 VP 3.20vo 805+vo50+vo+75i25=0;i=vo50Solving,vo=50 V;i=1 Aio=50 (75)(1)25=5 Ap75i=75iio=375 W:The dependent voltage source delivers 375 W to the circuit.P 3.21 3+vo200+vo+5i10+vo 8020=0;i=vo 8020a Solving,vo

35、=50 Vb ids=vo+5i10i=(50 80)=20=1:5 A:ids=4:25 A;5i=7:5 V:pds=(5i)(ids)=31:875 Wc p3A=3vo=3(50)=150 W(del)p80V=80i=80(1:5)=120 W(del)Xpdel=150+120=270 WCHECK:p200=2500=200=12:5 WProblems71p20=(80 50)2=20=900=20=45 Wp10=(4:25)2=10=180:625 WXpdiss=31:875+180:625+12:5+45=270 WP 3.22 aio=v2 v3502io+v1100

36、+v1 v225=0v2 v125+v2200+v2 v350v3 v250+v3 5io5+v3 38:520=0Solving,v1=50 V;v2=30 V;v3=2:5 Vb io=v2 v350=30 2:550=0:65 Ai3=v3 5io5=2:5 5(0:65)5=1:15 Aig=38:5 2:520=1:8 AXpdis=XpdevCalculateXpdevbecause we dont know if the dependent sources aredeveloping or absorbing power.Likewise for the independent

37、source.p2io=2iov1=2(0:65)(50)=65 W(dev)p5io=5ioi3=5(0:65)(1:15)=3:7375 W(dev)pg=38:5(1:8)=69:30 W(dev)Xpdev=69:3+65+3:7375=138:0375 W72CHAPTER 3.Techniques of Circuit AnalysisCHECKXpdis=2500100+900200+40025+(0:65)2(50)+(1:15)25+(1:8)2(20)=138:0375 W:Xpdev=Xpdis=138:0375 WP 3.23 a 5+v115+v1 v25=0v2 v

38、15+v230+v210+v2+5i30=0i=v1 v25Solving,v1=30 V;v2=15 V;i=3 A;io=15+1530=1 Ap5i=(15)(1)=15 W(del)p5A=5(30)=150 W(del):pdev=165 WbXpabs=(30)215+(15)230+(15)210+(3)2(5)+(1)2(30)=165 W:Xpdev=Xpabs=165 WP 3.24 i=v3 v44=235 2224=3:25 A30i=30(3:25)=97:5 Vv1+30i=v4v1=v4 30i=222 97:5=124:5 Vv3+v=250:v=250 235

39、=15 V3:2v=(3:2)(15)=48 Aig=250 124:52+250 2351=77:75 AProblems73p250V=250ig=250(77:75)=19;437:5 W(del)i30i i+v4=40+48=0i30i=i 222=40 48=3:25 5:55 48=50:3 Ap30i=(30i)i30i=(97:5)(50:3)=4904:25 W(dev)p3:2v=(3:2v)(v4)=(48)(22)=10;656 W(abs):Xpdev=19;437:5+4904:25=24;341:75 Wp10=v2110=(124:5)210=1550:025

40、 Wp2=(250 124:5)22=7875:125 Wp1=(250 235)21=225 Wp20=(235)220=2761:25 Wp4=(3:25)2(4)=42:25 Wp40=(222)240=1232:10 W:Xpdiss=10;656+1550:025+7875:125+225+2761:250+42:25+1232:1=24;341:75 WThus,Xpdev=Xpdiss;Agree with analystP 3.2574CHAPTER 3.Techniques of Circuit AnalysisNode equations:v120+v1 202+v3 v2

41、4+v380+3:125v=0v240+v2 v34+v2 201=0Constraint equations:v=20 v2v1 35i=v3i=v2=40Solving,v1=20:25 V;v2=10 V;v3=29 VLet igbe the current delivered by the 20 V source,thenig=20 (20:25)2+20 101=30:125 Apg(delivered)=20(30:125)=602:5 WP 3.2618+3i1+9i2 15+6i2+2i1=0;i2 i1=3Solving,i1=0:6 A;i2=2:4 Ap18V=18i1

42、=10:8 W(diss)p3=(0:6)2(3)=1:08 Wp2=(0:6)2(2)=0:72 Wp9=(2:4)2(9)=51:84 WProblems75p6=(2:4)2(6)=34:56 WXpdiss=99 Wvo=15i2 15=36 15=21 Vp3A=3vo=63 W(dev)p15V=15i2=36 W(dev)Xpdev=99 W=XpdissP 3.27100+5i1+15i2 15=05i1+15i2=115i2 i1=3;i2=i1+3;15i2=15i1+45:20i1=70i1=3:5 A;i2=6:5 Avo=15i2 15=97:5 15=82:5 Vp

43、100V=100i1=350 W(dev)p3A=3vo=247:5 W(dev)p15V=15i2=97:5 W(dev)Xpdev=Xpdis=695 WXpdis=(3:5)2(5)+(6:5)2(15)=695 W76CHAPTER 3.Techniques of Circuit AnalysisP 3.28 a Summing around the supermesh used in the solution to Problem 3.27 gives(10)+5i1+15i2 15=0i2=i1+3:i1=2 A;i2=1 Ap10V=10(2)=20 W(del)vo=15i2

44、15=0 Vp3A=3vo=0 Wp15V=15i2=15 W(del)Xpdiss=(2)2(5)+(1)2(15)=35 WXpdev=35 W=Xpdissb With 3 A current source replaced with a short circuiti1=2 A;i2=1 A:XPdiss=(2)2(5)+(1)2(15)=35 Wc A 3 A source with zero terminal voltage is equivalent to a short circuitcarrying 3 A.P 3.29 a40=50i1 45i264=45i1+50:5i2S

45、olving,i1=9:8 A;i2=10 Aia=i1=9:8 A;ib=i1 i2=0:2 A;ic=i2=10 Ab If the polarity of the 64 V source is reversed,we have40=50i1 45i264=45i1+50:5i2i1=1:72 Aandi2=2:8 Aia=i1=1:72 A;ib=i1 i2=1:08 A;ic=i2=2:8 AProblems77P 3.30600=25:6i1 16i2 5:6i3424=16i1+20i2 0:8i330=i3Solving,i1=35 A;i2=8 A;i3=30 Aa v30A=

46、0:8(i2 i3)+5:6(i1 i3)=0:8(8 30)+5:6(35 30)=10:4 Vp30A=30v30A=30(10:4)=312 W(abs)Therefore,the 30 A source delivers 312 W.b p600V=600(35)=21;000 W(del)p424V=424(8)=3392 W(abs)Therefore,the total power delivered is 21;000 Wc p4=(35)2(4)=4900 Wp3:2=(8)2(3:2)=204:8 Wp16=(35 8)2(16)=11;664 Wp5:6=(35 30)2

47、(5:6)=140 Wp0:8=(30+8)2(0:8)=387:2 WXpresistors=17;296 WXpabs=17;296+312+3392=21;000 W(CHECKS)78CHAPTER 3.Techniques of Circuit AnalysisP 3.31 a110+12=17i1 10i2 3i30=10i1+28i2 12i312 70=3i1 12i2+17i3Solving,i1=8 A;i2=2 A;i3=2 Ap110=110i1=880 W(del)p12=12(i1 i3)=120 W(del)p70=70i3=140 W(del):Xpdev=11

48、40 Wb p4=(8)2(4)=256 Wp10=(6)2(10)=360 Wp12=(4)2(12)=192 Wp2=(2)2(2)=8 Wp6=(2)2(6)=24 Wp3=(10)2(3)=300 W:Xpabs=1140 WP 3.32 a2400(i1 0:0025)+1500i1 150(i1 0:0025)=0Problems79:i1=1:5 mAi=i1 2:5=1:0 mAb vo=(0:0025)(400)+(0:001)(2400)=3:4 Vp2:5mA=3:4(2:5)=8:5 mW:p2:5mA(deliver)=8:5 mWc 150i=150(1:0 103

49、)=0:15 Vpdep source=150ii1=(0:15)(0:0015)=0:225 mWpdep source(absorbed)=0:225 mWP 3.33Mesh equations:25i1 5i2 2:5i3=075=5i1+12:5i2 7:5i3Constraint equations:i3=0:2vv=5(i2 i1)Solving,i1=3:6 A;i2=13:2 A;i3=9:6 A;v=48 Vvcs=125 v 2:5(i3 i1)=125 48 2:5(9:6 3:6)=62 Vpvc=(62)(9:6)=595:2 W(abs)p50V=50(i2 i3

50、)=50(13:2 9:6)=180 W(abs)80CHAPTER 3.Techniques of Circuit Analysisp125V=125i2=125(13:2)=1650 W(del)Thus,the total power developed is 1650 W.CHECK:p17:5=(3:6)2(17:5)=226:8 Wp5=(13:2 3:6)2(5)=460:8 Wp2:5=(9:6 3:6)2(2:5)=90 Wp7:5=(13:2 9:6)2(7:5)=97:2 W:Xpabs=226:8+460:8+90+97:2+180+595:2=1650 WP 3.34