《電路分析基礎(chǔ)》第2版

1、電路分析基礎(chǔ)各章習(xí)題參考答案第1章習(xí)題參考答案1(1)50W;(2)300V、25V,200V、75V;(3)2=12.5QRa=100Q,R4=37.5Q1- 2VA=8.5V，Vm=6.5V，VB=0.5V，VC=-12V，VD=-19V，Vp=-21.5V，UAB=8V，UBC=12.5，UDA=-27.5V1- 3電源(產(chǎn)生功率)：A、B元件；負(fù)載(吸收功率)：C、D元件；電路滿足功率平衡條件。1- 4(1)VA=100V,VB=99V,VC=97V,VD=7V,VE=5V,VF=1V,UAF=99V,UCE=92V,UBE=94V,UBF=98V,UCA=-3V；(2)VC=90V,

2、VB=92V,VA=93V,VE=-2V,VF=-6V,VG=-7V,UAF=99V,UCE=92V,UBE=94V,UBF=98V,UCA=-3V1- 5I0.18A,6度，2.7元6I=4A，I1=11A，I2=19A1- 7(a)U=6V,(b)U=24V,(c)R=5Q,(d)I=23.5A8(1)i6=-1A;(2)U4=10V,U6=3V;(3)P1=-2W發(fā)出，P2=6W吸收，P3=16W吸收，P4=-10W發(fā)出，P5=-7W發(fā)出，P6=-3W發(fā)出9I=1A,Us=134V,R7.8Q1- 10S斷開：Uab=-4.8V,Uao=-12V,Ubo=-7.2V;S閉合：Uab=-1

3、2V,Uao=-12V,Ubo=0V11支路3，節(jié)點(diǎn)2，網(wǎng)孔2，回路31- 12節(jié)點(diǎn)電流方程：(A)I1+I3-I6=0，(B)I6-I5-I7=0，(C)I5+I4-I3=0回路電壓方程：I6R6+US5+I5R5-Us3+13R3=0,-15R5-Us5+I7R7-Us4=0,-丨3R3+US3+US4+I1R2+I1R1=01-13Uab=11V,I2=0.5A,l3=4.5A,R32.4Q1- 14VA=60VVC=140VVD=90VUAC=-80VUAD=-30VUCD=50V15I1=-2AI2=3AI3=-5AI4=7AI5=2A第2章習(xí)題參考答案12.4Q5A2- 2(1)4

4、V2V1V;(2)40mA20mA10mA31.5Q2A1/3A1- 46Q36Q52A1A1- 61A72A2- 81A9I1=-1.4AI2=1.6AI3=0.2A2- 10I1=0AI2=-3AP1=0WP2=-18W11Ii=-1mA,I2=-2mA,E3=10V2- 12I1=6A,I2=-3A,I3=3A13I1=2A,I2=1A,I3=1A,I4=2A,I5=1A1- 14Va=12V,I1=-1A,I2=2A2-15Va=6V,I1=1.5A,I2=-1A,I3=0.5A2-16Va=15V,I1=-1A,I2=2A,I3=3A2-17I1=-1A,I2=2A2-18I1=1.

5、5A,I2=-1A,I3=0.5A2-19I1=0.8A,I2=-0.75A,I3=2A,I4=-2.75A,I5=1.55A2-20I3=0.5A2-21Uo=2V,Ro=4Q,I0=0.1A2-22I5=-1A2-23(1)I5=0A,Uab=0V;(2)I5=1A,Uab=11V2-24Il=2A2-25Is=11A,R0=2Q2-2618Q,-2Q,12Q2-27U=5V2-28I=1A2-29U=5V2-30I=1A2-3110V,180Q2- 32U0=9V,R0=6Q,U=15V第3章習(xí)題參考答案3- 150Hz,314rad/s,0.02s,141V,100V,120°

6、;2200V,141.4V2- 3u=14.1sin(314t-60°V4(1)®u1-贏2=120°(2)®1=-90°%=-210°,%1-屁=120°(不變)2- 5(1)U1=50._90V,U2=50一2.0V;(2)U3=1002sin(3t+45°)V,U4=100-2sin(®t+135°)V3- 6(1)i1=14.1sin(72°)A;(2)U2=300sin(360°)V7錯(cuò)誤：(1),，(4),(5)3- 8(1)R;(2)L;(3)C;(4)R9i=

7、2.82sin(10t-30°)A,Q40var2- 10u=44.9sin(3141-135°V,Q=3.18var11(1)I=20A;(2)P=4.4kW2- 12(1)I1.4A,I1.4-30A；(3)Q308var,P=0W;(4)i0.98sin(628t-30°)A13(1)I=9.67A,I=9.67450A,i=13.7sin(314t+150°)A;(3)Q=2127.4var,P=0W;(4)Ic=0A14(1)C=20.3尸;(2)Il=0.25A,lc=16A第4章習(xí)題參考答案4-1(a)，Y=0.2/_36.87S;(b)Z

8、=2.52L45I】,3- 2Y=(0.06-j0.08)S,R16.67Q,Xl=12.5Q,L0.04H3UR=60£0V,UL=80/90V,US=10053.13V3- 4卩=20/36.87A5Z=1002/45；：,卩=10A,UR=1000V,UL=125/90V,UC=25/90V4-6Y=0.25245S,U=4,2£0V,Ir=.2.0A,Il=022£90A,Ic=12、290A7ll=10、一245AUS=100乙90V1- 8(a)30V;(b)2.24A9(a)10V;(b)10A1- 10(a)10V;(b)10V11U=14.1V3

9、- 12Ul1=15V,Uc2=8V,Us=15.65V4-13Ux1=100V,U2=600V,X1=10Q,X2=20Q,X3=30Q3- 14Z=20、2/45門，l=2/45A,人之；2乙0A,l2=$2/90A,Uab=0V15(1)I=.$2a,Zrc=5.2，Z=5、i0;(2)R=10門，X10'J2- 16P=774.4W,Q=580.8var,S=968VA-17l1=5A,l2=4A4-18I1=1A,I2=2A,l=一5.26.565A,4-19Z=10",I=190A,Ur2=5,2135V,P=10W64-20a=5X10rad/s,p=1000Q

10、,Q=100,l=2mA,Ur=20mV,Ul=Uc=2V44-2130=10rad/s,p=100Q,Q=100,U=10V,Ir=1mA,Il=lc=100mA4-22L1=1H,L20.33H第5章習(xí)題參考答案4- 3M=35.5mH4301=1000rad/s,302=2236rad/s3- 5Z1=j31.4Q,Z2=j6.28Q6Zr=3+7.5Q4- 7M=130mH8“2二-2/45A2- 9U1=44.8V10M12=20mH,11=4A2- 11U2=220V,I1=4A12n=1.92- 13N2=254匝，N3=72匝14n=10,P2=31.25mW第6章習(xí)題參考答案

11、1(1)A相燈泡電壓為零，B、C相各位為220V3- 3Il=Ip=4.4A,Up=220V,Ul=380V,P=2.3kW4(2)Ip=7.62A,Il=13.2A3- 5A、C相各為2.2A,B相為3.8A6Ul=404V6-7Uan=202/47V8cos$=0.961,Q=5.75kvar6-9Z=33428.4門10(1)Ip=11.26A,Z=19.53/42.3°Q;(2)Ip=Il=11.26A,P=5.5kW5- 11Ul=391V12iA=222sin(t53.13)AiB=22.2sin(-，t-173.13)AiC=222sin(t66.87)A4- 13Uv

12、=160V14(1)負(fù)載以三角形方式接入三相電源(2)Nab=3.82.-15A，1仁=3.82也135A，仁=3.8.2105AIa=3.8*6/45A，Ib=3.8iQ165A，I.=3.8675A15L=110mH，C=91.9mF第7章習(xí)題參考答案1P=240W，Q=360var3- 2P=10.84W7-3(1)i(t)=4.7sin(100)3sin37A(2)I3.94A,U58.84V,P93.02W0MUmnoL1r227-4u2(t)sin(tarctan1)V,zR(LJz2R5- 5直流電源中有交流，交流電源中無直流6U1=54.3V,R=1Q,L=11.4mH;約為8

13、%,(L'=12.33mH)6- 7使總阻抗或總導(dǎo)納為實(shí)數(shù)(虛部為0)的條件為尺=&=氏=.L/C7-8C1=9.39jjF,C2=75.13疔4- 9L1=1H,L2=66.7mH10C1=10譏C2=1.25忻第8章習(xí)題參考答案6iL(0+)=1.5mA,ul(0+)=-15V5- 7i1(0+)=4A,i2(0+)=1A,ul(0+)=2V,)=3A,i2()=0,ul()=08i1(0+)=75mA,i2(0+)=75mA,i3(0+)=0,ul1(0+)=0,ul2(0+)=2.25V8-98-108-11*8-128-138-148-158-168-178-188-198-20*8-21*8-22ic(t2e-10tA4tUL(t)=6e_VuC(t)=10(1_e40t)V,iC(t)=56說*人500t貝uC(t)=115esin(866亠60)V10t10tUL(t)=12e一V,L(t)=2(1e)A1tUR(t)=Use下2CV,ur(3J-USe-V(1)t=0.1s,(2)uc(t)=10e-0tV,(3)t=0.1suC(t)=10_9e01V10t_iL(t)=5e一A(a) f(t)