计算机电路基础作业1.已知R1=2Ω、R2=3Ω、R3=9Ω、R4=6Ω,求图中电路的等效电阻RAB?R1R2R3R4ABC电路图解:R1与R2并联,R1//R2=2*3/2+3=1.2ΩR3与R4并联,R3//R4=9*6/9+6=3.6ΩRAB=(R1//R2)+(R3//R4)=1.2+3.6=4.8Ω2.已知R1=R2=2Ω、R3=3Ω、R4=5Ω,US=2V,IS=2A,用戴维南定理求图所示电路中的电流I,并计算电阻R4所消耗的功率。IISUSR3R4R2R1解:电流源的输出电流不变,Is就是总电流:I=Is*R3/(R3+R4)=2*3/(3+5)=0.75A;电流方向由Is决定。P=I*I*R4=2*2*5=20W3.分析图所示的电路,已知R1=R2=8Ω,R3=3Ω,R4=6Ω,US=18V,求电流I、I1、I2、I3、I4、I5?解:R总=R1/R2+R3/R4=1.5ΩI=Us/R总=12AI1=I*R2/(R1+R2)=6AI2=I-I1=6AI3=I*R4/(R3+R4)=8AI4=I-I3=4AI5=I1-I3=2AUSIR1R2R3R4I1I2I5I3I4ABC4.如图所示电路,已知:Us=9V,R1=3Ω,R2=6Ω,L=1H。在t=0时换路,即开关由1位置合到2位置,设换路前电路已经稳定,求换路后的初始值)0(),0(),0(21Luii。解:iL(0)=Us/R1=9/3=3Ai1(0)=iL(0)xR2/(R1+R2)=3x6/9=2Ai2(0)=iL(0)xR1/(R1+R2)=3x3/9=1AUL(0)=i2(0)R2=1x6=6VUsR1R212Si1i2iLLUL