Question

叠加定理求各支路电流，R1=5欧姆，R2=4欧姆，R3=5欧姆，R4=6欧姆，U=10伏，Is=2安培 。

Answer 1

U单独作用时，I6=0
I1=I3=U/(R1+R3)=1A;
I2=I4=U/(R2+R4)=1A;
I5=I1+I2=2A
Is单独作用时，I6=Is=2A
I2= - [Is*(R2*R4)/(R2+R4)]/R2= - 1.2A;
I4= [Is*(R2*R4)/(R2+R4)]/R4=0.8A;
I1= [Is*(R1*R3)/(R1+R3)]/R1=1A;
I3= - [Is*(R1*R3)/(R1+R3)]/R3= - 1A;
I5=I1+I2= - 0.2A
故：I1=2A, I2= - 0.2A, I3=0A, I4=1.8A, I5=1.8A, I6=2A