Question

如图电路所示,求电流I1,I2,I3,I4,I5,    要步骤,或提示一下方法

Answer 1

本题如用支路电流法,则需要列解六元一次方程组,手工很难完成.

用网孔电流法.
I2=I0-I1
I4=I0-I3
I5=I3-I1
只要在所画三个网孔列三个方程解出I0、I1、I3)即可
R1I1+R5(I1-I3)+R2(I1-I0)=0
R3I3+R4(I3-I0)+R5(I3-I1)=0
R2(I0-I1)+R4(I0-I3)=U
化简
(R1+R5+R2)I1-R5I3-R2I0=0
(R3+R4+R5)I3-R4I0-R5I1=0
(R2+R4)I0-R2I1-R4I3=U
化简
6I1-I3-2I0=0
8I3-5I0-I1=0
7I0-2I1-5I3=0.1
由第一式求出I3=6I1-2I0,代到第2、3式有
8(6I1-2I0)-5I0-I1=0
7I0-2I1-5(6I1-2I0)=0.1
化简
47I1-21I0=0
17I0-32I1=0.1
由第一式求出I1=0.4468I0,代到第2有
17I0-32×0.4468I0=0.1
2.702I0=0.1
I0=0.037A=37mA
I1=0.4468I0=16.5mA
I3=6I1-2I0=25.2mA
I2=I0-I1=20.5mA
I4=I0-I3=11.8mA
I5=I3-I1=8.7mA