题目

电阻星型连接与三角形连接的等效变换
U12=i1R1-i2R2,U23=i2R2-i3R3,i1+i2+i3=0
怎么变形计算得出 i1=(R3U12-R2U31)/(R1R2+R2R3+R3R1),我需要演算过程,自己算了半天也算不出来,

提问时间：2020-06-25

答案

I1=(U12+I2R2)/R1 ------1
I2=-(I1+I3) --------2
I3=(U31+I1R1)/R3 -------3
将2、3带入1
I1={U12-[I1+(U31+I1R1)/R3]R2}/R1
=(U12R3-I1R2R3-U31R2-I1R1R2)/R1R3
移向
I1+I1R3R2/R1R3+I1R1R2/R1R3=(U12R3-U31R2)/R1R3
I1(R1R2+R2R3+R1R3)/R1R3=(U12R3-U31R2)/R1R3
I1=(R3U12-R2U31)/(R1R2+R2R3+R3R1)