怎样用回路电流法求I?

如题所述

10i1=i-i2.........(1)

(2+3+5)i1-3i2-5i=0,10i1-3i2=5i,将(1)代入,i=-i2.........(2)

12=6i+2i1+4i2,将(2)代入,6=i1-i2........(3)

u=(5+6)i-5i1=12-(3+4)i2+3i1,12=11i-8i1+7i2=-8i1-4i2,-3=2i1+i2.........(4)

(3)+(4)          3=3i1,i1=1A,i2=-5A,i=5A。

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