节点电压法求各支路电流。

核心提示如图,a,b,c对应为节点1,2,3,Va=10vb: (1+0.5)Vb - 0.5Va=-i5,Va-3Vb=2i5.......(1)c: (1+0.5)Vc -Va=i5,3Vc-2Va=2i5.........(2)补: Vc-V

如图,a,b,c对应为节点1,2,3,Va=10v

b: (1+0.5)Vb - 0.5Va=-i5,Va-3Vb=2i5.......(1)

c: (1+0.5)Vc -Va=i5,3Vc-2Va=2i5.........(2)

补: Vc-Vb=5............(3)

解(1)(2) 3Vc+3Vb=3Va,Vc+Vb=10.....(4)

解(3)(4) 2Vb=5,Vb=2.5v,Vc=7.5v;

i4=(Va-Vc)=2.5A,i6=0.5(Va-Vb)=3.75A,i1=i4+i6=6.25A,i3=0.5Vc=3.75A

i2=Vb=2.5A,i5=i6-i2=1.25A。

 
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