次ぎもMillmanの定理の応用問題。



図の回路の電流分布をMillmanの定理を用いて求めよというもの。

AB間の電圧Eabがわかれば、各枝電流は容易に計算できる。

EabはMillmanの定理を用いて

Eab=ΣYi*Ei/ΣYi
=(16/4+0/(10/3+1/(1/4+1/8))+9/3+26/2)/(1/4+1/(10/3+1/(1/4+1/8))+1/3+1/2)
=(4+0+3+13)/(1/4+1/(10/3+8/3)+1/3+1/2)
=20/(1/4+1/6+1/3+1/2)
=20*12/(3+2+4+6)
=20*12/15
=4*4
=16 [V]

従ってそれぞれの枝電流は

I1=(Eab-16)/3
=(16-16)/3
=0 [A]

I2=Eab/(10/3+1/(1/4+1/8))
=16/(10/3+8/3)
=16*3/18
=16/6
=8/3
=2.67 [A]

I3=I2*8/(4+8)
=(8/3)*8/12
=8*8/3*12
=8*2/3*3
=16/9
=1.78 [A]

I4=I2*4/(4+8)
=(8/3)*4/12
=8*4/3*12
=8/3*3
=8/9
=0.889 [A]

I6=(Eab-9)/3
=(16-9)/3
=7/3
=2.33 [A]

I7=(Eab-26)/2
=(16-26)/2
=-10/2
=-5 [A]

I5=I6+I7
=7/3-10/2
=(7*2-10*3)/6
=(14-30)/6
=-16/6
=-8/3
=-2.67 [A]

ということになる。