ChIsl0-2 网络拓扑的基本概念 RI 29 +11 12 us +us RI 10V u2/R2 8 10Vu12gu2巾R2 89 分析图(a),(b)中的u1,i1,u2,i2? 10 =1(4)u4=l2=ly=10() R1+R,2+8 l4=R=2×1=2) i=n==5() l2=Ri2=8×1=8() =0=125(4) R28Ch1s10-2 分析图(a), (b)中的u1, i1, u2, i2? (A) R R u i i S 1 2 8 10 1 2 1 2 = + = + = = u u u (V ) 1 = 2 = s =10 (a) (b) 一. 网络拓扑的基本概念 u Ri 2 1 2(V) 1 = 1 1 =  = u R i 8 1 8(V) 2 = 2 2 =  = (A) R U i 5 2 10 1 1 1 = = = (A) R U i 1.25 8 10 2 2 2 = = =