解:(1)求短路电流lC a R=(R1R3)+(R2/R4) =5.89 SC R 3 12 4 2.07A b R58 x Us 1,=R31=10×2,07=1.38A R,+R,10+5 2=l4=I=1.035A SC=11-12 或:Isc=I4-l3 =1.38-1.035=0.345A 上一页下 一页上一页 下一页 解: (1) 求短路电流ISC 2 . 07A 5.8 S 12 = = = R U I 2 . 07 1 . 38 A 10 5 10 1 3 3 1  = + = + = I R R R I R =(R1 //R3 ) +( R2 //R4 ) = 5. 8 US a b + – I1 I4 ISC I3 I2 I 1 . 035 A 2 1 I2 = I4 = I = ISC = I1 – I2 =1. 38 – 1.035=0. 345A 或：ISC = I4 – I3