怎么求这个电路的Isc或Uoc呢？

将电源移至 1、1'；从左向右，6Ω电阻两端a1、a2等效电阻Ra = 9//6 = 9 * 6/(9+6) = 18/5Ω ；5Ω电阻两端b1、b2等效电阻 Rb = (Ra+7)//5 = (53/5) * 5/(53/5+5) = 265/78Ω；Ub = 5 * Rb/(Rb+2) = 5 * (265/78)/(265/78 + 2 ) = 1325/421 v；Ua = Ub * Ra/(Ra+7) = (1325/421) * (18/5)/(18/5+7) = 450/421 v；9Ω电阻电流I =Ua/9 = 50/421 A；由互易定理，电源在左侧，则Isc = I = 50/421 A；电源短路，1、1'端等效电阻r = (Rb+2)//10= (265/78+2)//10 = (421/78) * 10/(421/78+10) = 4210/1201 Ω；Uoc = Isc * r = (50/421) * (4210/1201) = 500/1201v 。