(1)求开路电压 U+U q 10×4/(4+6)+10×6/(4+6) 4+6=2V (2)求等效电阻Req Ra=4//6+6/14=4.89 q (3)Rx=1.29时, R R. I=Uoc/Reg +R)=0.333A R=5.2时, b 1=Uoc / Rea +R)=0.2Aa b + – 10V – + U2 + – U1 R I x I a b Uoc + – Rx Req (1) 求开路电压 Uoc = U1 + U2 = -104/(4+6)+10  6/(4+6) = -4+6=2V + Uoc _ (2) 求等效电阻Req Req=4//6+6//4=4.8 (3) Rx =1.2时， I= Uoc /(Req + Rx ) =0.333A Rx =5.2时， I= Uoc /(Req + Rx ) =0.2A