证明如下：R1+R2=RR1、R2的并联电阻为：R并=R1R2/(R1+R2) =R1(R-R1)/R =(R1R-R1²)/R =-(-R1R+R1²)/R =-(R²/4-R1R+R1²-R²/4)/R =-[(R/2-R1)²-R²/4]/R =[-(R/2-R1)²+R²/4]/R =-(R/2-R1)²/R+R/4上式当R/2=R1时有最小值R/4
