一道初中物理题如图所示，电源电压
设滑片P在端点b时电阻为Rab,I2=U/(Rab+RL),电压表示数U2=IRab=URab/(Rab+RL)P2=I2^2RL=RL*U^2/(Rab+RL)^2,则滑片P在中点c时电阻为Rab/2。 I=U/(Rab/2+RL),U1=IRab/2=(URab/2)/(Rab/2+RL)=URab/(Rab+2RL)U1:U2=URab/(Rab+2RL):URab/(Rab+RL)=(Rab+RL):(Rab+2RL)=2:3RL= :1P1=I1^2RL=RL*U^2/(Rab/2+RL)^2,P1/P2=[RL*U^2/(Rab/2+RL)^2]/[RL*U^2/(Rab+R...全部
设滑片P在端点b时电阻为Rab,I2=U/(Rab+RL),电压表示数U2=IRab=URab/(Rab+RL)P2=I2^2RL=RL*U^2/(Rab+RL)^2,则滑片P在中点c时电阻为Rab/2。
I=U/(Rab/2+RL),U1=IRab/2=(URab/2)/(Rab/2+RL)=URab/(Rab+2RL)U1:U2=URab/(Rab+2RL):URab/(Rab+RL)=(Rab+RL):(Rab+2RL)=2:3RL= :1P1=I1^2RL=RL*U^2/(Rab/2+RL)^2,P1/P2=[RL*U^2/(Rab/2+RL)^2]/[RL*U^2/(Rab+RL)^2]=[(Rab+RL)/(Rab/2+RL)]^2=[(Rab+RL)/2*(Rab+2RL)]^2=[(2/3)/2]^2=1/91)R-L:R-ab是1:12)P在c和b时,L消耗的电功率之比1:9。
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