解：P1=9W，P1'=4W根据公式P=U^2/R,可得P1/P1'=U^2/R1/(U-U2)^2/R1=9W/4W解得：U=3U2，U1=U-U2=2U2(1)S断开时，I=U1/R1=U2/R2，可得：R1/R2=U1/U2=2/1(2)P2=U2^2/R2=(U1/2)^2/(R1/2)=4W x 1/2=2W(3)S闭合与断开时，流过电阻R1的电流之比是:I/I'=(U/R1)/[U/(R1+R2)]=(R1+R2)/R1=(2R2+R2)/2R2=3/2答：电阻R1与R2的比值是2:1，S断开时，电阻R2消耗的电功率,2瓦，S闭合与断开时，流过电阻R1的电流之比是3:2....
