电源电压不变,电灯L的电阻值不变,开关S闭合.已知滑动变阻器的滑片P在上端

设电源电压为V;电灯电阻为RL;滑动变阻器ab间电阻为RP;
则滑片P在中点c时 :电压表 V1 = V*RP/[(RL+RP/2)*2]
则滑片P在中点b时 :电压表 V2 = V*RP/(RL+RP)
所以 V1:V2 = V*RP/[(RL+RP/2)*2]:V*RP/(RL+RP) = 3 :4
所以 RL:RP = 1 :2
则滑片P在中点c,电灯L的电功率 :P1= I^2*R = [V/(RL+RP/2)]^2 * RL
则滑片P在中点b,电灯L的电功率 :P2= I^2*R = [V/(RL+RP)]^2 * RL
因为 RL :RP = 1 :2 则RP = 2RL 所以
P1:P2 = [V/(RL+RP/2)]^2 * RL :[V/(RL+RP)]^2 * RL = 9:4

看不清楚图形，无法解答，请补充！