6V 3W灯泡的许用电流I1=P1/U1=3/6=0.5A,电阻R1=U1²/P1=6²/3=12Ω
6V 6W灯泡的许用电流I2=P2/U2=6/6=1A,电阻R2=U2²/P2=6²/6=6Ω
两灯泡都不允许超过许用电流,所以当其中一个灯泡正常发光时,电路总电流I=0.5A
此时总功率P = I²R = I²(R1+R2) = 0.5²×(12+6) = 4.5瓦