(1)∵P=
,U2 R
∴灯泡电阻:
R1=
=
U
P1额
=12Ω,R2=(6V)2
3W
=
U
P2额
=3Ω;(3V)2
3W
(2)电路电流:
I=
=U
R1+R2
=0.4A,6V 12Ω+3Ω
灯泡实际功率:
P1实=I2R1=(0.4A)2×12Ω=1.92W,P2实=I2R2=(0.4A)2×3Ω=0.48W;
(3)整个电路t=10s内消耗的电能:
W=UIt=6V×0.4A×10s=24J.
答:(1)L1的电阻为12Ω,L2的电阻为3Ω;
(2)两灯的实际功率各分别为1.92W、0.48W;
(3)整个电路在10s时间内消耗的电能为24J.