灯泡L 1 标有“12V 4W”,L 2 标有“6V 3W”,把这两盏灯串联接在电源上,闭合开关后有一只灯泡正常发光,则
1个回答

灯泡L 1的电阻R 1=

U 1 2

P 1 =

(12V) 2

4W =36Ω,额定电流I 1=

P 1

U 1 =

4W

12V =

1

3 A,灯泡L 2的电阻R 2=

U 22

P 2 =

(6V) 2

3W =12Ω,额定电流I 2=

P 2

U 2 =

3W

6V =0.5A,

所以串联时正常发光的是灯泡L 1,所以电源电压U=I 1(R 1+R 2)=

1

3 A×(36Ω+12Ω)=16V.

故选A.