标有“12V 6W”、“6V 6W”字样的灯L 1 和L 2 串联在某电源上,设灯丝的电阻不变,则灯L 1 和L 2 的实际

2025-01-02 09:38:36
推荐回答(1个)
回答1:


①∵P=
U 2
R

∴两只灯泡的电阻分别为R 1 =
U 1 2
P
=
(12 V) 2
6W
=24Ω,R 2 =
U 2 2
P
=
(6V) 2
6W
=6Ω;
∵P=I 2 R,
∴两灯串联时消耗的实际功率之比为
P 1实
P 2实
=
I 2 R 1
I 2 R 2
=
R 1
R 2
=
24Ω
=
4
1

②∵P=UI,
∴灯泡的额定电流分别为I 1 =
P
U 1
=
6W
12V
=0.5A,I 2 =
P
U 2
=
6W
6V
=1A,
要使灯泡安全工作,电路电流为I=I 1 =0.5A,
∴电源电压为U=I(R 1 +R 2 )=0.5A×(24Ω+6Ω)=15V.
故答案为:4:1;:15.