设电源的电压为U;
R1、R2串联时,电路中的电流为:
I=
U |
R1+R2 |
R1消耗的电功率为:
P=I2R1=(
U |
R1+R2 |
R1、R2并联时,R1消耗的电功率:
P1=
U2 |
R1 |
通过R2的电流为:
I2=
U |
R2 |
①②两式相比可得,
(
R1 |
R1+R2 |
9 |
16 |
解得R1=3R2,
代入①③式可得:
(
U |
R1+R2 |
U |
3R2+R2 |
1 |
4 |
U |
R2 |
1 |
4 |
解得:R2=3Ω,
R1=3R2=3×3Ω=9Ω.
答:
电阻R1和R2的阻值分别为9Ω、3Ω.
U |
R1+R2 |
U |
R1+R2 |
U2 |
R1 |
U |
R2 |
R1 |
R1+R2 |
9 |
16 |
U |
R1+R2 |
U |
3R2+R2 |
1 |
4 |
U |
R2 |
1 |
4 |