如图所示:电源电压恒定不变,R
1=24Ω,S闭合.甲、乙两表均为电压表时:U
甲=12V,U
乙=8V.
(1)求电源电压及R
2的阻值;
(2)当S断开时,甲、乙均为电流表时,求出I
甲、I
乙的示数;
(3)当S断开时,甲、乙均为电流表时,求出电阻R
1、R
2消耗的电功率P
1、P
2.

答案:
解:(1)由电路图知,电源电压U=U
甲=12V,
电阻R
1两端电压U
1=U﹣U
乙=12V﹣8V=4V,
电路电流I=

=

=

A,
电阻R
2的阻值R
2=

=

=48Ω;
(2)甲乙均为电流表,S断开时,两电阻串联,电流表甲测流过电阻R
2的电流,电流表乙测干路电流,
I
甲=I
2=

=

=0.25A,
I
1=

=

=0.5A,
I
乙=I
1+I
2=0.25A+0.5A=0.75A,
(3)电阻R
1的电功率P
1=UI
1=12V×0.5A=6W,
电阻R
2的电功率P
2=UI
2=12V×0.25A=3W,