(1)求电源电压及R2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R1、R2消耗的电功率P1、P2.答案:
解:(1)由电路图知,电源电压U=U甲=12V,电阻R1两端电压U1=U﹣U乙=12V﹣8V=4V,
电路电流I===A,电阻R2的阻值R2===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R2的电流,电流表乙测干路电流,I甲=I2===0.25A,
I1===0.5A,I乙=I1+I2=0.25A+0.5A=0.75A,
(3)电阻R1的电功率P1=UI1=12V×0.5A=6W,电阻R2的电功率P2=UI2=12V×0.25A=3W,
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