youyue1983
幼苗
共回答了21个问题采纳率:90.5% 举报
只闭上S1,则R1和R3串联,R2断开,电压表显示R1的电压
则P3=I1^2 *R3 U1/I1=R1
只闭上S2,则R2与R3串联,R1断开,电压表显示R2电压
则P′3=I2^2 *R3 U2/I2=R2
(1) P′3/P3=I2^2/(I1^2)=1/4 得I1/I2=1/2
(2) R1/R2=(U1/I1)/(U2/I2)
U2=2U1 I1/I2=1/2
得R1/R2=1/4
(3)
设电池电动势为e,则
e/(R1+R3)=I1 e/(R2+R3)=I2 I1/I2=1/2 R1/R2=1/4 P2=1.6=U2I2
可得 R3=2R1 e^2=14.4R1
都闭上后,R3短路,电路为R1R2并联.
并联电阻R=R1R2/(R1+R2)=0.8R1
则功率为:P=e^2/R=14.4/0.8=18W
1年前
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