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问题: 八年级物理题

电源电压一定,R1>R2.当只断开开关S3时,电路消耗的电功率为19.2W,电阻R1消耗的电功率为P1;当只闭合开关S3时,电路消耗的电功率为2.4W,电阻R3消耗的电功率为0.8W,电阻R1消耗的电功率为P1';则求P1:P1'

解答:

当只断开开关S3时,R1、R2并联接入电路,P1=U^2/R1 ,P2=U^2/R2
1/R=(R1+R2)/R1R2
P=U^2/R=U^2(R1+R2)/R1R2=19.2W
P1:P=U^2/R1: U^2(R1+R2)/R1R2=1:(R1+R2)/R2
P1=P*R2/(R1+R2)
当只闭合开关S3时,R1、R2和R3串联接入电路,
P'=U^2/(R1+R2+R3)=I^2(R1+R2+R3)=2.4W ,P3'=I^2 R3=0.8W
P3'/P'=R3:(R1+R2+R3)=0.8/2.4=1:3 ,R3=(R1+R2)/2
P1':P'=R1:(R1+R2+R3)=R1:3(R1+R2)/2
P1'=2P'*R1/3(R1+R2)

P:P'=U^2(R1+R2)/R1R2:U^2/(R1+R2+R3)=2(R1+R2)^2:3R1R2=19.2:2.4
R1=[10±(104)^1/2]R2/2
R1≈10R2(∵R1>R2)
P1:P1'=P*R2/(R1+R2):2P'*R1/3(R1+R2)=3PR2:2R1P'=3PR2:2P'810R2
=3P:20P'=3*19.2:20*2.4=6:5