初二物理题,请帮忙解答一下,谢谢
设电源电压为U
S闭合时:
R3被短路,R1、R2串联接入电路,V1表测量的是R1两端电压,V2表测量的是R2两端电压
则:I1=U/(R1+R2)
U1=I1R1=UR1/(R1+R2)
U2=I1R2=UR2/(R1+R2)
S断开时:
R1、R2、R3串联接入电路,V1表测量的是R1两端电压,V2表测量的是R2、R3两端电压
则:I2=U/(R1+R2+R3)=U/(R1+R2+4)
U1'=I2R1=UR1/(R1+R2+4)
U2'=I2*(R2+R3)=U(R2+R3)/(R1+R2+4)=U(R2+4)/(R1+R2+4)
已知:U1/U1'=3:2
即:[UR1/(R1+R...全部
设电源电压为U
S闭合时:
R3被短路,R1、R2串联接入电路,V1表测量的是R1两端电压,V2表测量的是R2两端电压
则:I1=U/(R1+R2)
U1=I1R1=UR1/(R1+R2)
U2=I1R2=UR2/(R1+R2)
S断开时:
R1、R2、R3串联接入电路,V1表测量的是R1两端电压,V2表测量的是R2、R3两端电压
则:I2=U/(R1+R2+R3)=U/(R1+R2+4)
U1'=I2R1=UR1/(R1+R2+4)
U2'=I2*(R2+R3)=U(R2+R3)/(R1+R2+4)=U(R2+4)/(R1+R2+4)
已知:U1/U1'=3:2
即:[UR1/(R1+R2)]/[UR1/(R1+R2+4)]=3/2
===> (R1+R2+4)/(R1+R2)=3/2
===> 2(R1+R2+4)=3(R1+R2)
===> 2(R1+R2)+8=3(R1+R2)
===> R1+R2=8…………………………………………………(1)
又,U2/U2'=9:10
即:[UR2/(R1+R2)]/[U(R2+4)/(R1+R2+4)]=9/10
===> [R2/8]/[(R2+4)/12]=9/10
===> 3R2/[2(R2+4)]=9/10
===> 30R2=18(R2+4)
===> 30R2=18R2+72
===> R2=6……………………………………………………(2)
联立(1)(2)得到:R1=2Ω。
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