题目内容
如图所示电路中,甲、乙两个毫安表的内阻均为6Ω,R3=12Ω,S断开时,AB之间电阻为3Ω,S闭合时,甲、乙两个毫安表的示数之比为1:2,求R1、R2的阻值各为多少?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218101735153.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241218101735153.jpg)
R1=3Ω,R2=1Ω
设S断开时AB间电阻为RAB,S闭合时,AB间电压为U,由欧姆定律得:
I甲=
……………………………………………① (2’)
I乙=
……………………………………………② (2’)
由题可知:I甲:I乙=1:2 ……………………………………………③ (2’)
RAB=
……………………………………………④ (2)
将RAB=3Ω,RA=6Ω,R3=12Ω,代方程,联立解得:
R1=3Ω (1’)
R2=1Ω (1’)
I甲=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121810188483.gif)
I乙=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121810204627.gif)
由题可知:I甲:I乙=1:2 ……………………………………………③ (2’)
RAB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121810220500.gif)
将RAB=3Ω,RA=6Ω,R3=12Ω,代方程,联立解得:
R1=3Ω (1’)
R2=1Ω (1’)
![](http://thumb2018.1010pic.com/images/loading.gif)
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