题目内容
有一个电流表G,内阻Rg=60Ω,满偏电流Ig=1mA。
(1)要把它改装为量程为0—3V的电压表,要串联多大的电阻?
(2)改装后的电压表的内阻是多少?
(1)要把它改装为量程为0—3V的电压表,要串联多大的电阻?
(2)改装后的电压表的内阻是多少?
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628339951.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626279811345.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628339951.png)
试题分析:把电流表G改装成电压表,电路图如图所示:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628667916.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628823840.png)
故电阻R分担的电压为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628979845.png)
由欧姆定律可求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241626279811345.png)
电压表的内阻等于Rg和R串联的总电阻即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162628339951.png)
点评:基础题,关键是明确串联电阻的分压作用,会求串联电阻阻值
![](http://thumb2018.1010pic.com/images/loading.gif)
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