题目内容
利用如入所示电路,测定电源电动势和小电阻r,已知R1=R2=R3=1Ω,电源内阻不计,当电键K断开时,电压表读数为0.8V;当电键K闭合时,电压表的读数为1V,该电源的电动势和小电阻分别为 V和 Ω。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412481144615386.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412481144615386.png)
2 0.5
根据题意可知,电键K断开时,为R1和R2串联,根据闭合电路欧姆定律有,
,当电键K闭合时,R1和R3并联然后再跟R2串联,则有
,联立可得该电源电动势为2V,小电阻为r=0.5Ω
故答案为:2 0.5
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124811477891.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124811508993.png)
故答案为:2 0.5
![](http://thumb2018.1010pic.com/images/loading.gif)
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