题目内容
如图所示,电源的电动势E=110V,电阻R1=21Ω.当电键S2断开时,电阻R1的电压U="105" V;当电键S2闭合时,电阻 R1的电功率是P=336W.求
(1) 当电键S2断开时R1上的电流I和电源的内电阻r;
(2)当电键S2闭合时R1上的电压和流过电动机的电流(可能用到的数据![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705659509.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705675320.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241227056913247.jpg)
(1) 当电键S2断开时R1上的电流I和电源的内电阻r;
(2)当电键S2闭合时R1上的电压和流过电动机的电流(可能用到的数据
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705659509.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705675320.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241227056913247.jpg)
(1)当电键S2断开时,通过R1的电流
(4分)
由闭合电路欧姆定律
,
得到电源内阻
(4分)
(2)当电键S2闭合时,由
得R1上的电压
(4分)
由闭合电路欧姆定律得,E=U’+I’r
代入数据解得流过电源的电流I’=26A.
由欧姆定律得流过R1的电流为
,
流过电动机的电流为I2= I’- I1="22A" (6分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705691669.gif)
由闭合电路欧姆定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705722464.gif)
得到电源内阻
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705737727.gif)
(2)当电键S2闭合时,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705753471.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705769784.gif)
由闭合电路欧姆定律得,E=U’+I’r
代入数据解得流过电源的电流I’=26A.
由欧姆定律得流过R1的电流为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824122705784523.gif)
流过电动机的电流为I2= I’- I1="22A" (6分)
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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