题目内容
如图所示电路中,电阻R1=8Ω.当电键S断开时,电压表V1的示数为5.7V,电流表的示数为0.75A,电源总功率是9W;当电键S闭合时,电压表V2的不数为4V.若电键断开和闭合时电源内部损耗的电功率之比是9:16,求电源的电动势和电阻R2、R3.,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412544398910575.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412544398910575.png)
ε=12V,R2=8Ω,R3=3.6Ω
:(1)电键S断开时 P="EI" 得到E=12V
(2)电键S断开时R3+R4=
=7.6Ω 电键断开和闭合时电源内部损耗的电功率之比是9:16 即
=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444083435.png)
而I1="0.75A" 故I2="1A" 电键S闭合时,R4=
=4Ω 故R3="7.6-4=3.6Ω"
(3)电键S断开时,I=
得r="0.4Ω" 电键S闭合时I=
得R2=8Ω
(2)电键S断开时R3+R4=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444020404.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444051464.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444083435.png)
而I1="0.75A" 故I2="1A" 电键S闭合时,R4=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444020404.png)
(3)电键S断开时,I=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444332743.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125444504991.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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