题目内容
( 10 分)如图所示,R3=0.5
,S断开时,两表读数分别为0.4A和2.4V, S闭合时,它们的读数分别变化了0.3A和0.3V
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241556349451974.png)
求
(1) R1、R2的阻值
(2) 电源的电动势和内阻(两表均视为理想表).
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155634882333.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241556349451974.png)
求
(1) R1、R2的阻值
(2) 电源的电动势和内阻(两表均视为理想表).
(1)
(2)E=2.8V;r=0.5Ω
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155635085526.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155635241552.png)
试题分析::(1)S断开时,R1=U1/I1=2.4/0.4=6Ω
S闭合时,依题意有:U2=2.4-0.3=2.1V
I2=I1+0.3=0.7A
∴IR1= U2/ R1=0.35A
IR2= I2-IR1=0.35A
∴R2= U2/IR2=6Ω
(2)S断开时,有:E= U1+ I1(R3+r)=2.4+0.4(0.5+r)
S闭合时,有:E= U2+ I2(R3+r)=2.1+0.7(0.5+r)
解上述方程组得E=2.8V;r=0.5Ω
点评:本题难度较小,本题在求解过程中要重点明确开关闭合和断开后电路中的串并联结构,熟悉闭合电路欧姆定律的内容
![](http://thumb2018.1010pic.com/images/loading.gif)
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