测电源电动势和内阻为什么用电流表的外接法误差大?

2025-04-02 18:52:27
推荐回答(1个)
回答1:

设电流表内阻rI 电压表内阻rU 电源内阻r 滑动变阻器阻值R 电源电动势E
1电流表内接
在一般的伏安法测电阻中,电流表测得了准确电流,但电压表测得的是电流表和滑动变阻器分的电压
U=I(R+ri)
但这个实验并非要测滑动变阻器电阻,所以这个误差反而精确的测定电源上所分得电压,只是电流表测定的是通过电源的电流减去电压表的电流
设电压表内阻与内接电流表支路合计电阻R1
I=(R1/R1+r)*E/ (ri+R) 而理想测量确是I=E/(R1+r)
通常由于要使电压表示数明显R常接近于电源内阻(通常不到1欧)这样的话rU>>ri+R从而R1=ri+R 从而误差很小
2电流表外接
电流表测得了准确电流(通过电源的电流)
而电压表测得的电压却未包含电流表分得的电压
设电压表与R并联所得电阻为R2
理想测量值:U=E*(R2+ri)/(R2+ri+r)
实际测量值: U=E*R2/(R2+ri+r)
刚才说了,R值通常接近ri 和 r,所以R2约为R,但注意此时ri已然不可忽略所以造成很大误差
明白了吗?

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