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A voltmeter with resistance 500Ω is used to measure the emf of a cell of internal resistance 4Ω. the percentage error in the reading the voltmeter will be

a) 0.2%
b) 0.8%
c) 1.4%
d) 2.2%


The answer is c)0.8
emf of a cell=500I+4I=504I
potential difference=500I
1-potential difference/emf of a cell=1-(500/504)=0.008
0 0 0
The percentage error in the voltmeter is 1-(500/504)=0.8%
1 1 1
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