The emf of a battery is 6.0 V and its internal resistance is `1.5 Omega`. Its potential differences is measured by a voltmeter of resistance `1000 Omega`. Calculate the percentage error in the reading of emf shown by voltmeter.

Explanation

Here, `epsilon = 6.0 V , r = 1.5 Omega`,
Resistance of voltmeter, `R= 1000 Omega`
Current in the circuit is `I = epsilon/R+r = (6V)/(1000) + 1.5 = 5.991 xx 10^(-3)A`
pot. Diff. across voltmeter
`=(5.991 xx 10^(-3))xx1000`
`=5.9991V`
Thus terminal potential difference of the cell,
V= pot. diff. across voltmeter = 5.991 V
% error `= (epsilon - V)/(V) xx 100 = (6-5.991)/(6) xx 100 = 0.15%`