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

Answer

`0.8%`

Explanation

Here `R = 500 Omega , r = 4 Omega`
`:. Delta epsilon = epsilon - V = 1r = (epsilon)/(R + r) xx r`
Percentage error `= ( Delta epsilon)/(epsilon) xx 100 = ((r )/( R + r)) xx 100`
`= ((4)/(500 + 4)) xx 100 ~~ 0.8 %`