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

Options

A. `0.2%`
B. `0.8%`
C. `1.4%`
D. `0.4%`

Answer

B

Explanation

(b) `I = (E)/(500 + 4) = (E)/(504)`
Reading of voltmeter
`= 500 I = (500 E)/(504)`
`%` Error ` = (E -(500 E)/(504))/(E) xx 100 = (400)/(504) = 0.8 %`