Two resistance of `400 Omega` and `800 Omega` are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance `10,000 Omega` is used to measure the potential difference across `400 Omega`. The error in measurement of potential difference in volts approximatley is

Options

A. 0.1
B. 0.02
C. 0.03
D. 0.05

Answer

D

Explanation

Before-
`V_(1)= 400/1200 xx 6 = 2V`
After-
`R_(eq) = (400 xx 10000)/10400 = 384.6 Omega`
`V_(2) = (384.6)/(800 + 384.6) xx 6`
`=1.945V =1.95 V`
Error `=2- 1.95 = 0.05 V`