The potential difference across the `100 Omega` resistance in the following circuit is measured by a voltmeter of `900 Omega` resistance. The percentage error made in reading the potential difference is

Options

A. `10/9`
B. 0.1
C. `1.0`
D. `10.0`

Answer

C

Explanation

Let the potential difference across `100 Omega` , before and after the voltmeter connection be `V_(1)` and `V_(2)`
`V_(1) = (100/110) xx V = (10 V)/11`
`V_(2) = ((900 xx 100)/1000)/(90 + 10) V = (90/100)V =(9V)/10`
`V_(1) - V_(2) =(10V)/11 -(9V)/10 = (100 V - 99V)/110 = V/110`
% Error ` =(V_(1) - V_(2))/V_(1) xx 100 =V/110 xx 11/(10 V) xx 100 = 1%`