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
D. 10

Answer

C

Explanation

(c ) Before connecting the voltemter, potential difference across `100 Omega` resistance
`V_(i) = (100)/((100 + 10)) xx V = (10)/(11) V`

Finally after connecting voltmeter across `100 Omega`
Equivalent resistance `(100 xx 900)/((100 + 900)) = 90 Omega`
Final potentail difference

`V_(f) = (90)/((90 + 10)) xxV = (9)/(10) V`
`%` error ` = (V_(i) - V_(f))/(V_(i)) xx 100`
`= ((10)/(11) V - (9)/(10) V)/((10)/(11) V) xx 100 = 1.0`