The resistance of a galvanometer is `10 Omega`. It gives full-scale deflections when `1 mA` current is passed. The resistance connected in series for converting it into a voltmeter of `2.5 V` will be

Options

A. `24.9 Omega`
B. `249 Omega`
C. `2490Omega`
D. `24900 Omega`

Answer

C

Explanation

Here: `I_(g) = 1 mA = 1 xx 10^(-3) A, G=10 Omega, V=2.5 V`
`V=I_(g)(G+R)` or `R=V/I_(g)-G`
Substituting the given values, we get,
`R=(2.5V)/(1 xx 10^(-3))A - 10 Omega = 2500 Omega - 10 Omega= 2490 Omega`