A galvanometer with `50` divisions on the scale has a resistance of `25 Omega`. A current of `2 xx 10^-4` A gives a deflection of one scale division. The additional series required to convert it into a voltmeter reading up to `25 V` is.

Options

A. `1200 Omega`
B. `1225 Omega`
C. `2475 Omega`
D. `2500 Omega`

Answer

C

Explanation

`i_g = 50 xx 2 xx 10^-4 = 10^-2 A, G = 25 Omega`
`V = i_g(R + G)`
`25 = 10^-2(R + 25) rArr R = 2475 Omega`.