The maximum current that can be measured by a galvanometer of resistance `40 Omega` is 10 mA . It is converted into a voltmeter that can read upto 50 V . The resistance to be connected in series with the galvanometer is ... (in ohm )

Options

A. `2010`
B. `4050`
C. `5040`
D. `4960`

Answer

D

Explanation

To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is
`R = V/(I_g) - G`
`= (50)/(10 xx 10^(-2)) - 40`
`= 5000 - 40 = 4960 Omega`