A galvanometer coil has a resistance of `12 Omega` and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V ?

Explanation

Here `R_G = 12 Omega , I_g = 3 mA = 3 xx 10^(-3)A` and voltmeter.s range V = 18 V
To convert galvanometer into a voltmeter of desired range, we should join a resistance R in series with the galvonometer, where
`R = V/(I_g) - R_G = 18/(3 xx 10^(-3)) - 12 = 6000 - 12 = 5988 Omega`.