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

Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, I_g = 3 mA = 3 x 10⁻³ A Range of the voltmeter is 0, which needs to be converted to 18 V.

Therefore ,V = 18 V

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:

R = V/I_g – G

= [18 /(3 x 10⁻³) ]-12 = 6000 -12 = 5988 Ω

Hence, a resistor of resistance 5988 Ω is to be connected in series with the galvanometer.