$A$ 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 $18\; V$?

(D) Resistance of the galvanometer coil,$G = 12\; \Omega$.
Current for which there is full-scale deflection,$I_{g} = 3\; mA = 3 \times 10^{-3}\; A$.
Range of the voltmeter,$V = 18\; V$.
To convert a galvanometer into a voltmeter,a high resistance $R$ must be connected in series with the galvanometer.
The formula for the series resistance is $R = \frac{V}{I_{g}} - G$.
Substituting the given values:
$R = \frac{18}{3 \times 10^{-3}} - 12$
$R = 6000 - 12 = 5988\; \Omega$.
Therefore,a resistor of $5988\; \Omega$ must be connected in series with the galvanometer.