A galvanometer coil has a resistance of $15 \Omega$ and the meter shows full scale deflection for a current of $4 mA$. To convert the meter into a voltmeter of range $0$ to $18 V$, the required resistance is

The galvanometer can be transformed into a voltmeter by applying a high resistance in series. If the series resistance is $R$, then total resistance in circuit $=\left(15+R\right) \Omega$

Now $15+R=\frac{V}{I}=\frac{18}{4\times {10}^{-3}}=\frac{18\times {10}^3}{4}=4.5\times {10}^3$

Therefore, $R=4500-15=4485 \Omega$