Physics - Current Electricity Question with Solution | TestHub

Let the resistance of the galvanometer be $G$ and its full-scale deflection current be $I_g$.

When converted into a voltmeter of range $V$, the series resistance $R_v$ is given by:
$V=I_g(G+R_v)\Rightarrow R_v=\frac{V}{I_g}-G$

When converted into a millivoltmeter of range $V^{\prime }$, the series resistance $R_{mv}$ is given by:
$V^{\prime }=I_g(G+R_{mv})\Rightarrow R_{mv}=\frac{V^{\prime }}{I_g}-G$

Since a millivoltmeter measures a smaller potential difference than a voltmeter, its range is smaller ($V^{\prime }<V$).

Therefore, $R_{mv}<R_v$. The series resistance of the millivoltmeter will be less.

Answer: less