When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is ' $n$ ' times…
When moving coil galvanometer (MCG) is converted into a voltmeter, the series resistance is ' $n$ ' times the resistance of galvanometer. How many times that of MCG the voltmeter is now capable of measuring voltage?
$n$
$\frac {n+1}{n}$
$n+1$
$n-1$
Solution
Series resistance is $\mathrm{n}$ times of galvanometer resistance: $\mathrm{R}_{\mathrm{s}}=\mathrm{nR}_{\mathrm{G}}$
Relation between voltage and current:
$\begin{array}{ll}
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{R}_{\mathrm{s}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{I}\left(\mathrm{nR}_{\mathrm{G}}+\mathrm{R}_{\mathrm{G}}\right) \\
\therefore & \mathrm{V}=\mathrm{IR}_{\mathrm{G}}(\mathrm{n}+1)
\end{array}$