Physics - Current Electricity Question with Solution | TestHub

Given that the galvanometer has $25$ divisions and the current required for deflection one division is $4\times 10^{-4}\text{A}$

current for full scale deflection $I=4\times 10^{-4}\times 25\text{A}$
$=10^{-2}\text{A}$

To use this galvanometer as a voltmeter for a maximum voltage of $2.5\mathrm{V}$, we need to add a resistance $R$ in series to the resistance of the galvanometer of $G=50\Omega$ such that,

$$\begin{gathered} I=\frac{2.5}{R+G} \\ \Rightarrow R=\frac{2.5}{10^{-2}}-50 \\ =200ohm \end{gathered}$$