Question

A galvanometer has a resistance of $60\Omega$ and has $50$ divisions. A current of $1mA$ is passed and it gives full-scale deflection. When shunt of $2.5\Omega$ is connected in parallel with the galvanometer, the deflection reduces, to $10$ divisions for the same current. Find the maximum current that can be measured with the galvanometer converted into an ammeter.

Answer 1

As the galvanometer has 50 divisions , the current is required to produce full scale deflection,

$\begin{array}{c}{I}_g=\frac{1}{10}\times 50=5mA\\ R_g=60\Omega \\ R_s=2.5\Omega \end{array}$

Let $I$ be the max current,

$\begin{array}{c}{I}_g=\frac{R_g}{R_g+R_s}\times I\\ I=\frac{R_g+R_s}{R_g}\times I_g=\frac{60+2.5}{2.5}\times 5=125mA\end{array}$