In a galvanometer, there is a deflection of 10 divisions per mA. The resistance of the galvanometer is \( 60\,\Omega \). If a shunt of \( 1\,\Omega \) is connected to the galvanometer and there are 100 divisions in all, on the scale of the galvanometer, the maximum current that this galvanometer can read is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 610 mA
A shunt in parallel with the galvanometer carries the extra current, so the pair works as an ammeter. The galvanometer and the shunt have the same potential difference across them.
The full-scale current of the galvanometer is the number of divisions divided by the deflection per mA.
\( \displaystyle I_g = \frac{100\ \text{divisions}}{10\ \text{divisions/mA}} = 10\,\mathrm{mA} \)
\( G = 60\,\Omega \)
\( S = 1\,\Omega \)
\( \displaystyle I_g\,G = (I - I_g)\,S \;\Rightarrow\; I = I_g\left(1 + \frac{G}{S}\right) \)
\( \displaystyle I = 10\,\mathrm{mA} \times \left(1 + \frac{60}{1}\right) = 610\,\mathrm{mA} \)