Q 12-04-214JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
The resistance of a galvanometer is $50\ \text{ohm}$ and the maximum current which can be passed through it is $0.002\ \text{A}$. What resistance must be connected to it in order to convert it into an ammeter of range $0$–$0.5\ \text{A}$?
Answer: (A) $0.2\ \text{ohm}$
A shunt $S$ in parallel carries the excess current:
$$S = \frac{I_gG}{I - I_g} = \frac{0.002\times50}{0.498} \approx 0.2\ \Omega$$
Solution by Sreeraj P, M.Sc Physics