Q 12-04-202JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
A moving coil galvanometer allows a full scale current of $10^{-4}\ \text{A}$. A series resistance of $2\times10^4\ \Omega$ is required to convert the galvanometer into a voltmeter of range $0$–$5\ \text{V}$. Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range $0$–$10\ \text{mA}$ is:
Answer: (C) $300\ \Omega$
Voltmeter: $5 = 10^{-4}(G + 2\times10^4)$, so $G = 3\times10^4\ \Omega$.
Ammeter: $I_gG = (I - I_g)S$:
$$S = \frac{10^{-4}\times3\times10^4}{10^{-2} - 10^{-4}} = \frac{3}{0.0099} \approx 300\ \Omega$$
Solution by Sreeraj P, M.Sc Physics