Q 12-04-075JEE MainJEE Main 2024 (5 Apr, Shift 2)Medium
A galvanometer of resistance $100\ \Omega$, when connected in series with $400\ \Omega$, measures a voltage of up to $10\ \text{V}$. The value of the resistance required to convert the galvanometer into an ammeter to read up to $10\ \text{A}$ is $x\times10^{-2}\ \Omega$. The value of $x$ is
Answer: (C) $20$
Full-scale current: $I_g = \dfrac{10}{100 + 400} = 0.02\ \text{A}$.
Shunt: $S = \dfrac{I_gG}{I - I_g} = \dfrac{0.02\times100}{10 - 0.02} \approx 0.2\ \Omega = 20\times10^{-2}\ \Omega$.
Solution by Sreeraj P, M.Sc Physics