Q 12-03-010NEETNEET 2023Top questionMedium
If the galvanometer $G$ does not show any deflection in the circuit shown, the value of $R$ is given by :
Answer: (C) $100\ \Omega$
No current flows through $G$, so the $2$ V cell carries no current. The potential difference across $R$ must then equal the emf of the $2$ V cell.
The $10$ V battery drives a current through $400\ \Omega$ and $R$ in series:
$$I = \frac{10}{400 + R}$$
$$IR = 2 \;\Rightarrow\; \frac{10R}{400 + R} = 2 \;\Rightarrow\; 10R = 800 + 2R \;\Rightarrow\; R = 100\ \Omega$$
Solution by Sreeraj P, M.Sc Physics