Q 12-02-102JEE MainJEE Main 2023 (8 Apr, Shift 1)Medium
In this figure the resistance of the coil of galvanometer G is $2\ \Omega$. The emf of the cell is $4$ V. The ratio of potential difference across $C_1$ and $C_2$ is
Answer: (B) $\dfrac45$
In the steady state no current flows through the capacitors. The current path is $A\to B$ ($6\ \Omega$) $\to$ G ($2\ \Omega$) $\to C\to D$ ($8\ \Omega$): $I=\dfrac{4}{16}=0.25$ A.
$V_{C_1}=V_A-V_C=0.25\times(6+2)=2$ V; $V_{C_2}=V_B-V_D=0.25\times(2+8)=2.5$ V.
Ratio $=\dfrac{2}{2.5}=\dfrac45$.
Solution by Sreeraj P, M.Sc Physics