A galvanometer $G$ of $2\ \Omega$ resistance is connected in the given circuit. The ratio of the charges stored in $C_1$ and $C_2$ is
Answer: (D) $\dfrac{1}{2}$
In the steady state no current flows through the capacitors. The only current path is $4\ \Omega \to G\,(2\ \Omega) \to 6\ \Omega$:
$$I = \frac{6}{4 + 2 + 6} = 0.5\ \text{A}$$
Take the left corner at $6\ \text{V}$ and the right corner at $0$. Then the top corner is at $6 - 0.5\times4 = 4\ \text{V}$ and the bottom corner at $4 - 0.5\times2 = 3\ \text{V}$.
$C_1$ (left to bottom): $V_1 = 6 - 3 = 3\ \text{V}$, $Q_1 = 4\times3 = 12\ \mu\text{C}$.
$C_2$ (top to right): $V_2 = 4 - 0 = 4\ \text{V}$, $Q_2 = 6\times4 = 24\ \mu\text{C}$.
$$\frac{Q_1}{Q_2} = \frac12$$
Solution by Sreeraj P, M.Sc Physics