Charge Flow Through Switch in RC Bridge Circuit

To determine the total amount of charge that flows from $\mathrm{Y}$ to $\mathrm{X}$ when the switch $\mathrm{S}$ is closed, we analyze the circuit before and after the switch is closed.

1. Initial State (Switch $\mathrm{S}$ is open)

Let the negative terminal of the $9\text{ V}$ battery be at a reference potential of $0\text{ V}$ and the positive terminal be at $9\text{ V}$.

• The potential of the left wire is $V_{\text{left}} = 9\text{ V}$.
• The potential of the right wire is $V_{\text{right}} = 0\text{ V}$.

Before the switch is closed, the two capacitors $C_1 = 3\text{ }\mu\text{F}$ and $C_2 = 6\text{ }\mu\text{F}$ are connected in series.
Since node $\mathrm{X}$ was isolated:

• The right plate of $C_1$ and the left plate of $C_2$ are connected together at node $\mathrm{X}$.
• The net initial charge at node $\mathrm{X}$ is:
  $q_{X, \text{initial}} = 0\text{ }\mu\text{C}$

2. Final State (Switch $\mathrm{S}$ is closed)

In the steady state after closing the switch, no current flows into or out of the capacitors. A steady current flows only through the resistive branch:
$I = \frac{V}{R_1 + R_2} = \frac{9\text{ V}}{3\text{ }\Omega + 6\text{ }\Omega} = 1\text{ A}$

The potential at junction $\mathrm{Y}$ is:
$V_{\mathrm{Y}} = V_{\text{left}} - I R_1 = 9\text{ V} - (1\text{ A})(3\text{ }\Omega) = 6\text{ V}$

Since switch $\mathrm{S}$ is closed, nodes $\mathrm{X}$ and $\mathrm{Y}$ are at the same electric potential:
$V_{\mathrm{X}} = V_{\mathrm{Y}} = 6\text{ V}$

Now, we calculate the charge on the plates connected to node $\mathrm{X}$:

• For capacitor $C_1 = 3\text{ }\mu\text{F}$:
  The potential difference across it is $V_{\text{left}} - V_{\mathrm{X}} = 9\text{ V} - 6\text{ V} = 3\text{ V}$.
  The charge on its right plate (connected to $\mathrm{X}$) is:
  $q_{1, \text{right}} = -C_1(V_{\text{left}} - V_{\mathrm{X}}) = -3\text{ }\mu\text{F} \times 3\text{ V} = -9\text{ }\mu\text{C}$

• For capacitor $C_2 = 6\text{ }\mu\text{F}$:
  The potential difference across it is $V_{\mathrm{X}} - V_{\text{right}} = 6\text{ V} - 0\text{ V} = 6\text{ V}$.
  The charge on its left plate (connected to $\mathrm{X}$) is:
  $q_{2, \text{left}} = +C_2(V_{\mathrm{X}} - V_{\text{right}}) = +6\text{ }\mu\text{F} \times 6\text{ V} = +36\text{ }\mu\text{C}$

Thus, the total net charge at node $\mathrm{X}$ in the final state is:
$q_{X, \text{final}} = q_{1, \text{right}} + q_{2, \text{left}} = -9\text{ }\mu\text{C} + 36\text{ }\mu\text{C} = +27\text{ }\mu\text{C}$

3. Charge Flow through the Switch

The net change in charge at node $\mathrm{X}$ must have entered via the switch from node $\mathrm{Y}$:
$\Delta q = q_{X, \text{final}} - q_{X, \text{initial}} = 27\text{ }\mu\text{C} - 0\text{ }\mu\text{C} = 27\text{ }\mu\text{C}$

Therefore, the total amount of charge that flows from $\mathrm{Y}$ to $\mathrm{X}$ is $27\text{ }\mu\text{C}$.

Correct Option: C