In the given figure, the charge stored in $$6\;\mu$$F capacitor, when points $$A$$ and $$B$$ are joined by a connecting wire is ______ $$\mu$$C.

When A and B are shorted, they become the same node.

step 1: identify final steady state

In DC steady state, capacitors act as open circuits.
So no current flows through capacitors, but they can hold charge.

step 2: reduce the circuit

Now A and B are same node.

So:

• top node = 9 V
• bottom node = 0 V (ground)
• A = B is some intermediate node

Between top and A:

6 Ω

Between A and bottom:

3 Ω

So it becomes a simple voltage divider.

step 3: find voltage at node A (= B)

$$V_A=\frac{3}{6+3}\times9=\frac{3}{9}\times9=3\text{ V}$$

step 4: voltage across 6 μF capacitor

Top plate is at 9 V, bottom plate at node B = 3 V

$$V_C=9-3=6\text{ V}$$

step 5: charge

$$Q=CV=6\mu F\times6=36\mu C$$