An ideal cell of emf 10V is connected in circuit shown in figure. Each resistance is $$2\Omega$$. The potential difference (in V) across the capacitor when it is fully charged is ___________.

Steady-state circuit conditions:

$$\text{Current through capacitor branch } I_C = 0 \implies I_{R5} = 0$$

$$\text{Potential at the right plate of the capacitor } V_{\text{right}} = 0\text{ V (connected to ground via } R_5\text{)}$$

Since $$I_C = 0$$, resistors $$R_1$$ and $$R_2$$ are in series. Let $$V_B$$ be the node voltage between $$R_2, R_3,$$ and $$R_4$$. The parallel combination of the upper branches connected to $$V_B$$ from the $$10\text{ V}$$ source is:

$$R_p = \frac{R_3 \times (R_1 + R_2)}{R_3 + (R_1 + R_2)} = \frac{2 \times (2 + 2)}{2 + 4} = \frac{8}{6} = \frac{4}{3}\ \Omega$$

Total circuit resistance connected to the source:  $$R_{\text{total}} = R_p + R_4 = \frac{4}{3} + 2 = \frac{10}{3}\ \Omega$$

Voltage at node $$V_B$$ via potential divider rule:  $$V_B = 10 \times \frac{R_4}{R_{\text{total}}} = 10 \times \frac{2}{10/3} = 6\text{ V}$$

Potential at the left plate of the capacitor ($$V_A$$), which lies between $$R_1$$ and $$R_2$$:

$$V_A = V_B + \left( \frac{R_2}{R_1 + R_2} \right) (10 - V_B) = 6 + \left( \frac{2}{2 + 2} \right) (10 - 6) = 8\text{ V}$$

Potential difference across the capacitor:  $$V_C = V_A - V_{\text{right}} = 8 - 0 = 8\text{ V}$$