Explanation

(c) A fully charged capacitor draws no current. Therefore, no current flows in arm $GHF$. So the $R$ of arm $HF$ is ineffective. The total resistance of the resistors in circuit
is $R^{'} = \frac{(R + R) \times R}{(R + R) \times R} + R$
$= \frac{(2 + 2) \times 2}{(2 + 2) + 2} + 2 = \frac{10}{3}\Omega$
Total current, $i = \frac{E}{R^{'}} = \frac{10}{(10/3)} = 3A$
In parallel circuit, the current divides in the inverse ratio of resistance, so current in arm $ABGD = 1A$ and current in arm $AD = 2A$
Potential difference between $G$ and $D$
$= V_{G} - V_{D} = 1 \times 2 = 2V$
Potential difference between $D$ and $F$
$= V_{D} - V_{F} = 3 \times 2 = 6V$
$\therefore\ \ V_{G} - V_{F} = \left( V_{G} - V_{D} \right) + (V_{D} - V_{F})$
$= 2 + 6 = 8\ V$
Potential difference across capacitor $= V_{G} - V_{F} = 8V$