In the given circuit when steady current flows, the potential drop across the capacitor is:

$V/3$

The correct answer is Option (2) → $V/3$

Current balance in the two resistive branches:

$ \frac{A + V - B}{R} + \frac{A + 2V - B}{2R} = 0 $

Multiply by $2R$:

$ 2(A + V - B) + (A + 2V - B) = 0 $

$ 3A + 4V - 3B = 0 $

$ B - A = \frac{4V}{3} $

Voltage across capacitor:

$ V_{C} = (A + V) - B $

$ V_{C} = V - \frac{4V}{3} $

$ V_{C} = -\frac{V}{3} $

Magnitude of drop: $ \frac{V}{3} $

The potential drop across the capacitor is $\frac{V}{3}$.