In the given circuit when steady current flows, the potential drop across the capacitor is:
Answer & explanation
Correct answer: option 2
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 $\frac{V}{3}$.