In the given circuit with steady current, the potential drop across the capacitor must be :

\frac{2 V}{3}

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\frac{V}{3}

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\frac{V}{2}

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V

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Solution

The correct option is B $\frac{V}{3}$
As the capacitors offer infinite resistance to steady current so no current will flow through capacitor branch. Now the equivalent circuit is shown in the figure and apply Kirchhoff's voltage law, we get
$2 V - V = 2 R I + R I \Rightarrow I = \frac{V}{3 R}$
Voltage across AB is $V_{A B} = V + R I = V + R (V / 3 R) = \frac{4 V}{3}$
also $V_{A B} = V + V_{C}$ where $V_{C} =$ potential across C.
or $(4 V / 3) = V + V_{C}$
$∴ V_{C} = \frac{4 V}{3} - V = \frac{V}{3}$

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