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(A) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch containing the capacitor $C$. Let the potent

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$V$

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(A) In a steady state,the capacitor acts as an open circuit,so no current flows through the middle branch containing the capacitor $C$. Let the potential at the right junction be $0$ and at the left junction be $V_L$. Applying Kirchhoff's loop law to the outer loop (top and bottom branches): The current $i$ flows through the top branch (resistance $R$) and the bottom branch (resistance $2R$). Using the loop equation: $V - iR - 2iR + 2V = 0$ $3V = 3iR$ $i = V / R$ Now,find the potential at the right junction $(V_A)$ and left junction $(V_L)$. Let the potential at the right junction be $0$. Then the potential at the left junction is $V_L = V - iR = V - (V/R)R = 0$. Alternatively,$V_L = 2V - i(2R) = 2V - (V/R)(2R) = 0$. So,the potential at the right junction is $0$ and at the left junction is $0$. The potential difference across the middle branch is the potential across the capacitor $C$ and the battery $V$. Since no current flows through the middle branch,the potential drop across the capacitor $V_C$ must balance the battery $V$ in that branch. Thus,$V_C = V$.

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

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