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(D) In a steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit consists of tw

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none of these

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(D) In a steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit consists of two parallel branches connected in series with the main line. The current $I$ flows through the first resistor $R$. At the junction before the second part of the circuit,the current splits. However,since the capacitor is in a steady state,the entire current $I$ must pass through the resistor $R$ that is in parallel with the capacitor branch.The potential difference across the capacitor is equal to the potential difference across the resistor $R$ that is in parallel with it.Since the current $I$ flows through this resistor $R$,the potential difference $V$ across it is $V = I R$.The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.Substituting $V = I R$ into the energy formula:$U = \frac{1}{2} C (I R)^2 = \frac{1}{2} C I^2 R^2$.

The capacitor shown in the figure is in a steady state. The energy stored in the capacitor is

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