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(C) In the steady-state condition,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor (Line $1$).The

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$10$

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(C) In the steady-state condition,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor (Line $1$).Therefore,the total current $i$ in the circuit flows only through the upper branch (Line $2$) containing the two $1\,\Omega$ resistors in series.The total resistance of the circuit is $R_{eq} = 1\,\Omega + 1\,\Omega + 0.5\,\Omega = 2.5\,\Omega$.The total current $i$ is given by $i = \frac{V}{R_{eq}} = \frac{2.5\,V}{2.5\,\Omega} = 1\,A$.The potential difference across the upper branch is $V_{upper} = i 	imes (1\,\Omega + 1\,\Omega) = 1\,A 	imes 2\,\Omega = 2\,V$.Since the capacitor branch is in parallel with the upper branch,the potential difference across the capacitor is equal to the potential difference across the upper branch,which is $2\,V$.The charge $Q$ on the capacitor is $Q = C 	imes V = 5\,\mu F 	imes 2\,V = 10\,\mu C$.

$A$ capacitor of capacitance $5\,\mu F$ is connected as shown in the figure. The internal resistance of the cell is $0.5\,\Omega$. The amount of charge on the capacitor plate is......$\mu C$.

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