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(A) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the circu

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$11 \, A$ and $3 \, \mu C$

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(A) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the circuit effectively consists of three resistors $(1 \, \Omega, 2 \, \Omega, 3 \, \Omega)$ connected in parallel across the $6 \, V$ battery.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$\frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6 + 3 + 2}{6} = \frac{11}{6} \, \Omega^{-1}$$R_{eq} = \frac{6}{11} \, \Omega$The total current $I$ drawn from the battery is:$I = \frac{V}{R_{eq}} = \frac{6}{6/11} = 11 \, A$Since the capacitor is in parallel with the battery,the potential difference across the capacitor is equal to the battery voltage,$V = 6 \, V$.The charge $Q$ on the capacitor is:$Q = C 	imes V = 0.5 \, \mu F 	imes 6 \, V = 3 \, \mu C$

In the given circuit diagram,the current through the battery and the charge on the capacitor,respectively,in the steady state are:

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