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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor and the $4 \Omega$ resis

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$11 	ext{ A}$ and $3 \mu	ext{C}$

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(C) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor and the $4 \Omega$ resistor. The circuit effectively consists of three resistors of $1 \Omega, 2 \Omega,$ and $3 \Omega$ connected in parallel across the $6 	ext{ V}$ battery.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$\frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6 + 3 + 2}{6} = \frac{11}{6} \Omega$Therefore,the current $I$ through the battery is:$I = \frac{V}{R_{eq}} = \frac{6}{11/6} = \frac{36}{11} 	ext{ A} \approx 3.27 	ext{ A}$.Wait,let's re-evaluate the calculation: $I = \frac{V}{R_{eq}} = 6 	imes \frac{6}{11} = \frac{36}{11} 	ext{ A}$.Looking at the options,there seems to be a discrepancy in the provided solution's calculation. Let's re-calculate: $I_1 = 6/1 = 6 	ext{ A}$,$I_2 = 6/2 = 3 	ext{ A}$,$I_3 = 6/3 = 2 	ext{ A}$. Total current $I = 6 + 3 + 2 = 11 	ext{ A}$.The charge on the capacitor is $Q = C 	imes V = 0.5 \mu	ext{F} 	imes 6 	ext{ V} = 3 \mu	ext{C}$.Thus,the current is $11 	ext{ A}$ and the charge is $3 \mu	ext{C}$.

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

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