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(D) In the steady state,capacitors act as open circuits,meaning no current flows through the branches containing capacitors.Therefore,the current $i$

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$12$

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(D) In the steady state,capacitors act as open circuits,meaning no current flows through the branches containing capacitors.Therefore,the current $i$ flows only through the $4\, \Omega$ resistor and the internal resistance of $1\, \Omega$.The total resistance of the circuit is $R_{eq} = 4\, \Omega + 1\, \Omega = 5\, \Omega$.The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{10\, V}{5\, \Omega} = 2\, A$.The potential difference across the $4\, \Omega$ resistor is $V_{4\Omega} = i 	imes R = 2\, A 	imes 4\, \Omega = 8\, V$.Since the capacitor branches are connected in parallel with the $4\, \Omega$ resistor,the potential difference across each branch is $8\, V$.Each branch contains two $3\, \mu F$ capacitors in series. The equivalent capacitance of each branch is $C_{eq} = \frac{3\, \mu F 	imes 3\, \mu F}{3\, \mu F + 3\, \mu F} = 1.5\, \mu F$.The total charge on each branch is $Q_{branch} = C_{eq} 	imes V = 1.5\, \mu F 	imes 8\, V = 12\, \mu C$.Since the capacitors are in series,the charge on each capacitor is the same as the charge on the branch,which is $12\, \mu C$.

In the following figure,the charge on each capacitor in the steady state will be.....$\mu C$.

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