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(B) At steady state,the capacitor acts as an open circuit. The circuit simplifies to a series combination of the $2 \ \Omega$ and $3 \ \Omega$ resisto

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$0.9 \ A$

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(B) At steady state,the capacitor acts as an open circuit. The circuit simplifies to a series combination of the $2 \ \Omega$ and $3 \ \Omega$ resistors in parallel,which is then in series with the $2.8 \ \Omega$ resistor and the $6 \ V$ battery.First,calculate the equivalent resistance of the parallel combination of $2 \ \Omega$ and $3 \ \Omega$ resistors:$R_p = \frac{2 	imes 3}{2 + 3} = \frac{6}{5} = 1.2 \ \Omega$Now,the total resistance of the circuit is:$R_{eq} = R_p + 2.8 \ \Omega = 1.2 \ \Omega + 2.8 \ \Omega = 4 \ \Omega$The total current $I$ flowing from the battery is:$I = \frac{V}{R_{eq}} = \frac{6 \ V}{4 \ \Omega} = 1.5 \ A$This current $I$ splits into two branches containing $2 \ \Omega$ and $3 \ \Omega$ resistors. Using the current divider rule,the current $I_2$ through the $2 \ \Omega$ resistor is:$I_2 = I 	imes \frac{3}{2 + 3} = 1.5 	imes \frac{3}{5} = 0.3 	imes 3 = 0.9 \ A$

In the given circuit,the current through $2 \ \Omega$ resistor is

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