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(B) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit simplifies t

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$0.9$

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(B) In the steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.The circuit simplifies to a $6\,V$ battery connected in series with a $2.8\,\Omega$ resistor and a parallel combination of $2\,\Omega$ and $3\,\Omega$ resistors.The equivalent resistance of the parallel part is $R_p = \frac{2 	imes 3}{2 + 3} = \frac{6}{5} = 1.2\,\Omega$.The total resistance of the circuit is $R_{eq} = 2.8\,\Omega + 1.2\,\Omega = 4.0\,\Omega$.The total current supplied by the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{4\,\Omega} = 1.5\,A$.Using the current divider rule,the current through the $2\,\Omega$ resistor is $I_2 = I 	imes \frac{3}{2 + 3} = 1.5 	imes \frac{3}{5} = 1.5 	imes 0.6 = 0.9\,A$.

In the circuit shown,the capacitance of the capacitor is $2\,\mu F$. Find the current flowing through the $2\,\Omega$ resistor in $A$.

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