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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. Therefore,the current $

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$60$

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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. Therefore,the current $I_2 = 0 \ A$.Since the capacitor branch is open,the current $I_1$ flows through the resistor $R_1$ and then through the resistor $R_3$. Thus,$I_1 = I_3$.The total resistance of the circuit is $R_{eq} = R_1 + R_3 = 4 \ \Omega + 6 \ \Omega = 10 \ \Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{10 \ V}{10 \ \Omega} = 1 \ A$.The voltage across the capacitor $V_c$ is equal to the voltage drop across the resistor $R_3$ because they are in parallel with the capacitor branch (considering the steady state where no voltage drops across $R_2$).$V_c = I_3 	imes R_3 = 1 \ A 	imes 6 \ \Omega = 6 \ V$.The charge stored in the capacitor is $q = C 	imes V_c = 10 \ \mu F 	imes 6 \ V = 60 \ \mu C$.

In the electrical circuit shown below,the amount of charge stored in the capacitor is . . . . . . $\mu C$.

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