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(C) In the steady state,the capacitor acts as an open circuit for $DC$. $1$. The $10\,\mu F$ capacitor is in series with the $DC$ source. In the stead

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$120\,\mu C, 0\,\mu C$

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(C) In the steady state,the capacitor acts as an open circuit for $DC$. $1$. The $10\,\mu F$ capacitor is in series with the $DC$ source. In the steady state,it gets fully charged to the source voltage of $12\,V$. Therefore,the charge on the $10\,\mu F$ capacitor is $Q_1 = C_1 V = 10\,\mu F 	imes 12\,V = 120\,\mu C$. $2$. The $24\,\mu F$ capacitor is connected in parallel with a $4\,\Omega$ resistor. In the steady state,the current through the circuit is zero because the $10\,\mu F$ capacitor blocks the $DC$ current. Since there is no current flowing through the $4\,\Omega$ resistor,the potential difference across it is $V_R = I 	imes R = 0 	imes 4 = 0\,V$. $3$. Since the $24\,\mu F$ capacitor is in parallel with the resistor,the potential difference across it is also $0\,V$. Thus,the charge on the $24\,\mu F$ capacitor is $Q_2 = C_2 V_2 = 24\,\mu F 	imes 0\,V = 0\,\mu C$. Therefore,the charges are $120\,\mu C$ and $0\,\mu C$ respectively.

In the given circuit,the charge on the capacitors of capacitance $10\,\mu F$ and $24\,\mu F$ in the steady state will be:

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