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(D) Initial charge on $2 \mu F$ capacitor: $Q_1 = C_1 V_1 = 2 \mu F 	imes 150 \ V = 300 \mu C$.Initial charge on $3 \mu F$ capacitor: $Q_2 = C_2 V_2

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all of the above

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(D) Initial charge on $2 \mu F$ capacitor: $Q_1 = C_1 V_1 = 2 \mu F 	imes 150 \ V = 300 \mu C$.Initial charge on $3 \mu F$ capacitor: $Q_2 = C_2 V_2 = 3 \mu F 	imes 120 \ V = 360 \mu C$.When the $1.5 \mu F$ capacitor is connected,let the charge on it be $x \mu C$. By charge conservation and Kirchhoff's Voltage Law,the potential difference across the $1.5 \mu F$ capacitor must equal the sum of potential differences across the other two capacitors in the loop.Let the new charges be $Q_1' = 300 - x$ and $Q_2' = 360 - x$.Applying $KVL$: $\frac{x}{1.5} = \frac{300 - x}{2} + \frac{360 - x}{3}$.Multiplying by $6$: $4x = 3(300 - x) + 2(360 - x) \implies 4x = 900 - 3x + 720 - 2x \implies 9x = 1620 \implies x = 180 \mu C$.Charge on $2 \mu F$ capacitor: $300 - 180 = 120 \mu C$.Since the initial charge on the $2 \mu F$ capacitor was $300 \mu C$ and it becomes $120 \mu C$,charge flows out of it. Similarly,charge flows out of the $3 \mu F$ capacitor. The net flow through point $A$ is from right to left to balance the potentials. Thus,all options are correct.

Two capacitors of $2 \mu F$ and $3 \mu F$ are charged to $150 \ V$ and $120 \ V$ respectively. The plates of the capacitors are connected as shown in the figure. $A$ discharged capacitor of capacity $1.5 \mu F$ is connected to the free ends of the wire. Then:

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