In the circuit shown in the figure,the potent | vedclass

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(D) The capacitors $3\,\mu F$ and $6\,\mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 3\,\mu F + 6\,\mu F = 9\,\mu F$.Now,thi

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$8$

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(D) The capacitors $3\,\mu F$ and $6\,\mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 3\,\mu F + 6\,\mu F = 9\,\mu F$.Now,this equivalent capacitor $C_p$ is in series with the $4.5\,\mu F$ capacitor.The total capacitance $C_{eq}$ of the circuit is given by $\frac{1}{C_{eq}} = \frac{1}{4.5} + \frac{1}{9} = \frac{2+1}{9} = \frac{3}{9} = \frac{1}{3}$.So,$C_{eq} = 3\,\mu F$.The total charge $Q$ supplied by the $12\,V$ battery is $Q = C_{eq} 	imes V_{total} = 3\,\mu F 	imes 12\,V = 36\,\mu C$.Since the $4.5\,\mu F$ capacitor is in series with the combination,the same charge $Q = 36\,\mu C$ flows through it.The potential difference $V_{4.5}$ across the $4.5\,\mu F$ capacitor is $V_{4.5} = \frac{Q}{C} = \frac{36\,\mu C}{4.5\,\mu F} = 8\,V$.

In the circuit shown in the figure,the potential difference across the $4.5\,\mu F$ capacitor is.......$V$.

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