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

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(C) The two capacitors $C$ and $2C$ are connected in series across the $10 \ V$ battery.The equivalent capacitance $C_{eq}$ of two capacitors in serie

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$40$

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(C) The two capacitors $C$ and $2C$ are connected in series across the $10 \ V$ battery.The equivalent capacitance $C_{eq}$ of two capacitors in series is given by:$C_{eq} = \frac{C 	imes 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C$Given $C = 6 \ \mu F$,the equivalent capacitance is:$C_{eq} = \frac{2}{3} 	imes 6 \ \mu F = 4 \ \mu F$The total charge $q$ drawn from the battery is:$q = C_{eq} 	imes V = 4 \ \mu F 	imes 10 \ V = 40 \ \mu C$Since the capacitors are in series,the same amount of charge $q$ is stored on each capacitor. Therefore,the charge stored in the capacitor of capacity $C$ is $40 \ \mu C$.

In the circuit shown in the figure,the battery is $10 \ V$ and $C = 6 \ \mu F$. The charge stored in the capacitor of capacity $C$ is......$\mu C$.

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