In the circuit shown in the figure,C = 6,mu F | vedclass

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(C) The two capacitors $C$ and $2C$ are connected in series across a potential difference of $10\,V$.In a series combination,the equivalent capacitanc

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$40$

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(C) The two capacitors $C$ and $2C$ are connected in series across a potential difference of $10\,V$.In a series combination,the equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C 	imes 2C}{C + 2C} = \frac{2C^2}{3C} = \frac{2}{3}C$Given $C = 6\,\mu F$,we have:$C_{eq} = \frac{2}{3} 	imes 6\,\mu F = 4\,\mu F$The total charge $q$ supplied by 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 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,$C = 6\,\mu F$. The charge stored in the capacitor of capacity $C$ is......$\mu C$.

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