Capacitors of capacities C_1, C_2 and C_3 are | vedclass

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(C) In a series combination,the equivalent capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.$\frac{1}{C} = \f

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$\frac{C_2 C_3 V}{C_2 C_3+C_1 C_3+C_1 C_2}$

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(C) In a series combination,the equivalent capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.$\frac{1}{C} = \frac{C_2 C_3 + C_1 C_3 + C_1 C_2}{C_1 C_2 C_3}$.Therefore,$C = \frac{C_1 C_2 C_3}{C_2 C_3 + C_1 C_3 + C_1 C_2}$.The total charge $Q$ stored by the combination is $Q = CV = \frac{C_1 C_2 C_3 V}{C_2 C_3 + C_1 C_3 + C_1 C_2}$.In a series combination,the charge on each capacitor is the same and equal to the total charge $Q$.Therefore,the potential difference across capacitor $C_1$ is $V_1 = \frac{Q}{C_1}$.Substituting the value of $Q$,we get $V_1 = \frac{C_1 C_2 C_3 V}{C_1 (C_2 C_3 + C_1 C_3 + C_1 C_2)} = \frac{C_2 C_3 V}{C_2 C_3 + C_1 C_3 + C_1 C_2}$.

Capacitors of capacities $C_1, C_2$ and $C_3$ are connected in series. If the combination is connected to a supply of $V$ volt,then the potential difference across capacitor $C_1$ is

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