Two capacitances of capacity C_1 and C_2 are | vedclass

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(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ of the series combination

Vedclass · Accepted Answer

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

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(C) When capacitors are connected in series,the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ of the series combination is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{C_1 + C_2}{C_1 C_2}$,so $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$.The total charge $Q$ stored in the combination is $Q = C_{eq} V = \frac{C_1 C_2 V}{C_1 + C_2}$.The potential difference $V_1$ across capacitor $C_1$ is given by $V_1 = \frac{Q}{C_1}$.Substituting the value of $Q$,we get $V_1 = \frac{1}{C_1} 	imes \left( \frac{C_1 C_2 V}{C_1 + C_2} ight) = \frac{C_2 V}{C_1 + C_2}$.

Two capacitances of capacity $C_1$ and $C_2$ are connected in series and a potential difference $V$ is applied across the combination. The potential difference across $C_1$ will be:

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