Capacitors of capacities C_1 and C_2 are conn | vedclass

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(B) When capacitors are connected in series, the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_

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

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(B) When capacitors are connected in series, the charge $Q$ on each capacitor is the same.The equivalent capacitance $C_{eq}$ 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 stored in the combination is $Q = C_{eq} V = \left( \frac{C_1 C_2}{C_1 + C_2} \right) V$.The potential difference across capacitor $C_2$ is $V_2 = \frac{Q}{C_2}$.Substituting the value of $Q$, we get $V_2 = \frac{1}{C_2} \left( \frac{C_1 C_2}{C_1 + C_2} \right) V = \frac{C_1 V}{C_1 + C_2}$.

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

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