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(B) When $n$ capacitors of capacitance $C$ are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \

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$C/3, 3V$

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(B) When $n$ capacitors of capacitance $C$ are connected in series,the equivalent capacitance $C_{	ext{eq}}$ is given by $\frac{1}{C_{	ext{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.Thus,$C_{	ext{eq}} = \frac{C}{3}$.In a series combination,the potential difference across each capacitor is $V$. Since the capacitors are identical,the total breakdown voltage $V_{	ext{total}}$ of the combination is the sum of the individual breakdown voltages.$V_{	ext{total}} = V + V + V = 3V$.Therefore,the combination has a capacitance of $C/3$ and a breakdown voltage of $3V$.

Three capacitors each of capacitance $C$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be

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