Three capacitors each of capacitance C and of breakdown voltage V are joined in series. The capacitance and breakdown voltage of the combination will beNEET 2009 (Screening)

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In series arrangement charge on each plate of each capacitor has same magnitude. The potential difference is distributed inversely in the ratio of capacitors ie, $V = V_{1} + V_{2} + V_{3} \dots$ Here, $V = 3 V$ The equivalent capacitance $C_{s}$ is given by
$\frac{1}{C _{s}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} + \dots C_{s} = \frac{C}{3}$

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