Three capacitors with capacitances of 1µF, 2µF and 3µF are connected in series. Each capacitor gets punctured, if a potential difference just exceeding 100 volt is applied. If the group is connected across 220 volt circuit then the capacitor most likely to puncture first is

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capacitor of capacitance 3µF

capacitor of capacitance 1µF

capacitor of capacitance 2µF

All three will be punctured at the same instant.

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{C_{eq}} = 1 + \frac{1}{2} + \frac{1}{3} \Rightarrow C_{eq} = \frac{6}{11} {μFQ}_{eq} = 120 μC \\ V_{1} = 120 V; V_{2} = 60 V; V_{3} = 40 V as V_{1} > 100 V \end{array}$

Hence 1µF = get punctured.

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