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A battery of emf 20V is connect to two capacitors 1µF and 3µF in series.1µF capacitor withstands a maximum of 9V and 3µF withstands a maximum voltage of 6V then

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1µF capacitors break

2µF capacitors break

no capacitor will break

both will break

answer is A.

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Detailed Solution

$\frac{V_{1}}{V_{2}} = \frac{C_{2}}{C_{1}} = \frac{3}{1}; V = V_{1} + V_{2} = 20$
1µF capacitor
$V_{1} = \frac{3}{4} \times 20 = 15>$ 9V so it breaks
3µF capacitor
$V_{2} = \frac{1}{4} \times 20 = 5 <$ 6V so it withstands

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