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Combination of Capacitors question from JEE Main 2017

Q82JEE Main 2017Difficult

Combination of Capacitors

A capacitance of 2 $\mu$ F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $\mu$ F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

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Answer: (c)

Each can take 300 V, so a series row must have $n$ with $300n \ge 1000$ : $n = 4$ .
One row: $\frac{1}{4}\ \mu\text{F}$ . Rows in parallel needed: $m\times\frac{1}{4} = 2 \Rightarrow m = 8$ .
Total $= n m = 4\times 8 = 32$
So the answer is (c) 32.

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