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JEE Main 2017 · Combination of Capacitors

Combination of Capacitors question from JEE Main 2017

Q82JEE Main 2017Difficult

Combination of Capacitors

A capacitance of 2 μ\muF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μ\muF 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)

  1. Each can take 300 V, so a series row must have nn with 300n≥1000300n \ge 1000: n=4n = 4.
  2. One row: 14 μF\frac{1}{4}\ \mu\text{F}. Rows in parallel needed: m×14=2⇒m=8m\times\frac{1}{4} = 2 \Rightarrow m = 8.
  3. Total =nm=4×8=32= n m = 4\times 8 = 32
  4. So the answer is (c) 32.
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