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JEE Main 2017 · 2 Apr · Shift 1, Physics question 15

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF 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 :

Choose one option
Answer and worked solution

Official answer: D

Each capacitor takes at most 300 V, so 1000 V needs 4 in series, giving 14\frac14 μF per string. For 2 μF, 8 strings are needed in parallel: 4×8=324\times8=32 capacitors.

Topics:Series and parallel combinations
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