question_answer

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:                            [JEE Main 2017]

A) 24

B) 32

C) 2

D) 16

Correct Answer: B

Solution :

[b] To hold 1 KV potential difference minimum four capacitors are required in series

\[\Rightarrow \]\[{{C}_{1}}=\frac{1}{4}\]for one series.

So for Ceq to be \[2\mu F,8\]parallel combinations are required.

\[\Rightarrow \] Minimum no. of capacitors \[=8\times 4=32\]