### JEE Main Solved Paper 2017 Question 29

##### Question: 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 Solved Paper-2017]

#### Options:

A) 24

B) 32

C) 2

D) 16

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

- 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 $