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 Solved Paper-2017

A)  24                                         

B)  32

C)  2                                            

D)  16

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 \[=8\times 4=32\]