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 :

 

  • Option 1)

    2

  • Option 2)

    16

  • Option 3)

    24

  • Option 4)

    32

 

Answers (2)

As we learnt in 

Series Grouping -

\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\cdots

- wherein

 

 \frac{1}{C_{equation}}= \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}

\frac{1}{C_{equation}}= \frac{4}{8}= \frac{1}{2}\; \; \! C_{equation}= 2\mu f

\therefore Total\ number\ of\ capacitance = 8\times 4= 32

 


Option 1)

2

Incorrect Option

Option 2)

16

Incorrect Option

Option 3)

24

Incorrect Option

Option 4)

32

Correct Option

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