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Question:

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:

32

2

16

24

Solution:

m × 300 > 1000
m > 10/3
orm = 4
For m = 4, capacitance of 1 branch is 1μF/4 = 0.25μF
For overall capacitance of 2μF, 8 such branches or equivalent is
Ceq = 8 × 0.25 = 2μF
Total capacitors required = 8 × 4 = 32