Question

electrician has large number of capacitors each of capacity 1 micro farad and each can with stand maximum voltage of 500 volts find minimum number of capacitors required to obtain a capacity of 2 microfarad across 3000 volts


*need explanation, rather than only answer*​

Answer 1

ANSWER:

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Explanation:

m×300>1000

m>

3

10

or m=4

For m=4, capacitance of 1 branch is

4

1μF

=0.25μF

For overall capacitance of 2μF, 8 such branches or equivalent

is C

eq

=8×0.25=2μF

Total capacitors required= 8×4=32