Question

An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer 1

answer : 18

if capacitors are connected in series then sum of potential actors each capacitor equal to potential actors circuit.

here, potential difference across circuit = 1kV = 1000V

and potential across each capacitor not more than 400V

so, minimum number of capacitors that must be connected in series in a row are n = 1kV/400V = 1000V/400V = 2.5 ≈ 3

capacitance of 3 capacitors in series in a row is 1/Ceq = 1/C + 1/C + 1/C

here, C = 1μF

so, 1/Ceq = 1/1μF + 1/1μF + 1/1μF

Ceq = 1/3 μF

minimum number of rows of 3 capacitors each to be connected in parallel to obtain net capacitance of 2μF are m = 2μF/(1/3 μF ) = 6

so the minimum number of capacitors required = m × n = 3 × 6 = 18