Question

Question 2.23: 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.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-89

Answer 1

if capacitors are connected in series then sum of potential actors each capacitor equal to potential actors circuit.
here, potential difference across circuit = 1kV
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 = 1 micro F
Ceq = 1/3 micro F

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

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