Question

A balanced bank of delta-connected capacitors is connected in parallel with the load which complex power associated with each phase of a balanced load is 146 191j kVA. The effect is to place a capacitor in parallel with the load in each phase. The line voltage at the terminals of the load thus remains at 2460 V . The circuit is operating at a frequency of 65 Hz . The capacitors are adjusted so that the magnitude of the line current feeding the parallel combination of the load and capacitor bank is at its minimum.

What is the size of each capacitor?

Answer 1

77.2805 μF

Step-by-step explanation:

Given data :

V = 2460 V

Q = 191 Kva

Calculate the size of Each capacitor

first step : calculate for the value of Xc

Q = V^2/ Xc

Xc ( capacitive reactance ) = V^2 / Q = 2460^2 / ( 191 * 10^3 ) = 31.683 Ω

Given that 1 / 2πFc = 31.683

∴ C ( size of each capacitor ) =
`(1)/(2\pi *65 *31.683)` = 77.2805 μF