Answer:
77.2805 μF
Explanation:
Given data :
V = 2460 V
Q = 191 Â Kva
Calculate  the size of Each capacitor
first step : calculate for the value of Xc Â
 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 ) = [tex]\frac{1}{2\pi *65 *31.683}[/tex]  =  77.2805 μF