1.Existing P.F of Load (tanQ1)(0.92)
2.Required P.F (Improvement in P.F)-(tanQ2)(0.99)
3.Supply Voltage - for reference(Voltage variation if any)
4.Load in KW (exam- availabl with you i.e 500KW)

KVAR= KW(tanQ1-tanQ2)
KVAR= 500(tan0.92- tan0.99)
KVAR = 500(0.2890)
KVAR = 141.75 KVAR (@ 142KVAR)
if you want to calculate by amp need following data
efficiency of motor, required P.F, KW,maganatising current
of motor,formula find in Electrical Machine Engineering

plz identify your systems existing power factor and
targeted power factor which u want to improve, then apply
this formula
KW X (Tan phi 1 - tan phi 2)

250 Kvar Capacitor bank is required for 500Kw load
(i.e,625Kva).
This is because the capacitor rating for any load is equal
to 40% of Kva.
i.e, Capacitor required=40%x Kva Load

Reactive power (KVAR) is increase due to run inductive load.as a result power factor is lagging. so have to use capacitive power to reduce this inductive or reactive load. generally we have to use 63% capacitive load to remove the reactive load.i.e.

active power=500KW
so 60% of this will be=(60*500/100) = 300 KVAR

so we have to use 300KVAr capacitive capacitor bank to remove this reactive power .

Is it mandatory statutory requirement of transformer oil
filtration annualy despite oil is withstanding required
dieelectrical strenth and all others parameters

we all know that in transformer, voltage and current are
inversely related i.e. Vp/Vs= Is/Ip....so in standard
CT..we have a ratio of 200/5...that means our voltage
should increase by ratio of 40..althoug that is not the
case...because that requires high insulation level.and
pratically not feasible......can anybody clarify my
ques.......thanx