The question requires more details.
First the "K" represents kilo or 1000.
The real power (kW) can be thought of as the x-axis and the
reactive power (kVar) from inductance or capacitance can be
thought of as the y-axis. The hypotenuse is the
Apparent power (KVA). The relationship of Real to apparent
power gives the Power Factor, where a power factor of 1.0
simply means that you only have real power, no reactive
power. The leading and lagging VARs from capacitance and
inductance, respectively can be added numerically since
they are on the same axis.
To ask, how muck kVars are required for a given kW doesn’t
make a lot of sense because you need to include the desired
If Pure Capacitor 90 Leads so The power factor is Unity(1)
*Sin(90)=1(But Reactive power is zero)
since I=Kvar/1.732*V*1 Eg. 1000/1.732*415*1=1.39A
other formula for Determine KW=KVAR/tan(pi)or KW=VIcos(PI)
Assume * cos(0.9)
since KW=KVAR/tan....Equ 1, take the cos invesr is equal to (0.9)=25.8419 ...Equ 2
they are more formulas
1.After synchronising the alternator to the grid,how we r
delivering power to the grid?if the i/p increases does the
speed hence frequency increases?
2.y the power factor is changing when exitation changes?what
is the relation with p.f and flux?