Load will have P, Q.
P1=VI cos pi
Q1= VI sin pi
Cos pi1= 0.6 or 0.7
To improve power factor to 0.9 (cos pi2)
Need to place capacitor in parallel with load.
Q2= VIsinpi2
Q2-Q1= Net reactive power to be supplied by capacitor=Q
Capacitor bank will have rating in V and Q.
The appropriate Q will be placed in parallel with Load to achive 0.9 powerfactor.
