A Power Factor Correction Example: Given a 240 Volt load of 32.5 KW at 0.65 power factor lagging (KVA-50), find the required capacitance to correct the pf to unity (pf-1), and find the % reduction in feeder current resulting from the pf correction. Feeder wm Source V 240V Load V-240 Power 32.5KW pf-0.65 KVA=KW | pf=50KVA VA-50000VA 1-KVA 1000/V-208.334 W-P*1000 32500W zmag V/I 1.15202 pf-W/VA=0.65 phai acosd(pf)-49.4584 = z=zmag (cosd (phai)+jsind(phai))=0.7488+ 0.8754 2 VAR=VA* sind (phai)=37997VARS Pfnew=1 With Unity pf, the new VA is equal to the power and the capacitance absorbs all the VARS VAnew-W-32500 Capacitance VARs-37997 xc V2/37997-1.51590 C=1/(2*pi*60*xc)=0.0017 Farad Inew-VAnew/V-135.14.4 Ireduction (1-Inew)*100/7-35%