Industrial sub-station with a 4 MW load has a capacitor of 2 MVAR to operate at 0.97 pf. What will be p.f if the capacitor goes out of service?

P.F without capacitor will be 0.8(lag)

Inductive MVAR with capacitor= (4/0.97)*sin(cosinv(0.97))=1MVAR

So, Inductive MVAR without capacitor=1+2=3MVAR

Apparent power=sqrt(4*4+3*3)=5MVA

P.F without capacitor=(4/5)=0.8