Capacitor range is actually given in Kvar, so to find that
range in kvar

As we know
Power factor = KW/KVA
for 65 Hp load
KW= 65HP*0.746 = 48.49KW
Considering Preent P.F=0.50
Target pf =0.99
Multiplying Factor by standard table= 1.59
Kvar= 48.49 KW * 1.59 = 77.09 Kvar

hence for the load of 65hp, capacitor range will be 77.09
Kvar.

as pf not giving ,taking pf 0.8
let pf of load =0.8
kw of load= 65 hp=65*0.746=48.5kw
kva =48.5/0.8=60.625kva
kvar =sqrt(60.625^2-48.5^2)=36.375kvar
if we want to improve pf to 0.95 then
improved kva= 48.5/0.95=51.052
improved kvar= sqrt(51.052^2-48.5^2)=15.9411
requard capacitor range = 36.375-15.94=20.43kvar.

first we should calculate the load(inductive or capacitive
or resistive).
in case of resistive load-----
in case of resistive load the, actual power factor will be
01(maximum),in this case the voltage will linior with
respecte to load current.
so tatal resistive load is(in example 65 hp),
so total load in kw = 65*0.746
=48.49.
As we know
Power factor = KW/KVA
for 65 Hp load(if pure resistive)

Considering Preent P.F=1
Target pf =0.99
Multiplying Factor by standard table= 1.56
Kvar= 48.49 KW * 1.56= 75.64

hence for the load of 65hp, capacitor range will be 75.64
Kvar.(in pure resistor).

in single phase we get 220 volt and in three phase 4420 v
why this
second Ques= in case three phase star conection our
neutrul becom disconect what happan
3 Q=== a three phase house load and have unbalce load on
different pahse diffrent current eg phase a 7A b 8A c
5A and voltage is 440 than how calculate it total load in
kw pl give formula thanks