Answer / r n seth

capacitor required is too be calculated from required power
factor to desired power factor.the formula is-

Req. Cap in KVAr= Tr. MVA* PF( tan cos-1(existinf PF- tan
cos-1(desired PF))

Answer / engr. tariq

Formula :

C = P (tan|old - tan| new) divided by Omega (w) and then
divided by Voltage (RMS) or Vrms.

theta (|) can be calculated by knowing cos theta OR power
factor. Example PF is 0.77 then Cos theta = 0.77 and Cos
inverse (0.77) = 39.646 = Theta OLD. Now

Now if i want to improve PF to 0.9 then theta = Cos inverse
0.99 = 25.841 = theta New.

After calculating theta put it in the Given formula to
calculate Capacitance required.

Answer / engbri

Another way i use is to take the PF and turn it into an
angular measurement in degrees. using j notation and the
magnitude of the wave you can calculate the size of
resistance, or real component and the size of the inductive
reactance or the -J component. Do this with a simple Polar
to rectangular conversion, if you cant your calculator
probably can. Then making the capacitive reactance equal to
the inductive reactance cancels it as they are oposites.
Then using the formula Xc = 1/2piFc transpose for c which
gives c =1/2piFXc. however as the cpacitor is taken as
ideal and with no resistive component this will never
reduce the PF to zero.

Answer / charu gupta

kVAr Rating = kW(tan A - tan B)
Where kVAr rating is the size of the capacitor needed,kW is
the average power drawn, tan A is the trigonometric ratio
for the present power facotr, and tan B is the
trigonometric ratio for the desired PF

Answer / engr. tariq

Please note that in answer # 5 omega (w) is equal to the
frequency of the system.

Answer / salman

to find above answer so we can directly go through
contract technical book this easiest way.

Answer / shah rukh khan

With the use of Calculator.. lol.. ;)

Answer / pavan vyas

as a thumb rule it should be 1/3 of load kw & theoretically
we can select from the chart

Answer / p.chandrasekharan nair

KVA LOAD*0.6

Answer / sachin gupta

transformer

why dont we connect the two 12 V DC batteries in parallel, for an inverter (UPS)? is there any harm ?

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