What the extra current is doing
At the defaults the shop takes 100 kW at a power factor of 0.78. Apparent power is 100 divided by 0.78, which is 128.2 kVA, and on a 480 volt three-phase feeder that is 154.2 amps. Correct to 0.95 and apparent power falls to 105.3 kVA and the current to 126.6 amps. The work being done never changed — the same motors turn the same shafts — but 27.6 amps have left the feeder.
The reactive part of that current is not consumed anywhere. It flows out to the motor windings, builds a magnetic field, collapses it and flows back, a hundred and twenty times a second on a 60 Hz supply. It does no work at either end and it heats the copper at both.
Heat falls faster than current
This is the part worth reading twice, and the page computes it rather than asserting it. Conductor loss goes as the square of the current, so a 17.9 percent reduction in current is a 32.6 percent reduction in heat. The ratio is exactly the square of the power factor ratio: 0.78 over 0.95 is 0.821, and 0.821 squared is 0.674, so 67.4 percent of the heat is left and 32.6 percent is gone.
Put a feeder resistance in and it becomes watts. At 0.05 ohms per conductor the three conductors were dissipating 3,567 W and now dissipate 2,405 W, which is 1,162 W saved continuously — 10,183 kWh a year if the load runs flat out all year, which no load does.
Wye and delta capacitors are not the same part
47.4 kVAR at 480 V and 60 Hz is 182 microfarads per phase if the three capacitors are connected in delta, and 545 microfarads per phase if they are connected in wye. The difference is exactly three, for the same reason the power difference on the wye and delta page is exactly three: kVAR goes as the square of the voltage across the capacitor, the delta capacitor sees 1.732 times the voltage, and 1.732 squared is 3.
The voltage rating diverges the same way. A delta capacitor lives at 480 V and a wye one at 277 V, so a bank assembled from the wrong cans is either badly undersized or badly overstressed. Neither of those announces itself on the day it is installed.
The bank is wrong at three in the morning
Fixed capacitors do not know what the load is doing. The 47.4 kVAR sized for a full shop exactly cancels the reactive power of the load once it falls to 59.0 kW, and below that the power factor goes leading. Leading power factor raises the bus voltage, is billed by some tariffs exactly as a lagging one is, and is genuinely unhelpful in front of a generator or a large drive.
That is the argument for a switched bank in steps, or for correction attached to individual motors so it comes and goes with them. Which is right for a particular site is not something arithmetic decides.
Where this stops
It stops at the capacitors. It does not size the cables to them, the switching device, the discharge arrangement or the protection, and it has no opinion on whether the existing feeder or panel can host them. A capacitor holds a charge after the supply is opened, which is the hazard people underestimate about this particular piece of equipment. And on a bus with a lot of drive and rectifier load, capacitors can resonate with the supply impedance at a harmonic frequency and draw far more current than any of this predicts. That is a power quality study, not a calculator.
Questions people ask
How many kVAR do I need to correct my power factor?
Multiply your real power in kW by the tangent of the angle you have now, do the same for the angle you want, and subtract. At 100 kW going from 0.78 to 0.95 that is 100 times 0.8023 minus 100 times 0.3287, which is 47.4 kVAR. The kW figure stays the same throughout — correction changes the reactive part and nothing else.
Why do wye capacitors need three times the capacitance of delta ones?
Because reactive power in a capacitor goes as the square of the voltage across it. A delta capacitor sits across the full line voltage and a wye one across the line-to-neutral voltage, which is 1.732 times smaller, so it needs 1.732 squared — three times — the capacitance to produce the same kVAR. Their voltage ratings differ by the same 1.732.
Does correcting power factor reduce my energy bill?
It reduces the current, so it reduces the heat lost in your own conductors, which is real but usually small. Whether it reduces the bill depends entirely on your tariff: if demand is billed in kVA or there is a separate power factor charge, correction moves that line; if demand is billed in kW, it does not. Read your own bill, and read a year of them.
What happens if I install too much capacitance?
The power factor goes past unity into leading. That raises the voltage on the bus, is billed the same way a lagging power factor is under many tariffs, and behaves badly in front of generators and drives. The page prints the load at which the bank you sized starts overcorrecting, which for the defaults is 59 kW, or 59 percent of the load it was sized for.
Can I put capacitors on a bus with variable frequency drives?
Not without somebody looking at the harmonics first. A capacitor bank and the supply impedance form a resonant circuit, and if that resonance lands near a harmonic the drives are producing, the capacitors draw far more current than their rating and fail early. It is a common failure mode on shop buses that look perfectly ordinary on paper, and it is a power quality study rather than a calculation.