A plant manager once showed me a utility bill with a penalty line for low power factor and asked a simple question: how much capacitor do we actually need? Not a rule of thumb, not a guess — a number we could defend. Here is the worked example I walked through with them.
The setup
A three-phase load draws 500 kW at 480 V, line-to-line, running at 0.75 power factor. The utility starts charging penalties below about 0.9 to 0.95 power factor, so the target is 0.95. Real power stays the same after correction — only the reactive part changes.
The formula
The capacitor rating needed is:
Qc = P x [tan(arccos PF1) - tan(arccos PF2)]
where P is real power, PF1 is the existing power factor and PF2 is the target. Apparent power is S = P / PF, and line current falls in the same ratio as S. This is the shunt-capacitor approach in IEEE 1036, with capacitor practice covered by IEC 60831.
The numbers
For 500 kW, 0.75 to 0.95:
- tan(arccos 0.75) = 0.882, and tan(arccos 0.95) = 0.329
- Qc = 500 x (0.882 - 0.329) = 276.6 kvar
- Apparent power falls from 666.7 kVA to 526.3 kVA
- Line current falls from 801.9 A to 633.1 A — a 21.1% reduction
- The capacitor bank itself draws about 332.7 A at 480 V
That 21% current reduction is the part people underestimate. It frees capacity in the transformer and cables feeding the load, and it cuts the I-squared-R losses along the way. The kvar number is what you buy; the current reduction is what the rest of the system feels.
Try it yourself
If you want to run your own numbers, the free Power Factor Correction Calculator from GridCircuits does this exact calculation. It shows its formula on the page and runs entirely in your browser — nothing is uploaded. Put in your kW, your existing power factor and your target, and it gives the capacitor kvar, the kVA before and after, and the current reduction.
Three things to check before you order
- Pick the next standard bank size up. The calculation gives 276.6 kvar; you buy a standard size at or above that, not a custom 276.6 kvar bank.
- Check for harmonic resonance. Capacitors can resonate with system inductance where drives or rectifiers are present, so a harmonic check belongs in the job, not after it.
- Do not over-correct. Correcting to a leading power factor at light load causes its own problems. Switched or automatic banks help when the load varies through the day.
Why this matters
Power factor correction is one of those jobs where the math is short and the consequences are long. A single worked example — 500 kW, 0.75 to 0.95, 276.6 kvar — turns a penalty line on a bill into a purchase order with a number behind it. Run the numbers first, then size the bank.
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