Power Factor Calculator
Price the correction — kvar of capacitors to move a load from the power factor it has to the one it should.
Power Factor Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Displacement PF — harmonics and distortion need filters
- One target; stepped banks interpolate in blocks
In short: A 50 kW load running at an 0.75 power factor draws 66.666667 kVA of apparent power; correcting to 0.95 takes 27.66165 kvar of capacitors and the same 50 kW reads 52.631579 kVA — 21.052632% of the current simply vanishes from the bill. The kW never moved; the burden did. That is the entire business case, and the utility's power-factor penalty is the deadline on it.
Formula
kvar = kW × (tan(acos PF₁) − tan(acos PF₂))
The power triangle: real power kW and reactive kvar make the apparent kVA, and the power factor is the ratio of work to burden. Capacitors supply kvar locally, so the utility only ships the work. The correction is the difference of tangents of the two phase angles, times the real power — a formula old enough to be on every nameplate sticker, quoted at 0.8 for a reason.
Worked Example
- Enter the real kW and the measured power factor.
- Pick the target — utilities typically stop penalizing near 95%.
- Read the kvar, then check the current that disappears.
Defaults: 27.66165 kvar to move 0.75 → 0.95. Drive the target to 100% and the bank reads 44.095855 kvar — chasing the last points to unity costs more bank than the first ninety did.
Strengths & Limits Of This Model
Where this engine is strong
- The before/after kVA prints the whole case on one line
- The tangent arithmetic shown, not hidden in a factor table
Where it stops
- Distorted waveforms need harmonic analysis first
Practical Use Cases
Penalty audits
the utility's PF line, defused
Bank sizing
kvar before the capacitor order
Capacity frees
kVA back out of an existing feeder
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Power Factor Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
What is power factor, in one sentence?
The fraction of current doing real work: 0.75 means three-quarters of the amps carry watts and the rest slosh between source and inductance, heating your wiring and the utility's on the way.
Why do utilities penalize a low power factor?
Because their copper carries your kVA, not your kW. A 0.75 plant makes them ship 33% more current than your metered energy justifies — the penalty is that extra copper, priced.
How is the kvar formula derived?
Reactive power is kW times the tangent of the phase angle, and the angle is arccosine of the power factor. The bank supplies the difference between the two tangents — 0.881917 at 0.75 minus 0.328684 at 0.95, times 50 kW, is the 27.66165 kvar on the default bill.
Why stop at 0.95 and not 1.00?
Because each added point costs more kvar than the last, past-unity banks resonate with the supply, and no utility rewards unity beyond their target. 95% is where the penalty ends and the payback flattens.
What does the plant gain besides the penalty?
Current: 66.666667 kVA falls to 52.631579 at the same kW — 21% fewer amps, which frees feeder and transformer capacity and shaves I²R losses by the square of that fraction.
Do motors or VFDs change this?
Induction motors and old lighting are the classic inductive loads — an unloaded motor is the worst offender. VFD front ends vary: older six-pulse drives distort (harmonics, displacement), which needs filters, not plain capacitors.
Where do the capacitors go?
At the load for maximum benefit (motor banks switch with the motor), or centrally at the main for penalty control only. Same kvar number; different share of the copper relieved.
Is a fixed bank ever wrong?
On lightly loaded feeders, yes — a fixed bank overshoots at low load and pushes the power factor leading. That is why automatic banks step in blocks, following the load the way this formula follows it.