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What Is Power Factor? The Current Your Watts Do Not Show

Watts describe the result. Volt-amps describe the burden — and conductor heating follows the square of the difference.

By Mohamed Zakrya

Updated · 9 min read

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Watts are the result; volt-amps are the burden What is power factor The meter bills one number. The wires carry the other. 1 · THE WORK Real power, in watts. What the meter bills. 800 W The job being done. 2 · THE BURDEN Apparent power, in volt-amps. 1,000 VA What the supply delivers. 3 · THE RATIO 800 ÷ 1,000 0.80 The rest is 600 VAR — a 3-4-5 triangle × 200. 4 · THE CURRENT I is set by VA, not W. × 1.2500 25% more current for exactly the same job. AND THEN IT GETS SQUARED Heating follows current squared 25 percent more current is not 25 percent more heat. Every ampere is in the squared term, so the conductor loss goes to 1.5625 — and voltage drop, following current, to 1.2500. 56.25% more heat · loss doubles at pf 0.7071 Who actually pays A residential meter bills real kilowatt-hours, and a demand charge stated in kW is billed on real power — so a poor power factor inflates neither of them. kVA demand or a pf penalty is what costs money This explains the physics and the billing. It does not size correction. No capacitors, no target power factor, no payback — those need the load profile, the harmonics and the actual tariff. The arithmetic here is the single-phase form the calculator carries; three-phase adds a root-three factor it does not.
Watts describe the result. Volt-amps describe the burden — and conductor heating follows the square of the difference.

An AC load has two power numbers. Real power, measured in watts, is the part converted into useful work — heat, light, motion, or whatever the load exists to produce — and it is the quantity behind the kilowatt-hours an ordinary residential meter records.

Apparent power, measured in volt-amps, describes what the supply has to deliver. The wires, the breaker, the transformer and everything else upstream carry current according to that larger figure, even where part of it produces no net work.

Power factor is real power divided by apparent power, so a load using 800 W while demanding 1,000 VA has a power factor of 0.80. The missing portion is not unaccounted energy; it is reactive power moving back and forth between the supply and the load.

That distinction has consequences a household bill never shows. Current is set by volt-amps rather than watts, and resistive heating rises with the square of current, so a modest increase in current creates a much larger increase in conductor loss.

The power triangle

Real, reactive and apparent power form a right triangle. Real power is one perpendicular side, reactive power is the other, and apparent power is the hypotenuse, so power factor compares the real-power side against that hypotenuse.

The power factor calculator starts with 800 W of real power and 1,000 VA of apparent power.

power factor = real power ÷ apparent power
power factor = 800 W ÷ 1,000 VA
power factor = 0.80

Reactive power comes from the same relationship. Square the apparent power, subtract the squared real power, and take the root.

reactive power = √(apparent power² − real power²)
reactive power = √(1,000² − 800²)
reactive power = √360,000
reactive power = 600 VAR

The check closes exactly, which is worth doing once. This is not a loose diagram or a figure softened by rounding — the three default values satisfy the identity to the digit.

800² + 600² = 1,000,000
1,000² = 1,000,000
800² + 600² = 1,000²

The proportions look familiar because 800, 600 and 1,000 are a 3-4-5 right triangle scaled by 200. That makes the default unusually good for seeing how the three powers relate without the arithmetic getting in the way.

Reactive power is energy that flows out from the source and returns during each AC cycle. It does no net work across a complete cycle, and the back-and-forth still requires current in the conductors and capacity through the whole supply.

Its unit is the volt-amp reactive, written VAR. The separate name keeps it distinguishable from watts even though both come from voltage and current, and apparent power takes the plain volt-amp because it describes the combined burden the hypotenuse represents.

The default is a 3-4-5 triangle scaled by 200 Three powers, one right angle Drawn to scale: 320 px is 800 W, 240 px is 600 VAR, and the hypotenuse lands on exactly 400 px. 800 W real what does the work, and what the meter bills 600 VAR reactive — out and back each cycle 1,000 VA apparent IT CLOSES EXACTLY 800² + 600² = 1,000,000 = 1,000² POWER FACTOR 800 ÷ 1,000 0.80 A 3-4-5 triangle scaled by 200 — which is why these particular defaults are so easy to check by eye.
800 W of real power and 600 VAR of reactive power close to 1,000 VA — a 3-4-5 triangle scaled by 200.

Current follows the volt-amps

The practical point is that current follows apparent power. In the single-phase relationship the calculator uses, current is volt-amps divided by volts, which means watts alone cannot establish a current unless the power factor is known too.

current = apparent power ÷ voltage
apparent power = real power ÷ power factor
current = real power ÷ (voltage × power factor)

Holding real power and voltage fixed exposes what power factor does. The unity-power-factor current becomes the reference, and the multiplier at any lower power factor is its reciprocal.

current ratio = 1 ÷ power factor
current ratio at 0.80 = 1 ÷ 0.80
current ratio at 0.80 = 1.2500

A ratio of 1.2500 means the conductors carry 25.0 percent more current than they would delivering the same real power at unity. Nothing about the useful output has increased; only the burden required to support it.

That extra current does no extra work. The load still converts 800 W into its result while the conductors, the breaker and the transformer accommodate the current belonging to 1,000 VA. Watts describe the result, volt-amps describe the burden, and only one of the two appears on a domestic meter.

This is also where a common misreading starts. Reactive power is not an extra block of watt-hours being consumed alongside the real power — it is a component of AC power flow that enlarges current and apparent power without adding any net work.

Loss follows current squared

Current is only the first half of the consequence. Resistive heating in a conductor varies with current squared, so relative loss is the square of relative current, and substituting the reciprocal of power factor gives the multiplier directly.

loss ratio = current ratio²
current ratio = 1 ÷ power factor
loss ratio = 1 ÷ power factor²

At a power factor of 0.80 the current multiplier is 1.2500, and squaring it gives 1.5625 — 56.25 percent more heat in the same conductors while the useful work is unchanged.

loss ratio = 1 ÷ 0.80²
loss ratio = 1.5625
extra loss = (1.5625 − 1.0000) × 100%
extra loss = 56.25%

That square is why power factor matters physically. A 25.0 percent rise in current does not produce merely 25.0 percent more heating: every ampere participates in the squared relationship, which turns the current increase into 56.25 percent more loss.

The ladder below holds real power at 800 W throughout. Apparent and reactive power move with the power factor, while the last three columns compare each row against a unity-power-factor load doing exactly the same useful work.

Power factorApparent power (VA)Reactive power (VAR)Current ratioLoss ratioMore loss
1.00800.00.01.0000×1.0000×0.00%
0.95842.1262.91.0526×1.1080×10.80%
0.90888.9387.51.1111×1.2346×23.46%
0.85941.2495.81.1765×1.3841×38.41%
0.801,000.0600.01.2500×1.5625×56.25%
0.701,142.9816.21.4286×2.0408×104.08%
0.601,333.31,066.71.6667×2.7778×177.78%

The pattern is not linear. Equal-looking steps in power factor do not produce equal steps in loss, because the reciprocal is being squared, and the lower the starting point the more sharply the loss responds to another reduction.

There is an exact milestone at which conductor loss doubles against unity power factor, and it falls at the reciprocal of the square root of two.

loss ratio = 1 ÷ power factor²
2 = 1 ÷ power factor²
power factor = 1 ÷ √2
power factor = 0.7071

Reactive power grows quickly too, relative to the fixed 800 W of real power. Setting it out as a share of that figure makes the changing balance easier to read.

Power factorReactive power (VAR)Reactive power ÷ real power
0.95262.90.329×
0.90387.50.484×
0.80600.00.750×
0.601,066.71.333×

At 0.60 the reactive power is 1.333 times the real power, so the circuit carries more reactive than real while the load still performs only the work the unchanged 800 W represents.

None of which implies that VAR can be added to watts. They are perpendicular components of the triangle, so apparent power comes from combining them geometrically, which is exactly why the calculator uses a square root rather than ordinary addition.

Current rises; loss rises squared The same 800 W, five power factors Both bars are multiples of what a power factor of 1.00 would need for identical useful work. PF CURRENT × CONDUCTOR LOSS × 1.00 1.0000 1.0000 0.90 1.1111 1.2346 0.80 1.2500 1.5625 — 56.25% more heat 0.70 1.4286 2.0408 0.60 1.6667 2.7778 loss doubles at pf 0.7071 current = 1 ÷ pf loss = 1 ÷ pf² Same watts, same job. Only the burden on the wire changes.
Current rises as one over power factor and conductor loss as one over its square, so 0.80 costs 25 percent more current and 56.25 percent more heat.

What it does to a voltage drop

Voltage drop follows current rather than current squared. With the cable, the length, the material and the voltage otherwise unchanged, the drop therefore moves by the current multiplier itself, so a power factor of 0.80 gives 1.2500 times the current and the same 1.2500 in drop.

voltage-drop ratio = current ratio
voltage-drop ratio = 1 ÷ power factor
voltage-drop ratio at 0.80 = 1 ÷ 0.80
voltage-drop ratio at 0.80 = 1.2500

The unchanged cable takes 25.0 percent more drop while supplying the same real power. That is a smaller penalty than the 56.25 percent in heating, and the difference between the two is the difference between a quantity proportional to current and one proportional to its square.

The division of labour between the two calculations is worth stating. Power factor supplies the current a load demands; the cable calculation then takes that current, with the conductor properties and the run length, and returns the drop. What size wire do I need covers that second step.

A wiring calculation based only on useful watts quietly assumes a power factor of one. Where the real figure is lower, that assumption understates the current, and therefore understates both the drop and the heating — with the heating understated by the larger squared amount.

One follows current, the other follows its square One power factor, two different penalties A power factor of 0.80 on an unchanged cable, carrying the same real power to the same load. THE CAUSE current × 1.2500 VOLTAGE DROP follows current × 1.2500 25.0% more drop CONDUCTOR LOSS follows current squared × 1.5625 56.25% more heat drop ∝ I loss ∝ I² The exponent is the whole difference between the two numbers. Power factor supplies the current. The cable calculation takes it from there.
Voltage drop follows current, so a 0.80 power factor costs 25 percent more drop on a cable that never changed.

Who actually pays for it

Residential meters bill real energy in kilowatt-hours. A poor power factor on a household appliance is therefore not something a homeowner can find as an added line of billed energy, nor reduce by correcting it — worth understanding, but not worth chasing as a way to lower a domestic bill.

The wires still carry the current that belongs to the apparent power, and every physical effect above remains real. The billing point is narrower than the physics: a residential kilowatt-hour total records real energy and not the enlarged volt-amp requirement.

A demand charge stated in kW is also billed on real power, so a poor power factor does not inflate a kW demand charge. The current may be higher and the apparent-power burden larger, and neither changes the meaning of a quantity expressly billed in kilowatts.

That point gets blurred because current, capacity and demand are usually discussed together, but the unit printed on the tariff is what settles it. What is a demand charge sets out how a kW demand structure is measured and billed.

Commercial billing makes power factor financially visible in two situations. The first is a tariff that bills demand in kVA rather than kW: since apparent power rises as power factor falls for fixed real power, a kVA measurement captures the larger burden directly.

The second is an explicit power-factor penalty applied below a threshold set by the tariff. Thresholds, measurement methods, averaging rules and penalty formulas all vary, so the tariff sheet is the only place to establish whether either provision applies to an account.

The right billing question is therefore not whether a low power factor always costs money. It is whether the applicable tariff bills kVA demand or carries a power-factor clause. Without one of the two, the physics still holds and the customer-facing charge may not move at all.

What this does not decide

These relationships do not select power-factor correction equipment. They do not size capacitors, set an installation target, or estimate cost and payback, all of which need the load profile, the switching conditions, the supply characteristics, the harmonic environment and the exact tariff.

The formulas here use the single-phase form the calculator carries. Three-phase arithmetic introduces a root-three factor and its own line quantities, so this current relationship should not be transplanted into a three-phase calculation unchanged.

A nameplate power factor is a rated figure at a stated load rather than a promise the equipment holds that value under every condition. The actual figure moves with the load and the operating state, and a lightly loaded motor generally runs at a worse power factor than its nameplate.

For a measured operating point the calculator answers a narrower question: it relates real, apparent and reactive power at that point. The power factor that falls out then gives the current multiplier, and its square gives the conductor-loss multiplier that watts alone never show.

Questions people ask

Why does a 0.80 power factor cost 56 percent more in conductor heating?

Because the current increase gets squared. Current is set by volt-amps rather than watts, so for a fixed amount of real work it scales as 1 divided by the power factor: at 0.80 that is 1.2500, or 25.0 percent more current. Resistive heating in a conductor varies with current squared, so the loss multiplier is 1 divided by 0.80 squared, which is 1.5625 — 56.25 percent more heat in the same wires while the useful output has not changed at all.

How much extra current does a low power factor draw?

The reciprocal of the power factor, for the same real power. Since apparent power is watts divided by power factor, and current is apparent power divided by volts, the current ratio is simply 1 divided by the power factor. At 0.90 that is 1.1111, at 0.80 it is 1.2500, at 0.70 it is 1.4286 and at 0.60 it is 1.6667. On the calculator default of 800 W and 1,000 VA, the conductors carry the current belonging to 1,000 VA even though only 800 W is doing work.

At what power factor does conductor loss double?

At 0.7071, which is 1 divided by the square root of two. Setting the loss multiplier 1 ÷ power factor squared equal to 2 and solving gives power factor = 1 ÷ √2 exactly. It is a useful marker because it sits between two figures people quote often: at 0.80 the loss is up 56.25 percent, and by 0.70 it has passed the doubling point at 2.0408, or 104.08 percent more. At 0.60 the multiplier reaches 2.7778, close to triple.

Does a poor power factor increase my kW demand charge?

No, and this is widely assumed the other way around. A demand charge stated in kilowatts is billed on real power, so the higher current and larger apparent-power burden do not change a quantity expressly measured in kW. What does make power factor cost money is a tariff that bills demand in kVA instead, since apparent power rises as power factor falls, or one that applies an explicit power-factor penalty below a stated threshold. The tariff sheet is where to check which, if either, applies.

Does power factor make voltage drop worse?

Yes, but by the current ratio rather than its square. Voltage drop is proportional to current, so at a power factor of 0.80 an otherwise unchanged cable takes 1.2500 times the drop — 25.0 percent more — while carrying the same real power to the same load. That is a smaller penalty than the 56.25 percent increase in conductor heating, and the gap between the two is exactly the difference between a quantity that follows current and one that follows current squared.

Can reactive power be added to real power?

No. Real power and reactive power are perpendicular components of the power triangle, so they combine geometrically rather than by ordinary addition, which is why apparent power comes from a square root. On the default figures, 800 W and 600 VAR give an apparent power of the square root of 800 squared plus 600 squared, which is 1,000 VA — not 1,400. Reactive power is also not an extra block of watt-hours being consumed; it flows out and back each cycle and does no net work.

Should I install power factor correction?

That is not a question this arithmetic can answer, and the guide deliberately does not. Selecting correction equipment, sizing capacitors, choosing a target power factor and estimating any payback all depend on the load profile, the switching conditions, the supply characteristics, the harmonic environment, equipment constraints and the exact tariff. What the calculator gives you is the relationship between real, apparent and reactive power at a measured operating point, plus the current and loss multipliers that follow from it.

Does this arithmetic work for three-phase?

Not as written. The relationships here use the single-phase form the calculator carries, and three-phase power arithmetic introduces a factor of the square root of three along with its own line and phase quantities. The power factor definition itself, real power divided by apparent power, still holds, but the current relationship should not be transplanted into a three-phase calculation unchanged. Note too that a nameplate power factor is a rated figure at a stated load, and a lightly loaded motor generally runs at a worse one.