Every electrical measurement you have ever taken of an AC circuit gives you two numbers: volts and amps. Multiplying them should give you the power the circuit is delivering. On a resistive load — a toaster, an electric fire, an incandescent lamp — it does. On a motor, a fluorescent lamp, a computer supply or an LED driver, it does not, and the gap between the two numbers is one of the most expensive inefficiencies in the whole subject.
What power factor measures
In an AC circuit the voltage and the current both oscillate at the same frequency, but not necessarily in step. The power factor is the cosine of the angle by which they are out of step — how closely the current peak lines up with the voltage peak:
PF = cos φ
It ranges from 0 to 1. A resistive load draws current in exact step with the voltage, so φ = 0 and PF = 1. A purely inductive load — an ideal coil — draws current a quarter cycle late, so φ = 90° and PF = 0. Most real loads sit somewhere in between, and a PF below 1 means current is flowing that does no useful work.
The reason is that an inductor stores energy in its magnetic field on the way up and has to release it on the way down. The current required to build the field is not the current that does the work, but the supply has to provide it anyway, every cycle, forever. A capacitor does the same thing with an electric field and releases it in the opposite half of the cycle, which is why capacitive and inductive loads can offset each other — the basis of compensation.
The power factor calculator takes active power, apparent power or the phase angle and returns the rest, so you can work in whichever units a meter gives you.
The three powers and the triangle
The relationships are geometric, and drawing them makes the vocabulary self-explanatory.
Real (active) power, P — measured in watts (W). This is the power doing useful work: turning a motor, heating an element, lighting a lamp. It is the only part that earns its keep.
Reactive power, Q — measured in volt-amperes reactive (var). This is the power that builds and collapses magnetic and electric fields. It does no work, it is genuinely consumed and regenerated in a cycle, and it loads the supply without producing anything.
Apparent power, S — measured in volt-amperes (VA). This is what the supply has to deliver, and it is simply the product of the two measured quantities:
S = V × I
Because the reactive part sits at 90° to the useful part, the three form a right-angled triangle:
S² = P² + Q²
and since PF = P ÷ S, the power factor is the cosine of the angle φ at the P–S corner, with Q opposite it. The two consequences that matter in practice are:
P = S × PF and Q = P × tan φ
The first is why the same supply delivers less real power as the power factor falls. The second is the sizing equation for compensation, and it is used in the worked example below.
Worked example 1: 30 kW at a power factor of 0.7
A three-phase installation on 400 V, 50 Hz, has a real power of 30 kW and a power factor of 0.7. Find the apparent power, the current, the reactive power, and then the current after compensation to 0.95.
Step 1 — apparent power. S = P ÷ PF = 30 ÷ 0.7 = 42.86 kVA
Step 2 — line current. For three phase, I = P ÷ (√3 × V × PF):
I = 30,000 ÷ (1.732 × 400 × 0.7) = 30,000 ÷ 485.0 = 61.9 A
At unity power factor the same 30 kW would need only 43.3 A. The extra 18.6 A does no work at all — it exists purely because the load is inductive.
Step 3 — the reactive power behind that. φ = cos-1(0.7) = 45.6°, so tan φ = 1.020:
Q = P × tan φ = 30 × 1.020 = 30.6 kvar
Step 4 — check the triangle. S = √(P² + Q²) = √(900 + 936.4) = √1836.4 = 42.86 kVA. ✓ Matches step 1 exactly, which is the check that the phase angle was computed correctly.
Step 5 — the penalty in plain terms. The installation needs 42.86 kVA of supply to deliver 30 kW of useful power. About 30% of the capacity being paid for exists only to shuttle energy back and forth inside fields. And because losses scale with I², that unused 30% is the most expensive part of the installation to carry.
Worked example 2: compensating to 0.95
The load is mostly motors, so capacitors can be installed in parallel to supply the reactive part locally and bring the supply's view of the power factor up to 0.95.
Step 1 — new apparent power. S = 30 ÷ 0.95 = 31.58 kVA
Step 2 — new line current. I = 30,000 ÷ (1.732 × 400 × 0.95) = 30,000 ÷ 658.2 = 45.6 A
Step 3 — the current reduction. From 61.9 A to 45.6 A — a fall of 16.3 A, or 26%. The real power is untouched at 30 kW; the supply is simply no longer carrying current that does nothing.
Step 4 — what that does to cable losses. Losses go as I², so the ratio is:
(45.6 ÷ 61.9)² = (0.7367)² = 0.54
Cable losses fall to 54% of their former value — a 46% reduction — for the same delivered power. This is the single most persuasive number in the whole subject, because it is a saving that recurs on every hour of every day for the life of the installation, and it comes entirely from reducing current.
Step 5 — how much capacitance. The target power factor has its own angle: φ₂ = cos-1(0.95) = 18.2°, so tan φ₂ = 0.329. The residual reactive power is Q₂ = 30 × 0.329 = 9.9 kvar. The capacitors must therefore supply:
Qc = Q₁ − Q₂ = 30.6 − 9.9 = 20.7 kvar
or, working directly from the two power factors, Qc = P × (tan φ₁ − tan φ₂) = 30 × (1.020 − 0.329) = 20.7 kvar. ✓
So a 21 kvar bank at 400 V is the answer. The power calculator will produce the same figures from the reactive power, if that is what the site survey gives you.
Where these power factors come from
Power factor is a property of the load, so knowing typical values tells you a great deal about what a piece of equipment will do to your supply.
- Resistive loads: 1.0. Heating elements, incandescent lamps, resistive water heaters. Nothing to correct.
- Induction motors: 0.75 to 0.85 at full load. And much worse at part load — a lightly loaded motor can sit below 0.4, which is why a bank sized for full load is switched out when the machines are idle. A motor is also the classic case for individual local compensation rather than site-wide correction.
- Fluorescent and discharge lighting: below 0.5 for older magnetic-ballast fittings, because the ballast is a large inductor. Magnetic ballasts are also the reason such fittings need a starter, and the reason replacing them with electronic or LED equivalents usually improves the site power factor for free.
- LED drivers: variable, and increasingly regulated. Good ones exceed 0.9; cheap ones can be well below. There is now a regulatory requirement in several markets for power factor above 0.9 for LED lamps above about 10 W, precisely because a large number of poor drivers would otherwise degrade the supply.
- Computers, TVs and switched-mode supplies: often 0.6 to 0.8. Their power supplies take current in narrow pulses; the true current is much higher than the average power justifies, and the gap shows up as a poor power factor.
- Transformers: 0.9 to 0.98 unloaded. A lightly loaded transformer is magnetising-current dominated and has a poor power factor, which is another reason not to oversize transformers.
Worth noting how the mix changes over time. A site that replaced its fluorescent lighting with LED and fitted new motors instead of old ones has probably improved its power factor substantially without doing anything deliberate. The voltage drop guide covers the other consequence of the current reduction, since lower current means lower drop as well as lower loss.
Harmonics and why a meter can disagree with theory
Everything above assumes a sinusoidal load, where the only deviation is the phase shift. Modern equipment breaks that assumption. Switched-mode power supplies, variable frequency drives, LED drivers, UPS units and most modern appliances draw current in narrow pulses rather than smooth waves. These are non-linear loads, and they generate harmonic currents at multiples of the fundamental frequency.
Harmonics matter to power factor in two ways. They add extra current that does no useful work, so the measured power factor falls even when the displacement angle is small. And because the harmonic currents superimpose on the fundamental, the peak current in the conductor is much higher than the RMS total would suggest — the crest factor — and peak current is what stresses conductors and drives cable sizing.
Capacitors need care here. A capacitor bank in a harmonic-rich supply can go into resonance with the network inductance, which amplifies voltage distortion rather than removing it. Detuned reactors are fitted in series with the capacitors specifically to prevent this, and they are not optional in a modern installation with substantial LED or drive load. Oversizing a bank on a site whose power factor is already poor is a common way to make the power quality worse rather than better.
The practical response to harmonics is a harmonic filter, not more capacitance, and the two problems are often confused. If the measured power factor is low and the current waveform is visibly distorted, a capacitor bank will not fix it. If the load is simply inductive and the waveform is clean, capacitors are exactly right.
Single phase versus three phase
For a single-phase load the current is straightforward:
I = P ÷ (V × PF)
A 2 kW heater at 230 V: at a power factor of 1.0 the current is 2000 ÷ 230 = 8.7 A. If the same real power were delivered at a power factor of 0.7, the current would be 2000 ÷ (230 × 0.7) = 12.4 A — a 43% increase in current for exactly the same useful output. Compensated to 0.95, it returns to 9.15 A, and the loss ratio is (9.15 ÷ 12.4)² = 0.54, the same 54% as the three-phase case. The ratio depends only on the power factors, not on the voltage or the phase count.
Three-phase compensation is normally done with a capacitor bank on a common bus, in stepped stages controlled by a power factor relay, so the bank can be switched out as the load falls. This matters because a fixed bank sized for full load will overcorrect at light load and push the power factor above 1.0, which means leading current — harmful to the supply, to the transformer and to the meter, and increasingly penalised.
Contactor sizes: at 400 V, one amp of capacitor current is 400 var = 0.4 kvar. A 21 kvar bank therefore draws about 52 A, which is a meaningful switching current and a meaningful cable. Distribution of stages — often split across the three phases to keep currents balanced — is part of the installation design, not an afterthought.
What it costs, and what to do about it
Three separate costs, which are often conflated:
1. Capacity you pay for and do not use. The installation needs 42.86 kVA to deliver 30 kW. If your tariff, your transformer rating or your supply agreement is expressed in kVA, that 43% overhead is billed directly. This is the most obvious cost and the easiest to see.
2. Cable losses, forever. I²R with I at 61.9 A instead of 45.6 A. On a long run this is a permanent operating cost that dwarfs any one-off installation saving, and it also increases conductor temperature, which shortens cable life.
3. Bill penalties. Many suppliers impose a reactive energy charge or a power factor adjustment on the monthly bill above a threshold, commonly 0.9 or 0.95. The watt calculator will not do this — it is a tariff question, and it needs your actual invoice to answer — but the direction is not in doubt: a lower power factor costs money wherever reactive power is charged or penalised.
The fix, in order of preference: reduce the reactive demand at the source first (replace magnetic-ballast fluorescents, use better drivers, avoid over-motoring), then install local compensation on the largest individual loads, then a site-wide stepped bank, and only then harmonic filters if the waveform is distorted. A capacitor bank bolted onto a site that has a harmonic problem is a partial fix that can make the power quality worse, which is why the diagnosis has to come before the equipment.
One last check worth doing before sizing anything: measure the power factor at different times of day and at different load levels, rather than at a single instant. A bank sized from a single measurement will be wrong for most of the year, and the most common fault in practice is a bank that overcorrects when the machines are off. An automatic stepped bank with a relay responds to the measurement continuously, which is why it is the standard solution rather than a fixed bank.
Frequently asked questions
What is a good power factor?
The closer to 1.0 the better, because every unit of reactive power is capacity you pay for and do not use. Electrical installations are commonly penalised or refused below 0.9, and utilities often require at least 0.95 at the point of supply for large loads. Motors typically arrive at 0.75 to 0.85, and industrial sites compensate to bring the site total above 0.95.
Will power factor correction actually reduce my electricity bill?
It reduces the bill only where your tariff penalises low power factor, or where you are billed on a maximum demand that the higher current raises. On a flat per-kWh tariff, correcting the power factor does not change the useful energy you consume, so the saving is zero. It reliably reduces losses in your own cables and frees up capacity in transformers and supply.
Why do LED lamps have such a poor power factor?
Because the driver is a switched-mode power supply, and switching a large capacitor's charge a few thousand times a second draws current in short pulses. Averaged over a cycle, the current peaks much higher than the power warrants, which shows up as a low measured power factor. Good drivers use active power factor correction to push it above 0.9, and regulations now require it.
How do I size a compensation bank for a three-phase load?
Calculate the reactive power at the present power factor, then the reactive power at the target, and install the difference: Qc = P × (tan φ1 − tan φ2). For 30 kW at 0.7 moving to 0.95, that is 30 × (1.020 − 0.329) = 20.7 kvar. Install it in automatic stages so the bank can be switched out when the load drops.