Power factor and the power triangle
What you’ll learn
- The three kinds of power: real (W), reactive (VAR) and apparent (VA).
- The power triangle, and power factor as P ÷ S or cos θ.
- Which loads make a lagging power factor, and which lead.
- Why low power factor costs current, capacity and money — and how capacitors correct it.
- The same arithmetic on three-phase, with √3.
Power factor is the share of the current you supply that actually does work. A motor draws some current to do real work and some just to build its magnetic field every cycle; power factor = real power ÷ apparent power, a number from 0 to 1. The lower it is, the more amps the supply, the transformer and the conductors must carry for the same useful output.
Three kinds of power
| Potencia | Symbol · unit | What it is |
|---|---|---|
| Real (true, active) | P · watts (W) | Turned into heat, light, motion. What the meter bills as kWh. |
| Reactive | Q · volt-amps reactive (VAR) | Borrowed and handed back each cycle by coils and capacitors. Does no work, but still flows in the wires. |
| Apparent | S · volt-amps (VA) | Volts × amps. What the supply has to deliver, and what heats the conductors. |
The power triangle
Real and reactive power are 90° apart — the same way resistance and reactance were in the last lesson — so they combine as the sides of a right triangle, with apparent power as the long side.
- S = V × IVA, single-phase
- P = S × PFwatts
- Q = √(S² − P²)VAR
- PF = P ÷ S = cos θ0 to 1
1. A single-phase motor load
- 240 V, 20 A, power factor 0.8 lagging.
- S = 240 × 20 = 4,800 VA.
- P = 4,800 × 0.8 = 3,840 W.
- Q = √(4,800² − 3,840²) = √(23,040,000 − 14,745,600) = √8,294,400 = 2,880 VAR.
- θ = cos⁻¹ 0.8 = 36.9°.
Lagging and leading
Anything with a magnetic circuit — induction motors, transformers, magnetic ballasts, welders — needs magnetizing current that lags the voltage: a lagging power factor. Capacitors do the opposite and make a leading one. Heaters and incandescent lamps are near unity. Most buildings lag, because most of what spins is a motor.
The supply has to deliver volt-amps, not just watts. At 0.8 power factor, every 100 W of work needs 125 VA of capacity — 25% more current in the service, the transformer and every conductor, and more heat lost in all of them. Utilities pass that cost on to large customers.
Correcting it with capacitors
Capacitors draw leading VARs. Put them across a lagging load and their VARs cancel part of the motor’s, so the supply only has to provide the difference. The watts don’t change — the motor does the same work — but the VA, and the current, drop.
2. Correcting 0.8 to 0.95
- Start: P = 3,840 W, Q = 2,880 VAR, 20 A at 240 V.
- Target 0.95: Snew = P ÷ 0.95 = 3,840 ÷ 0.95 = 4,042 VA.
- Qnew = √(4,042² − 3,840²) ≈ 1,262 VAR.
- Capacitor needed: 2,880 − 1,262 ≈ 1,618 VAR (about 1.6 kVAR).
- New current: 4,042 ÷ 240 ≈ 16.8 A, down from 20 A, for exactly the same work.
On three-phase
The triangle is the same; only apparent power changes. With line-to-line voltage and line current, S = √3 × VL × IL, and P = S × PF as before. (The three-phase lesson explains where √3 comes from.)
3. A 600 V motor (and its 480 V cousin)
- 600 V three-phase, 50 A per line, PF 0.85.
- S = 1.732 × 600 × 50 = 51.96 kVA.
- P = 51.96 × 0.85 = 44.17 kW. Q = 51.96 × √(1 − 0.85²) = 51.96 × 0.527 ≈ 27.4 kVAR.
- The same 50 A at 480 V (a common US system): S = 1.732 × 480 × 50 = 41.57 kVA, P = 35.33 kW.
Try one: 120 V, 8 A, 720 W measured — what is the power factor?
S = 120 × 8 = 960 VA. PF = 720 ÷ 960 = 0.75. Q = √(960² − 720²) = √403,200 ≈ 635 VAR.
En la obra
Tips from the job
- Motor nameplates often give a power factor at full load. It drops when the motor is lightly loaded — an oversized motor running near idle is a power-factor problem.
- Many utilities bill commercial and industrial customers on kVA demand or add a penalty below a set power factor. That is why correction banks exist.
- Correction capacitors store energy. Treat a bank as live until its discharge time has passed and you have metered it dead.
- A clamp meter reads amps; watts need a power meter that sees the phase angle. V × A on a motor gives VA, not W — see reading a motor nameplate.
Common mistakes
- Calling V × A “watts” on a motor load. It is volt-amps; watts need the power factor.
- Adding W and VAR straight to get VA. They are at right angles: S = √(P² + Q²).
- Thinking better power factor makes a motor do more work. It lowers the current; the work is the same.
- Forgetting √3 on three-phase, or using line-to-neutral volts with it.
Check yourself
8 questions. Tap an answer — it is marked straight away and the reason opens under it. Get 80% or better and the lesson is ticked off on the year page.
Not started yet.
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Power factor is the ratio of…
Answer and why
B. PF = P ÷ S = real power (W) ÷ apparent power (VA). It is also the cosine of the angle between voltage and current.
-
A 240 V load draws 20 A at a power factor of 0.8. How many watts?
Answer and why
B. S = 240 × 20 = 4,800 VA. P = S × PF = 4,800 × 0.8 = 3,840 W.
-
Same load: how many VARs?
Answer and why
B. Q = √(S² − P²) = √(4,800² − 3,840²) = √8,294,400 = 2,880 VAR. (Or S × sin θ, with θ = 36.9°.)
-
Which kind of load typically makes a lagging power factor?
Answer and why
C. Motors, transformers and magnetic ballasts need magnetizing current, which lags. A lightly loaded motor is among the worst. Heaters and incandescent lamps are close to unity; capacitors lead.
-
What is the usual fix for a low lagging power factor?
Answer and why
B. Capacitors draw leading VARs that cancel the motors’ lagging VARs, so less current has to come from the supply for the same real work.
-
Improving power factor on a load that still does the same work…
Answer and why
B. The watts stay the same; the VA — and therefore the amps the supply and conductors carry — go down. In the worked example, 20 A drops to about 16.8 A.
-
What is apparent power (S) for a three-phase load at 600 V line-to-line, 50 A per line?
Answer and why
B. S = √3 × VL × IL = 1.732 × 600 × 50 ≈ 51,960 VA ≈ 52 kVA.
-
A power factor of 1.0 (unity) means…
Answer and why
B. At unity the angle is 0°, cos 0° = 1, VARs are zero and every volt-amp does real work. A purely resistive load is unity.
Preguntas frecuentes
What is a good power factor?
Unity (1.0) is perfect. Most facilities aim to stay somewhere above about 0.9, and utilities that penalize low power factor set their own threshold in the rate schedule. Resistive loads are at or near 1; induction motors at full load are commonly in the 0.8s and worse when lightly loaded.
What is the difference between leading and lagging power factor?
Lagging means the current peaks after the voltage — inductive loads like motors and transformers. Leading means the current peaks before the voltage — capacitive loads. Correction adds the opposite kind to cancel the VARs.
Why are transformers and generators rated in kVA instead of kW?
Because their windings heat up from the current they carry, whatever the power factor of the load. The kVA rating limits the current; how many kW that turns into depends on the load.
Ver también
- Power factor (glossary) — the short definition.
- Reactance and impedance — the lesson before this one — the same triangle in ohms.
- Single-phase transformers — why they are rated in kVA.
- Leer la placa de datos de un motor
- Single & three phase formulas — printable.
- All second year lessons
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