Power factor from real and apparent power:
formula and worked example
Power factor says how much of the current you draw is doing work, and how much is just sloshing back and forth.
Calcylator Editorial Team
Updated · 5 min read
Why a load can draw more than it uses
With AC supply, some loads, motors and transformers above all, draw current that is partly out of step with the voltage. That portion of the current moves energy back and forth between the source and the load's magnetic fields. It does not do work, yet it still flows through the cables, the switchgear and the supplier's network.
Power factor is the ratio that tells you how bad this is. It compares the real power, which turns the shaft or heats the element, with the apparent power, the product of voltage and current. A ratio of 1 means all the current is useful, and anything lower means some is wasted effort.
- Real kW: the power doing useful work, in kilowatts.
- Apparent kVA: the product of RMS voltage and current, in kilovolt-amperes.
- Result: a ratio between 0 and 1, sometimes written as a percentage.
The formula and the power triangle
- real_kw:
- Real kw
- apparent_kva:
- Apparent kva
Real power, reactive power and apparent power form a right-angled triangle. Real power is the base, reactive power (kVAR) is the height and apparent power is the hypotenuse. Power factor is the cosine of the angle between base and hypotenuse, which is the same as base divided by hypotenuse.
That geometry gives a second relationship you will use often: apparent power squared equals real power squared plus reactive power squared. If you know any two sides you can find the third.
Worked example: 12 kW from 15 kVA
A workshop motor load is measured at 12 kW real power and 15 kVA apparent power. What is the power factor, and how much reactive power is involved?
- Divide real by apparent: 12 ÷ 15 = 0.8.
- Square the sides: 15² = 225 and 12² = 144.
- Find the reactive power: √(225 − 144) = √81 = 9 kVAR.
- State the result: a power factor of 0.8, lagging if the load is inductive.
Real kw
12
Apparent kva
15
Power factor from real and apparent power
0.8 ratio
Worked from the inputs shown: 12 kW of real power from 15 kVA apparent power.
The 9 kVAR is the reactive component. It is not paid for as energy by most small consumers, but it fills the cables and equipment capacity all the same.
Reading the figure and what it costs
Typical readings depend on the load. Induction motors running lightly can show a power factor well below 0.8, while heaters and filament lamps are close to 1. The ranges that suppliers treat as acceptable differ by country and tariff, so check the actual rule that applies to your connection.
- Current: for the same kW, current is proportional to 1 ÷ power factor, so 0.8 draws 25% more than unity.
- Capacity: a 15 kVA supply or generator can deliver only 12 kW at 0.8, but 15 kW at unity.
- Losses: cable loss is proportional to the square of current, so the 25% extra current gives about 56% more loss in the wiring.
- Charges: many commercial tariffs penalise or reward the power factor, which varies by supplier.
The numbers also show why a low power factor is a capacity problem before it is an energy problem: the meter's kWh reading is unaffected, but the kVA you must be able to supply is not.
What happens when the real power moves
Holding the apparent power fixed at 15 kVA and raising the real power by 10%, from 12 to 13.2 kW, lifts the power factor from 0.8 to 0.88. Of course, in a real installation the two quantities rise together, so a bigger load of the same type gives the same power factor.
| Scenario | Real kw | Calculated result |
|---|---|---|
| Original assumptions | 12 | 0.8 ratio |
| Real kw increased by 10% | 13.2 | 0.88 ratio |
The better way to improve it is to cut reactive power. Capacitors supply the reactive power locally so it does not have to come from the grid. For the 12 kW load, raising the power factor to 0.95 needs about 5 kVAR of capacitors, which reduces the apparent power from 15 to roughly 12.6 kVA.
Mistakes with power factor
- Mixing units: dividing kW by VA or W by kVA gives an answer off by a factor of 1,000.
- Using peak instead of RMS values, which makes the apparent power wrong.
- Reading a single instant: loads vary, and the average over a billing period is what is usually charged.
- Over-correcting with too many capacitors, which pushes the power factor leading and can raise the voltage.
A scenario: checking a generator size
A 15 kVA generator is meant to run the 12 kW load above. At 0.8 power factor it supplies exactly 12 kW, with no margin, and the engine may handle more than the alternator. Add a 3 kW load and the apparent power becomes about 18.8 kVA, beyond the generator rating.
The solution is not always a larger generator: correcting the power factor to 0.95 brings the same 12 kW down to 12.6 kVA, and the extra load fits. The check is quick and often cheaper than the larger set.
Common questions
What is power factor?
It is the ratio of real power, which does useful work, to apparent power, the product of voltage and current. A value of 1 means all the current is useful, while lower values mean part of it only circulates.
How do I calculate power factor from kW and kVA?
Divide real power in kW by apparent power in kVA. With 12 kW and 15 kVA, the power factor is 12 ÷ 15 = 0.8. Both numbers must refer to the same load and the same period.
Is a power factor of 0.8 good or bad?
Common for motor-heavy loads, but usually considered low. Many suppliers and standards encourage values above roughly 0.9, though the exact threshold and any charge vary by country and tariff.
Does a poor power factor increase my kWh consumption?
Not directly, since kWh meters record real energy. But it raises current, cable losses and the capacity you need, and some tariffs add a charge or a demand penalty based on kVA or kVAR.
How can I improve power factor?
Add capacitors near the inductive loads to supply reactive power locally, or use automatic correction panels for varying loads. Avoid running motors lightly loaded, and avoid over-correction, which causes a leading power factor.
Was this guide helpful?
Continue reading
View all blogsCable voltage drop DC: formula and worked example
Estimate DC cable voltage drop from current, one-way length and resistance per metre, with a 10 A, 25 m example and ways to bring the loss down.
5 min read
Fuse load headroom: formula and worked example
Express a load current as a percentage of the fuse rating, with a 12 A on a 16 A example, how to turn it into spare capacity and a safe loading limit.
5 min read
Air conditioner daily energy use: formula and worked example
Estimate an air conditioner's daily energy use in kWh from its average power and running hours, with a checked 1.4 kW example and a note on duty cycle.
5 min read




