Power factor comes up in every electrical science course and again whenever you look at motors, lighting or a large load. This post explains what it means, gives the formulas you need, and works through an example from motor test readings. The examples are taken word for word from our Electrical Science Calculations workbook, with the same working and rounding as the book.
True, apparent and reactive power
In a DC circuit, or an AC circuit with resistance only, volts times amps gives the power. Once an AC circuit has reactance (a motor winding, a transformer, some lighting control gear) the current and voltage are no longer in step, and only part of the volt-amps does useful work.
- True power P, in watts (W): the power that does the work. Only resistance takes true power.
- Apparent power S, in volt-amps (VA): supply voltage times supply current.
- Reactive power Q, in volt-amps reactive (VAr): the part that does no useful work.
The three form a right-angled triangle, so S² = P² + Q². Power factor is the ratio of true power to apparent power: pf = P ÷ S, which is also cos φ, where φ is the angle between voltage and current.
The formulas
- S = U × I
- P = U × I × pf
- pf = P ÷ S
- Q = √(S² − P²)
Worked example: motor test readings
Question (workbook Q127, Core): On test, a single-phase motor circuit gives these readings: wattmeter 1200 W, ammeter 7.5 A, voltmeter 230 V.
Calculate (a) the apparent power, (b) the power factor and (c) the reactive power.
Method and numbers
- (a) S = U × I = 230 × 7.5 = 1725 VA
- (b) pf = P ÷ S = 1200 ÷ 1725 = 0.70
- (c) Q = √(S² − P²) = √(1725² − 1200²) = 1239.2 VAr
Result
- (a) Apparent power = 1725 VA
- (b) Power factor = 0.70
- (c) Reactive power = 1239.2 VAr
A power factor of 0.70 means only 70% of the volt-amps is doing useful work. The supply and the cable still have to carry the full current.
Why power factor matters
For the same true power, a lower power factor means more current. More current means larger cables, more voltage drop and more heat lost in the wiring. Correction capacitors supply reactive power close to the load, so less current is drawn from the supply. This example shows the effect.
Question (workbook Q129, Core): A 4 kW single-phase load at 230 V has its power factor improved from 0.7 to 0.95.
Calculate the current before and after, and the reduction in current.
Method and numbers
- Before: I = P ÷ (U × pf) = 4000 ÷ (230 × 0.7) = 24.845 A
- After: I = P ÷ (U × pf) = 4000 ÷ (230 × 0.95) = 18.307 A
- Reduction = 24.845 − 18.307 = 6.54 A
Result
- Before = 24.845 A
- After = 18.307 A
- Reduction = 6.54 A
The load still takes the same true power, but the supply current drops by about a quarter.
Common slips
- Mixing up watts and volt-amps. A wattmeter reads true power; voltmeter times ammeter gives apparent power.
- Adding the apparent powers of different loads. True powers add and reactive powers add, but apparent powers do not.
- Forgetting to convert kW to W, or kVA to VA, before dividing by the voltage.
- Putting a power factor into a formula as a percentage. Use 0.7, not 70.
A note on the figures
Practice data. Every figure a question needs is given in the question. Physical constants such as resistivity, temperature coefficients, the acceleration due to gravity and the permittivity of free space are typical textbook values, given so the questions can be worked. Prices, ratings and measurements are invented for practice. Do not use any figure in this book for design, installation or certification.
Practise more
Electrical Science Calculations (£9.99, PDF) has 200 questions in 15 sections, graded Easy, Core and Challenging, each with a full worked answer. It runs from units and Ohm’s law through AC waveforms, reactance and impedance, power factor and power-factor correction, to three-phase, transformers, motors, mechanics and lighting.
The Electrician Study Bundle (£34.99) adds Inspection and Testing Calculations, the BS 7671 A4:2026 Practice Papers and the 2391-52 Inspection and Testing Practice Papers, so you get all four PDFs together.
Written by a City & Guilds 2391-51 qualified electrician.