Power Stations WORLDWIDETHE GLOBAL EDITION 01
LESSON 4 OF 6 · ORIGINAL EDUCATIONAL EXAMPLE

Power factor and load current

Calculate PF = P/S and see why a lower power factor increases current at fixed real power and voltage.

Learning goal: Separate true power factor from displacement power factor and avoid interpreting PF as efficiency.

Prepared by Power Stations Worldwide. Sources checked on 30 September 2026; not independently engineering peer reviewed.

The essentials

A ratio, not an efficiency

For a consuming load, true power factor is average real power divided by apparent power: PF = P/S. Efficiency instead compares useful output with input. An efficient machine can still have a non-unity power factor.

When cos φ is sufficient

For sinusoidal voltage and current, PF = cos φ, where φ is their phase angle. Lagging current is typical of an inductive load; leading current is typical of a capacitive load. With harmonic distortion, cos φ of the fundamental alone is displacement power factor, not necessarily true PF.

Same kW, more amperes

At fixed real power and voltage, reducing PF raises apparent power and current. Higher current increases I²R losses in a given resistance. Improving PF does not automatically reduce the load’s useful-energy demand or establish a particular electricity-bill saving.

Lagging current at PF = 0.80

Lagging current at PF = 0.80A voltage phasor points horizontally right. A current phasor of normalized length lags clockwise by 36.87 degrees; the marked angle has cosine 0.80. Lengths are normalized independently, not a comparison between volts and amperes.Voltage VCurrent Iφ = 36.87°cos φ = 0.80 · I lags V
Original phasor illustration with counterclockwise rotation convention. Angle φ ≈ 36.87°; lengths are independently normalized and do not compare voltage with current.
HYPOTHETICAL VALUES · NOT PLANT RECORDS

Worked example

20 kW at 230 V: PF = 0.80 versus PF = 1

Given

  • Single-phase electrical input P = 20 kW = 20,000 W
  • Voltage V = 230 V RMS
  • Compare a sinusoidal lagging load at PF = 0.80 with PF = 1

Formula toolbox

  • PF = P / S; for sinusoids PF = cos φ
  • Single-phase: S = P / PF; I = P / (V × PF)

Calculate step by step

  1. At PF = 0.80: S = 20 / 0.80 = 25 kVA; I = 20,000 / (230 × 0.80) ≈ 108.70 A.
  2. At PF = 1: S = 20 kVA; I = 20,000 / 230 ≈ 86.96 A.
  3. Current ratio = 108.70 / 86.96 ≈ 1.25: 25% more current at PF = 0.80.

108.70 A at PF 0.80 · 86.96 A at PF 1

Both cases transfer 20 kW of real power. This comparison does not prescribe capacitors or protection; harmonic resonance and local network conditions need separate assessment.

Results are rounded only for display. Reproduce the calculations using unrounded intermediate values.

Quick quiz

Choose one answer for each question, then check your answers. This is a practice check, not a qualification. Answers stay in this page only; refreshing or leaving resets them.

1. A consuming load takes 18 kW and 20 kVA. What is its true power factor?
2. At the same real power and supply voltage, what happens when PF decreases?

No answers checked yet.

Show answers and explanations (also works without JavaScript)
  1. 0.90. PF = P / S = 18 / 20 = 0.90. This ratio alone does not identify leading or lagging operation.
  2. Current increases. I is inversely proportional to PF for the stated model. Efficiency and PF are different quantities.

Curriculum connection: MIT 6.061 (Spring 2011). Its readings include 2007 notes; some equations use peak rather than RMS amplitudes. This lesson states its own value conventions and uses original diagrams, numbers and quiz questions. No MIT affiliation or endorsement is claimed.

Textbook references

Suggested technical background for this topic. These books are further reading, not proof that every explanation was checked against every edition. See the full bibliography and review scope.

  1. Fundamentals of Electric Circuits (7th ed.) — Alexander, C. K. & Sadiku, M. N. O., McGraw-Hill, 2021 · ISBN 9781260226409AC power and power-factor calculations.
  2. Electric Circuits (11th ed.) — Nilsson, J. W. & Riedel, S. A., Pearson, 2019 · ISBN 9780134746968Sinusoidal steady-state power and balanced three-phase loads.