You multiply an AC motor's voltage by its current and get 2,400. Is that 2,400 watts? For a single-phase load, RMS volts × RMS amps gives apparent power in volt-amperes (VA). To obtain real electrical input power in watts, you also need power factor. This original study example helps 313A candidates identify the requested quantity before calculating.
Which quantity does each formula describe?
| Quantity | Single-phase relationship | Unit |
|---|---|---|
| Apparent electrical power, S | VRMS × IRMS | VA |
| Real electrical input power, P | S × PF | W |
| Power factor, PF | P ÷ S | No unit |
Yokogawa's motor power guide, checked 2026-09-27, distinguishes these quantities. For a consuming load with power factor below one, watts are lower than VA. Keep the unit next to each intermediate answer so that the numerical product does not silently change its meaning.
How does power factor change a worked answer?
Suppose an educational example specifies a single-phase motor with 240 V RMS, 10 A RMS and a true power factor of 0.80. Find its real electrical input power.
- Calculate apparent power: 240 × 10 = 2,400 VA.
- Apply power factor: 2,400 × 0.80 = 1,920 W.
- Convert if requested: 1,920 ÷ 1,000 = 1.92 kW.
The 0.80 is supplied for this example, not a default for every motor. Omitting it produces 2,400 VA and overstates the requested watts by 480 W. If the question gives only voltage and current, you can calculate VA under these conditions; you cannot assign a numerical watt value without power factor or equivalent information.
Is power factor the same as efficiency?
No. Power factor compares real electrical power with apparent electrical power. Motor efficiency compares mechanical output with real electrical input. These ratios describe different steps.
Extend our example with an explicitly given efficiency of 0.85. Mechanical output is 1,920 W × 0.85 = 1,632 W. Multiplying 2,400 VA by efficiency directly would skip the electrical-input step. Write a three-line chain: 2,400 VA apparent input → 1,920 W real input → 1,632 W mechanical output.
For a reverse calculation, 1,632 ÷ 0.85 recovers 1,920 W real input. Dividing again by 0.80 recovers 2,400 VA. This is an arithmetic check, not a claim about the efficiency or power factor of a particular compressor.
When should you stop using this shortcut?
The worked formula above is single-phase. Identify the supply and requested quantity before using it for a three-phase problem. Yokogawa also distinguishes true power factor from displacement power factor: the cosine-of-angle relationship applies to sinusoidal waveforms. Drive waveforms need appropriate measurement methods.
Fluke's motor-analyzer documentation treats voltage, current, power, power factor and harmonic distortion as separate measurements. A current reading alone is not a complete power or motor-condition diagnosis. This article is calculation practice, not an instruction to open energized equipment or select wiring, protection or a replacement motor.
How can you practise without memorizing the numbers?
Try a new original example: a single-phase load has 120 V RMS, 5 A RMS and true power factor 0.75. Calculate VA first, then W. Explain the unit change before checking: 600 VA and 450 W. No efficiency is given, so mechanical output remains unknown.
In your notes, write four prompts: What supply? Which units? Is PF given? Is the question asking for electrical input or mechanical output? Compare this with the airflow formula example: both require identifying the variables and assumptions before multiplying.
Use 313A exam information and the current-standard guide to choose your official study outline. This example is not a recalled exam question and makes no claim that a specific question will appear. Continue with the free five-question sample, which is original, unofficial draft practice pending the existing certified-mechanic review.
Sources
- Yokogawa: Electric Motor Power Measurement and AnalysisReal/apparent power, true/displacement PF and efficiency definitions; checked 2026-09-27
- Fluke: 438-II Power Quality and Motor AnalyzerDistinct electrical power and motor measurement quantities; checked 2026-09-27
Examen Studio is independent and not affiliated with the Red Seal Program or Skilled Trades Ontario. This article is a study aid; on the job, follow current codes, regulations and manufacturer instructions.