Library / Electrical / ELE-20
Electrical
Real power, single-phase AC
Real power drawn by an AC load, including power factor.
- Ref
- ELE-20
- Rev
- v1
- Field
- Electrical
- Result
- P [kW]
P1 in watts, Vrms and Irms RMS voltage and current, φ the phase angle between them. cos φ is the power factor.
Only the part of the current that is in step with the voltage does useful work. The power factor tells you what fraction of the volt-amps becomes real power.
Calculator
- Apparent power S = VI2.3 kVA
- Reactive power Q1.003 kvar
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A low power factor means the supply must carry more current for the same real power, wasting cable and generator capacity.
- Phase angle φ and power factor are linked by pf = cos φ; a power factor of 0.9 corresponds to about 25.8°.
- Use RMS values. Peak values overstate power by a factor of two.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kW) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Irms RMS current | 1 to 100 A | 0.207 | 20.7 | 20.49 | +10% | +10 % |
| Vrms RMS voltage | 100 to 480 V | 0.9 | 4.32 | 3.42 | +10% | +10 % |
| cos φ Power factor | 0.5 to 1 | 1.15 | 2.3 | 1.15 | +undefined% | — |
Sweep of Power factor
| Power factor | Real power (kW) |
|---|---|
| 0.5 | 1.15 |
| 0.5625 | 1.294 |
| 0.625 | 1.438 |
| 0.6875 | 1.581 |
| 0.75 | 1.725 |
| 0.8125 | 1.869 |
| 0.875 | 2.013 |
| 0.9375 | 2.156 |
| 1 | 2.3 |
Sweep one variable
Variables and how each one contributes
Real power
What the equation gives you. Everything else feeds it.
RMS voltage
RMS supply voltage. Directly proportional.
RMS current
RMS line current. Directly proportional.
Power factor
1 for a pure resistor, lower for motors and inductive loads. Directly proportional: improving power factor raises real power at the same current.
Worked example with the values above
- Apparent power: S = VI = 230 × 10 = 2,300 VA.
- Real power: P = S cos φ = 2,300 × 0.9 = 2,070 W.
- That is 2.07 kW.
Assumptions and limits
- Sinusoidal voltage and current at the same frequency.
- RMS quantities.
- Single-phase supply.
Common mistakes
- Using peak instead of RMS values.
- Reading apparent power (VA) as real power (W).
- Forgetting the power factor.
Statistics and uncertainty
Power factor is the least certain input: meters typically measure it to ±1–2%, but loads vary with their duty. Errors combine as δP/P ≈ √[(δV/V)² + (δI/I)² + (δpf/pf)²].
Where it is used
- Generator and supply sizing.
- Motor load calculations.
- Energy bill estimation.
Maths you need
- RMS values: effective AC amplitude
- Phase angle: lead or lag of current