AC Power Basics
RMS values, phasors, real, reactive and apparent power, power factor and the power triangle.
Why AC, and what RMS means
Almost all mains power is alternating current (AC): the voltage swings sinusoidally, v = Vpeak sin(2π f t), at 50 or 60 Hz. AC won because a transformer can change its voltage cheaply, so power travels at high voltage and low current over long distances and is stepped down for use (see Transformers and Distribution). To compare an alternating supply with direct current we use the RMS (root-mean-square) value: the DC voltage that would heat a resistor equally. For a sine wave Vrms = Vpeak ÷ √2.
A “230 V” socket therefore swings to ±325 V (651 V peak to peak); a 120 V supply peaks at 170 V. Insulation and capacitors must be rated for the peak, and a cheap meter that reads the average and scales it is wrong for anything but a clean sine (use a true-RMS meter for drives, LED lamps and computers).
Phase, reactance and impedance
In a resistor the current follows the voltage exactly. In an inductor (a coil, a motor winding) the current lags the voltage by 90°; in a capacitor it leads by 90°. The opposition to AC current is the reactance: XL = 2π f L and XC = 1 ÷ (2π f C), in ohms. A resistor and a reactance combine at right angles into the impedance Z = √(R² + X²) with a phase angle φ = arctan(X ÷ R). Drawing voltages and currents as rotating arrows (phasors) turns the sums into triangles.
Worked example: a coil in series with a resistor
Worked example: R = 10 Ω and L = 0.1 H on 230 V, 50 Hz
- Reactance XL = 2π × 50 × 0.1 = 31.4 Ω.
- Impedance Z = √(10² + 31.4²) = 32.97 Ω, so the current is I = 230 ÷ 32.97 = 6.98 A, lagging the voltage by φ = 72.3°.
- Real power P = I² R = 487 W; reactive power Q = I² X = 1 529 var; apparent power S = V I = 1 605 VA; power factor = R ÷ Z = 0.303.
- Compare the same resistor on DC at 230 V: 5 290 W. The coil limits the current to a seventh, yet draws 1 605 VA from the supply for only 487 W of useful heat.
- A capacitor of 100 µF at 50 Hz has XC = 31.8 Ω. Put 0.1 H and 100 µF together and they resonate at f = 1 ÷ (2π√(L C)) = 50.3 Hz, where the two reactances cancel (the idea behind power-factor correction and also a hazard with harmonics, see Power Factor, Harmonics and Energy).
Real, reactive and apparent power
With a lagging or leading current the product V × I is not all useful work. Three powers describe it:
| Quantity | Symbol and unit | What it is | Formula (single-phase) |
|---|---|---|---|
| Real (active) power | P, watt (W) | the rate of doing work or producing heat; paid for | P = V I cos φ |
| Reactive power | Q, volt-ampere reactive (var) | energy swinging to and fro in magnetic and electric fields; does no work but loads the system | Q = V I sin φ |
| Apparent power | S, volt-ampere (VA) | the product of RMS voltage and current; what cables and transformers must carry | S = V I, S² = P² + Q² |
Power factor
power factor = P ÷ S = cos φ
Electricity suppliers bill reactive power or demand in kVA because their cables and transformers must carry it, and heat their windings with it. Power factor correction (capacitors near the load) is covered in Power Factor, Harmonics and Energy.
Worked example: the press motor
Worked example: the 11 kW motor of the hydraulic press
This guide follows the 11 kW, 400 V, 50 Hz, four-pole three-phase induction motor that drives the hydraulic press, at 91 percent efficiency and a power factor of 0.85.
- Electrical input = 11 000 ÷ 0.91 = 12.09 kW.
- Apparent power S = P ÷ pf = 12.09 ÷ 0.85 = 14.22 kVA; reactive power Q = √(S² − P²) = 7.49 kvar.
- Line current I = S ÷ (√3 × 400) = 20.5 A. At unity power factor the same work needs only 17.4 A: the reactive current is a pure overhead of 18 percent.
- Energy: running 4 000 hours a year at full load uses 48 352 kWh, about 5 802 a year at 0.12 per kWh. The kWh meter counts only real power.
Units and instruments
| Instrument | Measures | Note |
|---|---|---|
| True-RMS multimeter | V, I, sometimes frequency | essential for drives, LED lamps and switch-mode supplies |
| Clamp meter | current without breaking the circuit | clamp one conductor, not the whole cable |
| Power analyzer / energy meter | P, Q, S, power factor, kWh, harmonics | three-phase versions use three current clamps |
| Oscilloscope with isolated probe | waveforms | never ground a probe on a live mains conductor without isolation |
Convert between kW, hp and BTU/h in the Power converter (1 hp = 0.746 kW) and kWh and joules in the Energy converter.
Related
- Next: Three-Phase Systems.
- Electronics Fundamentals for resistors, capacitors and inductors.
- Ohm’s Law and Power Calculator and the Reactance and Resonance Calculator.
- Electrical Power: the whole guide.
FAQ
What is the difference between kW and kVA?
kW is real power that does work; kVA is apparent power that the wiring must carry. They are equal only at a power factor of 1. A 14.2 kVA load at 0.85 draws 12.1 kW.
Why does a motor have a power factor below 1?
Because the winding is an inductor: it needs a magnetizing current that stores and returns energy each cycle. A lightly loaded motor has a poorer power factor than a fully loaded one.
Is 230 V the peak or the RMS voltage?
RMS. The peak is 325 V.
Can reactive power be used?
It is needed to build magnetic fields, but it does no work and is not metered in homes. Industry is billed for it through kVA demand or power-factor penalties.
References
- Alexander, C. K., Sadiku, M. N. O., Fundamentals of Electric Circuits, McGraw-Hill (AC circuits and power).
- Wildi, T., Electrical Machines, Drives and Power Systems, Pearson; Glover, J. D., et al., Power System Analysis and Design, Cengage.
- IEC 60038 (standard voltages) and IEEE Std 1459 (definitions of power quantities).
Last reviewed: October 2026. Efficiency and power factor are typical for a four-pole 11 kW motor.