AC and three-phase power

The grid does not supply a steady voltage. It supplies a voltage that swings back and forth as a sine wave, 50 times per second in Algeria. Industrial motors use three of these waves at once. This lesson explains why.

The sine wave and its RMS value

A sine wave has a peak value $\hat{V}$ and a frequency $f$ (Hz). Because it spends most of its time below the peak, we describe it by its RMS (root mean square) value, the DC voltage that would heat a resistor equally:

V_\text{RMS} = \frac{\hat{V}}{\sqrt{2}}

Every voltage on a nameplate is RMS. A 230 V supply peaks at 325 V.

FoundationStart here if this is new to you

Picture a point going round a circle at constant speed. Its height, plotted against time, is a sine wave. The length of the arm is the peak value; one turn is one period. This rotating arm is called a phasor, and it is the easiest way to see two waves that are shifted in time: they are two arms at an angle to each other.

Power factor

In a motor the current does not rise and fall exactly with the voltage. Part of it only magnetises the iron: it flows in and back out every cycle and delivers no net energy. On the phasor diagram the current arm lags the voltage by an angle φ.

Real power P (kW) does the work: $P = V I \cos\varphi$ .
Apparent power S (kVA) is what the cables and transformers must carry: $S = V I$ .
Power factor is $\cos\varphi = P / S$ .

Try it: sine wave, phasor and power factor

Voltage peak325 V

Current peak20 A

Current phase φ-30°

V RMS: 229.8 V
I RMS: 14.1 A
Power factor cos φ: 0.866
Real power P: 2.81 kW
Reactive power Q: 1.62 kvar
Apparent power S: 3.25 kVA

Move the phase slider to −90°. The real power drops to zero even though current still flows. That current heats cables and does no work, which is why utilities charge for poor power factor.

Three-phase supply

Industrial supplies have three lines, each a sine wave shifted by 120° from the next. There are two voltages to know:

Line voltage $V_L$ , between two lines: 400 V in Algeria.
Phase voltage $V_\text{ph}$ , across one winding. It depends on how the windings are connected.

The real power of a balanced three-phase load is:

P = \sqrt{3}\, V_L\, I_L \cos\varphi

Predict first

The same motor windings are connected in star instead of delta, on the same 400 V supply. What happens to the line current at start?

It is unchangedIt drops to 1/√3 (58 %)It drops to 1/3 (33 %)It doubles

Try it: star or delta

Line voltage V_L400 V

Voltage per winding: 231 V
Current per winding: 23.1 A
Line current: 23.1 A
Torque vs delta: 33 %
Line current vs delta: 33 %

star: V_ph = V_L/√3, I_L = I_ph delta: V_ph = V_L, I_L = √3·I_ph

The rotating magnetic field

Put three coils around a circle, 120° apart, and feed each with one phase. Each coil makes a field that only pulses back and forth along its own axis. Added together, the three pulsing fields make one field of constant strength that rotates.

Try it: three coils make a rotating field

Resultant field: 1.50 × Î

Slow motion0.30×

Swap phases b and c

Synchronous speed at 50 Hz: 1,500 rpm
In rad/s: 157.1 rad/s
Rotation: forward

The field turns once per period for a two-pole machine. With more poles it turns more slowly. This is the synchronous speed:

n_s = \frac{120\, f}{p}

with $p$ the number of poles. At 50 Hz a 4-pole motor has $n_s = 1500$ rpm. Swap any two phases and the field turns the other way, which is how a contactor reverses a motor.

ExplorerGo deeper: derivations and open questions

Why exactly 1.5 times the peak? Write the phase currents as $\hat{I}\cos(\omega t)$ , $\hat{I}\cos(\omega t - 120°)$ , $\hat{I}\cos(\omega t - 240°)$ along axes at 0°, 120°, 240°. Summing as complex numbers gives $\tfrac{3}{2}\hat{I} e^{j\omega t}$ : a vector of fixed length $1.5\hat{I}$ turning at $\omega$ .

Why no neutral current? In a balanced system the three currents add to zero at every instant. Check it in the widget: the three coloured arrows always cancel along any one axis. That is why a three-phase motor needs no neutral wire.

Open question. A 2-pole motor on a VFD runs at 3000 rpm at 50 Hz. What output frequency gives 1800 rpm, and what voltage does a 400 V / 50 Hz V/Hz law apply there?