Power in three phases

There is a reason machines have three phases and not one, and it is not redundancy. A single-phase supply delivers power that swings between zero and twice its average, twice per electrical cycle. Three balanced phases deliver a constant.

Not approximately constant, and not constant on average. Constant at every instant, exactly. On a shaft that difference is the difference between smooth torque and torque that pulses at twice the electrical frequency — which the mechanical system feels, and which no control loop can remove because it is not an error.

How many phases carry current

One Two Three

The load Phase lag

Instantaneous power over one electrical cycle, at 24 V and 6 A peak. Each phase's contribution is drawn faintly; their sum is the heavy line. Switch to One and the sum is that single pulsing contribution. Switch back to Three and it flattens completely. Then drag Phase lag: the single-phase trace dips below zero — power flowing back out of the load — while the three-phase sum stays flat and simply sits lower.

One phase pulses to zero

With voltage and current at the same frequency, their product is

p (t) = \overset{v}{^} cos ω t \cdot \overset{}{^} cos (ω t - φ) = \frac{1}{2} \overset{v}{^} \overset{}{^} [cos φ + cos (2 ω t - φ)]

A constant, plus a cosine at twice the supply frequency. At 24 V and 6 A peak into a resistive load that is a mean of 72 W swinging between 0 W and 144 W.

Two things about that oscillating term are worth noticing, because they are easy to get backwards.

Its amplitude does not depend on the power factor. The swing is always $\frac{1}{2} \overset{v}{^} \overset{}{^}$ — 144 W peak-to-peak here — whatever the phase lag. Only the constant it rides on moves. So the ratio of ripple to mean is $2/ cos φ$ , which grows without limit as the load becomes reactive, while the ripple itself never changes at all.

It goes negative for any lag at all. The minimum is $\frac{1}{2} \overset{v}{^} \overset{}{^} (cos φ - 1)$ , which is below zero the moment $φ \neq = 0$ . There is no threshold to stay under. Unity power factor is the only case that merely touches zero:

Phase lag	Mean	Minimum	Ripple ÷ mean
0°	72 W	0 W	2.00
30°	62.4 W	−9.7 W	2.31
60°	36 W	−36 W	4.00

Three phases sum to a constant

Add three of those, each shifted by 120°. The constant terms add. The oscillating terms are three cosines at $2 ω$ spaced 240° apart — which is still a balanced set — so they sum to zero identically.

p (t) = \frac{3}{2} \overset{v}{^} \overset{}{^} cos φ

At the same 24 V and 6 A that is a flat 216 W, which is three times the 72 W one phase carried, delivered without a ripple to remove.

This is the same cancellation the space vectors page describes magnetically. There, three winding MMFs sum to one resultant that rotates without pulsing. Here, three powers sum to a constant. It is one fact seen from two sides, and the $3/2$ in both is the same $3/2$ .

Where the 3/2 in the dq expression comes from

Written in the two-axis quantities, the same power is

p = \frac{3}{2} (v_{α} i_{α} + v_{β} i_{β}) = \frac{3}{2} (v_{d} i_{d} + v_{q} i_{q})

Those two are equal because Park is a rotation and a rotation preserves a dot product — the frame you compute in cannot change the power.

The $3/2$ is there because this site uses the amplitude-invariant Clarke transform, where $i_{α}$ carries the phase peak. Two axes holding phase-peak values under-count three windings by exactly that ratio, so the power expression puts it back.

On the bench The power-invariant convention scales by $2/3$ instead, and then the power is simply $v_{d} i_{d} + v_{q} i_{q}$ with no factor — which is why some texts have the $3/2$ and others do not. Neither is wrong. Mixing them is, and the symptom is a torque constant out by 1.5 or by $1.5$ depending on which half of the calculation came from which book.

src/core is amplitude-invariant everywhere. If you take an equation from elsewhere, check which convention it assumes before trusting the factor.

What the machine cannot be given

A voltage identical on all three terminals delivers no power at all. With the star point floating the three currents must sum to zero, so

p_{c m} = v_{c m} (i_{a} + i_{b} + i_{c}) = 0

That is why space-vector PWM can inject a common-mode component to buy 15.5% more linear range and pay nothing for it. The injection is invisible to the machine — not approximately, not on average, but identically.

What a d-axis current costs

Copper loss in dq is

P_{c u} = \frac{3}{2} R_{s} (i_{d}^{2} + i_{q}^{2})

and the two axes appear at exactly the same weight. On the reference machine 6 A of $i_{q}$ costs 18.9 W. Six amps of $i_{d}$ costs the same 18.9 W — and on a non-salient machine produces no torque whatsoever.

That is the real price of field weakening, and the reason MTPA is worth the arithmetic: every amp you put on the d-axis is heating you are paying for, and the only question is whether it is buying you something.

What to take away

One phase delivers power that swings between zero and twice its mean, at twice the electrical frequency. Three balanced phases deliver a constant, exactly.
The single-phase ripple amplitude is independent of power factor; only the mean moves. The ratio is $2/ cos φ$ .
Single-phase instantaneous power goes negative for any phase lag at all. There is no threshold.
$p = \frac{3}{2} (v_{d} i_{d} + v_{q} i_{q})$ , and the $3/2$ is the amplitude-invariant convention rather than a fudge. The power-invariant convention has no factor.
Common-mode voltage delivers zero power into a floating star point, which is what makes SVPWM’s injection free.
Copper loss charges $i_{d}$ and $i_{q}$ at the same rate. A d-axis amp is heat unless it is buying voltage headroom or reluctance torque.