Three phases, and the wire that carries nothing

Until this page every source on this site has had a real value. That was never stated as a restriction because nothing needed it lifted: one source in a network sets the reference angle, and every phase in the answer is measured from it. Three sources is the first circuit where the phases between the sources are the subject, and a real number cannot carry one.

So the assembler was extended. A source’s value may now be a complex number, its imaginary half is carried in a second source vector beside the real one, and everything downstream of the solve is unchanged. The documentation for V has said “value is its phasor” since the first commit; this is the first circuit that took it at its word.

Why the sum is zero #

Three unit phasors at 0°, −120° and +120° sum to zero, and the reason is that they are the three cube roots of one. Their sum is the coefficient of the second-highest term in z³ − 1 = 0, which is absent, so the sum is exactly zero — an algebraic fact rather than a numerical coincidence.

Feed three identical impedances from them and each current is the same complex number divided by the same impedance, so the currents are also three phasors a third of a cycle apart and they sum to zero too. The neutral wire is therefore carrying nothing, and the solved network agrees: the measured sum comes out at 5.3×10⁻¹⁵ of a line current, which is the arithmetic’s own floor.

That is the entire economic argument for three-phase distribution. Three wires deliver three loads’ worth of power, and the fourth wire — the one that would be needed to return the current in a single-phase system — can be thin or, in some arrangements, absent.

The factor of √3, which is a geometry rather than a convention #

Two voltages are available in a star-connected supply and the ratio between them is the one number in this subject most often quoted and least often derived.

The voltage from any line to the star point is the source’s own, 230 V here. The voltage between two lines is the difference of two phasors a third of a cycle apart, and the difference of two unit phasors 120° apart has magnitude 2 sin 60° = √3. So the line-to-line voltage is 230 × √3 = 398 V, which is the 400 V that appears on every nameplate in Europe.

Nothing about that is a convention. It is the length of the third side of a triangle whose other two sides are equal and meet at 120°, and it is the same construction three voltages that close uses to show that a series circuit’s element voltages can each exceed the source. The arithmetic is identical; only the drawing is rotated.

The consequence is that a three-phase supply offers two voltages from the same four wires, and a load can be connected between a line and the neutral or between two lines. A load connected between two lines draws no current in the neutral at all — it has no connection to it — which is one reason larger equipment is usually wired that way, and one reason an installation’s neutral loading is decided by its small loads rather than its large ones.

The second route #

A claim that a wire carries nothing is a claim worth checking by a method that does not look at the wire, and one is available.

Any three phasors can be decomposed into three symmetrical components: a positive-sequence set rotating one way, a negative-sequence set rotating the other, and a zero-sequence set in which all three are identical. The zero-sequence component is the average of the three, and the neutral carries exactly three times it.

The routine that computes it takes the three line currents and produces the average with two complex rotations. Nothing in it touches the neutral branch, its resistance, or the solve that produced it beyond reading the three currents out. It agrees with the neutral current from the network to a part in 10⁹ at every imbalance on the slider.

For the balanced case both routes report zero, which is not a strong test — two routes agreeing on zero is easy. For the unbalanced case at thirty per cent they report the same 2.654 A, and that is a test: two different constructions of the same quantity, sharing only the solve.

The decomposition is worth a second look because it also says what kind of imbalance is present, which the neutral current on its own does not. A load imbalance that is purely a difference in magnitude produces zero- and negative-sequence components together; a set of currents that are equal in magnitude but wrongly spaced in angle produces a negative-sequence component and no zero-sequence one at all, and puts nothing in the neutral while still heating a motor connected between the lines. The neutral current is therefore a partial indicator: it detects one of the two ways a supply can be unbalanced and is silent about the other. A three-phase motor fed from a supply with a two per cent negative-sequence component runs measurably hotter for reasons that no neutral measurement can see, because the motor’s rotor is being driven backwards by a small counter-rotating field.

How quickly it stops being zero #

The interesting question is not whether the sum is zero when everything matches — it is, exactly — but how much mismatch it takes for the neutral to carry something that matters.

Bisecting on the solved network with one phase’s resistance scaled by 1 + x gives the answer: at x = 0.111, the neutral carries a tenth of a line current. Eleven per cent.

That number deserves a moment. Eleven per cent is not a fault condition. It is the ordinary difference between three circuits in an office with different numbers of things plugged into them, or three phases of a lighting installation where one corridor has more fittings than another. Nothing has gone wrong, nobody has made a mistake, and the neutral is carrying ten per cent of a line current.

Proportionality is what makes this awkward rather than dangerous. There is no cliff. A neutral sized at half a line conductor is fine up to roughly fifty per cent imbalance and the installation gives no warning as it approaches. The failure mode of an undersized neutral is a slow one — heat, then a loose connection, then a connection that arcs — and it is not detected by anything that monitors the three phases.

It is worth being explicit about what the eleven per cent is a property of, because the number moves. The bisection scales one phase’s resistance while holding the other two, so it describes a very particular kind of imbalance: one phase heavier, two equal. Imbalance distributed differently gives a different answer — one phase lighter and two equal is not the mirror image, because resistance and current are reciprocally related and a ten per cent lighter load does not draw ten per cent less current — and two phases displaced in opposite directions can produce a much larger neutral current for the same total departure from balance.

What survives all of those variations is the order of magnitude, and that is the useful part: the neutral current is roughly the arithmetic imbalance rather than something smaller. There is no averaging, no partial cancellation, no factor that makes a ten per cent difference produce a one per cent current. The three phasors sum to zero exactly when they are equal and to something of the order of their difference when they are not.

A worked case makes the scale concrete. Three twenty-ohm loads on 230 V draw 11.500 A each. Make one of them thirty per cent heavier — that is, raise its resistance by thirty per cent, so it draws 8.846 A instead — and the neutral carries 2.654 A, which is twenty-three per cent of a line current. The imbalance in the current was 2.654 A and the neutral carries 2.654 A: the whole of it, arriving in the conductor that was sized on the assumption that imbalance was the exception.

The harmonics that do not cancel #

There is a second mechanism, and it does not obey the proportionality above at all.

The cancellation argument rests on three phasors being a third of a cycle apart. Shift each of them by a third of a cycle and the set is unchanged, which is the symmetry that makes the sum vanish. Now consider the third harmonic of each phase’s current. Shifting the fundamental by a third of a cycle shifts the third harmonic by a whole cycle — that is, by nothing at all. The three third harmonics are therefore in phase with one another, and instead of cancelling in the neutral they add.

The same is true of the ninth, the fifteenth, and every odd multiple of three, which is why they are collected under the name triplen.

The consequence is one of the few genuinely counter-intuitive results in distribution: a perfectly balanced three-phase installation of non-linear loads can have a neutral current larger than any of its line currents. Not a fraction of one — larger. The fundamental cancels completely, the triplen harmonics add to three times their per-phase value, and if the third harmonic is a third of the fundamental the neutral carries as much as a line does.

None of that is visible to a measurement of the fundamental, or to a power factor computed as cos φ, or to any calculation that represents a current as a single phasor. It requires the current’s shape, which is the quantity the capacitor that was right once also turns on.

What the milliohm is doing there #

One detail of the netlist is a deliberate choice and worth defending, because it looks like a fudge.

The neutral is a one-milliohm resistor rather than a wire. That is not an approximation to a connection; it is the honest way to say “connected”, because a short circuit between two nodes is a constraint and not an element, and stamping it as an admittance of infinity is how a matrix acquires a row of nonsense. The assembler refuses a resistance of zero by name for exactly that reason.

A milliohm is four orders below the smallest load on the page, so it changes no digit that is printed. What it buys is that the figure which removes the neutral does so by making that resistance large rather than by deleting a line — the same circuit at two parameter values, with the same solver and the same checks, rather than two circuits that have to be trusted to correspond. The distinction matters whenever a limiting case is being drawn: a limit approached through a parameter can be checked at every point along the way, and a limit reached by editing the netlist cannot.

The star point, and what happens without a neutral #

Removing the neutral is modelled here by making its resistance large rather than by deleting a wire, which keeps it the same circuit at a different parameter value and keeps the solver’s refusals meaningful — a network with a node that has no path to ground is refused by name rather than returned as a large plausible number.

With the neutral gone and the load balanced, nothing changes: the star point sits at zero anyway, so a wire carrying no current can be removed with no effect. With the neutral gone and the load unbalanced, the star point moves, and it moves in the direction that makes things worse: the lightly loaded phase sees a higher voltage and the heavily loaded one a lower voltage. A three-phase supply without a neutral does not merely fail to correct imbalance; it converts a current imbalance into a voltage imbalance, which is why the neutral is present in every distribution arrangement that feeds single-phase loads.

The efficiency argument, restated #

That figure is worth revisiting in this context because it explains why three-phase distribution is shaped the way it is. A distribution system is not trying to extract the most power from a source; it is trying to lose the least in the wires. Every design decision follows: high voltage, so the current is small; three phases, so the return current is nothing; a low source impedance, so the voltage does not sag.

The neutral is where that logic has an exception, and this essay is about the size of it. Three phases make the return current zero under an assumption — balance — that holds well enough to be worth building on and not well enough to be built on carelessly. Eleven per cent of imbalance, or one rectifier per phase, is enough to make the wire that carries nothing carry something.