The half the neutral does not carry

Three phases, and the wire that carries nothing computed the neutral current of an unbalanced supply twice: once by solving the network and reading the current in the neutral branch, and once as three times the zero-sequence component of the three line currents. The two agree, and the second route never touches the neutral branch at all, so the agreement is real.

It is also one third of what that second route computed. Decomposing three phasors into symmetrical components produces three balanced sets, and the zero-sequence one — all three in phase, which is why the neutral carries their sum — is the only one that has ever been read. The positive-sequence set is the supply doing its job. The negative-sequence set is three balanced phasors going round the other way, and it is exactly the same size as the zero-sequence set in the case that ladder measured.

Nothing in the collection has looked at it, and no conductor anywhere reports it: the neutral carries the zero sequence and is blind to the negative one by construction, because three phasors 120° apart sum to nothing whichever way they rotate.

Two things come out of reading it. The equality between the two components is a theorem about the neutral rather than a fact about the numbers chosen, and it fails the moment the neutral is taken away — not by making the imbalance smaller but by moving all of it into the component nothing measures. And the negative-sequence voltage, which is the quantity a supply is actually held to, turns out to be the negative-sequence current divided by the ratio of the load impedance to the line impedance, which is to say divided by the thing an installation spends money to make large.

What is being decomposed, and what makes it exact #

Three phasors are three complex numbers, which is six real numbers, and the three symmetrical component sets are also six. The decomposition is a change of basis and nothing else:

$I_0 = \tfrac13(I_a + I_b + I_c)$, $I_+ = \tfrac13(I_a + \alpha I_b + \alpha^2 I_c)$ and $I_- = \tfrac13(I_a + \alpha^2 I_b + \alpha I_c)$, with $\alpha$ a rotation of 120°.

That is invertible and it is exact for any three phasors whatever — balanced, unbalanced, or three numbers picked at random. It assumes nothing about the circuit, which is what makes it worth checking against: a change of basis that failed to reproduce its own input would be an arithmetic error rather than a modelling one, and this collection’s habit is to make the instrument prove itself on the case where the answer is known before quoting it on the case where it is not — the calibration exact outside and wrong within is about, applied to a matrix rather than to a field.

There is one condition inside it and it is easy to lose, because it is carried by the word phasor rather than stated anywhere. Each of $I_a$, $I_b$ and $I_c$ has to be one complex number, which means one frequency. Everything below is a statement about a supply carrying sinusoids, and the last figure on this page is the case where that is false.

The reconstruction is the check rather than the decomposition, and the distinction is the point. A decomposition can be inspected and believed; a reconstruction can only be right or wrong, and it is wrong by 9.0 × 10⁻¹⁶ of a line current, which is where double-precision arithmetic runs out. That is the same floor the balanced neutral current sits at in the rung below — three currents cancelling to 5 × 10⁻¹⁵ amperes — and it is what entitles everything downstream to be quoted as a measurement.

The three sets are each perfectly balanced however unbalanced the thing they add up to, which is the whole content of the construction and is asserted in the figure rather than assumed: the largest and smallest magnitudes within each set differ by under 10⁻¹² of a line current at every position of the slider. The middle star turning the other way is not a drafting choice. It is what the arithmetic produces, and it is the reason a machine cares about a quantity no protective conductor carries.

Equal, and equal for a reason #

Vary one phase of a star load by a factor $1 + x$ with a good neutral, and the negative and zero sequences come out the same size. That is not a coincidence of the numbers chosen.

With the star points tied, each phase is independent: the two unchanged phases carry $V/Z$ at their own angles and the third carries $V/Z(1+x)$. The two unchanged ones sum to $-V/Z$ in both the zero-sequence and the negative-sequence combination, because $\alpha$ and $\alpha^2$ merely swap which of them is which, so both components come out as $\tfrac{V}{3Z}\cdot\tfrac{-x}{1+x}$ — the same number. The positive sequence gets $\tfrac{V}{3Z}(2 + \tfrac{1}{1+x})$ instead, and the ratio of the two is $x/(3 + 2x)$.

Two consequences of that are worth having in front of the figure rather than after it. The ratio is not proportional to the imbalance — it saturates, reaching 20.001 per cent at $x = 1$ where a linear reading would give a third — and it does not depend on the load resistance at all, only on the fraction by which one phase differs from the others. A 5 Ω installation and an 80 Ω one with the same percentage imbalance carry the same percentage of negative sequence.

The figure checks that expression against the solve at sixty imbalances and finds the worst disagreement at 3.0 × 10⁻⁵, which is the milliohm the neutral is modelled as rather than anything about the expression. It then bisects the network for the imbalance that puts the negative sequence at two per cent of the positive, gets 6.250 per cent, and compares that with what the expression inverts to. Two routes, and the second one never solves anything.

So the neutral current is an exact proxy for the negative sequence in this case — three times too large, and otherwise the same number. That is a comfortable position and it survives exactly one change to the installation.

Take the neutral away #

Removing the neutral does not remove the imbalance. It removes the path for one of the two components it produces, and the imbalance moves into the other one.

Three-wire distribution is what a motor circuit, a delta-fed load and most industrial supplies actually are, and the change is in the unfavourable direction: at full imbalance the negative sequence goes from 20.001 per cent of the positive to 25.000. The instrument that was measuring it — the current in the neutral — is now reading zero, correctly, on an installation whose negative sequence has gone up by a quarter.

That is the shape of defect this collection is mostly about, arriving as an absence: no reading is wrong, no conductor is overloaded, and the quantity that grew has stopped being observable at the one place anybody was watching.

The mechanism is a constraint rather than a resistance. Three currents whose sum is forced to zero have a zero-sequence component of zero by definition, so removing the neutral does not attenuate that component, it deletes the degree of freedom the component lived in. What was six real numbers of current is now four, the imbalance still has to go somewhere, and the only place left is the negative sequence. The star point moves instead — the load’s own neutral drifts away from the supply’s, which is what unbalances the voltages across the individual loads and is why a three-wire load with one arm open sees the other two in series across a line voltage rather than each across a phase voltage.

The one thing that adds #

Powers do not superpose. Two solves that add, and the one that does not is about exactly that: currents from separate sources add and the dissipation they produce does not, because a square has a cross term.

The line heating of an unbalanced three-phase load is the exception, and the exception is worth having because it turns the negative sequence from a diagnostic into a term in a bill.

The identity is $|I_a|^2 + |I_b|^2 + |I_c|^2 = 3(|I_0|^2 + |I_+|^2 + |I_-|^2)$, and it holds because the transformation is unitary up to the factor of three. So the copper loss in the three lines splits into one additive term per sequence with nothing left over, and the two terms that are not positive sequence produce no torque in any symmetrical machine on the supply. They are pure heat, in the cable, in the transformer and in the machine’s own windings, and the site’s own habit of pricing what a current costs rather than what it is called applies to them directly — the same argument the current that does no work makes about the reactive half, one dimension along.

The split is worth stating in the direction a designer meets it. Three ammeters on the lines give $|I_a|^2 + |I_b|^2 + |I_c|^2$ directly and therefore give the total conduction loss; what they cannot give without the decomposition is how much of it is doing anything. Balanced, the answer is all of it. At thirty per cent imbalance it is 98.6 per cent, and at full imbalance 92.6 — a loss of efficiency that appears in no power factor, no reactive power and no harmonic reading, and that is invisible to every instrument this field has so far pointed at a supply, because every current involved is at the fundamental and in phase with its own voltage — the true-power-factor argument of what a meter multiplies by has nothing to bite on here, and neither has a reactive-power calculation.

Seven and a half per cent of the conduction loss at full imbalance is not the headline number here, because it is a number somebody could in principle measure with three ammeters. The next one is not.

The unbalance the meter cannot see #

A supply is not held to a limit on current unbalance. It is held to a limit on voltage unbalance, usually two per cent, and a voltage unbalance is a current unbalance multiplied by the impedance between the source and the load. Make that impedance small — which is what a good supply is — and the meter reads nothing while the load does as it likes.

This is not a small correction. The current unbalance is set by the load and barely moves with the line: 19.94 per cent at a tenth of an ohm, 19.71 at half an ohm, 18.87 at two ohms, 17.39 at five. The voltage unbalance moves almost in proportion to the line resistance instead — 0.083, 0.410, 1.563 and 3.571 per cent at those same four values — so the ratio between the two runs from 240 down to under five, entirely on a quantity the load knows nothing about.

The open-phase case is worth pausing on because both halves of it are exact. Fifty per cent is a theorem: with one phase carrying nothing, the positive sequence is two thirds of a line current and the negative sequence is one third, whatever the parts are. And the divisor is nothing but the copper — one and a half times the load resistance divided by the line resistance, plus a half — checked against the solved network across a factor of ten in each of the two resistances and agreeing to a part in 10⁴.

So the two quantities are not merely different in size. They are the same quantity read through a divider, and the divider is the thing an installation spends money to make large. The negative sequence behaves here exactly as the reactive half of the power does in the energy a unity power factor doubles: the instrument everyone reads is a ratio, the ratio can be improved without improving anything the ratio was standing in for, and the quantity that decides the copper is on the other side of it. A supply engineered to hold its voltage — which is the far end that rises done well — is a supply engineered to make the negative sequence undetectable at the point of common coupling.

Where the meter starts to see it #

Reading that the other way round is more useful than reading it as a complaint. There is a line resistance at which an open phase does reach two per cent at the meter, and it can be bisected on the solve like any other edge: 1.224 Ω against a 20 Ω load, which is 6.12 per cent of it.

The two figures are the same load and differ only in the copper in front of it, which is worth saying plainly because the current curves are almost identical: 19.705 per cent against 18.868 at full imbalance, a difference of four per cent of itself. Everything that changed between them changed in the quantity a meter reads.

Which says the voltage unbalance factor measures the supply and not the load, and that reading it as a statement about the load is a category error with a factor of fifty in it. A weak supply feeding a mildly unbalanced load can fail the limit; a strong supply feeding a load with a phase missing can pass it. Both are the same instrument reporting the impedance in front of it, which is the same lesson the millivolts in the wire draws about a return conductor and what a network answers states in general: a measurement is a property of the circuit it is taken in.

The practical reading is short. A voltage unbalance inside the limit is evidence about the supply and almost none about the load, so an installation that wants to know whether its loads are balanced has to measure the currents, which costs three ammeters and a decomposition rather than one voltmeter at the board. And a voltage unbalance outside the limit is evidence about both at once, which is why it is a poor diagnostic and a reasonable regulation: it fires when the product is large, which is when something is actually being damaged.

What it does not say #

It does not say the neutral current is the wrong thing to measure. It sizes the neutral conductor, which is a real conductor with a real rating, and the neutral that carries more than a line shows it doing something no sequence argument here predicts at all — three balanced rectifier loads putting √3 times a line current down it, because the third harmonic of each phase is shifted by 360° and therefore not shifted.

That is the sharpest limit on everything above. Every number on this page is a single-frequency statement, and the commonest load in a modern installation is not a single frequency at all: the rectifier measured in the direct voltage that is a sawtooth draws its current in a pulse a fraction of a cycle wide, and a decomposition into three phasors has no representation for it. Sequence components can be computed harmonic by harmonic, which is a different and larger calculation, and nothing here does it.

It does not say either that negative sequence is bad in proportion to its size. What it costs depends on what is connected: a resistive load barely notices, while a machine presents a much lower impedance to a backwards-rotating field than to a forwards one, so a small negative-sequence voltage drives a disproportionate negative-sequence current into it. That amplification is a property of the machine and this collection has no machine in it, so the number is not quoted — only the current, the heat and the two routes to each. What can be said without a machine is the direction: a lower impedance to the negative sequence than to the positive means the current unbalance in a motor exceeds the voltage unbalance driving it, so the factor of fifty measured above is a lower bound on the disparity rather than an upper one.

It does not say the line resistance model is the whole of a real supply either. A distribution transformer’s impedance is mostly reactive, a cable’s is mostly resistive at 50 Hz, and a real feeder is both plus whatever else is on it. Replacing the half ohm of resistance with half an ohm of reactance was measured rather than argued about, and it barely moves anything: the open-phase voltage unbalance goes from 0.8265 per cent to 0.8333, and the divider from 60.50 to 60.00. What a reactive line changes is every angle in the solve; what it leaves alone is the ratio of impedances the whole argument is about.

And the milliohm neutral is a model. It is what makes “connected” expressible as an element rather than as a constraint, and it is why the equality between the negative and zero sequences comes out at 3.0 × 10⁻⁵ rather than at the arithmetic’s own floor, the way the recombination residual does.

The number worth carrying #

Fifty per cent in the current, 0.83 per cent at the meter, with one phase gone.

The habit that goes with it is about instruments rather than about supplies. When a quantity is measured through an impedance, the reading is the quantity divided by whatever that impedance is compared with the load — so a limit written on the reading is a limit on the product, and improving the supply improves the reading without improving anything the reading was meant to protect. The way to find out which is happening is to compute the quantity on the other side of the divider, which here costs one extra line of a decomposition the solve was already doing.