Circuit Response
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Series

Three-phase — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. The neutral of a three-phase supply with one phase 30% off. computed by solving, not by drawing. Balanced, the three line currents sum to 4.6e-16 of one of them and the neutral carries nothing. With one phase 30% heavier the neutral carries 2.65 A against a line current of 11.50 A. The neutral reaches a tenth of a line current at 11.1% imbalance.

    Three phases, and the wire that carries nothing

    Three sources a third of a cycle apart, feeding three equal loads, return a current of 5×10⁻¹⁵ amperes down the wire between the star points. That is zero, and the whole of three-phase distribution rests on it. What is worth measuring is how fast it stops being zero, and the answer is that an eleven per cent imbalance in one phase puts a tenth of a line current down a conductor often sized on the assumption that it carries none.

    part 1 · power
  2. Three balanced rectifier loads conducting 60°, and their neutral. computed by solving, not by drawing. The three phase currents are drawn faint and the neutral heavy. Balanced loads, identical in every respect, and the neutral carries 0.968 A against a line current of 0.559 A — a ratio of 1.7321, where √3 is 1.7321. The pulse trains are disjoint, so the neutral is their union and its mean square is three times one phase's. Rebuilding the same current from the multiples of three in one phase's spectrum gives 0.966 A, 0.13% away, by a route sharing only the waveform.

    The neutral that carries more than a line

    Three balanced loads draw currents summing to 5.3 × 10⁻¹⁵ amperes in the neutral. That is a theorem about sinusoids, and it uses nothing except that each current is a single frequency. A harmonic of order three is shifted by 360° between phases, which is no shift at all — so the third harmonics add, and for any conduction angle narrow enough that the pulse trains stay disjoint the neutral carries exactly √3 times a line current.

    part 2 · power
  3. The two sequences a neutral current says nothing about. computed by solving, not by drawing at 61 imbalances. Three 20 Ω loads on a 230 V, 50 Hz star supply, one of them raised by a fraction of itself, with the neutral in place. The zero-sequence current is the one the neutral carries three times and is the only one a neutral ammeter reads; the negative sequence is a balanced set of three phasors rotating the other way. At 30.0 per cent imbalance it is 0.8846 A against 10.6154 A of positive sequence, 8.333 per cent, against 8.333 per cent from x/(3 + 2x). Two per cent arrives at 6.250 per cent imbalance, bisected on the network.

    The half the neutral does not carry

    A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half a neutral ammeter reads; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.

    part 3 · power
  4. The harmonics of one phase of three balanced rectifier loads conducting 60°, each sorted into the sequence its three phasors form. computed by solving, not by drawing: each harmonic of the three phases decomposed into zero, positive and negative sequence. Every one is a single sequence to the arithmetic's floor: h1 positive, h3 zero, h5 negative, h7 positive, h9 zero, h11 negative, h13 positive, h15 zero. A balanced set shifts harmonic h by h × 120° between phases, so the third and its multiples are not shifted at all — zero sequence — the seventh is shifted as the fundamental is, and the fifth the other way round. Relative to the fundamental, the zero-sequence harmonics together are 72.5%, the negative 24.1% and the positive 16.7%.

    The sequence a harmonic belongs to

    Symmetrical components are defined on phasors, so they describe one frequency, and a rectifier load's current has many. Taken harmonic by harmonic, three balanced loads give a single sequence at every harmonic, to the arithmetic's floor: the third and its multiples zero sequence, the seventh, thirteenth and nineteenth positive, the fifth, eleventh and seventeenth negative — because harmonic h is shifted h × 120° between phases. The sequence decides where each harmonic can go. Connected line to line, the loads put no triplen in a line at all: 3.32 A of them circulates round the delta and never leaves it. Fed through a delta–wye transformer beside a second group, the fifth and seventh cancel in the supply and the eleventh and thirteenth remain, because the transformer turns the two sequences opposite ways.

    part 4 · power

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