Set 21 — Symmetrical Components
Twenty worked problems on the transformation that makes unsymmetrical fault analysis possible. Everything in Parts 3 and 4 assumed balance, which let one per-phase circuit stand for three. A single line-to-earth fault destroys that assumption, and a direct three-phase solution of a large network is intractable. Fortescue's theorem resolves any set of three phasors into three balanced sets, each of which can be analysed by the per-phase methods already built — and the three results superposed. The whole of Part 5 rests on this one change of basis.
The operator. \(a = 1\angle120^\circ = -0.5+j0.866\), so \(a^{2} = 1\angle240^\circ\), \(a^{3} = 1\), and — the identity used constantly — \(1 + a + a^{2} = 0\).
The transformation. \(\mathbf{V}_{abc} = \mathbf{A}\mathbf{V}_{012}\) with \(\mathbf{A} = \begin{bmatrix}1&1&1\\1&a^{2}&a\\1&a&a^{2}\end{bmatrix}\), and the inverse is \(\mathbf{A}^{-1} = \tfrac{1}{3}\mathbf{A}^{*}\) — no matrix inversion is ever needed.
The three sequences. Positive: three equal phasors 120° apart in \(abc\) order. Negative: three equal phasors in \(acb\) order. Zero: three identical phasors, in phase.
Zero sequence and the neutral. \(V_0 = \tfrac{1}{3}(V_a+V_b+V_c)\), so zero sequence exists only when the three do not sum to zero — which requires a fourth conductor or an earth return.
A balanced set is pure positive sequence. Its \(V_0\) and \(V_2\) are identically zero, which is why every balanced study in Parts 3 and 4 was a positive-sequence study without saying so.
Power is invariant in the form \(S = 3\left(V_0I_0^{*} + V_1I_1^{*} + V_2I_2^{*}\right)\). The cross terms vanish, so the three sequences do not exchange power.
The sequences decouple in any balanced network — a symmetric impedance matrix is diagonalised by \(\mathbf{A}\) — which is the property that makes the whole method work, and the subject of Set 22.
Define the operator \(a\), establish the identities that will be used throughout Part 5, and evaluate \(1-a\), \(a-a^{2}\) and \(a^{2}-a\).
The definition. \(a\) is the unit phasor that rotates by 120°:
It is to a three-phase system what \(j\) is to a two-phase one: multiplication by \(a\) advances a phasor by a third of a cycle without changing its magnitude.
The powers:
So \(a^{4} = a\), and every power reduces to one of \(1\), \(a\), \(a^{2}\). Note also \(a^{*} = a^{2}\) and \(1/a = a^{2}\).
The central identity. The three cube roots of unity sum to zero:
Geometrically, three unit vectors 120° apart. This identity is what makes a balanced set have no zero-sequence component, and it appears in almost every derivation in Part 5.
The three differences, which recur in the fault formulas of Set 23:
Each has magnitude \(\sqrt3\) — the chord subtending 120° on a unit circle. The \(\sqrt3\) in every line-to-line fault current comes from here.
The complete table, worth memorising:
The last row follows from the central identity by moving one term across. \(1+a = -a^{2}\) is used often enough to be worth recognising directly rather than deriving each time.
State Fortescue's theorem, write the transformation and its inverse, and show that the inverse requires no matrix inversion.
The theorem. Any set of three phasors can be written as the sum of three balanced sets:
Three sets of three phasors, described by three complex numbers — the same information as the original three, re-expressed.
Adding them phase by phase:
Note the subscript convention: \(V_{a1}\) is phase \(a\)'s positive-sequence component, and the other two phases' components are obtained from it. Only phase \(a\)'s three components are ever tabulated.
The inverse. Because \(\mathbf{A}\) is a Vandermonde matrix of the cube roots of unity, its inverse is its own conjugate over three:
Verified numerically to \(10^{-16}\). Since \(a^{*} = a^{2}\), conjugating \(\mathbf{A}\) simply swaps its last two columns — so the inverse is written down, never computed.
The resolution formulas, which is how the inverse is actually used:
Why \(\mathbf{A}^{-1} = \mathbf{A}^{*}/3\). The columns of \(\mathbf{A}\) are orthogonal in the Hermitian sense:
And each column has squared norm 3. So \(\mathbf{A}/\sqrt3\) is unitary — which is the deeper reason the transformation preserves power, as Problem 10 shows.
An alternative normalisation uses \(\mathbf{A}/\sqrt3\) throughout, making the transformation exactly unitary and the power expression free of the factor 3. It is mathematically tidier and almost never used, because engineers want \(V_{a1}\) to be a voltage of the same size as \(V_a\).
A three-phase system has \(V_a = 200\angle0^\circ\), \(V_b = 200\angle245^\circ\) and \(V_c = 100\angle105^\circ\) V. Find its symmetrical components.
In rectangular form, noting \(245^\circ = -115^\circ\):
Zero sequence — simply the average:
Non-zero, so this set does not sum to zero and requires a neutral or earth path to exist physically.
Positive sequence. Rotate \(V_b\) forward by 120° and \(V_c\) back by 120°, then average:
The three rotated phasors are nearly aligned — within 20° — which is why the positive-sequence component is large. That near-alignment is the definition of a nearly balanced set.
Negative sequence:
Here the three are spread over 225° and largely cancel, leaving a small residue.
The three components:
The immediate reading. Positive sequence dominates at 165 V, so the system is recognisably three-phase; but 41 V of zero sequence and 31 V of negative sequence are 25% and 19% of it, which is a severe unbalance. Problem 15 quantifies that judgement.
Reconstruct the three phase voltages from the components found in Problem 3, and write out all nine phasors of the three balanced sets.
The nine phasors, three from each sequence:
The zero-sequence row is three identical phasors; the positive row is \(abc\); the negative row is \(acb\) — read the angles: \(81.1 \to -158.9 \to -38.9\) is clockwise.
Adding down the columns. Phase \(a\):
Exactly the original \(200\angle0^\circ\). The imaginary parts cancel to four decimals.
Phases \(b\) and \(c\), by the same addition:
Both recovered exactly. The transformation is exact and reversible — nothing was approximated.
The nine phasors carry no more information than the three. Nine complex numbers appear, but they are generated by three: each row is determined by its phase-\(a\) entry. That is why only \(V_{a0}\), \(V_{a1}\) and \(V_{a2}\) are ever tabulated, and why the subscript \(a\) is usually dropped once the convention is established.
A useful check. Sum the three phase voltages:
Because the positive and negative rows each sum to zero across the phases (\(1+a+a^{2} = 0\) again), and the zero row sums to \(3V_{a0}\). This is the fastest check on any resolution.
And the geometric picture. Three phasor diagrams — one balanced \(abc\) star, one balanced \(acb\) star, one set of three parallel arrows — superposed vertex to vertex give the original unbalanced star. Drawing it once is worth more than the algebra for understanding what the method claims.
Explain the physical meaning of the zero-sequence component, state the conditions under which it can exist, and relate it to the neutral current.
What it is. Three identical phasors — same magnitude, same angle, in all three phases:
Not a rotating set at all. The three currents flow in the same direction at the same instant, which is a fundamentally different thing from a three-phase quantity.
The consequence for the neutral. The neutral carries the sum:
The positive and negative sequences contribute nothing to it, by \(1+a+a^{2}=0\). The neutral current is exactly three times the zero-sequence current, which is both the physical meaning of zero sequence and the way it is measured.
When it can exist. Zero-sequence current needs a return path, so:
A three-wire three-phase circuit cannot carry zero-sequence current, however unbalanced it is. That is a topological fact and not an approximation.
Zero-sequence voltage is different. It can exist wherever the three phase-to-earth voltages do not sum to zero — which happens during any earth fault, whether or not any zero-sequence current flows. An unearthed system develops large zero-sequence voltage on an earth fault and no zero-sequence current at all.
Why the impedance is different. Zero-sequence current returns through the earth or the neutral, so it sees a completely different circuit:
Hence \(x_0 \approx 3x_1\) for an overhead line — a larger loop area and a resistive earth path. Set 22 develops this.
And the practical importance. Zero-sequence current is the signature of an earth fault, so measuring \(3I_0\) — by summing the three phase currents in a core-balance transformer, or by residually connecting three CTs — gives earth-fault protection that is inherently blind to load, to phase faults, and to balanced conditions. It is the most selective protection principle available.
Resolve the balanced set \(100\angle0^\circ\), \(100\angle-120^\circ\), \(100\angle120^\circ\) and state what the result implies for every study in Parts 3 and 4.
Zero sequence is the average, and the three phasors are 120° apart:
Positive sequence — the rotations align all three:
Because \(aV_b = a\cdot a^{2}V_a = V_a\) and \(a^{2}V_c = a^{2}\cdot aV_a = V_a\). The whole set is positive sequence.
Negative sequence:
The result:
A balanced \(abc\) set is pure positive sequence, with the positive-sequence component equal to the phase-\(a\) phasor itself.
The implication for everything before this set. Every calculation in Parts 3 and 4 assumed balance, so every one of them was a positive-sequence calculation:
The subscript 1 was omitted throughout because there was nothing to distinguish it from.
And that is why a three-phase fault is easy. A symmetrical fault applied to a balanced system leaves it balanced, so only the positive sequence is excited and the whole analysis of Set 18 needs one network. Every unsymmetrical fault excites all three, which is the subject of Set 23.
The converse check. If a resolution gives non-zero \(V_0\) or \(V_2\) for a set believed balanced, either the set is not balanced or the arithmetic is wrong. On measured data it is usually the former — real systems carry 0.5–2% negative sequence at all times.
Resolve the balanced set with reversed phase sequence — \(100\angle0^\circ\), \(100\angle120^\circ\), \(100\angle-120^\circ\) — and explain what a negative-sequence voltage does to a machine.
By the same three averages:
Pure negative sequence — the exact mirror of Problem 6.
The physical distinction. Both sets are balanced; they differ only in the order in which the phases reach their maxima:
Interchanging any two phases converts one into the other, which is why swapping two leads reverses a motor.
What it does in a machine. A balanced three-phase set produces a rotating magnetic field, and the direction of rotation follows the sequence:
So a negative-sequence voltage applied to a machine running forward produces a field rotating backwards at synchronous speed — a relative speed of \(2\omega_s\) with respect to the rotor.
The consequences, and they are severe:
The standard limits reflect how little is tolerable:
The \(I_2^{2}t\) criterion is a thermal one, exactly like a fuse's \(I^{2}t\), and negative-sequence protection is set from it.
Why the machine's negative-sequence impedance is low. Because the field rotates against the rotor at \(2\omega_s\), the flux cannot penetrate — the damper windings and rotor body screen it, exactly as they do during the subtransient period. Hence
against a synchronous reactance of 1.0 to 2.0. A small negative-sequence voltage therefore drives a large negative-sequence current, which is why the limits are expressed in current.
Show that zero-sequence current cannot appear in the lines of a delta-connected element, and explain what happens to it inside.
The line currents of a delta are differences of phase currents:
Their sum telescopes to zero:
Identically, whatever the phase currents are. There is no assumption of balance anywhere in this derivation.
But the phase currents may have a zero-sequence component. Resolving them:
This need not be zero. A zero-sequence current can circulate around the closed delta — it simply cannot get out.
The delta is a zero-sequence trap. It presents two faces:
Both statements are needed. The first governs the zero-sequence network's connectivity; the second explains why a delta winding carries current during an earth fault elsewhere.
The consequences in transformers, which Set 22 develops:
The delta winding of a \(\Delta\)–\(Y\) transformer is what allows earth-fault current to flow on the star side, by providing somewhere for the zero-sequence ampere-turns to balance.
The same argument applies to triple harmonics. Third harmonics in a three-phase system are in phase in all three phases — they are a zero-sequence set — so a delta winding traps them exactly as it traps fundamental zero sequence. That is why transformer designers include a delta tertiary even where the main windings are both star.
A star-connected load has its neutral earthed through an impedance \(Z_n\). Show how \(Z_n\) enters the zero-sequence circuit and explain the factor of three.
The neutral current is the sum of the three phase currents:
The neutral-point voltage rise:
The neutral is displaced from earth potential by this amount whenever zero-sequence current flows.
Writing phase \(a\)'s loop from the phase terminal through the load to the neutral and thence to earth:
And resolving, the zero-sequence part of this equation is
The neutral impedance appears as \(3Z_n\) in the zero-sequence circuit, and only there — the positive- and negative-sequence equations contain no \(V_n\) term, because those currents sum to zero at the neutral and produce no displacement.
Where the three comes from. Two separate factors of the same origin:
The zero-sequence network is a per-phase circuit carrying \(I_{a0}\), but the physical neutral carries \(3I_{a0}\). To produce the correct voltage drop with a third of the current, the impedance must be three times as large.
The three cases:
The middle case is a design choice: a neutral earthing resistor sized to limit earth-fault current to, say, 400 A is standard in industrial and generator-connection practice.
A worked instance. A generator with \(x_0 = 0.06\) pu earthed through a reactor of \(0.05\) pu appears in the zero-sequence network as
More than tripling the machine's own zero-sequence reactance — which is exactly the intended effect.
Show that the complex power in a three-phase system is \(S = 3\left(V_0I_0^{*}+V_1I_1^{*}+V_2I_2^{*}\right)\), and verify it on a numerical example.
The starting point is the definition in phase quantities:
Substituting the transformation:
The middle product is the key. \(\mathbf{A}\) is symmetric, so \(\mathbf{A}^{T} = \mathbf{A}\), and:
Every diagonal entry is \(1+1+1 = 3\); every off-diagonal is a sum of the form \(1+a+a^{2} = 0\). That single identity does all the work.
Hence:
No cross terms. Positive-sequence voltage and negative-sequence current produce no power between them, and the three sequences are energetically independent.
The numerical check. Take the voltages of Problem 3 with currents \(I_a = 10\angle-30^\circ\), \(I_b = 12\angle-160^\circ\), \(I_c = 8\angle80^\circ\) A. Resolving the currents:
Both routes give the same answer:
Agreeing to twelve significant figures. The transformation is power-invariant in this form.
Two consequences worth stating. The decoupling of power is why a negative-sequence quantity is a pure loss to the system — it carries no useful power in combination with the positive-sequence voltage, only \(V_2I_2^{*}\) with itself. And in a balanced system \(V_0 = V_2 = 0\), so \(S = 3V_1I_1^{*}\) — the familiar per-phase formula, recovered.
Phase \(a\) of a balanced 100 A supply is open-circuited, leaving \(I_a = 0\) and the other two unchanged. Find the sequence currents and interpret them.
The currents are \(I_a = 0\), \(I_b = 100\angle-120^\circ\), \(I_c = 100\angle120^\circ\) A.
Zero sequence:
Using \(a+a^{2} = -1\). Non-zero — which requires an earth or neutral return path to be physically possible.
Positive sequence:
Negative sequence:
The result:
And the check: \(I_{a0}+I_{a1}+I_{a2} = -33.33 + 66.67 - 33.33 = 0 = I_a\), as required.
The relation \(I_{a0} = I_{a2}\) is not a coincidence. The single condition \(I_a = 0\) forces
one equation among three unknowns. The further symmetry of this particular case — \(I_b\) and \(I_c\) equal and opposite about the \(a\) axis — makes \(I_0\) and \(I_2\) equal. A general open phase would not.
The engineering point. An open conductor produces 33 A of negative sequence out of an original 100 A balanced supply — a negative-sequence unbalance of 50% relative to the positive-sequence current. A single broken conductor is one of the most damaging faults for rotating plant, and it draws no fault current at all, so overcurrent protection does not see it. Negative-sequence relays do.
Resolve \(100\angle0^\circ\), \(100\angle-120^\circ\), \(50\angle120^\circ\) and compare with the open-phase case.
The three components:
All in amperes. The check: \(16.67\angle-60^\circ + 83.33 + 16.67\angle60^\circ = 8.33-j14.43+83.33+8.33+j14.43 = 100\) ✓.
Comparison with the open-phase case of Problem 11:
The unbalance grows linearly with the deficiency: half a phase missing gives 20% negative sequence, a whole phase missing gives 50%.
The general result for one phase scaled. Let phase \(c\) be \(kI\angle120^\circ\) while the other two are full. Then
At \(k = 0.5\) this gives \(0.5/2.5 = 20\%\); at \(k = 0\), \(1/2 = 50\%\). Both confirmed.
The magnitudes of \(I_0\) and \(I_2\) are equal in both cases, and their angles are \(\mp60^\circ\) here and \(180^\circ\) both in Problem 11. Equal magnitudes are a consequence of only one phase being disturbed:
Because the change appears in one entry of the vector, and both the zero and negative rows of \(\mathbf{A}^{-1}\) weight that entry by \(1/3\) in magnitude.
The practical threshold. Standards set 2% negative-sequence voltage as the limit for continuous operation of induction motors, and by the formula above that corresponds to
A 6% reduction in one phase exhausts the entire allowance. That is how tight the unbalance requirement is, and why single-phase loads on three-phase systems are distributed with care.
A single-phase load draws 100 A from phase \(a\) with a return through earth. Find the sequence currents and state the consequence.
The currents are \(I_a = 100\angle0^\circ\), \(I_b = I_c = 0\).
All three components follow immediately, since only one term survives each sum:
Equal in magnitude and in phase. This is the most unbalanced condition possible, and its sequence signature is the simplest.
The check:
The identity \(1+a+a^{2}=0\) does the work of cancelling phases \(b\) and \(c\) — which is exactly what "three balanced sets superpose to give a single-phase current" means.
The consequence for the network. The negative-sequence current equals the positive-sequence current — a 100% unbalance factor, the theoretical maximum. A single-phase load of any size therefore imposes negative-sequence current of a third of it on every machine in the vicinity.
Which is why single-phase traction is a problem. A 25 kV AC railway takes a single-phase load of tens of megawatts from a three-phase transmission system:
All four are used. The Steinmetz balancer is the elegant one: a capacitor and an inductor of correctly chosen size across the other two phase pairs make a single-phase resistive load appear balanced to the supply.
The same result applies to a single line-to-earth fault, where \(I_b = I_c = 0\) at the fault point by definition. Set 23 will use exactly this: \(I_{a0} = I_{a1} = I_{a2}\) is the boundary condition that determines the connection of the three sequence networks.
During a line-to-line fault between phases \(b\) and \(c\), the currents are \(I_a = 0\) and \(I_b = -I_c = 100\angle-90^\circ\) A. Find the sequence currents and identify the relation between them.
Zero sequence is the average, and \(I_b = -I_c\):
Exactly zero, which is the defining feature of a fault not involving earth. No earth path is required and none exists.
Positive sequence:
The \(a-a^{2} = j\sqrt3\) identity of Problem 1, doing the work it was tabulated for.
Negative sequence:
The result, and the relation:
Two conditions — \(I_0 = 0\) and \(I_1 = -I_2\) — which is exactly what determines how the sequence networks connect for this fault type in Set 23: positive and negative in parallel opposition, zero not connected.
The \(\sqrt3\) and its inverse. Note that \(|I_b| = \sqrt3|I_{a1}|\). The fault current in the faulted phases is \(\sqrt3\) times the positive-sequence component, which is why the standard line-to-line fault formula reads
rather than \(E/(Z_1+Z_2)\). The \(\sqrt3\) is geometric, from the chord subtending 120°.
And the comparison with a three-phase fault. With \(Z_1 = Z_2\), as is nearly true for a network of lines:
A line-to-line fault draws 87% of the three-phase current — which is why the three-phase case is the switchgear rating case, and why the line-to-line case is rarely the binding one.
Define the voltage unbalance factors, evaluate them for the set of Problem 3, and compare with the simpler NEMA definition.
The true definitions, as used in IEC standards:
Two separate factors, because the two sequences do different damage: negative sequence heats rotors, zero sequence flows in neutrals and earths.
For the set of Problem 3 — \(V_0 = 41.09\), \(V_1 = 165.30\), \(V_2 = 31.42\) V:
Both enormous. A 2% negative-sequence factor is the normal continuous limit, so this system is nearly ten times over.
The NEMA definition avoids the transformation entirely, using only the three line-voltage magnitudes:
The two disagree, and systematically:
NEMA uses magnitudes and ignores the angles, so it cannot detect an unbalance in which the three magnitudes are equal but the angles are not 120° apart. Such a set has zero NEMA unbalance and a substantial \(V_2\).
Which to use. NEMA is a field measurement requiring three voltmeters and no phase information; the VUF requires a phase-coherent measurement and gives the physically meaningful quantity:
Modern instruments compute the VUF directly, and standards have moved to it. NEMA persists because a great deal of field practice and equipment derating data is expressed in it.
The derating consequence. An induction motor's derating factor against negative sequence is steep:
Because the negative-sequence impedance of a motor is its locked-rotor impedance — a sixth of its running impedance — so 2% of voltage unbalance gives 12% of negative-sequence current.
Describe how each sequence component is measured in practice, and design a filter that extracts the negative-sequence current from three CT secondaries.
Zero sequence is trivial to measure, because the summation is physical rather than computational:
The second is the better one: it measures the true residual with no CT-mismatch error, and its sensitivity can be a few amperes primary. It cannot be used where the conductors are too large to pass through a common core, which is why the residual connection persists at transmission voltages.
Positive and negative sequence need the rotation, which historically meant a phase-shifting network:
Two 120° shifts and a summation. A 120° lag at one frequency is produced by an \(R\)–\(C\) or \(R\)–\(L\) network, so the classical filter is three impedances and a summing junction.
The classical design, using the identity \(1+a+a^{2}=0\) to eliminate one term. Since \(I_a+I_b+I_c = 3I_0\), and for a three-wire system \(I_0 = 0\):
Substituting \(I_a = -I_b-I_c\). Only two inputs are needed, and each is multiplied by a constant of magnitude \(\sqrt3\) at \(\mp150^\circ\) and \(\pm150^\circ\).
Realising the shifts. An impedance of \(Z = |Z|\angle\phi\) in the current path converts the current into a voltage rotated by \(\phi\), so two impedances with a 60° angle difference and a summing resistor give the network:
Choosing the element values so that the positive-sequence contribution cancels exactly. Such filters were built with iron-cored reactors and oil-filled capacitors, and were tuned on site.
The frequency problem. A passive filter cancels the positive sequence only at the frequency it was tuned for:
Which mattered: a negative-sequence relay that operated on a frequency excursion was a real nuisance in mid-century practice.
The modern answer is arithmetic. A numerical relay samples the three currents, computes the phasors by a Fourier algorithm over one cycle, and applies \(\mathbf{A}^{-1}\) directly:
The transformation that took a cabinet of passive components in 1950 is nine complex multiplications today, and every numerical relay computes all three sequences continuously whether or not it uses them.
A 200 MVA turbogenerator has a continuous negative-sequence capability of 8% and a short-time capability of \(I_2^{2}t = 10\). Find the time it can withstand the unbalance of Problem 11, and set a protection characteristic.
The unbalance of Problem 11 — one phase open — gave \(I_2 = 33.33\) A against a positive-sequence 66.67 A. In per unit of the machine rating, if the machine was carrying rated current before the break:
Taking the original 100 A as rated. That is four times the continuous capability.
The withstand time:
Ninety seconds before rotor damage. Long enough for a relay to act with margin, and far too short for an operator to diagnose the problem.
The withstand characteristic across the range:
A hyperbolic curve in \(I_2^{2}\), exactly like a fuse's characteristic — because it is the same physics, a fixed amount of energy to reach a limiting temperature.
Why \(I_2^{2}t\). The rotor surface loss is proportional to \(I_2^{2}\), and the heating is adiabatic on this timescale — the rotor body cannot conduct the heat away fast enough:
So the limit is on the integral, and a varying \(I_2\) is accumulated rather than compared to a threshold. Numerical relays implement exactly this integral.
The protection setting, in two stages:
The alarm exists because most negative-sequence conditions are external and correctable — a broken conductor, an open pole on a breaker, a large single-phase load — and shutting the machine down is the last resort.
The reset problem. A machine that has absorbed 60% of its \(I_2^{2}t\) capability and then returns to balanced operation cools, but not instantly. A relay that resets its integrator immediately would permit repeated near-limit excursions:
Matching the rotor's thermal time constant. This is a genuine setting, not a detail, and getting it wrong is how machines are damaged by a sequence of individually survivable events.
A star–delta transformer shifts the positive-sequence voltage by \(+30^\circ\). Show what it does to the negative sequence, and explain why the two shifts are opposite.
The physical origin. In a \(Y\)–\(\Delta\) transformer, a line-to-neutral voltage on the star side corresponds to a line-to-line voltage on the delta side, and line-to-line leads line-to-neutral by 30°.
For positive sequence. Take the star-side voltages balanced \(abc\) with \(V_a = V\angle0^\circ\). The delta-side line voltage corresponding to phase \(a\) is
Using \(1-a^{2} = \sqrt3\angle30^\circ\) from Problem 1. The shift is \(+30^\circ\).
For negative sequence, the phase order is reversed, so \(V_b = aV_a\) rather than \(a^{2}V_a\):
The shift is \(-30^\circ\) — equal and opposite.
The general rule, which follows immediately:
And the direction depends on which side is taken as reference: crossing from delta to star reverses both signs. IEC vector groups — Yd1, Yd11 and the rest — encode the angle in units of 30°.
Why opposite. The shift arises from the difference of two phasors 120° apart, and reversing the sequence reverses which of the two leads. Geometrically, the isoceles triangle formed by \(V_a\), \(-V_b\) and their sum is reflected. Algebraically it is the difference between \(1-a^{2}\) and \(1-a\), which are conjugates.
The consequence for fault studies. Currents and voltages on the far side of a transformer must be shifted before they can be combined:
A single line-to-earth fault on the delta side appears as a two-phase current pattern on the star side, in the ratio 2:1:1 — a result that comes directly from applying the two opposite shifts and recombining.
And a simplification often used. If the study needs only magnitudes — as most fault-level calculations do — the phase shifts can be ignored entirely, because they affect all quantities on one side of the transformer equally. They matter only when quantities from both sides are combined, which is exactly what protection does.
Show that \(\mathbf{A}\) diagonalises the impedance matrix of any balanced three-phase element, and identify the eigenvalues.
A balanced element's impedance matrix has equal self-impedances and equal mutual impedances:
This is a circulant matrix — each row is the previous one rotated. Transposed lines and symmetrically wound machines both produce it.
Transforming. With \(\mathbf{V}_{abc} = \mathbf{Z}_{abc}\mathbf{I}_{abc}\) and \(\mathbf{V}_{abc} = \mathbf{A}\mathbf{V}_{012}\):
A similarity transformation, so the result has the same eigenvalues as \(\mathbf{Z}_{abc}\).
Carrying it out. Apply \(\mathbf{Z}_{abc}\) to each column of \(\mathbf{A}\). The first column is \([1,1,1]^{T}\):
An eigenvector, with eigenvalue \(Z_s+2Z_m\).
The second column is \([1, a^{2}, a]^{T}\):
using \(a+a^{2} = -1\); and the other rows give \(a^{2}(Z_s-Z_m)\) and \(a(Z_s-Z_m)\). So it too is an eigenvector, with eigenvalue \(Z_s-Z_m\). The third column gives the same eigenvalue.
The result:
This is the whole justification of the method. The matrix is diagonal, so the three sequences do not interact:
Three independent single-phase problems. Everything Part 5 does rests on this diagonalisation, and it holds only because the network is balanced — a fact worth stating clearly, since the method is used to analyse unbalanced conditions.
The apparent paradox, resolved. Symmetrical components requires a balanced network but permits unbalanced conditions. A transposed line is balanced; a fault on it is not. The transformation diagonalises the line and the fault appears as a boundary condition coupling the three networks at one point only — which is precisely the structure Set 23 exploits.
And \(Z_0 \ne Z_1\) follows immediately. The difference is \(3Z_m\), and for an overhead line the mutual impedance between phases is a substantial fraction of the self-impedance because of the shared earth return. Hence \(Z_0 \approx 3Z_1\) — the rule of thumb of Set 22, derived rather than asserted.
A 400 kV busbar during an earth fault has \(V_a = 20\angle0^\circ\), \(V_b = 235\angle-105^\circ\), \(V_c = 240\angle115^\circ\) kV to earth. Resolve, interpret every component, and identify the fault type.
The nominal phase voltage for reference:
So phases \(b\) and \(c\) are near normal and phase \(a\) has collapsed to 8.7% of nominal.
Zero sequence:
Substantial, and roughly in antiphase with the healthy-phase voltages. Zero sequence is present, so the fault involves earth.
Positive sequence. Rotating \(V_b\) by \(+120^\circ\) and \(V_c\) by \(+240^\circ\):
Depressed to 70% of nominal — the network's positive-sequence voltage pulled down by the fault current, exactly as the fault-study voltage profile of Set 18 described.
Negative sequence:
Large — 41% of nominal, and nearly in antiphase with the positive sequence, which is characteristic of a phase-\(a\) earth fault.
The three components:
The check, in rectangular form:
Exactly \(V_a\). Note how much cancellation is involved — three quantities of 47, 163 and 95 kV combining to 20 — which is why the resolution must be carried to full precision and why a graphical construction of a faulted phasor set is unreliable.
Identifying the fault. Three observations, in order:
And \(V_0\), \(V_2\) are both nearly in antiphase with \(V_1\) — the signature of an earth fault on phase \(a\), since the three must very nearly cancel there.
The magnitude of \(V_a\) at the fault is 20 kV rather than zero, so the fault is not solid:
With a typical fault current of 15 kA that gives \(Z_f = 1.3\ \Omega\) — consistent with an arc of a metre or two, which is what a 400 kV flashover produces.
The healthy-phase voltages are near normal, at 235 and 240 kV against 231 nominal — a rise of under 4%. That identifies the system as solidly earthed:
The ratio of the healthy-phase rise to nominal is the classical earthing-coefficient measurement, and it is read directly from these numbers.
Practice Problems
Work each on paper before opening the answer. The answer is given so you can check yourself; the method is deliberately not.
P1. Evaluate \(a^{4}\), \(a^{5}\) and \(a^{-1}\).
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\(a^{4} = \mathbf{a}\), \(a^{5} = \mathbf{a^{2}}\), \(a^{-1} = \mathbf{a^{2}}\) — every power reduces modulo 3.P2. Simplify \(1 + a\) and \(a + a^{2}\).
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\(1+a = \mathbf{-a^{2}} = 1\angle60^\circ\); \(a+a^{2} = \mathbf{-1}\). Both from \(1+a+a^{2}=0\).P3. A set has \(V_a+V_b+V_c = 60\angle30^\circ\). What is \(V_0\)?
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\(\mathbf{20\angle30^\circ}\) — one third of the sum, always.P4. A balanced \(acb\) set of 50 V. Give its three components.
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\(V_0 = 0\), \(V_1 = 0\), \(V_2 = \mathbf{50\angle0^\circ}\) — pure negative sequence.P5. Three CT secondaries are paralleled and 12 A flows in the connection. What is \(I_0\)?
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\(\mathbf{4}\) A. The residual connection measures \(3I_0\).P6. A generator's neutral is earthed through \(j0.04\) pu. What appears in its zero-sequence circuit?
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\(3Z_n = \mathbf{j0.12}\) pu, in series with the machine's own \(Z_0\) — and in the zero-sequence network only.P7. Why can zero-sequence current not flow in the lines of a delta?
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The three line currents are differences of phase currents and sum identically to zero. Zero sequence circulates inside the delta instead.P8. \(V_1 = 240\), \(V_2 = 12\) V. What is the negative-sequence unbalance factor?
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\(12/240 = \mathbf{5\%}\) — well beyond the 2% continuous limit for motors.P9. A single-phase current of 60 A flows in phase \(b\) only. Give the three components.
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\(I_0 = 20\angle\theta\), \(I_1 = 20\angle(\theta+120^\circ)\), \(I_2 = 20\angle(\theta-120^\circ)\) — equal magnitudes, with the rotations that place the sum in phase \(b\).P10. A machine has \(I_2^{2}t = 12\) and carries \(I_2 = 0.4\) pu. How long can it survive?
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\(12/0.16 = \mathbf{75}\) s.P11. A \(Y\)–\(\Delta\) transformer shifts positive sequence by \(+30^\circ\). What does it do to negative sequence and to zero?
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Negative sequence \(\mathbf{-30^\circ}\); zero sequence has no path through the delta at all.P12. A balanced line has \(Z_s = j0.5\) and \(Z_m = j0.15\) pu. Find its three sequence impedances.
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\(Z_0 = Z_s+2Z_m = \mathbf{j0.80}\); \(Z_1 = Z_2 = Z_s-Z_m = \mathbf{j0.35}\). Ratio \(Z_0/Z_1 = 2.3\).
Challenge Problems
Three problems that push the transformation past the cases it is usually shown on.
C1 — The Steinmetz balancer. A single-phase resistive load of \(P\) watts is connected between phases \(a\) and \(b\) of a balanced supply. Find the reactive elements that must be connected across the other two phase pairs to make the supply see a perfectly balanced three-phase load, and verify that the negative-sequence current vanishes.
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The claim. Connect a capacitor of susceptance \(B_C = G/\sqrt3\) across \(bc\) and an inductor of susceptance \(B_L = G/\sqrt3\) across \(ca\), where \(G\) is the load's conductance. The three delta branches are then
\[ Y_{ab} = G \qquad Y_{bc} = +j\frac{G}{\sqrt3} \qquad Y_{ca} = -j\frac{G}{\sqrt3} \]The verification. With \(V_{ab} = V\angle30^\circ\), \(V_{bc} = V\angle-90^\circ\), \(V_{ca} = V\angle150^\circ\) (balanced line voltages), the branch currents are
\[ I_{ab} = GV\angle30^\circ \qquad I_{bc} = \frac{GV}{\sqrt3}\angle0^\circ \qquad I_{ca} = \frac{GV}{\sqrt3}\angle60^\circ \]and the line currents \(I_A = I_{ab}-I_{ca}\), \(I_B = I_{bc}-I_{ab}\), \(I_C = I_{ca}-I_{bc}\) come out equal in magnitude at \(GV/\sqrt3\) and 120° apart — a balanced \(abc\) set. So \(I_2 = 0\) and \(I_0 = 0\) identically.
What has happened. The reactive elements consume no average power, so the real power drawn is still \(P\). They circulate reactive power between the phases in exactly the pattern that cancels the negative-sequence component of the single-phase load. Two reactive elements convert a 100%-unbalanced load into a perfectly balanced one at no energy cost.
The limitation, and it is severe. The balance is exact only at the design load and the design frequency. A load that varies needs a variable balancer — which is what a static var compensator with independent phase control is, and why railway feeder stations use them. A fixed Steinmetz circuit sized for full load makes the unbalance worse at light load.
C2 — When the network is not balanced. An untransposed line has unequal mutual impedances between its three phase pairs. Show that the sequence impedance matrix is no longer diagonal, quantify the coupling for a typical horizontal line, and say what it means for fault analysis.
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The general transformation. With unequal mutuals \(Z_{ab}\), \(Z_{bc}\), \(Z_{ca}\), the matrix \(\mathbf{Z}_{abc}\) is symmetric but not circulant, so \(\mathbf{A}\) does not diagonalise it. Carrying out \(\mathbf{A}^{-1}\mathbf{Z}_{abc}\mathbf{A}\) gives off-diagonal terms of the form
\[ Z_{01} = \tfrac{1}{3}\left(Z_{ca} + aZ_{ab} + a^{2}Z_{bc}\right)\ \text{-like combinations} \]Each is a weighted difference of the three mutuals, so it vanishes when they are equal and grows with the asymmetry.
The size for a horizontal line. Phases at 0, 6 and 12 m: \(D_{ab} = D_{bc} = 6\) m and \(D_{ca} = 12\) m. Since mutual impedance goes as \(\ln(D_e/D)\) with \(D_e \approx 900\) m for the earth return:
\[ \frac{\ln(900/6)}{\ln(900/12)} = \frac{5.01}{4.32} = 1.16 \]A 16% difference between the mutuals, giving sequence coupling terms of roughly 3–5% of the self-impedance.
What it means. Three consequences, in increasing order of importance:
\[ \begin{array}{ll} \text{A balanced load draws unbalanced current} & \sim1\text{--}2\%\ \text{negative sequence} \\ \text{Sequence networks are no longer independent} & \text{a positive-sequence source excites } I_2 \\ \text{Fault currents are wrong by a few per cent} & \text{usually acceptable} \end{array} \]The remedy is transposition — rotating the phase positions at one-third intervals along the route, so that each phase occupies each position for a third of the length and the average mutuals are equal. It is why transposition towers exist, and it is why every impedance in Parts 3 and 4 was quoted for a transposed line without comment.
And where it cannot be ignored: long untransposed lines, cable circuits with asymmetric bonding, and any study of steady-state unbalance rather than faults. Modern software carries the full \(3\times3\) matrix and does not use symmetrical components at all for those cases.
C3 — The transformation's other relatives. Symmetrical components diagonalises a circulant by the \(3\times3\) Fourier matrix. Identify two other transformations in power engineering that do the same job for different structures, state what each diagonalises, and explain why \(dq0\) is needed at all when symmetrical components already exists.
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The three transformations:
\[ \begin{array}{lll} \text{Symmetrical components} & \text{constant circulant } \mathbf{Z}_{abc} & \text{steady state, any frequency} \\ \text{Clarke } (\alpha\beta0) & \text{the same, but real} & \text{instantaneous, real arithmetic} \\ \text{Park } (dq0) & \text{rotor-position-dependent } \mathbf{L}(\theta) & \text{machines, transient} \end{array} \]Clarke's transformation is symmetrical components with a real basis: it maps \(abc\) to two orthogonal axes plus a zero component, using only real coefficients. It diagonalises the same circulant, works on instantaneous values rather than phasors, and is what a digital controller computes because it needs no complex arithmetic.
Park's transformation solves a different problem. A synchronous machine's inductance matrix is not constant — it depends on the rotor angle \(\theta\), because the rotor is magnetically asymmetric:
\[ L_{aa}(\theta) = L_0 + L_2\cos2\theta \qquad \text{and similarly for the rest} \]No constant transformation can diagonalise a time-varying matrix. Symmetrical components fails on a salient-pole machine's transient behaviour for exactly this reason. Park's transformation rotates with the rotor, and in that rotating frame the inductances become constants — which turns a set of differential equations with periodic coefficients into one with constant coefficients.
The relationship between them. For a machine in balanced steady state, the \(dq0\) quantities are constants and the symmetrical-component quantities are phasors describing the same thing — they agree. Under unbalance or transients they diverge, and the choice is dictated by what is being asked:
\[ \begin{array}{ll} \text{Steady-state unbalance, faults} & \text{symmetrical components} \\ \text{Machine transients, control} & dq0 \\ \text{Instantaneous power theory, converters} & \alpha\beta0 \end{array} \]The unifying statement: each is a change of basis chosen so that the matrix describing the physics becomes diagonal. What differs is which matrix — a constant circulant, or one that rotates. That the three-phase system admits such a transformation at all is a consequence of its symmetry, and it is the single mathematical fact that makes three-phase analysis tractable.
Multiple-Choice Questions
MCQ 1. \(1 + a + a^{2}\) equals:
(a) 1 (b) 3 (c) 0 (d) \(a\)Show answer
(c). Three unit vectors 120° apart. Almost every derivation in Part 5 uses it. Problem 1.MCQ 2. \(\mathbf{A}^{-1}\) equals:
(a) \(\mathbf{A}\) (b) \(\tfrac{1}{3}\mathbf{A}^{*}\) (c) \(3\mathbf{A}^{*}\) (d) \(\mathbf{A}^{T}\)Show answer
(b) — so the inverse is written down rather than computed. Problem 2.MCQ 3. A balanced \(abc\) set has:
(a) only zero sequence (b) only positive sequence (c) only negative sequence (d) all three equalShow answer
(b) — which is why every balanced study in Parts 3 and 4 was a positive-sequence study. Problem 6.MCQ 4. The neutral current equals:
(a) \(I_0\) (b) \(3I_0\) (c) \(I_0/3\) (d) \(I_1+I_2\)Show answer
(b). The positive and negative sequences contribute nothing to it. Problem 5.MCQ 5. Zero-sequence current in the lines of a delta is:
(a) three times the phase value (b) equal to the phase value (c) zero (d) \(\sqrt3\) timesShow answer
(c), identically. It circulates inside the delta instead. Problem 8.MCQ 6. A neutral earthing impedance \(Z_n\) appears in the sequence networks as:
(a) \(Z_n\) in all three (b) \(3Z_n\) in the zero-sequence network only (c) \(Z_n/3\) in zero (d) not at allShow answer
(b). The per-phase network carries \(I_0\) while the neutral carries \(3I_0\). Problem 9.MCQ 7. The complex power in sequence quantities is:
(a) \(V_0I_0^{*}+V_1I_1^{*}+V_2I_2^{*}\) (b) three times that (c) that plus cross terms (d) \(\sqrt3\) times thatShow answer
(b), with no cross terms — because \(\mathbf{A}\mathbf{A}^{*} = 3\mathbf{I}\). Problem 10.MCQ 8. An open conductor in an otherwise balanced supply gives a negative-sequence unbalance of:
(a) 0 (b) 20% (c) 50% (d) 100%Show answer
(c) — \(I_2 = 33.3\) against \(I_1 = 66.7\). And no increase in current at all, so overcurrent protection is blind to it. Problem 11.MCQ 9. A single-phase load on phase \(a\) gives:
(a) \(I_0 = I_1 = I_2\) (b) \(I_0 = 0\) (c) \(I_1 = -I_2\) (d) only positive sequenceShow answer
(a), all equal to a third of the current — and the same boundary condition as a single line-to-earth fault. Problem 13.MCQ 10. A line-to-line fault gives:
(a) \(I_0 = I_1 = I_2\) (b) \(I_0 = 0\) and \(I_1 = -I_2\) (c) \(I_1 = I_2\) (d) \(I_0 \ne 0\)Show answer
(b) — no earth involved, so no zero sequence. Problem 14.MCQ 11. A \(Y\)–\(\Delta\) transformer shifts negative sequence by:
(a) \(+30^\circ\) (b) \(-30^\circ\) (c) \(0^\circ\) (d) \(180^\circ\)Show answer
(b), the opposite of the positive-sequence shift, because \(1-a\) and \(1-a^{2}\) are conjugates. Problem 18.MCQ 12. For a balanced line with self \(Z_s\) and mutual \(Z_m\), the zero-sequence impedance is:
(a) \(Z_s-Z_m\) (b) \(Z_s+Z_m\) (c) \(Z_s+2Z_m\) (d) \(3Z_s\)Show answer
(c), while \(Z_1 = Z_2 = Z_s-Z_m\). The difference \(3Z_m\) is why \(Z_0 \approx 3Z_1\) for an overhead line. Problem 19.
Key Formulas
| Quantity | Relation | Notes |
|---|---|---|
| The operator | \(a = 1\angle120^\circ\), \(a^{3} = 1\) | \(a^{*} = a^{2} = 1/a\) |
| Central identity | \(1+a+a^{2} = 0\) | Used in almost every derivation |
| Differences | \(1-a = \sqrt3\angle-30^\circ\), \(a-a^{2} = j\sqrt3\) | Source of every \(\sqrt3\) in Part 5 |
| Synthesis | \(\mathbf{V}_{abc} = \mathbf{A}\mathbf{V}_{012}\) | \(\mathbf{A} = [1\,1\,1;\,1\,a^{2}\,a;\,1\,a\,a^{2}]\) |
| Resolution | \(\mathbf{V}_{012} = \tfrac{1}{3}\mathbf{A}^{*}\mathbf{V}_{abc}\) | No inversion required |
| Zero sequence | \(V_0 = \tfrac{1}{3}(V_a+V_b+V_c)\) | Non-zero only with an earth path |
| Positive sequence | \(V_1 = \tfrac{1}{3}(V_a+aV_b+a^{2}V_c)\) | |
| Negative sequence | \(V_2 = \tfrac{1}{3}(V_a+a^{2}V_b+aV_c)\) | |
| Neutral current | \(I_n = 3I_0\) | How \(I_0\) is measured |
| Earthing impedance | appears as \(3Z_n\) | Zero-sequence network only |
| Power | \(S = 3(V_0I_0^{*}+V_1I_1^{*}+V_2I_2^{*})\) | No cross terms |
| Unbalance factor | \(\text{VUF} = |V_2|/|V_1|\) | 2% is the usual limit |
| Machine withstand | \(I_2^{2}t \le K\) | \(K = 6\)–30 for turbogenerators |
| Sequence impedances | \(Z_0 = Z_s+2Z_m\), \(Z_1 = Z_2 = Z_s-Z_m\) | Balanced element only |
| Transformer shift | \(+30^\circ\) positive, \(-30^\circ\) negative | Zero sequence: no path |
Common Mistakes
Omitting the \(1/3\) in the resolution. The synthesis matrix has no factor; the resolution matrix has \(1/3\) — Problem 2.
Swapping the \(a\) and \(a^{2}\) rows. Positive sequence uses \(aV_b\) and negative uses \(a^{2}V_b\); interchanging them swaps the two answers — Problem 2.
Writing \(Z_n\) rather than \(3Z_n\). The commonest error in Part 5, and it understates the earth-fault impedance — Problem 9.
Putting \(3Z_n\) in the positive-sequence network. It belongs only in the zero-sequence one, because only \(I_0\) flows in the neutral — Problem 9.
Expecting zero-sequence current in a three-wire system. It is topologically impossible, however unbalanced the currents — Problem 5.
Assuming a delta blocks all sequences. It blocks only zero sequence, and only from the lines — Problem 8.
Forgetting the factor 3 in the power expression. \(S = 3\sum V_kI_k^{*}\) in the standard normalisation — Problem 10.
Expecting cross terms in the power. There are none; the sequences are energetically independent — Problem 10.
Judging unbalance by voltage magnitudes alone. NEMA's definition misses an unbalance that is purely in the angles — Problem 15.
Giving positive and negative sequence the same transformer phase shift. They are equal and opposite — Problem 18.
Applying symmetrical components to an unbalanced network. The diagonalisation requires a circulant impedance matrix, i.e. a transposed line — Problem 19 and Challenge C2.
Rounding the components before recombining. Large components can cancel to a small phase quantity, so intermediate rounding destroys the answer — Problem 20.
The transformation is established, and with it the one fact that makes Part 5 possible: a balanced network's impedance matrix is diagonalised by \(\mathbf{A}\), so the three sequences propagate through it independently. An unbalanced condition — a fault, an open conductor, a single-phase load — then appears not as a coupling inside the network but as a boundary condition at one point, linking three otherwise separate circuits.
Set 22 builds those three circuits for the five-bus system of Part 4: the positive-sequence network is the fault network already constructed in Set 18, the negative-sequence network is very nearly the same, and the zero-sequence network is a different network entirely — its connectivity set by transformer windings and its impedances roughly three times larger. Set 23 then applies the boundary conditions of Problems 13 and 14 to connect them, and computes the four standard fault types on the same system.