19.2 Symmetrical Components & Unbalanced Fault Analysis
Key Takeaways
- Any set of three unbalanced phasors resolves into positive-, negative- and zero-sequence sets, with Va0 = (Va + Vb + Vc)/3 and the operator a = 1∠120°.
- A three-phase balanced fault uses the positive-sequence network only; single line-to-ground connects all three sequence networks in series and gives Ia1 = Ea/(Z1 + Z2 + Z0).
- Line-to-line faults connect positive and negative sequence networks in parallel with no zero-sequence involvement, so LL fault current is about 86.6% of the three-phase value.
- Zero-sequence current requires a return path: a delta winding traps it and an ungrounded wye blocks it entirely, which is why transformer connection determines ground-fault current.
- Faults ranked by typical frequency of occurrence are single line-to-ground (about 70–80%), line-to-line, double line-to-ground, then three-phase (about 5%), which is nevertheless usually the most severe.
19.2 Symmetrical Components & Unbalanced Fault Analysis
Balanced three-phase faults are the easy case and account for only about 5% of real faults. The remaining 95% are unbalanced, and the only practical way to analyse them by hand is Fortescue's method of symmetrical components, a core component of Enhanced TOS topic K, Power System Analysis (PRBEE Resolution No. 40, s. 2024 — 9.00% of the examination, 20 of the 100 Electrical Engineering items).
1. The Symmetrical Component Transformation
Any three unbalanced phasors decompose into three balanced sets:
- Positive sequence (1) — three equal phasors 120° apart in the normal rotation $a$–$b$–$c$; this is the only set present in a healthy balanced system;
- Negative sequence (2) — three equal phasors 120° apart in the reversed rotation $a$–$c$–$b$; it produces a counter-rotating field that heats generator rotors, which is why negative-sequence relays exist;
- Zero sequence (0) — three equal phasors in phase; it can flow only where a neutral or ground return path exists.
The operator $a = 1\angle 120° = -0.5 + j0.866$ satisfies
Synthesis (components to phase quantities):
Analysis (phase quantities to components):
Two consequences are examined repeatedly:
- In a three-wire system (no neutral), $I_a + I_b + I_c = 0$, so zero-sequence current is zero by definition;
- In a four-wire system the neutral carries $I_n = 3I_{a0}$ — the neutral current is three times the zero-sequence current, not equal to it.
2. Sequence Networks
Each sequence has its own network with its own impedance.
| Element | $Z_1$ | $Z_2$ | $Z_0$ |
|---|---|---|---|
| Synchronous generator | $X_d''$ | $\approx X_d''$ | Smallest; typically $0.15$–$0.6 \times X_1$ |
| Transformer | $Z_T$ | $Z_T$ | $Z_T$, but path depends on winding connection |
| Transmission line | $Z_1$ | $Z_1 = Z_2$ | $2$–$3.5 \times Z_1$ (ground return path) |
For static equipment — lines, cables, transformers — $Z_1 = Z_2$ always, because a static device cannot tell the direction of phase rotation. For rotating machines $Z_2 \ne Z_1$ in general, though $Z_2 \approx X_d''$ is the usual working assumption.
Zero-sequence line impedance is substantially higher than positive sequence because the return path is the earth and any overhead ground wires, enclosing a much larger flux loop.
Transformer Zero-Sequence Connection Rules
This table is the single most examinable item in the topic:
| Connection | Zero-sequence path |
|---|---|
| Grounded wye – grounded wye | Continuous through the transformer |
| Grounded wye – delta | Flows on the wye side, circulates and is trapped in the delta; does not pass through |
| Delta – delta | Blocked on both sides; no external zero-sequence path |
| Ungrounded wye – anything | Blocked; no zero-sequence current at all |
| Grounded wye – delta (as grounding bank) | Provides a deliberate zero-sequence source for ground-fault detection |
A delta winding is a zero-sequence trap: it gives circulating third-harmonic and zero-sequence currents somewhere to go, which is why a delta tertiary is added to large wye-wye autotransformers.
3. Fault Types and Sequence Network Interconnection
| Fault type | Network connection | Fault current expression |
|---|---|---|
| Three-phase (balanced) | Positive sequence only | $I_f = \dfrac{E_a}{Z_1}$ |
| Single line-to-ground (SLG) | $Z_1$, $Z_2$, $Z_0$ in series | $I_{a1} = \dfrac{E_a}{Z_1+Z_2+Z_0}$, $I_a = 3I_{a1}$ |
| Line-to-line (LL) | $Z_1$, $Z_2$ in parallel; no zero sequence | $I_{a1} = \dfrac{E_a}{Z_1+Z_2}$, $I_b = -j\sqrt{3},I_{a1}$ |
| Double line-to-ground (DLG) | $Z_1$ in series with $Z_2 \parallel Z_0$ | $I_{a1} = \dfrac{E_a}{Z_1 + \dfrac{Z_2Z_0}{Z_2+Z_0}}$ |
| Open conductor | Series (rather than shunt) network connection | Produces negative-sequence current without high fault current |
Fault-current arithmetic worth memorising: with $Z_1 = Z_2$, the line-to-line fault current is
so an LL fault is always about 86.6% of the three-phase value. A single line-to-ground fault can exceed the three-phase value when $Z_0 < Z_1$, which happens on solidly grounded systems close to a grounded-wye transformer — a genuine engineering result and a favourite examination trap for candidates who assume the three-phase fault is always the worst case.
Faults ranked by frequency of occurrence: SLG 70–80%, LL about 15%, DLG about 10%, and three-phase only about 5%.
4. System Grounding
| Grounding method | Ground-fault current | Transient overvoltage | Typical application |
|---|---|---|---|
| Solidly grounded | Very high | Low | LV distribution, most utility MV |
| Low-resistance grounded | Limited to 100–1000 A | Moderate | Industrial MV systems |
| High-resistance grounded | Limited to about 5–10 A | Moderate | Continuous-process plants |
| Ungrounded | Capacitive only | High (arcing ground fault) | Legacy systems; increasingly avoided |
Neutral grounding impedance $Z_n$ appears in the zero-sequence network as $3Z_n$ — three times, because the total zero-sequence current $3I_{a0}$ flows through it. Forgetting the factor of three is the standard error in grounded-neutral fault problems.
Solved Board Exam Examples
Example 1: Resolving Unbalanced Line Currents
A four-wire feeder carries $I_a = 10\angle 0°$ A, $I_b = 10\angle 230°$ A and $I_c = 10\angle 130°$ A. Find the zero-sequence current and the neutral current.
Solution.
Converting to rectangular form: $10\angle 0° = 10 + j0$; $10\angle 230° = -6.428 - j7.660$; $10\angle 130° = -6.428 + j7.660$.
The neutral carries three times this value:
Had the feeder been three-wire with no neutral or ground return, $I_a + I_b + I_c$ would be forced to zero and no zero-sequence component could exist at all — a useful sanity check on any set of measured currents.
Example 2: Single Line-to-Ground Fault Current
At a bus, $Z_1 = j0.15$ pu, $Z_2 = j0.15$ pu and $Z_0 = j0.05$ pu on a 100 MVA, 13.8 kV base. Find the SLG fault current in amperes and compare it with the three-phase fault.
Solution. For an SLG fault the sequence networks are in series:
The three-phase fault gives $1.0/0.15 = 6.667$ pu $= 27.89$ kA. Because $Z_0 < Z_1$, the ground fault is 28% more severe than the three-phase fault — precisely the case that defeats the "three-phase is always worst" assumption.
Example 3: Line-to-Line Fault Ratio
For a machine with $Z_1 = Z_2 = j0.20$ pu, find the ratio of line-to-line fault current to three-phase fault current.
Solution.
For the LL fault, $I_{a1} = \dfrac{1.0}{j(0.20+0.20)} = -j2.5$ pu, and the faulted phase current is
This √3/2 ratio holds whenever $Z_1 = Z_2$, independent of the actual impedance values.
In a four-wire grounded system, what is the relationship between the neutral current and the zero-sequence current of phase a?
A grounded-wye to delta transformer feeds a load. What happens to zero-sequence current that flows in the grounded-wye winding?
At a particular bus Z1 = Z2 = j0.10 pu and Z0 = j0.30 pu. How does the single line-to-ground fault current compare with the three-phase fault current?