3.3 Symmetrical Components & Fault-Current Paths

The Three Sequence Networks

Quick Answer: Any set of three unbalanced phasors can be decomposed into three balanced sets: positive sequence (normal rotation, ABC), negative sequence (reverse rotation, ACB), and zero sequence (three in-phase phasors that add in the neutral/ground path).

C.L. Fortescue's 1918 theorem is the foundation of modern fault analysis. Each sequence has its own network impedance:

• Z1 (positive sequence): the impedance seen by a balanced positive-sequence set — the normal load impedance, including source, transformer, and line.
• Z2 (negative sequence): the impedance seen by a negative-sequence set. For static equipment (transformers, lines) Z2 ≈ Z1. For rotating machines Z2 is different and usually smaller.
• Z0 (zero sequence): the impedance seen by three in-phase currents that return through neutral or ground. Z0 depends strongly on transformer winding connection and grounding — a delta winding blocks zero sequence, a wye with grounded neutral passes it.

A balanced three-phase system contains only positive sequence. Unbalanced conditions (L-G fault, open phase, unbalanced load) introduce negative and/or zero sequence.

Which Networks Each Fault Type Uses

Fault Magnitude Relationships

Assume the common simplification Z1 = Z2 = Z (true for static equipment) and compare to the three-phase fault current I3φ = V / Z1:

• Three-phase: I3φ = V / Z (the reference).
• L-L fault: ILL = V / (Z1 + Z2) = V / (2Z). Compared to I3φ: ILL / I3φ = Z / (2Z) × (note L-L uses L-L voltage). The standard result is ILL ≈ 0.866 × I3φ (√3/2 of the three-phase fault) when Z1 = Z2.
• L-G fault (Z0 = Z1): ILG = 3V / (3Z) = V / Z = I3φ. L-G equals the three-phase fault when Z0 = Z1.
• L-G fault (Z0 = 3 × Z1, typical impedance-grounded): ILG = 3V / (Z + Z + 3Z) = 3V / (5Z) = 0.6 × V/Z = 0.6 × I3φ. Lower than 3φ.
• L-G fault on an ungrounded system: Z0 is essentially infinite (only capacitive charging current returns), so ILG ≈ 0 — only a few amps of charging current. This is why ungrounded systems keep operating on a single ground fault but alarm instead of tripping.

The key insight: L-G fault magnitude depends almost entirely on Z0, which depends on transformer connection and grounding choice. Three-phase and L-L faults do not involve Z0.

The Fault-Current Path in a Wye System

On a solidly grounded wye system (e.g., 480Y/277 V with neutral-ground bond at the service), a line-to-ground fault drives current from the source transformer secondary, through the phase conductor, through the fault to ground/neutral, back through the equipment grounding conductor (EGC) to the service neutral-ground bond, and through the neutral to the source. The path is metallic and low-impedance, so L-G fault current is high — high enough to operate a standard overcurrent relay (51) or instantaneous element (50).

On an impedance-grounded system (covered in 3.4) the neutral-ground bond includes a resistor or reactor, so the same path carries only a limited current (typically 5-400 A). The system alarms but does not trip on the first ground fault — useful where continuity of service matters.

Why This Matters for NETA Work

• Relay coordination: A 51/51N relay must be set to trip for L-G faults in its zone but to wait (coordinate) for a downstream device to clear first. Knowing whether the system is low-Z or high-Z grounded tells you whether to expect a high-magnitude L-G fault or a low-magnitude alarm condition.
• CT selection: CT ratios and accuracy classes (e.g., C100, C200, C400, C800) are chosen so the CT does not saturate during the maximum fault current. If you underestimate L-G fault current because you forgot the system is solidly grounded, the CT saturates and the relay misoperates.
• Differential (87) relays: Through-fault stability depends on CT matching and on the actual fault current. An L-G through-fault on a grounded-wye transformer produces zero-sequence current that can fool a differential scheme unless the CTs are wye-connected with the zero sequence accounted for.
• Breaker interrupting duty: The breaker must interrupt the maximum fault at its location. Three-phase faults often set the duty on solidly grounded systems, but L-G faults can exceed 3φ on systems with very low Z0 (e.g., solidly grounded generator neutrals).

Exam Trap

A common Level 2 trap is to assume L-G fault current is always lower than 3φ fault current. It is not always lower. On a solidly grounded system with Z0 < Z1 (e.g., a generator with a solidly grounded neutral), L-G fault current can exceed the 3φ fault current by up to ~1.5×. The relationship depends entirely on Z0. When a question gives a grounding configuration, use it — do not default to "L-G is smaller."