Lesson 10 of 10

Symmetrical Components & Fault Calculation

A three-phase fault is easy to calculate: the network stays balanced, so one per-phase circuit is enough. But most real faults involve only one or two phases, and then the network is no longer balanced. In 1918 Charles Fortescue showed how to turn one unbalanced problem into three balanced ones. That method — symmetrical components — is still how every short-circuit study and every Fault Analysis Lab result on this platform is calculated.

The idea: three balanced sets

Any set of three unbalanced phasors can be written as the sum of three balanced sets, each easy to analyse on its own:

Positive sequence (1)

Three equal phasors 120° apart in the normal rotation a–b–c. A healthy, balanced system contains only this component.

Negative sequence (2)

Three equal phasors 120° apart in the reverse rotation a–c–b. It appears whenever the system is unbalanced and heats rotating machines.

Zero sequence (0)

Three equal phasors exactly in phase. It can only flow if there is a return path through earth or a neutral conductor.

Using the operator a = 1∠120° (so a² = 1∠240°), the phase voltages are rebuilt from the sequence components, and the components are extracted from the phase voltages:

The same relations hold for currents.

Sequence networks

The trick works because, in a network that is itself symmetrical, each sequence current only produces voltage drops of the same sequence. Each component therefore sees its own impedance network:

Positive-sequence Z₁

The normal per-phase impedances used in power flow — with generators represented by their subtransient reactance X″d and internal voltage.

Negative-sequence Z₂

For lines, cables and transformers equal to Z₁. For rotating machines somewhat different, but of the same order as X″d. There are no sources in this network.

Zero-sequence Z₀

Depends strongly on earthing and transformer connections. A delta winding blocks zero-sequence current from passing through; an unearthed neutral blocks it entirely; a neutral earthed through impedance Zₙ adds 3Zₙ to the path.

Connecting the networks for each fault type

Each fault type imposes conditions at the fault point that tell you how to connect the three networks. With V_f the pre-fault voltage and Z_f the fault impedance:

Three-phase (3φ)

Only the positive-sequence network is involved: I_f = V_f / Z₁.

Single line-to-ground (SLG)

All three networks in series: I₁ = I₂ = I₀ = V_f / (Z₁ + Z₂ + Z₀ + 3Z_f), and the fault current is I_a = 3I₀.

Line-to-line (LL)

Positive and negative networks in parallel, no zero sequence: I₁ = −I₂ = V_f / (Z₁ + Z₂), giving a phase current of magnitude √3·|I₁|.

Double line-to-ground (LLG)

The positive network in series with the negative and zero networks in parallel: I₁ = V_f / (Z₁ + Z₂ ∥ (Z₀ + 3Z_f)).

A worked example

Take V_f = 1.0 p.u. at a bus close to a solidly earthed transformer, with Z₁ = Z₂ = j0.20 p.u. and Z₀ = j0.10 p.u., and a bolted fault (Z_f = 0):

Here the earth fault is larger than the three-phase fault — the case flagged in the faults lesson. Earth the same transformer through a resistor or reactor so that Z₀ rises to j0.60 p.u., and the earth-fault current falls to 3 / 1.0 = 3.0 p.u. This is exactly why network operators choose an earthing method: it sets the size of the most common fault.

How it is done in practice

Real short-circuit studies follow a standard so that results are comparable. In Europe that is IEC 60909, which replaces detailed pre-fault load flow with an equivalent voltage source cU_n/√3 at the fault location — the voltage factor c (typically 1.1 for maximum currents in high-voltage networks) covers the spread of real operating voltages — and then uses exactly the sequence networks described here to find the initial symmetrical short-circuit current, the peak current and the breaking current.