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Symmetrical components

A three-phase fault loads the three phases equally, and one impedance per phase describes the network. An unbalanced fault, between two phases or between a phase and earth, does not. The method of symmetrical components replaces the unbalanced system by three balanced ones, each with its own impedance of the network. This topic explains the three sequence networks as the design's short-circuit study builds them, and how the fault equations of IEC 60909-0:2016 combine them.

Read first: Per-unit quantities

1Three balanced systems

The currents of the three phases at a fault can be written as the sum of three balanced sets. The positive-sequence set has the phase order of the network. The negative-sequence set has the reverse order. The zero-sequence set is three equal currents in phase, which can only flow where a path to earth or to a neutral exists. Their sum to earth is three times the zero-sequence current, $3 I_{0}$ , the earth-fault current from which the earth mat's grid current follows (Grid current).

Each set sees its own impedance of the network at the fault location: the positive-sequence impedance $Z_{1}$ , the negative-sequence impedance $Z_{2}$ and the zero-sequence impedance $Z_{0}$ . The rotation operator $a = e^{j 2 π /3}$ turns a phasor by a third of a period, and the fault equations use it to recover the phase currents from the sets (Short-circuit currents by IEC 60909-0, Eq. (8)).

2The sequence networks of the design

The design's study builds $Z_{1}$ and $Z_{0}$ at the high-voltage (HV) bus and at a medium-voltage (MV) section from the elements below. It takes $Z_{2} = Z_{1}$ at both buses, as for a network far from a generator (Short-circuit currents by IEC 60909-0, The equivalent voltage source).

Element	Positive sequence $Z_{1}$	Zero sequence $Z_{0}$
Grid feeder	from the three-phase fault level and the grid's X/R ratio	from the single-phase fault level, at the same X/R ratio
Overhead line	$R_{1} + j X_{1}$ per km, times the length over the parallel circuits	$R_{0} + j X_{0}$ per km, the same way
ICT, earthed star with a delta tertiary	$K_{T} Z_{T}$	$k_{0} K_{T} Z_{T}$ , an earth path at the HV bus
ICT with a delta HV winding (Dyn11)	$K_{T} Z_{T}$	no earth path at the HV bus
MV neutral earthing resistor	none	$3 R_{NGR}$ in series
Converter plant	a current source $k I_{r}$ added to the three-phase current	none
Generator unit	$Z_{S}$ in parallel with the grid	through the generator transformer's earthed star

The grid's zero-sequence impedance follows from the single-phase fault level $I_{k 1 Q}^{''}$ by Eq. 52 (Short-circuit currents by IEC 60909-0, Eq. (9)). With $Z_{2} = Z_{1}$ at the remote bus, Eq. 52 gives $∣ Z_{Q 0} ∣ = 3 c U_{n Q} / I_{k 1 Q}^{''} - 2∣ Z_{Q 1} ∣$ . When the two fault levels are equal, as on Auto, $Z_{Q 0}$ equals $Z_{Q 1}$ .

The zero-sequence impedance of an interconnecting transformer (ICT) is the ratio $k_{0}$ times its positive-sequence impedance, 0.9 by default. A closed delta tertiary carries the zero-sequence current that the earthed HV star draws, so the transformer is an earth path at the HV bus. A delta HV winding carries none to the HV bus (Earthing of the neutral, The design's transformer neutrals).

where
$Z_{1}$ , $Z_{2}$ , $Z_{0}$	positive-, negative- and zero-sequence impedance at the fault location	Ω
$I_{0}$	zero-sequence current	A
$a$	rotation operator, $e^{j 2 π /3}$
$I_{k 1 Q}^{''}$	single-phase fault level of the grid at the remote bus	kA
$Z_{Q 1}$ , $Z_{Q 0}$	positive- and zero-sequence impedance of the grid feeder	Ω
$k_{0}$	ratio of the transformer's zero-sequence to positive-sequence impedance
$R_{NGR}$	resistance of the neutral earthing resistor	Ω

3How the fault types use the sequences

The fault equations of IEC 60909-0:2016 connect the sequence networks at the fault (Short-circuit currents by IEC 60909-0, The initial symmetrical currents). A three-phase fault involves the positive sequence only. A line-to-line fault connects the positive and negative networks in series, with no zero sequence. A line-to-earth fault connects the three networks in series:

I_{k 1}^{''} = \frac{3 c U _{n}}{∣ Z _{1} + Z _{2} + Z _{0} ∣}

(1)

IEC 60909-0:2016 §7.5, Eq. 52

With $Z_{2} = Z_{1}$ , the denominator is $∣2 Z_{1} + Z_{0} ∣$ . The ratio of $Z_{0}$ to $Z_{1}$ then decides how the line-to-earth current compares with the three-phase current. Where $Z_{0}$ is smaller than $Z_{1}$ at a similar angle, the line-to-earth current is the larger. Where $Z_{0}$ is larger, as on a MV section earthed through a neutral earthing resistor (NGR), it is the smaller.

A line-to-line fault with $Z_{2} = Z_{1}$ draws $3 /2$ of the three-phase current, 86.6 %, for any network. At the default project's HV bus the line-to-line current is 18.32 kA, 86.6 % of the network's three-phase current of 21.15 kA.

Example 1Sequence impedances of the default 220 kV project

Given the default 220 kV / 33 kV project: the grid at 40.0 kA three-phase and 40.0 kA single-phase, X/R 15, one line, 2 ICTs with $k_{0}$ = 0.9.

Grid by Eq. 4 and Eq. 52: $Z_{Q 1}$ = 0.232 Ω + j3.485 Ω, and $Z_{Q 0}$ = 0.232 Ω + j3.485 Ω, equal to $Z_{Q 1}$ since the two levels are equal.

Line: $Z_{L 1}$ = 0.28 Ω + j3.10 Ω, and $Z_{L 0}$ = 2.00 Ω + j10.00 Ω.

ICTs: $Z_{T 1}$ = 0.92 Ω + j36.75 Ω and $Z_{T 0}$ = 0.83 Ω + j33.07 Ω for one ICT, and $Z_{T 0} / n$ = 0.41 Ω + j16.54 Ω for 2 in parallel.

HV bus, the grid path in series with the line, in parallel with the ICTs' earth paths for the zero sequence:

Z_{1} Z_{0} = Z_{Q 1} + Z_{L 1} = 0.512 + j 6.585 Ω = (Z_{Q 0} + Z_{L 0}) ∥ \frac{Z _{T 0}}{n} = (2.232 + j 13.485) ∥ (0.41 + j 16.54) = 0.758 + j 7.464 Ω

Line-to-earth current by Eq. (1):

∣2 Z_{1} + Z_{0} ∣ = 20.711 Ω, I_{k 1}^{''} = \frac{3 \times 1.10 \times 220}{20.711} = 20.24 kA

Comparison at the HV bus, without the plant's infeed of 0.83 kA:

Fault	Sequence networks	Current
Three-phase	positive	21.15 kA
Line-to-line	positive and negative	18.32 kA
Line-to-line-to-earth, to earth	positive, negative and zero	19.40 kA
Line-to-earth	positive, negative and zero	20.24 kA

Result $∣ Z_{0} ∣$ is 1.136 times $∣ Z_{1} ∣$ , so $Z_{0}$ is the larger, and the line-to-earth current of 20.24 kA is below the three-phase current of 21.15 kA. On a MV section, $Z_{0}$ = 57.18 Ω + j0.74 Ω against $Z_{1}$ = 0.032 Ω + j0.975 Ω, and the line-to-earth current is 1097 A.

4The X/R ratio of a line-to-ground fault

The ANSI / IEEE set calculates the breaker duty by IEEE C37.010-2016, on the same sequence networks with $Z_{2} = Z_{1}$ . Its X/R ratio comes from separate reductions of the reactances and the resistances. A line-to-ground fault combines them as the current does, $(2 X_{1} + X_{0}) / (2 R_{1} + R_{0})$ (Short-circuit duty by IEEE C37.010, Eq. (3)).

In the 230/34.5 kV example at 60 Hz, the HV bus has $R_{1}$ = 0.501 Ω, $X_{1}$ = 6.412 Ω, $R_{0}$ = 0.369 Ω and $X_{0}$ = 7.599 Ω. The three-phase X/R ratio is 12.8 and the line-to-ground ratio 14.9. On a MV section the NGR's resistance enters $R_{0}$ , and the line-to-ground ratio falls to 0.048.

Sources

[1]IEC 60909-0:2016, Short-circuit currents in three-phase AC systems, Part 0: Calculation of currents. §7.5, Eq. 52.

[2]IEEE C37.010-2016, IEEE Application Guide for AC High-Voltage Circuit Breakers.

CitationSubstationLayout Reference, Symmetrical components, revision 1, 8 Nov 2026, https://substation.solarlayout.app/library/reference/symmetrical-components

Symmetrical components

1Three balanced systemsCopy link to Three balanced systems

2The sequence networks of the designCopy link to The sequence networks of the design

3How the fault types use the sequencesCopy link to How the fault types use the sequences

4The X/R ratio of a line-to-ground faultCopy link to The X/R ratio of a line-to-ground fault

1Three balanced systems

2The sequence networks of the design

3How the fault types use the sequences

4The X/R ratio of a line-to-ground fault