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General rules of electrical installation design
Connection to the MV utility distribution network
Connection to the LV utility distribution network
MV and LV architecture selection guide for buildings
LV Distribution
Protection against electric shocks and electrical fires
Sizing and protection of conductors
LV switchgear: functions and selection
Overvoltage protection
Energy Efficiency in electrical distribution
Power Factor Correction
Power harmonics management
Characteristics of particular sources and loads
PhotoVoltaic (PV) installation
Residential premises and other special locations
ElectroMagnetic Compatibility (EMC)


In a 3-phase installation Isc at any point is given by:

Isc = \frac{U_{20} }{\sqrt 3 Z_T} where

U20 = phase-to-phase voltage of the open circuited secondary windings of the power supply transformer(s).
ZT = total impedance per phase of the installation upstream of the fault location (in Ω)

Method of calculating ZT

Each component of an installation (MV network, transformer, cable, busbar, and so on...) is characterized by its impedance Z, comprising an element of resistance (R) and an inductive reactance (X). It may be noted that capacitive reactances are not important in short-circuit current calculations.

The parameters R, X and Z are expressed in ohms, and are related by the sides of a right angled triangle, as shown in the impedance diagram of Figure G35.

Fig. G35Impedance diagram

The method consists in dividing the network into convenient sections, and to calculate the R and X values for each.

Where sections are connected in series in the network, all the resistive elements in the section are added arithmetically; likewise for the reactances, to give RT and XT.

The impedance (ZT) for the combined sections concerned is then calculated from Z_T= \sqrt {R_T\ ^2 + X_T\ ^2}

Any two sections of the network which are connected in parallel, can, if predominantly both resistive (or both inductive) be combined to give a single equivalent resistance (or reactance) as follows:

Let R1 and R2 be the two resistances connected in parallel, then the equivalent resistance R3 will be given by:

R_3 =\frac {R_1 \times R_2}{R_1 + R_2} or for reactances X_3 = \frac{X_1 \times X_2}{X_1 + X_2}

It should be noted that the calculation of X3 concerns only separated circuit without mutual inductance. If the circuits in parallel are close togother the value of X3 will be notably higher.

Determination of the impedance of each component

Network upstream of the MV/LV transformer

(see Fig. G36)

The 3-phase short-circuit fault level PSC, in kA or in MVA[1] is given by the power supply authority concerned, from which an equivalent impedance can be deduced.

Psc U20 (V) Ra (mΩ) Xa (mΩ)
250 MVA 420 0.07 0.7
500 MVA 420 0.035 0.351

Fig. G36The impedance of the MV network referred to the LV side of the MV/LV transformer

A formula which makes this deduction and at the same time converts the impedance to an equivalent value at LV is given, as follows:

Za = \frac{U_{20}\ ^2}{Psc}


Za = impedance of the MV voltage network, expressed in milli-ohms
U20 = phase-to-phase no-load LV voltage, expressed in volts
Psc = MV 3-phase short-circuit fault level, expressed in kVA

The upstream (MV) resistance Ra is generally found to be negligible compared with the corresponding Xa, the latter then being taken as the ohmic value for Za. If more accurate calculations are necessary, Xa may be taken to be equal to 0.995 Za and Ra equal to 0.1 Xa.

Figure G36 gives values for Ra and Xa corresponding to the most common MV[2] short-circuit levels in utility power-supply networks, namely, 250 MVA and 500 MVA.


(see Fig. G37)

The impedance Ztr of a transformer, viewed from the LV terminals, is given by the formula:

Ztr = \frac {U_{20}\, ^2}{Sn} \times \frac{Usc}{100} in milli-ohms


U20 = open-circuit secondary phase-to-phase voltage expressed in volts
Sn = rating of the transformer (in kVA)
Usc = the short-circuit impedance voltage of the transformer expressed in %

The transformer windings resistance Rtr can be derived from the total load-losses as follows:

Pcu=3In^2  so that Rtr=\frac{Pcu \times 10^3}{3In^2} in milli-ohms


Pcu = total load-losses in watts
In = nominal full-load current in amps
Rtr = resistance of one phase of the transformer in milli-ohms (the LV and corresponding MV winding for one LV phase are included in this resistance value).

Xtr = \sqrt {Ztr^2 - Rtr^2}

Note: for an approximate calculation, in the absence of more precise information on transformer characteristics, Cenelec 50480 suggests to use the following guidelines:

  • if U20 is not known, it may be assumed to be 1.05 Un
  • in the absence of more precise information, the following values may be used: Rtr = 0.31 Ztr and Xtr = 0.95 Ztr

Example: for a transformer of 630kVA with Usc=4% / Un = 400V, approximate calculation gives:

  • U20 = 400 x 1.05 = 420V
  • Ztr = 4202 / 630 x 4% = 11 mΩ
  • Rtr = 0.31 x Ztr = 3.5 mΩ and Xtr = 0.95 x Ztr = 10.6 mΩ
Rated Power kVA) Oil-immersed Cast-resin
Usc (%) Rtr (mΩ) Xtr (mΩ) Ztr (mΩ) Usc (%) Rtr (mΩ) Xtr (mΩ) Ztr (mΩ)
100 4 37.9 59.5 70.6 6 37.0 99.1 105.8
160 4 16.2 41.0 44.1 6 18.6 63.5 66.2
200 4 11.9 33.2 35.3 6 14.1 51.0 52.9
250 4 9.2 26.7 28.2 6 10.7 41.0 42.3
315 4 6.2 21.5 22.4 6 8.0 32.6 33.6
400 4 5.1 16.9 17.6 6 6.1 25.8 26.5
500 4 3.8 13.6 14.1 6 4.6 20.7 21.2
630 4 2.9 10.8 11.2 6 3.5 16.4 16.8
800 6 2.9 12.9 13.2 6 2.6 13.0 13.2
1,000 6 2.3 10.3 10.6 6 1.9 10.4 10.6
1,250 6 1.8 8.3 8.5 6 1.5 8.3 8.5
1,600 6 1.4 6.5 6.6 6 1.1 6.5 6.6
2,000 6 1.1 5.2 5.3 6 0.9 5.2 5.3

Fig. G37Resistance, reactance and impedance values for typical distribution 400 V transformers (no-load voltage = 420 V) with MV windings ≤ 20 kV


The resistance of busbars is generally negligible, so that the impedance is practically all reactive, and amounts to approximately 0.15 mΩ/metre[3] length for LV busbars (doubling the spacing between the bars increases the reactance by about 10% only).

In practice, it's almost never possible to estimate the busbar length concerned by a short-circuit downstream a switchboard.

Circuit conductors

The resistance of a conductor is given by the formula:

Rc=\rho \frac{L}{S}


ρ = the resistivity of the conductor material at the normal operating temperature

ρ has to be considered:

  • at cold state (20°C) to determine maximum short-circuit current,
  • at steady state (normal operating temperature) to determine minimum short-circuit current.

L = length of the conductor in m
S = c.s.a. of conductor in mm2

20 °C PR/XLPE 90 °C PVC 70 °C
Copper 18.51 23.69 22.21
Alu 29.41 37.65 35.29

Fig. G38Values of ρ as a function of the temperature, cable insulation and cable core material, according to IEC60909-0 and Cenelec TR 50480 (in mΩ mm2/m).

Cable reactance values can be obtained from the manufacturers. For c.s.a. of less than 50 mm2 reactance may be ignored. In the absence of other information, a value of 0.08 mΩ/metre may be used (for 50 Hz systems) or 0.096 mΩ/metre (for 60 Hz systems). For busways (busbar trunking systems) and similar pre-wired ducting systems, the manufacturer should be consulted.


At the instant of short-circuit, a running motor will act (for a brief period) as a generator, and feed current into the fault.

In general, this fault-current contribution may be ignored. However, if the total power of motors running simultaneously is higher than 25% of the total power of transformers, the influence of motors must be taken into account. Their total contribution can be estimated from the formula:

Iscm = 3.5 In from each motor i.e. 3.5m In for m similar motors operating concurrently.

The motors concerned will be the 3-phase motors only; single-phase-motor contribution being insignificant.

Fault-arc resistance

Short-circuit faults generally form an arc which has the properties of a resistance. The resistance is not stable and its average value is low, but at low voltage this resistance is sufficient to reduce the fault-current to some extent. Experience has shown that a reduction of the order of 20% may be expected. This phenomenon will effectively ease the current-breaking duty of a CB, but affords no relief for its fault-current making duty.

Recapitulation table

(see Fig. G39)

Parts of power-supply system R (mΩ) X (mΩ)
DB422317.svg Supply network
Figure G34
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\frac {Ra}{Xa} = 0.1 Xa = 0.995 Za

\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Za = \frac{ U_{20}\, ^2}{Psc}

Figure G35
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Rtr = \frac {Pcu \times 10^3 }{3In\, ^2} where

 In = \frac {Sn \times 10^3}{U_{20} \sqrt 3}
Rtr is often negligible compared to Xtr for transformers > 100 kVA

\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Xtr = \sqrt {Ztr^2 - Rtr^2} with
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Ztr = \frac {U_{20}\, ^2}{Sn}\times \frac{Usc}{100}
Circuit-breaker Not considered in practice
Busbars Negligible for S > 200 mm2, below use the formula:
 \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}R = \rho \frac{L}{S} [a]
Xb = 0.15 mΩ/m
Circuit conductors[b] \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}R = \rho\frac{L}{S} [a] Cables: Xc = 0.08 mΩ/m
Motors See Motors (often negligible at LV)
Three-phase maximum
circuit current in kA
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Isc= \frac{U_{20} }{\sqrt 3 \sqrt{ R_T\, ^2 + X_T}\, ^2}

[a] ρ = resistivity at 20°C
[b] If there are several conductors in parallel per phase, then divide the resistance of one conductor by the number of conductors. The reactance remains practically unchanged.
U20: Phase-to-phase no-load secondary voltage of MV/LV transformer (in volts).
Psc: 3-phase short-circuit power at MV terminals of the MV/LV transformers (in kVA).
Pcu: 3-phase total losses of the MV/LV transformer (in watts).
Sn: Rating of the MV/LV transformer (in kVA).
Usc: Short-circuit impedance voltage of the MV/LV transfomer (in %).

RT : Total resistance. XT: Total reactance

Fig. G39Recapitulation table of impedances for different parts of a power-supply system

Example of short-circuit calculations

(see Fig. G40)

LV installation R (mΩ) X (mΩ) RT (mΩ) XT (mΩ) \definecolor{bgblue}{RGB}{65,193,232}\pagecolor{bgblue}Isc= \frac{420}{\sqrt 3 \sqrt {R_T\, ^2 + X_T\, ^2} }
DB422318.svg MV network

Psc = 500 MVA

0.035 0.351
Transformer 20 kV / 420 V
Pn = 1000 kVA
Usc = 5%
Pcu = 13.3 x 103 watts
2.35 8.5
Single-core cables
5 m copper
4 x 240 mm2/phase
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Rc= \frac {18.51}{4}\times \frac {5}{240} = 0.10 Xc = 0.08 x 5 = 0.40 2.48 9.25 Isc1 = 25 kA
Main circuit-breaker Not considered in practice
Busbars 10 m Not considered in practice
Three-core cable
100 m
95 mm2 copper
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Rc= {18.51}\times \frac {100}{95} = 19.5 Xc = 100 x 0.08 = 8 22 17.3 Isc3 = 8.7 kA
Three-core cable
20 m
10 mm2 copper
final circuits
\definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}Rc= {18.51}\times \frac {20}{10} = 37 Xc = 20 x 0.08 = 1.6 59 18.9 Isc4 = 3.9 kA

RT : Total resistance. XT: Total reactance. Isc : 3-phase maximum short-circuit current
Calculations made as described in figure G36

Fig. G40Example of maximum short-circuit current calculations for a LV installation supplied at 400 V (nominal) from a 1000 kVA MV/LV transformer


  1. ^ Short-circuit MVA: \definecolor{bggrey}{RGB}{234,234,234}\pagecolor{bggrey}\sqrt 3 EL Isc where:
    • EL = phase-to-phase nominal system voltage expressed in kV (r.m.s.)
    • Isc = 3-phase short-circuit current expressed in kA (r.m.s.)
  2. ^ up to 36 kV
  3. ^ For 50 Hz systems, but 0.18 mΩ/m length at 60 Hz