IEC & BS 7671 calculators

Prospective fault current calculator

Turn loop impedance readings into prospective short-circuit and earth fault currents, estimate three-phase levels, and check the breaking capacity of the protective devices.

Calculated with electrician.wiki · https://electrician.wiki/calculators/prospective-fault-current/

Result · BS 7671

PFC 958 A ≤ 6 kA · OK

With Z(L–N) = 0.24 Ω and an earth loop of 0.35 Ω, the prospective short-circuit current is 958 A and the earth fault current 657 A. The 958 A maximum is within the 6 kA breaking capacity.

Prospective fault current (the larger): 958 A
PSCC, line–neutral: 958 A
PEFC, line–earth: 657 A
Minimum earth fault current (cmin 0.95): 624 A
Breaking capacity: 6 kA
Device rating: Enough

Show the work

1. Prospective short-circuit current (line–neutral)
$I_{psc} = \frac{U_{0}}{Z_{L–N}} = \frac{230 V}{0.24 Ω} = 958.3 A$
2. Prospective earth fault current (line–earth)
$I_{pef} = \frac{U_{0}}{Z_{e}} = \frac{230 V}{0.35 Ω} = 657.1 A$
3. Highest fault current the devices must break
$I_{pf} = 958.3 A$
BS 7671 practice compares the value at nominal voltage, as loop testers display it.
4. Breaking capacity is enough
$958.3 A \leq 6,000 A$
6 kA: MCB or RCBO marked 6000 (Icn)
5. Minimum earth fault current (for disconnection checks)
$I_{ef,min} = \frac{c_{\min} \cdot U_{0}}{Z_{e}} = \frac{0.95 \cdot 230 V}{0.35 Ω} = 624.3 A$
cmin = 0.95 (IEC 60909-0; BS 7671 Cmin). Use the earth fault loop impedance calculator to compare it with the device’s Ia.

References: BS 7671 Regulation 434.5.1 (2018): breaking capacity not less than the prospective fault current · ENA ER P23 (via the IET): typical maximum Ze; 16 kA design value · IEC 60909-0 (2016): voltage factors cmax and cmin

U0 = 230 V nominal; the fault itself has negligible impedance (a bolted fault).
BS 7671 practice compares the value at nominal voltage. IEC 60909 would add a 1.10 voltage factor for a supply allowed to run 10 % high: 1.05 kA.
Loop impedances are taken as measured, with no allowance for conductor temperature: warmer conductors would lower the fault current slightly.
Motor contribution and the asymmetric (peak) current are ignored; for those, and for large installations, use a full IEC 60909 study.

Report a problem with this result

How to use the prospective fault current calculator

Enter the two loop impedances a loop tester measures at the point you care about, usually the origin: Z(L–N), between line and neutral, and the earth fault loop impedance, which at the origin is Ze. The calculator turns each into a current at the nominal voltage, takes the larger as the prospective fault current (PFC), and checks it against the breaking capacity of the device you choose. For a three-phase supply it estimates the three-phase fault from the single-phase reading. In BS 7671 mode you can skip the readings and use the UK design values instead.

PSCC, PEFC and PFC

A bolted fault between line and neutral drives the prospective short-circuit current (PSCC); a fault from line to earth drives the prospective earth fault current (PEFC). Each is the voltage divided by the impedance of its loop:

I_{psc} = \frac{U_{0}}{Z_{L–N}}

I_{pef} = \frac{U_{0}}{Z_{e}}

The prospective fault current you record and design for is the larger of the two. On a TN-C-S supply they are often close; on a TN-S supply the earth path through the cable sheath usually makes the PEFC smaller; on a TT supply the PEFC is tiny. With the default readings, 0.24 Ω and 0.35 Ω, the PSCC is 958 A and the PEFC 657 A. Most loop testers display exactly this: the nominal voltage over the measured impedance.

Why it matters: breaking capacity

Every device that may have to interrupt a fault must be able to break the largest current that can flow where it is installed (BS 7671 Regulation 434.5.1). Beyond that rating, a breaker’s contacts can weld or the case can rupture. Circuit-breakers to BS EN 60898 show their rated short-circuit capacity Icn in a box, in amperes: 6000 for 6 kA, 10000 for 10 kA. Some also quote a service capacity Ics, the level after which the breaker is still fit for use, which can be lower.

UK consumer units add a twist. A consumer unit to BS EN 61439-3 is designed as an assembly, with its devices and outer enclosure, to withstand a prospective fault current of up to 16 kA in domestic and similar premises (NICEIC). That conditional rating is why 6 kA breakers are normal in a house whose supply could deliver more than 6 kA. With the UK design value of 16 kA, a 6 kA breaker on its own fails the check, while the same breaker in a 16 kA consumer unit passes.

Typical UK supplies

UK distributors declare the most Ze their supplies will present, and the IET recommends designing with those figures from the Energy Networks Association’s Engineering Recommendation P23. For the prospective short-circuit current the IET recommends assuming 16 kA at the origin of a single-phase supply up to 100 A where the service cable is longer than 2 m, unless the distributor gives a different figure. Measure on site whenever you can: real values are usually lower, but an installation near a substation can be higher.

Typical maximum Ze declared by UK distributors for single-phase supplies up to 100 A, and the earth fault current each implies at 230 V.
Earthing	Typical max Ze	PEFC at 230 V	Note
TN-C-S (PME)	0.35 Ω	657 A	combined neutral and earth in the supply cable
TN-S	0.8 Ω	288 A	separate earth, usually the cable sheath
TT	21 Ω	11 A	distributor’s part only; your earth electrode adds to it

Three-phase supplies: × 2 or × √3?

A single-phase loop tester measures one line and the neutral. Splitting that loop into the line path ZL (transformer winding plus line conductor) and the neutral path ZN, the reading and the three kinds of fault are:

I_{L–N} = \frac{U_{0}}{Z_{L} + Z_{N}}

I_{3φ} = \frac{U_{0}}{Z_{L}}

I_{L–L} = \frac{\sqrt{3} \cdot U_{0}}{2 \cdot Z_{L}}

If the neutral path has the same impedance as the line path, the symmetrical three-phase fault is exactly twice the line–neutral reading, and a line-to-line fault is √3 times it, which is (√3/2) ≈ 0.87 of the three-phase value, the relation IEC 60909-0 gives. The IET On-Site Guide takes the first, simpler route: double the single-phase reading for a three-phase supply. It usually errs high, because the neutral path doesn’t include the transformer winding, so ZN is smaller than ZL. Multiplying by √3 instead answers a different question, the line-to-line fault, and understates the worst case by about 13 %. A reduced or very long neutral reverses the margin, so measure between lines where you can.

One line–neutral loop reading of 0.24 Ω, three answers. The On-Site Guide’s doubling gives the three-phase fault, the largest; √3 gives a line-to-line fault.

Voltage factors: cmax and cmin

The supply isn’t always at exactly 230 V. IEC 60909-0, the standard for short-circuit calculations, scales the voltage with a factor c: for low-voltage networks, cmax = 1.10 where the voltage may run 10 % high (1.05 for a 6 % tolerance) when you want the largest current, and cmin = 0.95 when you want the smallest. In IEC mode the calculator applies cmax to the breaking-capacity check; our data has those cmax figures from a single source, so the result says so. BS 7671 practice compares the value at nominal voltage, and the UK design value is already a maximum. The minimum earth fault current, cmin × U0 / Ze, is the figure that matters for disconnection: 624 A for the default readings, which you can carry into the earth fault loop impedance calculator.

Worked examples

Readings at a consumer unit

Z(L–N) = 0.24 Ω and Ze = 0.35 Ω at the origin of a TN-C-S supply, with 6 kA breakers.

1. Prospective short-circuit current (line–neutral)
$I_{psc} = \frac{U_{0}}{Z_{L–N}} = \frac{230 V}{0.24 Ω} = 958.3 A$
2. Prospective earth fault current (line–earth)
$I_{pef} = \frac{U_{0}}{Z_{e}} = \frac{230 V}{0.35 Ω} = 657.1 A$
3. Highest fault current the devices must break
$I_{pf} = 958.3 A$
4. Breaking capacity is enough
$958.3 A \leq 6,000 A$
5. Minimum earth fault current (for disconnection checks)
$I_{ef,min} = \frac{c_{\min} \cdot U_{0}}{Z_{e}} = \frac{0.95 \cdot 230 V}{0.35 Ω} = 624.3 A$

Designing without readings

A new single-phase supply with no measurements yet: the UK design value against a 16 kA consumer unit.

1. Short-circuit current (UK design value)
PSCC = 16 kA
2. Earth fault current with the typical TN-C-S Ze
$I_{pef} = \frac{U_{0}}{Z_{e}} = \frac{230 V}{0.35 Ω} = 657.1 A$
3. Highest fault current the devices must break
$I_{pf} = 16,000 A$
4. Breaking capacity is enough
$16,000 A \leq 16,000 A$
5. Minimum earth fault current (for disconnection checks)
$I_{ef,min} = \frac{c_{\min} \cdot U_{0}}{Z_{e}} = \frac{0.95 \cdot 230 V}{0.35 Ω} = 624.3 A$

For design, a UK single-phase supply up to 100 A is taken as 16 kA, so devices need a 16 kA rating or more: this one qualifies. The earth fault current with a typical TN-C-S Ze of 0.35 Ω is 657 A.

A three-phase board near a substation

A 0.05 Ω line–neutral loop at a three-phase distribution board feeding 10 kA breakers.

1. Prospective short-circuit current (line–neutral)
$I_{psc} = \frac{U_{0}}{Z_{L–N}} = \frac{230 V}{0.05 Ω} = 4,600 A$
2. Prospective earth fault current (line–earth)
$I_{pef} = \frac{U_{0}}{Z_{e}} = \frac{230 V}{0.12 Ω} = 1,917 A$
3. Three-phase fault, estimated from the single-phase reading
$I_{3φ} = 2 \cdot I_{psc} = 2 \cdot 4,600 A = 9,200 A$
4. Highest fault current the devices must break
$I_{pf} = 9,200 A$
5. Breaking capacity is enough
$9,200 A \leq 10,000 A$
6. Minimum earth fault current (for disconnection checks)
$I_{ef,min} = \frac{c_{\min} \cdot U_{0}}{Z_{e}} = \frac{0.95 \cdot 230 V}{0.12 Ω} = 1,821 A$

With Z(L–N) = 0.05 Ω and an earth loop of 0.12 Ω, the prospective short-circuit current is 4.6 kA and the earth fault current 1.92 kA; a three-phase fault could reach about 9.2 kA. The 9.2 kA maximum is within the 10 kA breaking capacity.

The IEC 60909 maximum

The same consumer unit readings, with cmax = 1.10 for a supply allowed 10 % above nominal: 1.05 kA.

1. Prospective short-circuit current (line–neutral)
$I_{psc} = \frac{U_{0}}{Z_{L–N}} = \frac{230 V}{0.24 Ω} = 958.3 A$
2. Prospective earth fault current (line–earth)
$I_{pef} = \frac{U_{0}}{Z_{e}} = \frac{230 V}{0.35 Ω} = 657.1 A$
3. Maximum fault current with cmax = 1.10 (IEC 60909-0)
$I_{k,max} = 1.1 \cdot 958.3 A = 1,054 A$
4. Breaking capacity is enough
$1,054 A \leq 6,000 A$
5. Minimum earth fault current (for disconnection checks)
$I_{ef,min} = \frac{c_{\min} \cdot U_{0}}{Z_{e}} = \frac{0.95 \cdot 230 V}{0.35 Ω} = 624.3 A$

With Z(L–N) = 0.24 Ω and an earth loop of 0.35 Ω, the prospective short-circuit current is 958 A and the earth fault current 657 A. The 1.05 kA maximum is within the 6 kA breaking capacity.

Common mistakes

Recording only the earth fault current. The PFC is the larger of PSCC and PEFC; on many supplies that is the short-circuit value.
Using √3 for a three-phase supply. That gives the line-to-line fault; the three-phase fault is larger. At the default reading the difference is 257 A.
Checking 6 kA breakers against 16 kA without the consumer unit’s rating. The conditional rating belongs to the tested assembly, not to loose devices in a different enclosure.
Measuring away from the device. PFC falls along a circuit; a distribution board 30 m from the origin sees less than the origin.
Adding a voltage factor to a design value. The distributor’s or the IET’s figure is already a maximum.

The three-phase example above, at 9.2 kA, would exceed 6 kA devices but suits 10 kA ones. Size the cables with the cable size calculator and check that their protective conductors survive the fault with the cpc size calculator. The US equivalent of this calculation is the available fault current; for a supply transformer’s own contribution, see the transformer calculator.

Method and verification

Currents are U0 over the loop impedance, with the IEC 60909 factors applied where the mode calls for them. The UK typical Ze values and the 16 kA design figure come from the IET’s summary of ENA ER P23, confirmed by NICEIC and ELECSA; the 16 kA consumer unit rating from NICEIC. Tests reproduce the typical values (657 A for 0.35 Ω, 288 A for 0.8 Ω) and check that the three-phase estimate and the line-to-line value keep their √3/2 ratio.

A first estimate for single installations; motor contribution, peak current and the X/R ratio need a full IEC 60909 study. Updated 2026-10-14. Corrections log.

Questions people ask

What is prospective fault current?

The current that would flow in a fault of negligible impedance at a given point: the larger of the prospective short-circuit current (line to neutral) and the prospective earth fault current (line to earth). Protective devices there must be able to break it.

How do you calculate PFC from Ze?

Divide the voltage to earth by Ze: 230 V / 0.35 Ω = 657 A. That is the earth fault current; do the same with the line–neutral loop impedance for the short-circuit current, and record the larger.

How do I find the three-phase prospective fault current?

From a single-phase reading, the IET On-Site Guide doubles the line–neutral value. Multiplying by √3 gives the line-to-line fault instead, about 13 % lower than the three-phase fault. Measuring between lines is better still.

Can I fit 6 kA MCBs if the PFC is above 6 kA?

In a UK consumer unit to BS EN 61439-3, yes, up to its 16 kA conditional rating, because the assembly was tested as a whole. Outside such an assembly, the device’s own breaking capacity must cover the PFC.

What PFC should I assume for a UK house?

16 kA for a single-phase supply up to 100 A with a service cable over 2 m, unless the distributor states otherwise (IET). Measured values are usually lower; record what you measure.

What is the difference between PSCC and PEFC?

PSCC is the line-to-neutral fault current, set by Z(L–N); PEFC is the line-to-earth fault current, set by Ze or Zs. The PFC is the larger. The smaller earth fault value also decides how fast devices disconnect.