Prospective Fault Current Calculator (PFC)

The highest current a fault could produce, from a loop impedance, a transformer nameplate or your meter readings — checked against the breaking capacity of the device that has to interrupt it.

Prospective fault current is the highest current that could flow if a fault occurred at a given point, and every protective device there has to be able to interrupt it. It is Ohm’s law on the fault loop: I = U ÷ Z. A 230 V supply with a loop impedance of 0.35 Ω gives 657 A. From a transformer nameplate instead, the fault at the terminals is the rated current divided by the per-unit impedance — a 500 kVA 400 V unit at 5% delivers 722 ÷ 0.05 = 14.4 kA. From meter readings, take the greater of the line-neutral and line-earth figures; on three-phase the symmetrical fault is exactly twice the highest line-neutral value, because a line-neutral fault sees two conductors and a three-phase fault sees one. UK distributors typically declare up to 16 kA at a domestic origin.

Formula
I = U ÷ Z
From a transformer
Isc = I_rated ÷ (Z% ÷ 100)
Three-phase from L-N
× 2, exactly
PFC
the greater of PSCC and PEFC
UK domestic declared max
16 kA
Typical consumer unit
6 kA
Regulation
BS 7671 434.1, 612.11

Inputs

Three routes to the same number. The transformer route gives the worst case at the supply; the impedance route gives it at any point you know Z for.

V

Line to earth for an earth fault, line to neutral for a short circuit.

Ω

Zs for an earth fault, or the line-neutral loop for a short circuit.

Icn on a BS EN 60898 breaker, Icu on a moulded-case device. 6 kA is the usual domestic consumer unit.

s

Only used to report the let-through energy for the adiabatic check.

Results

Prospective fault current657.1 A
Breaking capacity needed
1.5 kA
Margin
5.343 kA
Let-through energy (A²s)
4.318e+4

6.00 kA of breaking capacity covers a 657 A fault, with 5.34 kA to spare. The device can interrupt the worst case at this point.

657 A from 230 V across 0.35 Ω. This is Ohm's law on the fault loop — the same impedance the loop impedance calculator reports, read as a current instead of a resistance.

Smallest standard breaking capacity that covers it: 1.50 kA. Never round down — the rating is the current the device is tested to interrupt safely, not a target to design close to.

Let-through energy 4.32e+4 A²s at 0.1 s. That is the number the adiabatic check takes — it decides the minimum conductor that survives the fault, which is a separate requirement from the device being able to break it.

A prospective fault current is a worst case at one point. It falls as you move away from the source, because every conductor between adds impedance — so the figure at the origin is not the figure at the end of a final circuit, and using the origin figure everywhere oversizes but never undersizes.

Worked examples

A domestic origin, PME

TN-C-S at the declared 0.35 Ω. This is the earth fault current at the cutout of an ordinary house.

657 A — comfortably inside a 6 kA consumer unit.

A 500 kVA substation on the doorstep

The worst case at the transformer terminals, before any cable adds impedance. This is the figure that decides the main switchgear.

14.4 kA — twenty times rated current, and far past a 6 kA device.

Three-phase readings, and the doubling rule

A 3 kA line-to-neutral reading on a three-phase board. The symmetrical three-phase fault is twice it, and taking the reading at face value would understate the duty by half.

6 kA, not 3 kA. Exactly at the limit of a 6 kA device.

The 16 kA declared supply

What a distributor declares at a domestic origin, against the 6 kA consumer unit everyone fits. The apparent contradiction, and its resolution.

Exceeds the device — and is normally acceptable under 434.5.1. See the note.

Far from the source

The same house, but at the end of a final circuit where Zs is 1.05 Ω rather than 0.35 Ω. Distance is what makes fault current manageable.

219 A — a third of the figure at the origin.

Reviewed by Salamot Hok, electrician with 10+ years · Last reviewed 25 August 2026

Engineering guidance, not a code sign-off. Verify against the governing standard before relying on this result for safety-critical or code-compliance work.

What it is, and why the number has to be big enough to worry about

When to use this: selecting a consumer unit or a distribution board; justifying a 6 kA board on a supply declared at 16 kA; sizing switchgear from a transformer nameplate; recording a PFC reading on a certificate; or feeding the I²t figure into a conductor check.

Prospective fault current is the current that would flow if a fault of negligible impedance happened at a given point — a bolted fault, with nothing limiting it but the impedance of the supply and the conductors leading to it.

I = U ÷ Z

That is the whole calculation. Ohm’s law applied to the fault loop. The difficulty is never the arithmetic; it is knowing which impedance and which voltage belong to the fault you are actually worried about.

It matters because every protective device has to be able to interrupt it. A device asked to break more current than it is rated for does not politely fail to trip. It can fail destructively — welded contacts, an arc it cannot extinguish, a fault that stays live behind a device that has already given up.

Three names for two things

  • PSCC — prospective short-circuit current. Line to neutral, or line to line.
  • PEFC — prospective earth fault current. Line to earth, through the protective conductor.
  • PFC — whichever of those two is greater, because that is the one the device has to survive.

On a TN-C-S supply the two are usually close together. On TN-S the earth path is longer and the short-circuit figure normally wins. Measure both; record the higher.

From a transformer nameplate

The supply-side calculation, and the one that produces the alarming numbers.

Isc = Irated ÷ (Z% ÷ 100)

The impedance voltage on a transformer’s plate — typically 4% to 6% on a distribution unit — is defined as the percentage of rated voltage you need to apply to drive rated current through a short-circuited secondary. Invert it and you have, directly, how many times rated current a bolted fault at the terminals produces.

A 500 kVA 400 V three-phase transformer is rated 722 A. At 5% impedance a fault at its terminals draws 14.4 kA — twenty times rated current. At 4% it draws 18 kA. A stiffer supply means a bigger fault, which is worth sitting with for a moment: the better the supply, the harder the switchgear has to work.

The three-phase doubling rule — and why it is exact

Take the highest line-to-neutral reading and double it. This gets taught as a rule of thumb and it is not one; it falls out of counting conductors.

A line-to-neutral fault travels out along the line conductor and back along the neutral. It sees two conductors in series:

ILN = U0 ÷ (ZL + ZN) = U0 ÷ 2Z

A three-phase symmetrical fault has no return through the neutral at all — the three phase currents are 120° apart and sum to zero at the fault — so each phase sees only its own impedance:

I3ph = U0 ÷ Z

Divide one by the other and the factor is exactly 2, whenever the neutral has the same impedance as the line. Where the neutral is reduced — a half-size neutral on a sub-main — the line-neutral loop is 3Z rather than 2Z, the true factor is 3, and doubling understates the fault. Measure line-to-line directly if that applies.

The contradiction everyone runs into

A UK distributor declares up to 16 kA at the origin of a domestic supply. The consumer unit you fit is rated 6 kA. Every house in the country is wired this way, and on the face of it every one of them is non-compliant.

They are not, and the reason is Regulation 434.5.1: a device may have a breaking capacity below the prospective fault current where an upstream device provides backup protection.

The distributor’s cutout fuse — normally a 100 A BS 88-3 — is current-limiting. On a severe fault it clears within the first few milliseconds, long before the current reaches its prospective peak, so the energy that actually arrives at the breaker behind it is a fraction of what an unrestricted 16 kA fault would deliver. The pair together has a conditional short-circuit rating well above the breaker’s own rating.

Fault current falls with distance

Every conductor between the source and the fault adds impedance, so the further into the installation you go, the smaller the fault:

PointZsFault current
Origin, PME supply0.35 Ω657 A
End of a 30 m final circuit1.05 Ω219 A

Same house, a third of the current. Using the origin figure everywhere is conservative — it oversizes breaking capacity but never undersizes it — which is why it is the normal design practice and why nobody calculates PFC circuit by circuit.

The trade-off nobody mentions

Low loop impedance is what makes a circuit disconnect quickly. Low loop impedance is also what produces a large fault current. The two requirements pull in opposite directions:

  • Loop impedance wants Z low, so the device operates in time.
  • Breaking capacity wants the fault current low, which means Z high.

A large prospective fault current is therefore usually a sign of a good supply, not a bad design. It is the reason both calculations exist and the reason neither can be done without the other.

What this does not settle

Breaking capacity is about whether the device can interrupt the fault. It says nothing about whether the conductor survives it for however long that takes. That is the adiabatic check, S = √(I²t) ÷ k, and the I²t figure this page reports is exactly what it takes as input.

Two requirements, two failure modes, one fault current feeding both. A circuit can have entirely adequate switchgear and a protective conductor that will not survive the fault the switchgear is about to clear.

How to use this calculator

  1. Choose the route

    From a loop impedance if you know or measured Z at the point in question. From a transformer nameplate for the worst case at the supply. From meter readings if you are on site with a tester.

  2. For the impedance route, use the right voltage and the right loop

    Line to earth with Zs for an earth fault; line to neutral with the line-neutral loop for a short circuit. The prospective fault current is whichever of the two is larger.

  3. For the transformer route, read the impedance voltage off the plate

    Usually 4–6% on a distribution transformer. It is the percentage of rated voltage needed to drive rated current through a shorted secondary, so its reciprocal is how many times rated current a bolted fault produces.

  4. For readings, take the greater — and double on three-phase

    The device has to survive whichever fault happens, so use the larger of the line-neutral and line-earth figures. On a three-phase supply the symmetrical fault is exactly twice the highest line-neutral reading.

  5. Check the breaking capacity

    The device’s rating must cover the fault. If it does not, either fit a larger device or establish that backup protection from an upstream current-limiting device applies — which needs the manufacturer’s tested combination, not a calculation.

  6. Pass the energy to the conductor check

    The device being able to break the fault is one requirement. The conductor surviving it is another, and that takes the I²t figure into the adiabatic equation.

Frequently asked questions

What is prospective fault current?

The highest current that would flow if a fault of negligible impedance occurred at a given point — a bolted fault, with nothing limiting it but the impedance of the supply and the conductors leading to it. It matters because every protective device has to be able to interrupt it. A device asked to break more current than its rating does not simply fail to trip: it can fail destructively, and the fault stays live behind it.

What is the difference between PFC, PSCC and PEFC?

PSCC is the prospective short-circuit current — a fault between line and neutral, or between lines. PEFC is the prospective earth fault current — line to earth, through the protective conductor. PFC is simply whichever of the two is greater, because that is the one the device has to be able to handle. On a TN-C-S supply the two are usually close; on TN-S the earth path is longer and the short-circuit figure normally wins.

How do you calculate fault current?

I = U ÷ Z. The voltage driving the fault divided by the impedance of the loop it flows around. For an earth fault that impedance is Zs, so a 230 V supply with Zs of 0.35 Ω gives 657 A. That is the whole calculation — the difficulty is never the arithmetic, it is knowing which impedance and which voltage apply to the fault you are worried about.

How do you calculate fault current from a transformer?

Divide the rated secondary current by the per-unit impedance. The impedance voltage on the nameplate — 5% on a typical distribution transformer — is defined as the percentage of rated voltage needed to push rated current through a short-circuited secondary, so its reciprocal is how many times rated current a bolted fault produces. A 500 kVA 400 V three-phase transformer is rated 722 A, and at 5% delivers 722 ÷ 0.05 = 14.4 kA at its terminals. A 4% transformer of the same size delivers 18 kA: a stiffer supply means a bigger fault.

Why is the three-phase fault current twice the line-to-neutral reading?

Because of how many conductors each fault path uses, and it is exact rather than a rule of thumb. A line-to-neutral fault travels out along the line conductor and back along the neutral, so it sees two conductors in series — twice the impedance. A three-phase symmetrical fault has no return through the neutral at all, because the three phase currents sum to zero at the fault, so each phase sees only its own impedance. Equal line and neutral impedances therefore give exactly double the current. If the neutral is reduced, the true factor is larger than two and doubling understates the fault.

How can a 6 kA consumer unit be legal on a 16 kA supply?

Because the device does not have to break the fault alone. BS 7671 Regulation 434.5.1 permits a device with a breaking capacity below the prospective fault current where an upstream device provides backup protection. The distributor’s cutout fuse — normally a 100 A BS 88-3 — is current-limiting: it clears a severe fault so quickly that the energy actually reaching the breaker behind it is far below what an unrestricted 16 kA fault would deliver. The pair has a conditional short-circuit rating higher than the breaker’s own. That rating comes from tested combinations the manufacturer publishes and cannot be calculated, which is why no calculator will give it to you.

What fault current should I assume at a domestic supply?

UK distributors typically declare a maximum of 16 kA at the origin of a domestic single-phase supply. That is a worst case they undertake not to exceed, not a measurement, and the figure at an actual house is usually a small fraction of it — the earth fault current at a 0.35 Ω PME origin is under 700 A. Design against 16 kA unless you have measured otherwise, but do not be surprised when the meter reads far less.

Does fault current change further along a circuit?

Yes, substantially, and always downward. Every metre of conductor between the source and the fault adds impedance, so the current falls as you move away. The same house that has 657 A of earth fault current at the cutout has around 219 A at the end of a final circuit with Zs of 1.05 Ω. Using the origin figure everywhere is conservative — it oversizes the breaking capacity but never undersizes it — which is why it is the normal design practice.

What does the prospective fault current test measure?

The tester injects a small known load, measures the resulting voltage drop, calculates the loop impedance and reports U ÷ Z as a current. Take the reading at the origin, on each line conductor, both line-to-neutral and line-to-earth, and record the highest. On a three-phase supply, double the highest line-to-neutral figure to get the symmetrical three-phase fault. It is the same measurement as a loop impedance test presented in different units.

Does a big fault current mean my design is wrong?

No — it usually means the supply is good. Low impedance produces a large prospective fault current and also makes it easy to satisfy the disconnection time requirement, since a large fault current trips the device fast. The two pull in opposite directions: you want low impedance for disconnection and can end up needing higher breaking capacity because of it. That trade-off is the reason both calculations exist.

What else does the fault current affect?

The conductor. Breaking capacity is about whether the device can interrupt the fault; the adiabatic check is about whether the conductor survives it for the time that takes. Both are required, they fail in different ways, and the fault current is the input to both. The let-through energy reported here — I²t — is what the adiabatic equation takes.

Sources and further reading

How this calculator is checked

Salamot Hok, Technical reviewer

Technical reviewer

Electrician · 10+ years of installation work in Bangladesh and the wider South Asian region

He reads the result the way an installer would: are the defaults values people actually meet, does the warning fire where you would stop and think, and is the answer something you could buy and fit? The code figures themselves come from the published standards cited below, not from him — that boundary is set out on his profile.

  • The maths lives in a pure function with its own test suite, asserted against worked examples from published references and standards. A calculator does not ship until those tests pass.
  • 4 sources cited by name and linked, so any figure on the page can be traced back to the document it came from.
  • Last reviewed . Review dates are advanced only when the page is actually re-read, never to look fresh.
  • Unusable input returns no answer. Where the inputs do not describe a real design, the calculator says so and withholds the number rather than printing a plausible-looking wrong one.

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