Earth Fault Loop Impedance Calculator (Zs)

Zs = Ze + (R1 + R2) against the maximum permitted for the protective device, to BS 7671 Chapter 41 — with the maximum derived from U0 × Cmin ÷ Ia rather than looked up, and the cold-measurement rule applied separately.

Earth fault loop impedance is the total impedance of the path a fault current takes from a line conductor back to the source through earth: Zs = Ze + (R1 + R2), where Ze is the external part supplied by the distributor and R1 and R2 are the line conductor and protective conductor of the circuit, which are in series and so add. The circuit complies when Zs is low enough that the device disconnects in time: Zs ≤ U0 × Cmin ÷ Ia, which for a 230 V supply is 218.5 ÷ Ia. For a circuit breaker Ia is a fixed multiple of the rating — 5× for Type B, 10× for Type C, 20× for Type D — so a 32 A Type B allows 218.5 ÷ 160 = 1.37 Ω. A cold measurement is compared against 0.75 × that figure, because the table assumes the conductors are at operating temperature.

Formula
Zs = Ze + (R1 + R2)
Compliance
Zs ≤ U0 × Cmin ÷ Ia
At 230 V
218.5 ÷ Ia
Ia — Type B / C / D
5× / 10× / 20× In
Max Zs — 32 A Type B
1.37 Ω
RCD limit
Zs ≤ 50 ÷ IΔn
Cold measurement
0.75 × tabulated

Inputs

What are you doing

Designing calculates Zs from the supply and the cable. Verifying compares a meter reading against the cold-measurement limit, which is a different number.

External impedance Ze

Twin and earth pairings: 1/1, 1.5/1, 2.5/1.5, 4/1.5, 6/2.5, 10/4, 16/6 mm².

Twin and earth has a reduced cpc, which is why it usually contributes more to R1+R2 than the line conductor does.

m

From the board to the furthest point of the circuit.

Design at the operating temperature: that is when resistance is highest and the circuit is hardest to disconnect.

Breaker type

Sets Ia: 5 × In, 10 × In or 20 × In. A Type C permits exactly half the impedance of a Type B at the same rating.

V

Line to earth. 230 V in the UK.

Results

Loop impedance Zs1.05 ΩZs = Ze + (R1 + R2)
Maximum permitted Zs
1.366 Ω
Maximum for a cold test
1.024 Ω
Margin
315.3 mΩ
R1 + R2
700.3 mΩ
R1 + R2 (mΩ/m)
23.34
Fault current
219 A
Device operates at
160 A

Ze taken as 0.35 Ω — PME. 0.35 Ω is the declared maximum. Low Ze makes loop impedance easy to satisfy — and is why PME installations rarely fail on Zs and often fail on something else. A declared maximum is a design assumption, not a measurement, and the real figure is usually lower. Measure it before you rely on a marginal result.

1.05 Ω is within the 1.37 Ω maximum for a 32 A Type B — 5 × In, with 0.32 Ω of margin. The circuit disconnects in time.

The circuit contributes 0.700 Ω of the 1.05 Ω total, 67% — so the cable is what decides this, and cable is what you can change.

(R1 + R2) is 23.34 mΩ/m for 2.5/1.5 mm² at 70 °C — 19.51 mΩ/m at 20 °C multiplied by 1.196. Both conductors carry the fault current in series, so their resistances add.

Maximum Zs derived as U0 × Cmin ÷ Ia = 230 × 0.95 ÷ 160 A. Ia is 5 × In because it is the magnetic element that operates, not the thermal one — which is also why a breaker has one Zs figure rather than separate ones for 0.4 s and 5 s.

If you test this circuit, a cold reading must be at or below 1.02 Ω — 0.75 × the tabulated maximum. The older rule of thumb was 0.8; if you learned that figure, this is why it moved.

Prospective earth fault current 219 A. Check the protective device is rated to break it — a loop impedance low enough to disconnect quickly is also a loop impedance that lets a lot of current through.

Satisfying Zs is one requirement. The protective conductor must also survive the fault, which is the adiabatic check, and be at least the size Table 54.7 requires where that route is used.

Worked examples

32 A ring on 2.5/1.5, 30 m, PME

The commonest domestic circuit in Britain. TN-C-S supply, 2.5 mm² twin and earth with a 1.5 mm² cpc, checked at the conductor operating temperature.

Zs 1.05 Ω against a 1.37 Ω maximum — passes with 0.32 Ω in hand.

The same circuit on a Type C breaker

Nothing about the cable changes. The Type C needs ten times its rating to trip instantly rather than five, so the permitted impedance halves.

Maximum falls to 0.683 Ω. The identical circuit now fails by 0.37 Ω.

A long lighting circuit on TN-S

A 6 A Type B lighting circuit in 1.0/1.0, 60 m from the board, on a TN-S supply with its higher declared Ze. Small breakers permit huge impedances.

Zs 3.40 Ω against 7.28 Ω — passes comfortably despite the length.

TT with a 30 mA RCD

An earth electrode at 100 Ω, which no overcurrent device could ever operate through. The RCD is judged on touch voltage instead: Zs ≤ 50 ÷ IΔn.

Zs 100.7 Ω against a 1667 Ω limit — passes easily. On a breaker it would be hopeless.

Verifying a test result

A meter reading of 1.15 Ω on that 32 A Type B ring. The tabulated maximum is 1.37 Ω, so it looks fine — but the comparison for a cold measurement is a different number.

Cold limit is 1.02 Ω. 1.15 Ω FAILS, despite being under the tabulated 1.37 Ω.

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 the loop is, and why its impedance decides everything

When to use this: designing a circuit and needing to know whether it will disconnect in time before you install it; checking a test result at the end of a job; working out why a long run to an outbuilding will not comply; or deciding whether a Type C breaker is going to cause a problem.

When a line conductor touches earth, the current has to get back to the transformer. It goes out along the line conductor, through the fault, back along the protective conductor, and home through the distributor’s network. That round trip is the earth fault loop, and its impedance is the only thing standing between the supply voltage and an unlimited fault current.

Zs = Ze + (R1 + R2)

Ze is everything outside your installation — the distributor’s cable and transformer. R1 is your circuit’s line conductor and R2 its protective conductor. They add rather than combining any other way because the fault current flows through both of them one after the other: they are in series.

The impedance matters because it sets the fault current, and the fault current is what makes the protective device operate. A high loop impedance means a small fault current, which means a slow disconnection, which means a dangerous touch voltage sitting on exposed metal for longer than a person can safely be in contact with it.

The compliance test, and where the numbers come from

Zs ≤ U0 × Cmin ÷ Ia

  • U0 — nominal line-to-earth voltage. 230 V in the UK.
  • Cmin — 0.95. The supply is not guaranteed to be at 230 V when the fault happens, and a low supply voltage drives less current through the same impedance, so it takes longer to trip the device. Designing at the nominal voltage is optimistic by exactly that 5%.
  • Ia — the current that makes the device operate within the required time.

For a 230 V supply the numerator is always 218.5, so the whole of the breaker table is 218.5 ÷ Ia. And for a circuit breaker, Ia is a fixed multiple of the rating, because it is the magnetic element that operates:

TypeIa32 A deviceMax Zs
B5 × In160 A1.37 Ω
C10 × In320 A0.68 Ω
D20 × In640 A0.34 Ω

Why a breaker has one figure and a fuse has two

A circuit breaker has two tripping mechanisms. The thermal element handles overloads slowly; the magnetic element handles short circuits in about ten milliseconds. Once the fault current reaches 5 × In on a Type B, the magnetic element operates essentially instantly — so whether the regulation asked for 0.4 s or 5 s makes no difference at all to the impedance you need. One column covers both.

A fuse has no magnetic element. The current needed to clear it in 0.4 s is genuinely larger than the current needed in 5 s, so it gets two columns. That current comes off a measured time/current curve with no closed form, which is why fuse figures are not built into this calculator: there is nothing to derive. Select “fuse or other”, read Ia off the curve for your device, and the rest of the arithmetic is done for you.

The mistake that passes circuits which should fail

This is the single most common error in this subject, and it happens at the end of the job rather than the start.

The tabulated maximum assumes the conductors are at their operating temperature — 70 °C for thermoplastic. That is deliberate: hot copper is more resistive, the loop impedance is higher, and the circuit is at its hardest to disconnect. Compliance has to hold in that condition, not in the easiest one.

You test a circuit cold. Room temperature, nothing loaded, resistance about 20% lower than it will be in service. Compare that reading against the hot limit and you will pass circuits that do not actually comply.

BS 7671 Appendix 14 handles it with a rule of thumb: a cold measurement should be at or below 0.75 × the tabulated maximum. For a 32 A Type B that is not 1.37 Ω but about 1.02 Ω — and a reading of 1.15 Ω, which looks comfortably inside the table, fails.

Where R1 + R2 actually comes from

Add the resistance per metre of the two conductors and multiply by the length. A 2.5 mm² conductor is 7.41 Ω/km and a 1.5 mm² is 12.1 Ω/km, so 2.5/1.5 twin and earth is 19.51 mΩ/m at 20 °C. Multiply by 1.20 for 70 °C.

Notice the split. The cpc is 12.1 of that 19.51 — 62% of the total, from the conductor that carries no load current at all. Twin and earth is sold with a reduced cpc, and that reduced cpc is what dominates the loop impedance of most domestic circuits. It is also why increasing the cpc is usually the most effective single change when a circuit fails on Zs.

Ze, and how much to trust it

SystemDeclared max ZeWhat it means
TN-C-S (PME)0.35 ΩCombined neutral and earth. Low impedance, rarely the problem.
TN-S0.8 ΩSeparate earth, usually the cable sheath.
TTYour own electrode. No typical value exists.

Those are declared maxima, not measurements — worst cases the distributor undertakes not to exceed. The real figure is usually a good deal lower. On a marginal design that difference decides the answer, and it is worth measuring rather than assuming.

Why TT needs an RCD, and it is not a matter of preference

A TT earth electrode is commonly tens or hundreds of ohms. A 32 A Type B breaker needs the entire loop under 1.37 Ω. At 100 Ω the fault current is about two amps — the breaker would sit there indefinitely while the fault stayed live.

An RCD is judged on something else entirely: Zs ≤ 50 ÷ IΔn, which comes from the 50 V touch voltage limit rather than from disconnection time. A 30 mA device permits 1,667 Ω. That is three orders of magnitude more headroom, and it is the entire reason TT installations are workable.

The RCD covers fault protection. The circuit still needs an overcurrent device for overload and short circuit, and that device still has its own requirements.

What passing does not prove

A compliant Zs means the device disconnects within the required time. Two other things have to be true of the same protective conductor and neither is checked here:

  • It has to survive the fault — the adiabatic check, S = √(I²t) ÷ k. A conductor can be low-resistance enough to trip the device quickly and still be too small to carry that current for the time it takes.
  • Where the Table 54.7 route is used, it has to meet that table’s minimum size regardless of what any calculation says.

How to use this calculator

  1. Choose designing or verifying

    They compare against different numbers. A design uses the tabulated maximum with the conductors at operating temperature. A test result is compared against 0.75 of that figure, because you measured the conductors cold and they will be hotter in service.

  2. Establish Ze

    The declared maxima are 0.35 Ω for TN-C-S and 0.8 Ω for TN-S. They are design assumptions rather than measurements, and the real figure is usually lower. TT has no typical value at all — the electrode is yours and must be measured.

  3. Describe the circuit

    Line conductor, protective conductor and length. Twin and earth has a reduced cpc, and because both conductors carry the fault current in series their resistances add — the smaller cpc usually contributes more than the line conductor does.

  4. Design at the operating temperature

    70 °C for thermoplastic, 90 °C for thermosetting. That is when the conductors are most resistive and the circuit is hardest to disconnect, which is the condition compliance has to hold in.

  5. Pick the device

    The breaker type sets Ia — 5 × In for Type B, 10 × for Type C, 20 × for Type D. For a fuse, read Ia off its time/current curve and enter it; fuse figures are not built in because that curve has no formula.

  6. Read the verdict, and remember what it does not cover

    Passing on Zs means the device disconnects in time. The protective conductor still has to survive the fault — that is the adiabatic check — and meet the Table 54.7 minimum where that route is used.

Frequently asked questions

What is earth fault loop impedance?

The total impedance of the path a fault current takes from a line conductor, through the fault to earth, and back to the source. It has two parts: Ze, the external portion belonging to the distributor, and (R1 + R2), the line conductor and protective conductor of your circuit. Zs = Ze + (R1 + R2). It matters because it is what limits the fault current, and the fault current is what makes the protective device operate. A high loop impedance means a small fault current, which means a slow disconnection, which means a dangerous touch voltage persisting for longer than it should.

How do you calculate maximum Zs?

Zs ≤ U0 × Cmin ÷ Ia. U0 is the nominal line-to-earth voltage, 230 V in the UK; Cmin is 0.95, allowing for the supply sitting at the bottom of its tolerance; and Ia is the current that makes the device operate within the required disconnection time. For a 230 V supply the numerator is always 218.5, so the whole of the breaker table is 218.5 ÷ Ia. A 32 A Type B needs 160 A to trip instantly, so its maximum Zs is 218.5 ÷ 160 = 1.37 Ω.

What is the max Zs for a 32 A Type B breaker?

1.37 Ω. A Type B trips its magnetic element at up to 5 × its rating, so a 32 A device needs 160 A, and 218.5 ÷ 160 = 1.37 Ω. The same 32 A device in Type C needs 320 A and permits 0.68 Ω; in Type D it needs 640 A and permits 0.34 Ω. If you are comparing a test result rather than designing, the figure to beat is 0.75 of that, about 1.02 Ω.

Why is the measured Zs limit lower than the table?

Because the table and your meter are describing conductors at different temperatures. The tabulated maximum assumes the conductors are at their operating temperature — 70 °C for thermoplastic — since that is when their resistance is highest and the circuit is hardest to disconnect. You test a circuit cold, at room temperature, when its resistance is about 20% lower. Comparing a cold reading against the hot limit would pass circuits that will not actually comply in service. BS 7671 Appendix 14 handles this with a rule of thumb: a cold measurement should be at or below 0.75 × the tabulated maximum.

Was the rule of thumb not 0.8?

It was, and a lot of people learned it that way. The current edition uses 0.75. The change came with the introduction of the Cmin factor of 0.95 into the maximum-Zs calculation — the tabulated maxima themselves came down by 5%, and the correction factor was restated at the same time. If you are working from an older set of figures, check which basis they are on before mixing them with a modern table.

Why does a breaker have one Zs value rather than one for 0.4 s and one for 5 s?

Because it is the magnetic element that operates, not the thermal one, and it operates in about 10 milliseconds. Once the fault current reaches 5 × In on a Type B the breaker trips essentially instantly, so whether the requirement was 0.4 s or 5 s makes no difference to the impedance you need. A fuse is different: it has no magnetic element, the current required to clear it in 0.4 s is genuinely larger than the current required in 5 s, and its tables have two columns for that reason.

What is Ze and what values should I use?

Ze is the external earth fault loop impedance — everything outside your installation, measured at the origin with the main bonding disconnected. UK distributors declare maximum values of 0.35 Ω for TN-C-S (PME) and 0.8 Ω for TN-S. Those are worst cases they undertake not to exceed, not measurements: the real figure is usually a good deal lower, and on a marginal design it is worth measuring rather than assuming. TT has no declared value because the earth electrode is yours.

Why does a TT installation need an RCD?

Because its loop impedance is far too high for any overcurrent device to operate through. A TT earth electrode is commonly tens or hundreds of ohms; a 32 A Type B breaker needs the whole loop under 1.37 Ω. The fault current would be a couple of amps, and the breaker would sit there indefinitely. An RCD is judged on a different criterion entirely — Zs ≤ 50 ÷ IΔn, from the 50 V touch voltage limit — which for a 30 mA device permits 1,667 Ω. That is what makes TT workable.

How do I calculate R1 + R2?

Take the resistance per metre of the line conductor and of the protective conductor, add them, and multiply by the circuit length. They add because the fault current flows out along one and back along the other — they are in series. A 2.5 mm² conductor is 7.41 Ω/km and a 1.5 mm² is 12.1, so 2.5/1.5 twin and earth is 19.51 mΩ/m at 20 °C. Then multiply by the temperature factor: 1.20 for 70 °C. Note how much of that total is the cpc — the reduced protective conductor in twin and earth contributes more than the line conductor does.

My Zs is too high. What are my options?

In rough order of cost: increase the cpc, since it usually dominates R1 + R2; shorten the run; use a Type B device rather than a Type C or D, which multiplies the permitted impedance by two or four; or provide fault protection by RCD, which raises the limit by orders of magnitude. Check first whether the supply or the cable dominates — if Ze is most of your Zs, changing the cable has limited effect and the conversation is with the distributor.

Does passing the Zs check make the circuit compliant?

No. It proves the protective device disconnects within the required time. The protective conductor also has to survive the fault current for that duration without damage, which is the adiabatic check, and where the Table 54.7 route is used it must meet that table’s minimum size regardless. Those are separate calculations with separate failure modes, and a circuit can pass this one and fail either of them.

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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