IEC & BS 7671 calculators

Earth fault loop impedance (Zs) calculator

Work out Zs for a circuit, the maximum its protective device allows for automatic disconnection, the longest run that complies, and the most a test should read.

Calculated with electrician.wiki · https://electrician.wiki/calculators/earth-fault-loop-impedance/

Code / standard

Earthing system

Circuit

Cable

Line conductor

Cable length

For a ring, the full length of the ring.

Protective device

Device rating In

Circuit type

Measured Zs (optional)

Leave at 0 to skip the test comparison.

Result · BS 7671

Zs 1.05 Ω ≤ 2.19 Ω · Pass

With Ze = 0.35 Ω and 30 m of 2.5/1.5 mm² twin and earth, Zs is 1.05 Ω: within the 2.19 Ω maximum for a B20 at 0.4 s (BS 7671), so it disconnects in time. A test near 10 °C should read no more than 1.75 Ω.

Zs (design, at operating temperature): 1.05 Ω
Maximum Zs (B20, 0.4 s): 2.19 Ω
R1 + R2: 0.702 Ω
Ia (operating current): 100 A
Earth fault current (0.95 × U0 / Zs): 208 A
Longest run for this device: 78.4 m
Most a test should read at 10 °C (0.8 × max): 1.75 Ω
…at 20 °C: 1.82 Ω

Ze = 0.35 Ω is the typical maximum UK distributors declare for TN-C-S supplies (ENA ER P23). Many supplies measure lower; use the measured value when you have it.

Show the work

1. R1 + R2 per metre at 20 °C (IEC 60228 maximum resistance)
$r_{20} = 7.41 mΩ/m + 12.1 mΩ/m = 19.51 mΩ/m$
2.5 mm² line + 1.5 mm² protective conductor, copper.
2. Correct to the conductor’s operating temperature (70 °C)
$m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2$
BS 7671 uses α = 0.004 per °C. The IET applies the same 1.20 factor for 70 °C cables.
3. R1 + R2 for the run
$R_{1} + R_{2} = r_{20} \cdot L \cdot m = 0.01951 Ω/m \cdot 30 m \cdot 1.2 = 0.7024 Ω$
4. Earth fault loop impedance
$Z_{s} = Z_{e} + (R_{1} + R_{2}) = 0.35 Ω + 0.7024 Ω = 1.052 Ω$
Ze is the distributor’s typical maximum for TN-C-S; measure it on site.
5. Required disconnection time: 0.4 s
$t = 0.4 s$
Final circuit with socket-outlets up to 63 A, TN system (BS 7671 Regulation 411.3.2 and Table 41.1).
6. Operating current Ia (B20)
$I_{a} = 100 A$
5 × In for type B, its instantaneous trip
7. Maximum Zs (BS 7671 Regulation 411.4.4)
$Z_{s,max} = \frac{C_{\min} \cdot U_{0}}{I_{a}} = \frac{0.95 \cdot 230 V}{100 A} = 2.185 Ω$
Cmin = 0.95 allows for the supply voltage running below 230 V (Appendix 3).
8. Compare: Zs is within the maximum
$1.052 Ω \leq 2.185 Ω$
9. Most a test should read (80 % rule, conductors near 10 °C)
$Z_{s,test} = 0.8 \cdot 2.185 Ω = 1.748 Ω$
At 20 °C multiply by 1.04 (IET GN3 Table A7 basis): 1.82 Ω.

References: BS 7671 Regulation 411.4.4 (2018): Zs × Ia ≤ U0 × Cmin · BS 7671 Table 41.1 and Regulation 411.3.2 (2018): disconnection times · BS 7671 Table 41.3 (2018): circuit-breakers · BS 7671 Appendix 3 (2018): Cmin and the 0.8 rule for measured values · IEC 60228: conductor resistance at 20 °C

Conductor resistance is the IEC 60228 maximum at 20 °C, raised to 70 °C; a real cable usually measures a little lower.
U0 = 230 V and Cmin = 0.95, as BS 7671 Tables 41.2 to 41.4 assume.
Reactance is ignored, which is fine for cables up to about 35 mm²; parallel earth paths (bonding, metal pipes) are not credited.

New to this? How TN-S, TN-C-S and TT earthing systems work

Report a problem with this result

How to use the Zs calculator

Enter Ze, the external earth fault loop impedance: either the distributor’s typical maximum for the earthing system or the value you measured at the origin. Then describe the circuit: twin and earth (the protective conductor size follows from the line size) or any pair of conductor sizes, the length (for a ring final, the whole ring), the protective device and the type of circuit. The calculator works out R1 + R2 at the conductor’s operating temperature, adds Ze, and compares Zs with the maximum that still makes the device disconnect in the time BS 7671 or IEC 60364 requires. It also gives the longest run that complies and the highest reading a test should show. Enter a measured Zs to check it against that figure.

How to calculate earth fault loop impedance, step by step

Get Ze. Measure it at the origin, or use the distributor’s typical maximum for the earthing system (the calculator lists them).
Work out R1 + R2. Add the line and protective conductor resistances per metre (mΩ/m at 20 °C) and multiply by the length. For a ring final circuit, use a quarter of the end-to-end value.
Correct to operating temperature. Conductors are tested cold but must work hot: multiply R1 + R2 by the factor for the cable’s maximum operating temperature (see the section on conductor temperature below).
Add them: Zs = Ze + (R1 + R2).
Compare with the maximum. Zs must not exceed Cmin × U0 / Ia (Cmin = 0.95 in BS 7671), where Ia is the current that makes the device disconnect in the required time.
Check measured readings at 80 %. A cold test reading should be no more than 80 % of the tabulated maximum.

What earth fault loop impedance is

When a line conductor touches an earthed metal casing, the fault current flows round a loop: out from the supply transformer along the line conductor, through the circuit’s line conductor to the fault, back along the circuit protective conductor (cpc) to the main earthing terminal, and home through the supply’s earth path. Zs is the impedance of that whole loop. The lower it is, the more current flows and the faster the fuse or circuit-breaker opens. Automatic disconnection of supply, the main protection against electric shock in most installations, depends on it.

Zs is the whole loop: Ze from the transformer to the main earthing terminal and back, plus the circuit’s line and protective conductors.

The loop splits naturally into the part outside the installation and the circuit itself:

Z_{s} = Z_{e} + (R_{1} + R_{2})

R1 + R2 and conductor temperature

R1 and R2 come from the conductor resistance per metre (the maximum values IEC 60228 allows at 20 °C) times the length. For 2.5/1.5 mm² twin and earth that is 19.51 mΩ/m. But a fault can happen when the circuit is fully loaded and its conductors are at their maximum operating temperature, 70 °C for PVC, and copper’s resistance rises with temperature. BS 7671 uses a coefficient of 0.004 per °C, so the correction from 20 °C is:

m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2

That 1.20 is the multiplier IET Wiring Matters applies in its worked example. For 90 °C thermosetting cable the same formula gives 1.28; we haven’t cross-checked that figure against a published table, so the calculator warns when it uses it. In a ring final circuit the furthest point is fed both ways round, so R1 + R2 there is a quarter of the end-to-end value: a 60 m ring of 2.5/1.5 mm² adds only 0.35 Ω.

Maximum Zs: Cmin · U0 / Ia

Each device needs a minimum current Ia to disconnect within the required time. Zs must be low enough to deliver it even when the supply voltage sags. BS 7671 Regulation 411.4.4 writes the condition as Zs × Ia ≤ U0 × Cmin, with Cmin = 0.95 for a public supply, so:

Z_{s} \leq \frac{C_{\min} \cdot U_{0}}{I_{a}}

For circuit-breakers to BS EN 60898 and RCBOs to BS EN 61009, Ia is the top of the instantaneous tripping band: 5 × In for type B, 10 × In for type C and 20 × In for type D. A B32 therefore needs 160 A, and its maximum Zs is 0.95 × 230 / 160 = 1.366 Ω, the 1.37 Ω printed in BS 7671 Table 41.3. Where 5 s is allowed, B and C devices keep the same Ia, but a type D only needs 10 × In: a D50 allows 0.22 Ω at 0.4 s and 0.44 Ω at 5 s.

Computed from Zs(max) = 0.95 × 230 V / Ia for 0.4 s, with Ia = 5, 10 or 20 × In. The test value is 80 % of the type B maximum; the last column is the longest 2.5/1.5 mm² twin-and-earth radial on a type B device with Ze = 0.35 Ω.
In	Type B	B, test (80 %)	Longest 2.5/1.5 radial (B)	Type C	Type D
6 A	7.28 Ω	5.83 Ω	296 m	3.64 Ω	1.82 Ω
10 A	4.37 Ω	3.50 Ω	172 m	2.19 Ω	1.09 Ω
16 A	2.73 Ω	2.19 Ω	102 m	1.37 Ω	0.68 Ω
20 A	2.19 Ω	1.75 Ω	78.4 m	1.09 Ω	0.55 Ω
32 A	1.37 Ω	1.09 Ω	43.4 m	0.68 Ω	0.34 Ω
40 A	1.09 Ω	0.87 Ω	31.7 m	0.55 Ω	0.27 Ω
50 A	0.87 Ω	0.70 Ω	22.4 m	0.44 Ω	0.22 Ω

Fuses have curves rather than a single multiple. The calculator uses fuse operating currents only where two independent publications agree, for example a BS 88-3 63 A fuse (about 600 A for 0.4 s, from BS 7671 Appendix 3, as NICEIC worked it). For any other fuse it asks you to read Ia from the curve and enter it, rather than lean on a single unconfirmed table.

Disconnection times

0.4 s for final circuits in a TN system at 230 V: up to 63 A with socket-outlets, or up to 32 A supplying only fixed equipment (BS 7671 Table 41.1 and Regulation 411.3.2).
5 s for distribution circuits and larger final circuits in TN systems.
0.2 s and 1 s for the same two cases in a TT system, which an overcurrent device seldom achieves; an RCD does.

IEC 60364-4-41 sets the same times and, since its 2017 amendment, the same limits: the original 2005 text applied the short time to final circuits up to 32 A, and the amendment raised that to 63 A for circuits with socket-outlets. BS 7671:2018 adopted the change through the European harmonisation document, so a 40 A socket circuit needs 0.4 s in both modes.

The 80 % rule for measured values

A loop test is made with the circuit cold, not at 70 °C, so a reading that only just meets the maximum would fail once the circuit warms up. The rule of thumb in BS 7671 Appendix 3, the IET’s Guidance Note 3 and the On-Site Guide is that a measurement should not exceed 80 % of the tabulated maximum. The factor is no accident: the resistance ratio from 10 °C to 70 °C is 1.20 / 0.96 = 1.25, and 1/1.25 = 0.8. Guidance Note 3 gives correction factors for testing at other temperatures; at 20 °C the allowance rises by 1.04×, so a B20 circuit tested on a warm day may read up to 1.82 Ω instead of 1.75 Ω.

Skipping the temperature correction is the classic design error, and IET Wiring Matters (May 2024) shows why. 1.5/1.0 mm² twin and earth, 77 m long, with Ze = 0.35 Ω: uncorrected, Zs = 2.68 Ω, apparently inside the 2.73 Ω of a B16; corrected, Zs = 3.14 Ω and the circuit fails.

TT systems and RCDs

On a TT supply the fault current returns through the soil between your earth electrode and the distributor’s, and Ze is often tens or hundreds of ohms. No circuit-breaker trips fast on a few amperes, so fault protection comes from an RCD. The condition (Regulation 411.5.3) is that the touch voltage stays within 50 V: RA × IΔn ≤ 50 V, where RA is the electrode plus protective conductor resistance. The calculator uses Zs in its place, which always exceeds RA and so errs safe. For a 30 mA RCD the limit is 1,667 Ω (BS 7671 Table 41.5), so the 85 Ω electrode in the example below passes easily. A high but compliant value still deserves attention, because electrode resistance rises when the ground dries or freezes.

Worked examples

A 20 A radial on a TN-C-S supply

30 m of 2.5/1.5 mm² twin and earth on a B20, with the distributor’s typical Ze for PME.

1. R1 + R2 per metre at 20 °C (IEC 60228 maximum resistance)
$r_{20} = 7.41 mΩ/m + 12.1 mΩ/m = 19.51 mΩ/m$
2. Correct to the conductor’s operating temperature (70 °C)
$m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2$
3. R1 + R2 for the run
$R_{1} + R_{2} = r_{20} \cdot L \cdot m = 0.01951 Ω/m \cdot 30 m \cdot 1.2 = 0.7024 Ω$
4. Earth fault loop impedance
$Z_{s} = Z_{e} + (R_{1} + R_{2}) = 0.35 Ω + 0.7024 Ω = 1.052 Ω$
5. Required disconnection time: 0.4 s
$t = 0.4 s$
6. Operating current Ia (B20)
$I_{a} = 100 A$
7. Maximum Zs (BS 7671 Regulation 411.4.4)
$Z_{s,max} = \frac{C_{\min} \cdot U_{0}}{I_{a}} = \frac{0.95 \cdot 230 V}{100 A} = 2.185 Ω$
8. Compare: Zs is within the maximum
$1.052 Ω \leq 2.185 Ω$
9. Most a test should read (80 % rule, conductors near 10 °C)
$Z_{s,test} = 0.8 \cdot 2.185 Ω = 1.748 Ω$

The IET example: a long lighting run that fails

77 m of 1.5/1.0 mm² on a B16 with Ze = 0.35 Ω, as worked in IET Wiring Matters, May 2024.

1. R1 + R2 per metre at 20 °C (IEC 60228 maximum resistance)
$r_{20} = 12.1 mΩ/m + 18.1 mΩ/m = 30.2 mΩ/m$
2. Correct to the conductor’s operating temperature (70 °C)
$m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2$
3. R1 + R2 for the run
$R_{1} + R_{2} = r_{20} \cdot L \cdot m = 0.0302 Ω/m \cdot 77 m \cdot 1.2 = 2.79 Ω$
4. Earth fault loop impedance
$Z_{s} = Z_{e} + (R_{1} + R_{2}) = 0.35 Ω + 2.79 Ω = 3.14 Ω$
5. Required disconnection time: 0.4 s
$t = 0.4 s$
6. Operating current Ia (B16)
$I_{a} = 80 A$
7. Maximum Zs (BS 7671 Regulation 411.4.4)
$Z_{s,max} = \frac{C_{\min} \cdot U_{0}}{I_{a}} = \frac{0.95 \cdot 230 V}{80 A} = 2.731 Ω$
8. Compare: Zs exceeds the maximum
$3.14 Ω > 2.731 Ω$
9. Most a test should read (80 % rule, conductors near 10 °C)
$Z_{s,test} = 0.8 \cdot 2.731 Ω = 2.185 Ω$

With Ze = 0.35 Ω and 77 m of 1.5/1 mm² twin and earth, Zs is 3.14 Ω: over the 2.73 Ω maximum for a B16 at 0.4 s (BS 7671), so it may not disconnect in time. A test near 10 °C should read no more than 2.19 Ω.

A ring final circuit

A 60 m ring of 2.5/1.5 mm² twin and earth on a B32.

1. R1 + R2 per metre at 20 °C (IEC 60228 maximum resistance)
$r_{20} = 7.41 mΩ/m + 12.1 mΩ/m = 19.51 mΩ/m$
2. Correct to the conductor’s operating temperature (70 °C)
$m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2$
3. R1 + R2 at the far point of a ring: a quarter of the end-to-end value
$R_{1} + R_{2} = \frac{r_{20} \cdot L \cdot m}{4} = \frac{0.01951 Ω/m \cdot 60 m \cdot 1.2}{4} = 0.3512 Ω$
4. Earth fault loop impedance
$Z_{s} = Z_{e} + (R_{1} + R_{2}) = 0.35 Ω + 0.3512 Ω = 0.7012 Ω$
5. Required disconnection time: 0.4 s
$t = 0.4 s$
6. Operating current Ia (B32)
$I_{a} = 160 A$
7. Maximum Zs (BS 7671 Regulation 411.4.4)
$Z_{s,max} = \frac{C_{\min} \cdot U_{0}}{I_{a}} = \frac{0.95 \cdot 230 V}{160 A} = 1.366 Ω$
8. Compare: Zs is within the maximum
$0.7012 Ω \leq 1.366 Ω$
9. Most a test should read (80 % rule, conductors near 10 °C)
$Z_{s,test} = 0.8 \cdot 1.366 Ω = 1.093 Ω$

With Ze = 0.35 Ω and 60 m of 2.5/1.5 mm² twin and earth (ring), Zs is 0.70 Ω: within the 1.37 Ω maximum for a B32 at 0.4 s (BS 7671), so it disconnects in time. A test near 10 °C should read no more than 1.09 Ω.

A TT supply with a 30 mA RCD

A rural property with an 85 Ω electrode reading at the origin.

1. R1 + R2 per metre at 20 °C (IEC 60228 maximum resistance)
$r_{20} = 7.41 mΩ/m + 12.1 mΩ/m = 19.51 mΩ/m$
2. Correct to the conductor’s operating temperature (70 °C)
$m = 1 + 0.004 \cdot (70 °C - 20 °C) = 1.2$
3. R1 + R2 for the run
$R_{1} + R_{2} = r_{20} \cdot L \cdot m = 0.01951 Ω/m \cdot 20 m \cdot 1.2 = 0.4682 Ω$
4. Earth fault loop impedance
$Z_{s} = Z_{e} + (R_{1} + R_{2}) = 85 Ω + 0.4682 Ω = 85.47 Ω$
5. Maximum Zs with an RCD (TT): Zs × IΔn ≤ 50 V
$Z_{s,max} = \frac{50 V}{I_{Δn}} = \frac{50 V}{0.03 A} = 1,667 Ω$
6. Compare: Zs is within the maximum
$85.47 Ω \leq 1,667 Ω$

With Ze = 85 Ω and 20 m of 2.5/1.5 mm² twin and earth, Zs is 85.47 Ω: within the 1,667 Ω limit for a 30 mA RCD (Zs × IΔn ≤ 50 V).

BS 7671 and IEC 60364

The physics is the same; the condition isn’t quite. BS 7671 multiplies U0 by Cmin = 0.95 to allow for a supply running below its nominal voltage. IEC 60364-4-41 states the condition as Zs × Ia ≤ U0 with no factor, so the same B32 is allowed 1.44 Ω in IEC mode against 1.37 Ω in BS 7671 mode. IEC mode also lets you set U0 (for 220 V, 230 V or 240 V systems); the UK fuse types and the ring final option are BS 7671 only.

Common mistakes

Comparing a cold reading with the tabulated maximum. Use 80 % of it, corrected for the test temperature.
Forgetting the temperature correction in design. R1 + R2 at 20 °C understates the loop by about a sixth.
Using the wrong time. A 32 A socket circuit needs 0.4 s; a sub-main may use 5 s, which matters for type D devices and fuses.
Relying on parallel paths. Metal pipes and bonding lower a measured Zs, but they can be removed; the design should work without them.
Assuming the typical Ze. Distributors declare maxima of 0.35 Ω (TN-C-S) and 0.80 Ω (TN-S); a real supply may be lower, but measure before relying on it.

A compliant Zs also sets the fault current your protective conductor must survive: check the cpc with the adiabatic equation. To size the cable in the first place, use the cable size calculator; for breaking capacity, the prospective fault current calculator.

Method and verification

Maximum Zs values are computed from Cmin · U0 / Ia rather than copied, and the tests confirm they reproduce every published figure we could check: B32 1.37 Ω and B16 2.73 Ω (IET), B6 7.28 Ω (NICEIC), D50 0.22 Ω and 0.44 Ω (IET). Conductor resistances are the IEC 60228 maxima confirmed by three manufacturers, and the IET Wiring Matters example above is a test case.

Assumes copper or aluminium conductors without significant reactance and no parallel earth paths. Updated 2026-10-14. Corrections log.

Questions people ask

What is the maximum Zs for a B32 MCB?

1.37 Ω for 0.4 s disconnection: 0.95 × 230 V divided by Ia = 5 × 32 = 160 A. A test reading with the circuit cold should not exceed about 1.09 Ω.

How do you calculate Zs?

Add Ze, the loop impedance outside the installation, to R1 + R2, the resistance of the circuit’s line and protective conductors. Work R1 + R2 out from the resistance per metre times the length, multiplied by 1.20 for 70 °C PVC cable.

What is the 80 % rule for earth fault loop impedance?

Measured values are taken with cold conductors, so they should be no more than 80 % of the tabulated maximum, which assumes conductors at their operating temperature. Guidance Note 3 gives factors to adjust for the temperature during the test.

What is a typical Ze in the UK?

Distributors declare maxima of 0.35 Ω for TN-C-S (PME) supplies and 0.80 Ω for TN-S. For TT supplies the distributor’s part is up to 21 Ω, and your own electrode adds to it.

What Zs do I need with an RCD on a TT system?

The touch voltage must stay within 50 V: Zs × IΔn ≤ 50 V, so up to 1,667 Ω for a 30 mA RCD. In practice aim far lower, because an electrode’s resistance changes with the seasons.

Why is my measured Zs higher than calculated?

Loose terminations, a long or undersized cpc, extra joints, or a higher Ze than assumed. Lower readings than calculated are often parallel earth paths through bonding and pipework, which shouldn’t be relied on.