How to calculate maximum Zs: worked examples

When you test a circuit you need two figures: its earth fault loop impedance Z_s, and the most its protective device will allow. This post shows how to work out that maximum and check a circuit against it. The three examples are taken word for word from our Inspection and Testing Calculations workbook, with the same working and rounding.

What maximum Zs means

Z_s is the impedance of the whole fault loop: the supply, the line conductor out to the fault, and the protective conductor back. The higher it is, the lower the fault current. If the fault current is too low, the protective device will not operate within the required disconnection time, so every device has a maximum Z_s.

The circuit figure is Z_s = Z_e + (R₁ + R₂). The maximum comes from the device.

The formula

Z_s(max) = (U₀ × C_min) ÷ I_a

• U₀ is the nominal voltage to earth. The examples use 230 V.
• C_min is the minimum voltage factor, which allows for the supply voltage varying. IET Wiring Matters (issue 105, May 2025) gives it as 0.95 for a low voltage supply provided under the ESQCR.
• I_a is the current that makes the device operate within the required time. It comes from the device data.

With these figures the top line is always 230 × 0.95 = 218.5, so each maximum is 218.5 divided by I_a.

A note on the figures

These examples come from the workbook, so its copyright-page note applies to them too:

Practice data only. All figures in the questions, including maximum earth fault loop impedances, resistances per metre, cable ratings, voltage drop figures, correction factors, diversity allowances and time limits, are supplied for calculation practice. Conductor resistances, voltage-drop figures and some cable current ratings used in the questions are typical published values for copper cable, given so the questions can be worked. No figure in this book may be used for design, installation, inspection, testing or certification. Always check against your own copy of the current edition of BS 7671, the relevant IET guidance and the manufacturer’s data.

Example 1: the maximum Zs for a small device

Question (workbook Q54, Core): A small protective device operates within the required time when the fault current is at least 30 A (I_a, given).

• Use Z_s(max) = (U₀ × C_min) ÷ I_a, with U₀ = 230 V and C_min = 0.95

Calculate the maximum Z_s to 3 decimal places.

Method and numbers

1. Z_s(max) = (230 × 0.95) ÷ 30 = 7.283 Ω

Result

• Z_s(max) = 7.283 Ω

One division gives the answer.

Example 2: does the circuit comply?

Question (workbook Q53, Core): A protective device operates within the required disconnection time when the fault current is at least 200 A (this is I_a, taken from the device data given).

• Use Z_s(max) = (U₀ × C_min) ÷ I_a, with U₀ = 230 V and C_min = 0.95; this maximum applies with the conductors at operating temperature
• Circuit: Z_e = 0.35 Ω, (R₁ + R₂) at 20 °C = 0.80 Ω
• Multiplier from 20 °C to operating temperature: 1.20

Calculate the maximum Z_s to 4 decimal places, then the circuit Z_s at operating temperature, and say whether the circuit complies.

Method and numbers

1. Z_s(max) = (230 × 0.95) ÷ 200 = 1.0925 Ω
2. Hot (R₁ + R₂) = 0.80 × 1.20 = 0.960 Ω
3. Z_s = 0.35 + 0.960 = 1.31 Ω
4. 1.31 Ω > 1.0925 Ω, so the circuit does not comply.

Result

• Z_s(max) = 1.0925 Ω
• Z_s at operating temperature = 1.31 Ω
• The circuit does not comply (1.31 Ω > 1.0925 Ω)

Maximum Z_s figures apply with the conductors at operating temperature, but (R₁ + R₂) is measured cold, so the working multiplies it by 1.20 first. The hot Z_s is over the maximum, so the circuit fails.

Example 3: changing the protective device

Question (workbook Q62, Challenging): An existing circuit has Z_s = 1.21 Ω at conductor operating temperature (worked out from its test readings, given). Its protective device (with I_a = 160 A) is to be replaced with a different device whose I_a = 200 A.

• Use Z_s(max) = (U₀ × C_min) ÷ I_a, U₀ = 230 V, C_min = 0.95; this maximum applies with the conductors at operating temperature

Calculate the maximum Z_s for each device (4 d.p.) and say whether the circuit complies with each.

Method and numbers

1. Z_s(max) existing = 218.5 ÷ 160 = 1.3656 Ω
2. Z_s(max) replacement = 218.5 ÷ 200 = 1.0925 Ω
3. 1.21 Ω ≤ 1.3656 Ω, so it complies with the existing device.
4. 1.21 Ω > 1.0925 Ω, so it does not comply with the replacement device.

Result

• Existing device Z_s(max) = 1.3656 Ω
• Replacement device Z_s(max) = 1.0925 Ω
• It complies with the existing device (1.21 Ω ≤ 1.3656 Ω)
• It does not comply with the replacement device (1.21 Ω > 1.0925 Ω)

The Z_s itself has not changed. The replacement device needs a larger current to operate in time, so its maximum Z_s is lower and the same circuit no longer complies. Check Z_s against the new device whenever a device is changed for one with a different I_a.

Common slips

• Leaving out C_min and dividing 230 by I_a. That gives a maximum that is too high.
• Using I_a for a different device type or rating.
• Comparing a cold reading straight with the maximum. Correct (R₁ + R₂) to operating temperature first, or use the rule-of-thumb limit from the workbook’s temperature section.
• Rounding the maximum up, or the circuit figure down, so a circuit that is just over the limit appears to pass.

Practise more

The free sampler has ten questions from ten sections of the workbook with full worked answers, including another maximum Z_s question.

Inspection and Testing Calculations (£9.99, PDF) has 200 questions in 14 sections, graded Easy, Core and Challenging, each with a full worked answer.

Written by a City & Guilds 2391-51 qualified electrician.