Earth Fault Loop and RCD Formulas

Earth fault loop formulas: the loop impedance from the supply and the cable, the fault current, the maximum impedance for a breaker and the limit with an RCD.

The formulas

Ze is the impedance of the supply side of the loop in ohms, L the cable length in metres, S1 and S2 the line and protective conductor areas in mm², ρ the resistivity in Ω·m (copper 1.7241e-8 at 20 °C), α the temperature coefficient (0.00393 for copper), U0 the voltage from line to earth, Ia the current that makes the device trip in the required time (for a type B breaker, the usual rule is 5 × In) and IΔn the rated residual current of the RCD. The 50 V limit for the touch voltage is the usual value for normal conditions and the trip times and the limit come from the standard you work to, so use your own. The resistance is taken at the working temperature because the loop must trip with hot cables.

Worked example: a 25 m final circuit

Ze = 0.35 Ω, 2.5 mm² line conductor and 1.5 mm² protective conductor, copper, 230 V, protected by a 20 A type B breaker.

1. R1 + R2 = 1.7241e-8 × 25 × (1 / 2.5 + 1 / 1.5) × 10⁶ = 0.4598 Ω at 20 °C
2. Hot (70 °C): 0.4598 × (1 + 0.00393 × 50) = 0.5501 Ω
3. Zs = 0.35 + 0.5501 = 0.9001 Ω
4. If = 230 / 0.9001 = 255.5 A
5. The breaker needs 5 × 20 = 100 A, so Zs(max) = 230 / 100 = 2.3 Ω, and the circuit passes

Worked example: earth resistance with a 30 mA RCD

1. RA ≤ 50 / 0.03 = 1,667 Ω

Largest earth resistance by RCD rating (50 V limit)

Frequently asked questions

What is the loop impedance?

The total impedance of the path of a fault current: from the source through the line conductor to the fault and back through the protective conductor and the earth path.

Why calculate with the cable hot?

The resistance of copper rises with temperature, and a fault current flows when the cable is already warm, so the cold value would give a result that is too good.

Does an RCD replace a good earth?

No. An RCD trips on a small leakage and makes a high earth resistance acceptable for shock protection, but the earth is still needed to give the RCD a path and the fault current.

Which supply type is this for?

The loop formulas are for a TN system where the fault current returns through the protective conductor. For a TT system use the earth resistance and the RCD rule instead.

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