Cable Sizing

Cable Fault Calculation Worked Example

Cable fault calculation worked example using complex impedance for phase and earth fault currents.

Updated June 4, 2026

This cable fault calculation worked example shows how cable impedance and source fault level can be combined to estimate fault current at the load end of a low-voltage cable. Complex impedance is used so that both magnitude and phase angle are retained.

Problem

Calculate the load-end phase and earth fault currents for a 400 V three-phase circuit, then review the effect of adding an external circuit protective conductor in parallel with the cable armour return path.

Given data

  • Cable: 185 mm2 single-core copper XLPE cable, 52 m long
  • Installation: flat formation, spaced one cable diameter apart
  • Positive-sequence cable impedance, Z1 = 0.00759 + j0.00726 Ω
  • Zero-sequence cable impedance, Z0 = 0.02486 + j0.00549 Ω
  • Source phase fault level = 10 kA at 0.25 power factor
  • Source earth fault level = 9.8 kA at 0.25 power factor

Step 1: calculate phase voltage

Uo = 400 / √3 = 230.9 V

Step 2: convert source fault levels to source impedance

The source fault level is treated as a complex current. At 0.25 power factor, the sine component is:

sin φ = sin(cos⁻¹ 0.25) = 0.968

For the phase fault source:

Ik3 = 10000 × 0.25 – j(10000 × 0.968) = 2500 – j9680 AZe3 = Uo / Ik3 = 230.9 / (2500 – j9680)Ze3 = 0.0058 + j0.0224 Ω

For the earth fault source:

Ik0 = 9800 × 0.25 – j(9800 × 0.968) = 2450 – j9486 AZe0 = Uo / Ik0 = 230.9 / (2450 – j9486)Ze0 = 0.0059 + j0.0228 Ω

Step 3: calculate phase fault current at the load end

The total phase fault impedance is the source impedance plus the positive-sequence cable impedance.

Zt3 = Ze3 + Z1Zt3 = (0.0058 + j0.0224) + (0.00759 + j0.00726)Zt3 = 0.01339 + j0.02966 Ω

Using Cmax = 1.05 and Cmin = 0.95:

If3,max = Cmax × Uo / Zt3 = 1.05 × 230.9 / Zt3 = 7.46 kA at 0.41 pfIf3,min = Cmin × Uo / Zt3 = 0.95 × 230.9 / Zt3 = 6.75 kA at 0.41 pf

Step 4: calculate earth fault current at the load end

The total earth fault loop impedance includes the source earth fault impedance, positive-sequence cable impedance and zero-sequence cable impedance.

Zt0 = Ze0 + Z1 + Z0Zt0 = (0.0059 + j0.0228) + (0.00759 + j0.00726) + (0.02486 + j0.00549)Zt0 = 0.03835 + j0.03555 Ω If0,max = 1.05 × 230.9 / Zt0 = 4.64 kA at 0.73 pfIf0,min = 0.95 × 230.9 / Zt0 = 4.20 kA at 0.73 pf

For the BS 7671 earth fault loop impedance check using Cmin = 0.95, the earth fault current is 4.20 kA and the loop impedance magnitude is 0.0523 Ω.

Step 5: add an external protective conductor

The original example then adds an external protective conductor return path in parallel with the cable armour return path.

Zcpc = (0.495e-3 + 0.135e-3) × 50 = 0.0315 ΩZ2 = (Zcpc × Z0) / (Zcpc + Z0)|Z2| = 0.0142 Ω

The earth fault current increases because the return path impedance is reduced:

  • Maximum earth fault current = 5.77 kA at 0.65 pf
  • Minimum earth fault current = 5.22 kA at 0.65 pf

For background on the method, see fault calculations, earth fault loop impedance and IEC 60909 fault calculations.

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