Choosing cable size for a long run to a welder outlet
A customer wants a 15 A single-phase outlet installed for a portable welder at the far end of a 42 m run from the main switchboard. The welder draws a continuous 12 A with an inrush of about 60 A for a fraction of a second at strike. The cable will be run in thermal insulation in a ceiling space for 20 m and then clipped to a wall in free air for the remaining 22 m. The available cable is 2.5 mm² copper with a tabulated current-carrying capacity of 24 A in free air and 18 A when surrounded by insulation. The earth loop impedance at the switchboard is 0.35 Ω and the maximum permitted for the circuit breaker is 2.4 Ω. The electrician must select a cable and protective device that will work reliably.
1. Determine whether 2.5 mm² cable is adequate, showing the voltage drop and derating reasoning.
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First derate for installation conditions: the worst section governs, so the cable is limited to 18 A where it passes through insulation. A 15 A or 16 A breaker protects the cable, so the current-carrying requirement is satisfied. Voltage drop: for 2.5 mm² copper single-phase, resistance is about 7.4 mΩ/m per conductor, so the loop resistance over 42 m is 2 × 42 × 0.0074 ≈ 0.62 Ω. At 12 A the drop is 0.62 × 12 ≈ 7.5 V, or about 3.2% of 230 V. That exceeds the commonly applied 3% limit for a final subcircuit but is within the 5% total installation limit if the rest of the installation drops less than 1.8%. If the welder is the only load and the supply is healthy, 2.5 mm² may be acceptable; otherwise a 4 mm² cable reduces the drop to about 2% and is the safer choice. The answer must state the limit being applied, because the decision hinges on it.
2. Check earth fault loop impedance and confirm whether the breaker will trip within the required time.
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Earth fault loop impedance at the outlet equals the supply value plus the circuit conductor impedance: 0.35 + 0.62 ≈ 0.97 Ω for 2.5 mm². This is well below the 2.4 Ω maximum, so the breaker will operate magnetically within the required disconnection time. With 4 mm² the loop impedance is about 0.35 + 0.39 ≈ 0.74 Ω, also compliant. The inrush of 60 A is not a fault current but may cause nuisance tripping of a Type B breaker; a Type C or D device may be needed, and the manufacturer's let-through data should be checked. The decision is therefore not just 'does it trip' but 'does it trip fast enough without tripping on start-up', which is why the breaker curve must match the load.
3. The customer says the run must go underground for the last 8 m through a garden bed. Explain how this changes the cable selection.
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Underground installation changes the derating and mechanical protection requirements. Direct-buried cable must be rated for burial and protected by a suitable enclosure or conduit at the transition; the thermal resistivity of the soil and the burial depth affect current-carrying capacity, so the tabulated value for buried installation must be used rather than the free-air value. If the cable passes through the garden bed it may need additional mechanical protection such as conduit or a warning tape at the correct depth. The voltage drop calculation is unchanged in length but the total run is now 50 m, which raises the drop to about 8.9 V (3.9%) on 2.5 mm² - reinforcing the case for 4 mm². The electrician should also confirm the earthing and bonding at the outlet and record the as-installed loop impedance.