An electrician uses a 50 m extension cord (2.5 mm² Cu) for a 2 kW floodlight at 230 V. Calculate the voltage drop and assess whether it is acceptable per NEN 1010 §525.

Question

Accepted Answer

ΔU ≈ 6.1 V (2.7%) — acceptable, because within the 4% limit of 9.2 V — NEN 1010 §525: Voltage drop calculation for extension cord: Current: I = P/U = 2000 W / 230 V = 8.70 A. Cable resistance (forward + return = 2× length): R = ρ × (2L) / A = 0.0175 Ω·mm²/m × (2 × 50 m) / 2.5 mm² = 0.70 Ω. Voltage drop: ΔU = I × R = 8.70 A × 0.70 Ω = 6.09 V ≈ 6.1 V. Percentage: ΔU% = 6.1 / 230 × 100% = 2.65%. Limit §525: for lighting and final circuits ≤ 4% of Un = 0.04 × 230 = 9.2 V. Conclusion: 6.1 V < 9.2 V → acceptable. Note: extension cords DO fall under §525 when used as part of the fixed installation or structurally (not occasionally). Option B (12 V, 5.2%) is wrong — this would require a shorter cord or thicker conductor (4 mm²). Options D and A are wrong: §525 applies to all final circuits including extension cords.

Answer

The 4% limit applies to fixed installations and not to mobile tools on an extension cable

Answer

ΔU ≈ 12 V (5.2%) — not acceptable, because above the 4% limit of 9.2 V

Answer

ΔU ≈ 6.1 V (2.7%) — acceptable, because within the 4% limit of 9.2 V

Answer

Extension cables are exempt from the voltage-drop requirement of §525 because they are used temporarily

Become an electrician in the Netherlands — learn NEN 1010 and NEN 3140 in your own language

What do you want to do?

Question of the day

Featured guides

Fuse utilisation category gG / aM

Inspection report, risk class & liability (NEN 3140)

Key exam terms

Ready for your NEN 3140 VP exam?