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November 10, 2026 at 9:55 am #5524
MEP Center Engineering
KeymasterStarter question — a frequently asked question posted by the MEP Center team to open this forum. Share your own experience below.
An outdoor lighting final circuit at 230 V single-phase carries 10 A at unity power factor to fittings about 60 m from the distribution board. The designer used 2.5 mm² copper PVC cable with a C16 MCB, because the current is low.
- What voltage-drop limit applies to a final circuit, and is it measured from the board or from the origin of the installation?
- Is 2.5 mm² acceptable here?
- Are there other checks that long final circuits commonly fail?
November 10, 2026 at 9:55 am #5525MEP Center Engineering
KeymasterThe usual limits are counted from the origin of the installation, not from the final board: IEC 60364-5-52 (Annex G) recommends 3 % for lighting and 5 % for other uses when supplied from a public LV network, and 6 % and 8 % from a private transformer. At 60 m, 2.5 mm² fails the voltage-drop check and also the fault-loop check with a C16.
Step 1: Set the budget
Subtract the drop already used upstream. If the sub-main to this board uses 1 %, the lighting final circuit has 2 % left of a 3 % total. NEC treats this as an informational note (3 % for a branch circuit, 5 % for feeder plus branch), and many specifications make it mandatory.
Step 2: Calculate
Single-phase: ΔU = 2 × L × I × (R cos φ + X sin φ). For small cables at cos φ = 1, X can be ignored. Use R at the operating temperature (70 °C for PVC): R70 = R20 × (1 + 0.00393 × 50).
Cable R at 70 °C (Ω/km) ΔU at 10 A, 60 m (V) ΔU (%) Total with 1 % upstream 2.5 mm² 8.87 10.6 4.63 5.63 % (fails) 4 mm² 5.52 6.6 2.88 3.88 % (fails) 6 mm² 3.69 4.4 1.92 2.92 % (passes) The circuit needs 6 mm², even though 2.5 mm² carries 10 A easily.
Step 3: Check fault-loop impedance (often forgotten)
Automatic disconnection in a TN system needs Zs × Ia ≤ U0. A C16 must trip magnetically at 10 × 16 = 160 A, so Zs ≤ 230 / 160 = 1.44 Ω. Use conductor resistance at operating temperature, or apply a 0.8 factor to values at 20 °C, not both.
- 2.5 mm² with a 2.5 mm² protective conductor: loop R = 2 × 0.060 × 8.87 = 1.06 Ω. With Ze = 0.40 Ω, Zs = 1.46 Ω, which fails.
- 6 mm²: 0.44 + 0.40 = 0.84 Ω, which passes.
- Alternatives are a B16 (Zs ≤ 2.88 Ω) or RCD protection. Outdoor circuits usually need a 30 mA RCD anyway; see RCD Types AC, A, F and B.
Practical points
- Use the real cable route length, including vertical drops and the loop through fittings, not the straight-line distance.
- For distributed loads (lamps spaced along the route) the drop is lower than with the whole load at the end. Calculate it segment by segment rather than halving it blindly.
- For motors, check the starting voltage drop as well (often limited to 10 to 15 % at the motor terminals during starting). See Motor Starting: DOL, Star-Delta, Soft Starter or VFD.
- LED drivers tolerate a wide voltage range, but emergency luminaires and control gear may not. Do not relax the limit on that basis.
Common mistakes
- Measuring 3 % from the final board instead of from the origin.
- Using resistance at 20 °C for voltage drop and then calling the result conservative.
- Upsizing phase conductors but leaving a small protective conductor, so Zs still fails.
For deeper study: Electrical Power Systems Design course.
Check the edition of the code that applies to your project.
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