Undersized Submain to a Beach-Front Extension
A two-storey guest house in Worthing has a single-phase 100 A service. The owner adds a 30 m submain from the main distribution board to a new first-floor distribution board supplying four air-conditioning units (each rated 12 A at 230 V, continuous), ten 13 A socket-outlets, and twenty LED downlights (5 W each). The electrician proposes 6 mm² two-core PVC-insulated copper cable clipped direct, protected by a 40 A miniature circuit-breaker. The measured loop impedance at the main board is 0.35 ohm. The owner reports that when all four air-conditioners start together the 40 A breaker trips after several minutes, and the voltage at the far board drops noticeably.
1. Assess whether the 40 A breaker and 6 mm² cable are appropriate for the connected load, showing your demand calculation.
Read the worked answer
Diversity must be applied deliberately, not by guesswork.
- Air-conditioning: 4 × 12 A = 48 A. Because the units are continuous and compressor starting is the reported symptom, treat the full 48 A as the design current for this circuit; no diversity is justified for a single grouped load of this type.
- Socket-outlets: 10 points. Applying a first 100% plus 40% of the remainder approach to a 13 A per-point allowance gives roughly 13 A + 0.4 × (9 × 13 A) ≈ 60 A, but for a guest-house submain the realistic after-diversity socket demand is far lower; a defensible figure is 20–25 A.
- Lighting: 20 × 5 W = 100 W ≈ 0.5 A, negligible.
Even taking the conservative socket figure of 20 A, the diversified demand is about 48 + 20 + 0.5 ≈ 68.5 A. That exceeds the 40 A breaker, so it will trip on sustained overload exactly as reported. The 6 mm² clipped-direct cable has a tabulated capacity around 47 A - adequate for 40 A but not for 68 A. The correct fix is to recalculate the submain for at least 70 A, select a cable with sufficient tabulated capacity (typically 16 mm² for this installation method), and coordinate the protective device with that capacity. The voltage drop must then be checked separately against the 4% (about 9.2 V) limit for a 230 V installation.
2. The measured earth-fault loop impedance at the main board is 0.35 ohm. Explain what must be verified at the remote board and why the breaker tripping does not prove the circuit is safe.
Read the worked answer
The main-board reading of 0.35 ohm is only the starting point. The submain adds conductor resistance over 30 m, and the remote board will see a higher loop impedance. Using the relationship Zs = Ze + (R1 + R2), and estimating 16 mm² line and 16 mm² circuit protective conductor at roughly 1.15 milliohms per metre each, the added impedance is about 30 × 2 × 0.00115 ≈ 0.07 ohm, giving an expected Zs near 0.42 ohm at the remote board.
For a 40 A type B breaker, the maximum permitted Zs for a 0.4 s disconnection is approximately 1.15 ohm (based on the tabulated multiplier applied to the rated current), so the expected value is comfortably within limits - but this must be confirmed by measurement, not calculation alone. For a type C device the limit is roughly 0.57 ohm, which is tighter and must be checked if the breaker is changed.
Crucially, a breaker tripping on overload says nothing about earth-fault disconnection. A high-resistance protective conductor, a loose termination, or a missing CPC could leave the metalwork live while the breaker still trips on current. The electrician must measure Zs at the remote board and at the furthest point of each final circuit, confirm the disconnection time is met, and verify that the main protective bonding to the water and gas services is present and of adequate size.