BS 7671 Wire Sizing Guide: UK Wiring Regulations Explained

What Is BS 7671?

BS 7671 is published by the Institution of Engineering and Technology (IET) and the British Standards Institution (BSI). It is not a law itself, but compliance is effectively mandatory because the Electricity at Work Regulations 1989 and Building Regulations Part P reference it as the means of meeting statutory safety requirements. Electricians in the UK must hold a qualification that demonstrates competence in applying BS 7671, such as the City & Guilds 2382 certificate.

Cable Sizing Methodology

BS 7671 Appendix 4 outlines a six-step process: (1) Determine the design current Ib from the load. (2) Select the rated current In of the protective device so that In ≥ Ib. (3) Determine the tabulated current-carrying capacity It required, accounting for correction factors: It ≥ In / (Ca × Cg × Ci × Cf), where Ca = ambient temperature, Cg = grouping, Ci = thermal insulation, Cf = semi-enclosed fuse factor (0.725). (4) Select a cable from Appendix 4 tables with Iz ≥ It. (5) Check voltage drop. (6) Verify fault current withstand using the adiabatic equation.

Appendix 4 Current-Carrying Capacity Tables

BS 7671 Appendix 4 provides tables for different cable types and installation methods. Table 4D1A covers single-core PVC cables (copper) in conduit: 1.5 mm² = 17.5 A, 2.5 mm² = 24 A, 4 mm² = 32 A. Table 4D2A covers multicore PVC cables (copper) clipped direct: 1.5 mm² = 19.5 A, 2.5 mm² = 27 A, 4 mm² = 36 A, 6 mm² = 46 A. The values differ by installation method because heat dissipation varies. Tables also cover XLPE/EPR, mineral-insulated, and armoured cables.

Voltage Drop Requirements

BS 7671 Regulation 525.1 limits voltage drop to 3% for lighting and 5% for other circuits, measured from the origin of the installation to the current-using equipment. Appendix 4 provides voltage drop per ampere per metre (mV/A/m) for each cable size. For 2.5 mm² two-core PVC cable, the value is 18 mV/A/m. The voltage drop formula is: ΔU = mV/A/m × Ib × L / 1000. For a 20 A load over 25 m: ΔU = 18 × 20 × 25 / 1000 = 9.0 V = 3.9% of 230 V. This exceeds 3% for lighting, requiring a 4 mm² cable (11 mV/A/m → 5.5 V = 2.4%).

RCD Protection Requirements

BS 7671 requires 30 mA RCD protection for all socket outlets rated up to 32 A (Regulation 411.3.3), all circuits in bathrooms and shower rooms, cables concealed in walls at a depth less than 50 mm, and circuits supplying mobile equipment outdoors. The 18th Edition significantly expanded RCD requirements compared to earlier editions. RCD protection complements but does not replace correct cable sizing — the protective conductor must still be sized for fault current.

Practical Example: Ring Final Circuit

A ring final circuit serving 13 A socket outlets in a UK home uses 2.5 mm² twin-and-earth cable protected by a 32 A MCB. Because it is a ring (two paths for current), the effective current-carrying capacity is doubled. The maximum floor area served is 100 m². Voltage drop calculation for a ring uses half the route length. For a 40 m ring with a 13 A load at the midpoint: effective length = 20 m, ΔU = 18 × 13 × 20 / 1000 = 4.68 V = 2.0%. This is within the 5% limit for power circuits.

Adiabatic Equation for Fault Protection

BS 7671 Regulation 434.5.2 requires that the protective device disconnects the fault before the cable is damaged. The adiabatic equation is: t ≤ (k × S / I)², where t is disconnection time in seconds, k is a material constant (115 for copper PVC), S is conductor cross-section in mm², and I is fault current in amps. For a 2.5 mm² cable with a 1 kA fault: t ≤ (115 × 2.5 / 1000)² = 0.083 s. The 32 A MCB must disconnect within 83 ms at 1 kA, which a Type B MCB achieves.

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