Cable sizing to BS 7671 follows a five-step process: determine the design current, select the protective device, apply correction factors, choose the cable from Appendix 4, and verify voltage drop. This guide covers every step with worked examples, reference methods, and the adiabatic check.
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Written and reviewed by Andrew Moore, founder of Elec-Mate, against BS 7671:2018+A4:2026, IET Guidance Note 3 and the IET On-Site Guide.
Five steps: (1) calculate the design current Ib; (2) select a protective device with In ≥ Ib; (3) apply the correction factors — ambient (Ca), grouping (Cg), thermal insulation (Ci) and BS 3036 fuse (Cf) — so the tabulated rating It ≥ In ÷ (Ca×Cg×Ci×Cf); (4) choose a cable whose current-carrying capacity from the Appendix 4 table for the correct reference method meets that figure; (5) verify voltage drop is within limits (3% for lighting, 5% for other uses per Appendix 4) and run the adiabatic check on the CPC (Reg 543.1.3).
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Correct cable sizing is one of the most fundamental design tasks in electrical installation. A cable that is too small will overheat under load, potentially causing a fire or damaging the insulation. A cable that is too large wastes material and money. BS 7671:2018+A4:2026 provides a systematic process for selecting the correct cable size based on the load, the installation conditions, and the protective device.
The process uses the current-carrying capacity tables in Appendix 4 of BS 7671, combined with correction factors that account for the specific installation conditions. The result is a cable that can safely carry the design current under the worst-case conditions it will encounter, while keeping the voltage drop within acceptable limits.
While the process can be done manually using the tables in the brown book, it involves multiple lookups and calculations that are easy to get wrong. This is why many electricians use a cable sizing calculator — and why Elec-Mate has built every Appendix 4 table and correction factor into its cable sizing calculator.
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The design current is the maximum current the cable must carry in normal service. For single-phase circuits: Ib = P / (V x cos phi). For three-phase circuits: Ib = P / (root 3 x VL x cos phi). For a resistive load (heaters, immersion, electric shower), the power factor is 1.0. For motors and fluorescent lighting, the power factor is typically 0.8-0.85. For domestic circuits with a fixed rating (e.g., 32A ring circuit, 20A radial), the design current is simply the expected maximum load.
Choose a protective device with a rated current In greater than or equal to Ib. Standard MCB ratings are 6, 10, 16, 20, 25, 32, 40, 50, and 63A. Standard RCBO ratings are the same. The protective device must also be appropriate for the load type: Type B for general circuits (trips at 3-5 times rated current), Type C for motor loads (5-10 times), and Type D for high inrush loads like transformers (10-20 times).
The correction factors account for conditions that reduce the cable's ability to dissipate heat. Divide the protective device rating by the product of all applicable correction factors to get the minimum tabulated current rating:
It = In / (Ca x Cg x Ci x Cf)
Find the appropriate Appendix 4 table for the cable type and installation method (reference method). Select a cable with a tabulated current-carrying capacity Iz greater than or equal to It. This ensures the cable can carry the full load current even under the worst-case derating conditions.
Calculate the voltage drop using VD = mV/A/m x Ib x L / 1000 and compare against the BS 7671 limits: 3% for lighting (6.9V from 230V) and 5% for other circuits (11.5V from 230V). If the voltage drop exceeds the limit, increase the cable size and recalculate.
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The reference method describes how the cable is installed. This is critical because the installation method determines how effectively the cable can dissipate heat — and therefore its current-carrying capacity. A cable clipped directly to a surface (Method C) can dissipate heat much better than the same cable enclosed in an insulated wall (Method A), so it has a higher current rating.
Lowest ratings. The cable is surrounded by insulation on all sides, severely restricting heat dissipation. Common in domestic rewires where cables run through insulated stud walls.
Slightly better than Method A as the conduit or trunking is mounted on a surface, allowing some heat dissipation from the exterior.
Good heat dissipation. Cable is clipped directly to a wall, ceiling, or other surface. The most common method for domestic T&E cable clipped to joists in accessible locations.
Used for SWA cables buried in the ground. The thermal conductivity of the soil affects the rating. Common for supplies to outbuildings, garages, and EV chargers.
Excellent heat dissipation with air circulation on all sides. High current ratings. Common in commercial and industrial installations.
Similar to Method E but mounted vertically. Slightly different ratings due to convection effects on vertical runs.
Cable supported on brackets or cleats spaced away from the surface. Free air circulation gives the highest current ratings.
Important: If a cable run uses multiple installation methods along its route, you must use the worst-case (lowest rating) reference method for the entire cable length. For example, if a cable is clipped direct (Method C) for most of its run but passes through an insulated wall for 2 metres (Method A), the Method A rating applies.
Four correction factors may apply to a cable sizing calculation. Each factor is a decimal less than or equal to 1.0 that increases the required tabulated current by reducing the divisor:
The Appendix 4 current ratings assume an ambient temperature of 30C. If the ambient temperature is higher, the cable cannot dissipate heat as effectively, and the rating must be reduced.
| Ambient Temp | Ca (70C PVC) | Ca (90C XLPE) |
|---|---|---|
| 25C | 1.03 | 1.02 |
| 30C | 1.00 | 1.00 |
| 35C | 0.94 | 0.96 |
| 40C | 0.87 | 0.91 |
| 45C | 0.79 | 0.87 |
| 50C | 0.71 | 0.82 |
When cables are grouped together, they heat each other up and cannot dissipate heat as effectively. The grouping factor depends on the number of circuits and the arrangement:
| No. of Circuits | Bunched | Single Layer (Touching) |
|---|---|---|
| 1 | 1.00 | 1.00 |
| 2 | 0.80 | 0.85 |
| 3 | 0.70 | 0.79 |
| 4 | 0.65 | 0.75 |
| 6 | 0.57 | 0.72 |
| 9 | 0.50 | 0.70 |
Cables in contact with or enclosed in thermal insulation require significant derating. If the cable is totally surrounded by thermal insulation for more than 0.5 metres: Ci = 0.5 (a massive derating — the cable can carry only half its normal current). If the cable touches insulation on one side only: Ci = 0.89. This is one of the most commonly missed correction factors, particularly in loft spaces where insulation is being retrofitted around existing cables.
If the circuit is protected by a BS 3036 semi-enclosed (rewirable) fuse, apply Cf = 0.725. This accounts for the higher fusing factor of rewirable fuses compared to MCBs. For MCBs, RCBOs, and BS 88 HRC fuses, Cf = 1.0 (no correction needed). BS 3036 fuses are rarely installed in new work but may still be encountered in existing installations.
Enter the ambient temperature, number of grouped circuits, insulation conditions, and fuse type — Elec-Mate applies Ca, Cg, Ci, and Cf automatically.
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Start free trialLoad: Domestic ring final circuit, design current 20A (typical mixed load)
Protective device: 32A Type B MCB
Conditions: Clipped direct (Method C), 30C ambient, not grouped, not in thermal insulation
Correction factors: Ca = 1.0, Cg = 1.0, Ci = 1.0, Cf = 1.0
It = 32 / (1.0 x 1.0 x 1.0 x 1.0) = 32A
From Table 4D5A (XLPE 90°C T&E), 2.5mm² clipped direct (Method C) has Iz = 30A per conductor. Since this is a ring circuit, both legs share the load, so 2.5mm² is adequate. Voltage drop for a 50m ring with 20A load: VD = (18 x 20 x 25) / 1000 = 9.0V (within 11.5V limit). Result: 2.5mm² T&E.
A4:2026 — AFDD (Reg 421.1.7): BS 7671:2018+A4:2026 Regulation 421.1.7 recommends the installation of arc fault detection devices (AFDDs) on AC final circuits of a fixed installation to mitigate the risk of fire from arc fault currents. For domestic ring final and radial circuits, consider AFDD requirements at the design stage. Note the wording is recommendatory rather than mandatory.
Load: Kitchen socket radial, design current 18A
Protective device: 20A Type B RCBO
Conditions: Enclosed in insulated wall (Method A) for 3m, then clipped direct — use Method A (worst case), 30C ambient, grouped with 1 other circuit, not in thermal insulation
Correction factors: Ca = 1.0, Cg = 0.80, Ci = 1.0, Cf = 1.0
It = 20 / (1.0 x 0.80 x 1.0 x 1.0) = 25A
From Table 4D5A, 2.5mm2 T&E Method A has Iz = 20A — not enough. 4mm2 T&E Method A has Iz = 27A — sufficient (27 ≥ 25). Result: 4mm2 T&E.
Load: First-floor lighting, design current 5A
Protective device: 6A Type B MCB
Conditions: Method C where clipped to joists, but touching loft insulation on one side for 6m. 35C in loft space. Grouped with 1 other circuit.
Correction factors: Ca = 0.94, Cg = 0.80, Ci = 0.89, Cf = 1.0
It = 6 / (0.94 x 0.80 x 0.89 x 1.0) = 6 / 0.669 = 8.97A
From Table 4D5A, 1.0mm² T&E Method C has Iz = 16A — sufficient. 1.5mm² also works with Iz = 20A. Voltage drop for 1.5mm² with 5A over 18m: VD = (29 x 5 x 18) / 1000 = 2.61V (within 6.9V lighting limit). Result: 1.5mm² T&E (standard for lighting).
A4:2026 — RCD on domestic lighting (Reg 411.3.4): BS 7671:2018+A4:2026 Regulation 411.3.4 requires that, within domestic (household) premises, AC final circuits supplying luminaires shall be provided with additional protection by an RCD with a rated residual operating current not exceeding 30 mA. A 6A MCB alone is no longer sufficient for a new domestic lighting circuit — an RCBO or RCD upstream is mandatory for compliance with A4:2026.
After selecting the cable based on current-carrying capacity and voltage drop, there is one final check: verifying that the cable can withstand fault current without damage. This uses the adiabatic equation:
k2S2 ≥ I2t
k = Cable factor (115 for PVC/copper line conductor, 143 for PVC/copper CPC)
S = Cross-sectional area of the conductor in mm2
I = Prospective fault current in amps
t = Disconnection time of the protective device in seconds
If k2S2 is greater than or equal to I2t, the cable can withstand the fault current. If not, the cable must be upsized. In practice, this check rarely fails for domestic installations because the prospective fault current is relatively low and the disconnection times are fast. However, it is an essential check on commercial and industrial installations where fault levels can be significantly higher.
Site verification — Zs and the 0.8 factor: When verifying earth fault loop impedance (Zs) on site during initial verification, the measured value must satisfy Zs(measured) < 0.8 × the maximum tabulated Zs value. The 0.8 multiplier (per BS 7671 Reg 411.4.4 / 411.5.4 and Reg 253.0) accounts for the fact that conductor resistance increases with temperature — a cable at full load is hotter than at the ambient conditions when measured. Measured Zs values that pass at ambient but exceed 80% of the tabulated maximum will fail under load; always apply the 0.8 factor before recording compliance.
For more detail, see the adiabatic equation calculator guide.
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Try it free for 7 daysAppendix 4 of BS 7671 contains the current-carrying capacity and voltage drop tables for all standard cable types and installation methods. Understanding which table to use is the first step in using them correctly.
Start with Table 4A2: Before reading any current-carrying capacity table, consult Table 4A2 (referenced in BS 7671 Reg 521.3 and Reg 125.8). Table 4A2 maps each physical installation method to the correct reference method and current-carrying capacity table — for example, T&E clipped direct maps to Reference Method C and Table 4D1A (PVC 70°C) or 4D5A (XLPE 90°C); singles in conduit on a wall map to Reference Method B and Table 4D4A. Selecting the wrong table — even with the right reference method — will give incorrect Iz values.
Each current-carrying capacity table has multiple columns corresponding to different reference methods. Ensure you read the correct column for your installation method. The voltage drop columns in each table provide the mV/A/m values for both single-phase and three-phase circuits.
Follow this process to size any cable to BS 7671:2018+A4:2026 using the Appendix 4 tables and correction factors.
Determine the design current (Ib) from the load. For single-phase: Ib = P / (V x cos phi). For three-phase: Ib = P / (root 3 x VL x cos phi). For domestic circuits with known ratings (32A ring, 20A radial), use the expected maximum load current.
Choose a protective device with rated current In greater than or equal to Ib. Standard MCB ratings: 6, 10, 16, 20, 25, 32, 40, 50, 63A. Select the appropriate type: Type B for general, Type C for motors, Type D for high inrush loads.
Determine the applicable correction factors: Ca (ambient temperature from Table 4B1), Cg (grouping from Table 4C1), Ci (thermal insulation — 0.5 if fully enclosed, 0.89 one side), and Cf (0.725 for BS 3036 fuses, 1.0 for MCBs). Calculate It = In / (Ca x Cg x Ci x Cf).
Find the correct Appendix 4 table for your cable type and installation method (reference method A through G). Select a cable with tabulated current-carrying capacity Iz greater than or equal to It. Use the worst-case reference method if the cable route includes multiple installation methods.
Calculate voltage drop: VD = mV/A/m x Ib x L / 1000. Compare against 3% for lighting (6.9V) or 5% for other circuits (11.5V from 230V). If voltage drop exceeds the limit, increase the cable size. Then check the adiabatic equation (k2S2 >= I2t) to verify the cable can withstand fault current. If both checks pass, the cable is correctly sized.
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