How to Size Cables to BS 7671
The complete step-by-step guide to cable sizing using BS 7671:2018+A4:2026 Appendix 4 tables. Current carrying capacity, correction factors, voltage drop verification, and fault current withstand — with worked examples.
The 6-Step Cable Sizing Process
Determine the Design Current (Ib)
The design current is the current the cable must carry in normal service. Calculate it from the load:
Single-phase: Ib = P ÷ (V × cosφ)
Three-phase: Ib = P ÷ (√3 × VL × cosφ)
For a 7.4kW electric shower on a 230V single-phase supply (power factor 1.0): Ib = 7,400 ÷ (230 × 1.0) = 32.2A
Select the Protective Device (In)
Choose a protective device with rated current In ≥ Ib. Standard MCB ratings: 6, 10, 16, 20, 25, 32, 40, 50, 63A.
For our 32.2A shower: the next standard rating up is 40A Type B MCB.
Tip: Select the MCB type based on the load. Type B for general circuits (trips at 3-5× rated current). Type C for motors (5-10×). Type D for transformers and high inrush loads (10-20×).
Apply Correction Factors & Calculate It
This is where most mistakes happen. You must account for conditions that reduce the cable's ability to dissipate heat:
It = In ÷ (Ca × Cg × Ci × Cc)
Ca — Ambient Temperature (Table 4B1): At 30°C = 1.0. At 35°C = 0.94. At 40°C = 0.87. At 45°C = 0.79.
Cg — Grouping (Table 4C1): 2 circuits = 0.80. 3 circuits = 0.70. 4 circuits = 0.65. 6 circuits = 0.57. 9 circuits = 0.50.
Ci — Thermal Insulation: Cable touching insulation on one side = 0.89. Fully enclosed in insulation (over 0.5m) = 0.50.
Cc — Semi-enclosed Fuses: 0.725 for BS 3036 fuses. 1.0 for MCBs and RCBOs (no correction needed).
For our shower: installed alone (Cg = 1.0), in a 35°C loft (Ca = 0.94), touching insulation one side (Ci = 0.89), MCB protection (Cc = 1.0):
It = 40 ÷ (0.94 × 1.0 × 0.89 × 1.0) = 40 ÷ 0.8366 = 47.8A
Select Cable from Appendix 4
Find a cable with current-carrying capacity Iz ≥ It (47.8A) from the correct table for your installation method and cable type.
For twin and earth in an enclosed space (Reference Method A), from Table 4D5A column 6:
| Cable Size | Iz (Method A) | Suitable? |
|---|---|---|
| 6mm² | 32A | No (32 < 47.8) |
| 10mm² | 43A | No (43 < 47.8) |
| 16mm² | 57A | Yes (57 ≥ 47.8) |
Result: 16mm² twin and earth for this installation. However, if using Reference Method C (clipped direct), 10mm² may suffice as Iz = 52A for that method.
Common Mistake: Using the wrong installation method. Reference Method A (enclosed in insulated wall) has much lower ratings than Method C (clipped direct). Check the actual route the cable will take — use the worst-case method for any portion of the run.
Verify Voltage Drop
Check the cable selection meets BS 7671 voltage drop limits: 3% for lighting (6.9V from 230V) and 5% for power (11.5V from 230V). Use the cable sizing calculator to automate these checks.
VD = mV/A/m × Ib × L ÷ 1000
For 10mm² T&E (clipped direct), the mV/A/m value from Appendix 4 is 4.4. With Ib = 32.2A and a 12m cable run:
VD = 4.4 × 32.2 × 12 ÷ 1000 = 1.7V (0.74% of 230V)
1.7V is well within the 5% power limit (11.5V). Compliant.
Verify Fault Current Withstand
The final check ensures the cable can withstand a short circuit without damage. Use the adiabatic equation:
k²S² ≥ I²t
k = 115 (PVC/copper line conductor) | S = cable CSA in mm² | I = prospective fault current | t = disconnection time
For 10mm² cable, fault current 3kA, 40A MCB disconnecting in 0.1s:
k²S² = 115² × 10² = 13,225 × 100 = 1,322,500
I²t = 3,000² × 0.1 = 9,000,000 × 0.1 = 900,000
1,322,500 ≥ 900,000 Compliant
Worked Example Summary
Load: 7.4kW electric shower
Design current (Ib): 32.2A
Protective device: 40A Type B MCB
Conditions: 35°C loft, touching insulation one side
Required It: 47.8A
Cable selected: 10mm² T&E (clipped direct, Iz = 52A)
Voltage drop: 1.7V (0.74%) — compliant
Fault withstand: k²S² > I²t — compliant
Skip the Manual Calculations
Elec-Mate's cable sizing calculator does all 6 steps in seconds. Every BS 7671 table is built in — just enter your load, conditions, and cable run length.
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Appendix 4 current-carrying capacity tables, correction factor tables, and mV/A/m values — no flipping through the brown book.
Automatic Derating
Enter ambient temperature, grouping, and insulation conditions. The calculator applies Ca, Cg, Ci, and Cc automatically.
Voltage Drop Verification
Checks your cable selection against BS 7671 voltage drop limits: 3% for lighting (6.9V) and 5% for power (11.5V) from a 230V supply.
Fault Current Check
Verifies the cable can withstand fault current using the adiabatic equation (k²S² ≥ I²t) so your design is fully compliant.
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Frequently Asked Questions
What is the cable sizing process to BS 7671?
Cable sizing to BS 7671 follows four steps: (1) Determine the design current (Ib) from the load. (2) Select a protective device with rated current In ≥ Ib. (3) Calculate the tabulated current rating It by dividing In by all applicable correction factors (ambient temperature, grouping, thermal insulation, semi-enclosed fuse). (4) Select a cable from BS 7671 Appendix 4 with current-carrying capacity Iz ≥ It. Then verify voltage drop and fault current withstand. Elec-Mate’s cable sizing calculator does all of this automatically with every BS 7671 table built in.
What are the BS 7671 correction factors for cable sizing?
There are four correction factors: Ca (ambient temperature — from Table 4B1, e.g. 0.87 at 40°C for 70°C PVC), Cg (grouping — from Table 4C1, e.g. 0.70 for 3 circuits touching on a surface), Ci (thermal insulation — 0.5 if fully surrounded, 0.89 if one side only), and Cc (semi-enclosed fuse factor — 0.725 for BS 3036 fuses). These multiply together: It = In ÷ (Ca × Cg × Ci × Cc).
What are the voltage drop limits in BS 7671?
BS 7671 limits voltage drop to 3% for lighting circuits and 5% for all other circuits, measured from the origin of the installation. From a 230V single-phase supply, that’s 6.9V for lighting and 11.5V for power. For three-phase 400V supplies, it’s 12V for lighting and 20V for power. Voltage drop is calculated as: VD = mV/A/m × Ib × L ÷ 1000, where mV/A/m comes from Appendix 4 tables.
How do I check fault current withstand for a cable?
Use the adiabatic equation: k²S² ≥ I²t, where k is the cable factor (115 for PVC/copper line conductor, 143 for PVC/copper CPC), S is the cross-sectional area in mm², I is the prospective fault current in amps, and t is the disconnection time of the protective device in seconds. If k²S² is greater than or equal to I²t, the cable can withstand the fault. Elec-Mate calculates this automatically.
What cable size do I need for a 32A ring circuit?
A standard domestic ring final circuit uses 2.5mm² twin and earth cable with a 32A Type B MCB. This is suitable for a maximum floor area of 100m². The cable has a current-carrying capacity of 27A per leg (Reference Method C), but because it’s a ring, both legs share the load. Always check voltage drop for longer cable runs and apply derating factors if cables are grouped or in thermal insulation.
Do I need to derate cables in thermal insulation?
Yes. If a cable is enclosed in thermal insulation for more than 0.5m, you must apply correction factor Ci. For cables totally surrounded by thermal insulation: Ci = 0.5 (a massive derating). For cables touching insulation on one side only: Ci = 0.89. This is one of the most commonly missed derating factors and can lead to undersized cables overheating. BS 7671 Regulation 523.9 covers this requirement.
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