Size cables correctly to the 18th Edition IET Wiring Regulations. Automatic correction factors, voltage drop calculation, fault current verification, and all Appendix 4 tables built in.
Professional cable sizing with BS 7671 compliance validation
The cable is sized on In, not on Ib (Reg 433.1.1: Ib ≤ In ≤ Iz). A BS 3036 semi-enclosed fuse also applies Cf = 0.725 per Appendix 4 §5.1.1(c)(i).
Standard: 30°C
Affects current rating
Lighting: 3%, Power: 5%
1.0 = 100% simultaneous load
Voltage drop, App 4 §6.2. Resistive 1.0, inductive 0.8-0.9
Enter Ze to check Zs against BS 7671 Tables 41.2–41.4 and the CPC against the adiabatic equation. The calculator steps the cable up a size if it fails.
Measured, or the declared maximum: TN-C-S 0.35 Ω, TN-S 0.80 Ω. For a sub-board, use Zs at that board.
Empty: Design Current (A), Cable Length (m), Ze or Zdb (Ω).
Cable sizing depends on multiple factors beyond current rating alone:
Always consult relevant electrical codes and standards for your specific application.
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Cable sizing is the process of selecting the correct cross-sectional area (CSA) of conductor for an electrical circuit. It is one of the most fundamental design decisions in any electrical installation. Get it wrong, and the consequences range from nuisance tripping and poor equipment performance to cable overheating, insulation failure, and fire.
BS 7671:2018 (the IET Wiring Regulations, 18th Edition) sets out the methodology for cable sizing in Section 523, with the current carrying capacity tables in Appendix 4 and the voltage drop data in Appendix 4, Section 6.4. The process involves calculating the design current, selecting a protective device, determining the installation method, applying correction factors for environmental conditions, selecting a cable with sufficient current carrying capacity, and then verifying voltage drop and fault current withstand.
The cable must be large enough to carry the full load current without its temperature exceeding the rated value for its insulation (70 degrees C for thermoplastic/PVC, 90 degrees C for thermosetting/XLPE). It must also limit voltage drop to within the permitted values, and it must be able to survive the thermal effects of a short circuit for the time it takes the protective device to disconnect.
For working electricians, cable sizing calculations are a daily requirement. Whether you are wiring a domestic kitchen circuit, running a submain to an outbuilding, or designing a three-phase distribution board for a commercial fit-out, you need to verify the cable size against BS 7671 before you install it. The Elec-Mate cable sizing calculator does this in seconds, right on your phone. Use it alongside the voltage drop calculator and the cable derating calculator for a complete circuit design check.
BS 7671 requires you to consider five key factors when selecting a cable size. Each factor can independently determine the minimum cable CSA, so all five must be checked. The governing factor (the one that requires the largest cable) determines the final cable size.
The maximum sustained current the circuit will carry in normal service. This is your starting point. For simple resistive loads, Ib = P / V. For motor loads, use the full load current. For discharge lighting, multiply lamp watts by 1.8. The protective device rating (In) must be equal to or greater than Ib.
The cable must have a tabulated current carrying capacity (Iz) that, after correction factors are applied, is at least equal to the protective device rating (In). The required tabulated value is calculated as It = In / (Ca x Cg x Ci x Cf). You then select the smallest cable from the Appendix 4 tables with Iz greater than or equal to It.
Regulation 525.202 requires that the voltage drop from the origin of the installation to any socket-outlet or fixed equipment terminal does not exceed the values in Appendix 4, Section 6.4. The standard limits are 3% for lighting circuits (6.9V on a 230V supply) and 5% for other circuits (11.5V on a 230V supply). Voltage drop is calculated using the mV/A/m values from Appendix 4, Section 6.4, multiplied by the design current and cable length.
Under short circuit conditions, the cable must be able to withstand the heat generated by the fault current for the duration it takes the protective device to disconnect. This is checked using the adiabatic equation from Regulation 434.5.1 (for live conductors) and 543.1 (for protective conductors): minimum CSA = square root of (I squared t) / k, where I squared t is the energy let-through of the device and k is a constant for the conductor and insulation type (115 for copper/PVC, 143 for copper/XLPE).
The total earth fault loop impedance (Zs) must be low enough for the protective device to operate within the required disconnection time (0.4 seconds for final circuits, 5 seconds for distribution circuits). A longer cable run increases the circuit impedance, so in some cases you may need to increase the cable size (or the CPC size) to achieve an acceptable Zs value. Note: the tabulated maximum Zs values in BS 7671 assume conductors at maximum operating temperature. On-site measured Zs values (taken at ambient) must be multiplied by the 0.8 factor from GN3 Appendix 3 before comparing to the tabulated limits — an installation signed off without this correction may fail when conductors reach full operating temperature.
The tabulated current carrying capacities in Appendix 4 of BS 7671 assume a set of reference conditions: an ambient temperature of 30 degrees C (for cables in air), a single circuit with no grouping, and no thermal insulation in contact with the cable. When real-world conditions differ from these reference conditions, correction factors must be applied.
If the ambient temperature is above 30 degrees C, the cable can carry less current. Tables 4B1 and 4B2 provide the correction factors. For example, at 40 degrees C with 70 degree C thermoplastic insulation, Ca = 0.87. At 35 degrees C, Ca = 0.94. For installations below 30 degrees C (e.g., underground at 20 degrees C), Ca is greater than 1, slightly increasing the permitted capacity.
When multiple circuits are grouped together (e.g., in trunking, conduit, or on a cable tray), they share heat and each cable can carry less current. Tables 4C1 through 4C5 provide grouping factors based on the number of circuits and the arrangement. For example, 3 circuits in a single layer on a tray gives Cg = 0.79, while 6 circuits in conduit gives Cg = 0.57.
Cables surrounded by thermal insulation cannot dissipate heat effectively. If a cable is totally surrounded by thermally insulating material for more than 0.5 metres, a derating factor of 0.5 applies (Regulation 523.9). For cables touching insulation on one side, reference method 100 (formerly 101/102) applies, which is built into the table values. The calculator selects the correct approach based on your input.
When the protective device is a BS 3036 semi-enclosed (rewirable) fuse, an additional factor of 0.725 must be applied. This is because rewirable fuses have a fusing factor of approximately 2 (they may not blow until the current reaches twice their rated value), so the cable must be rated higher to allow for this. This factor is not applied for MCBs, RCBOs, or cartridge fuses, which have tighter operating characteristics.
The way a cable is installed has a significant effect on its ability to dissipate heat, and therefore on its current carrying capacity. BS 7671 Table 4A2 defines a series of reference methods that cover the most common installation arrangements. When you use the cable sizing tables in Appendix 4, you must select the column that corresponds to your reference method.
The most common reference methods encountered in UK domestic and commercial installations are:
Choosing the wrong reference method is one of the most common cable sizing errors. The Elec-Mate calculator presents the reference methods with clear descriptions and illustrations, so you select the right one every time.
Everything you need to size cables to BS 7671 on your phone or tablet, with no internet connection required.
All current carrying capacity tables from Appendix 4 and voltage drop data from Appendix 4, Section 6.4 are embedded in the calculator. No need to carry the book.
Enter the ambient temperature, grouping arrangement, and insulation conditions. The calculator applies Ca, Cg, Ci, and Cf automatically.
Calculates voltage drop for your route length and design current. Flags results that exceed the 3% lighting or 5% general limit.
Verifies the selected cable can withstand the prospective fault current using the adiabatic equation from BS 7671 Regulation 434.5.1.
Supports all standard reference methods from Table 4A2 including clipped direct, conduit, trunking, cable tray, and underground ducts.
Covers thermoplastic (PVC) and thermosetting (XLPE/LSF) cables, singles, flat twin and earth, SWA, and flexible cables.
Cable sizing is just one of 70 electrical calculators. Also includes voltage drop, max demand, diversity, conduit fill, earth rod, and more.
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Follow these seven steps to select the correct cable size for any circuit, using the methodology set out in BS 7671:2018. The Elec-Mate calculator handles steps 2 through 7 automatically once you enter the circuit parameters.
Calculate the design current (Ib) of the circuit. For resistive loads, divide the power in watts by the supply voltage (Ib = P/V). For motor loads, use the full load current from the manufacturer. For discharge lighting, multiply the lamp wattage by 1.8 to account for control gear losses and harmonic currents.
Choose a protective device with a nominal rating (In) equal to or greater than the design current. The device must also be appropriate for the type of circuit — for example, a Type B MCB for resistive loads, Type C for motor or transformer inrush, or Type D for very high inrush currents.
Determine the reference method from BS 7671 Table 4A2 that matches how the cable will be installed. Common methods include Method A (conduit in insulated wall), Method B (conduit on wall), Method C (clipped direct), and Reference Method 100 for thermally insulating material on one side. The method determines which column of the current carrying capacity tables to use.
Apply all relevant correction factors. Ca for ambient temperature (from Table 4B1/4B2), Cg for grouping (from Table 4C1 to 4C5), Ci for thermal insulation (from Regulation 523.9), and Cf for semi-enclosed fuses (0.725). Calculate the required tabulated current carrying capacity: It = In / (Ca x Cg x Ci x Cf).
Look up the appropriate Appendix 4 table for your cable type (e.g., Table 4D5 for 70 degree C thermoplastic flat cable, or Table 4E1A for 90 degree C thermosetting singles). Find the smallest cable size with a tabulated Iz value equal to or greater than your calculated It value.
Calculate the voltage drop using the mV/A/m values from Appendix 4, Section 6.4 for your selected cable size and installation method. Multiply by design current and route length, then divide by 1000. Ensure the result does not exceed 3% for lighting or 5% for other circuits, as required by Regulation 525.202. If it exceeds the limit, increase the cable size and re-check.
Confirm the selected cable can withstand the let-through energy of the protective device under fault conditions. The adiabatic equation is: S = square root of (I squared t) / k, where S is the minimum conductor cross-sectional area in mm squared, I squared t is the energy let-through of the protective device, and k is a constant for the conductor and insulation type (from BS 7671 Regulation 434.5.1 for live conductors and Regulation 543.1 for protective conductors). The cable CSA must be at least equal to S.
A domestic ring final circuit supplying socket outlets in a kitchen, protected by a 32A Type B MCB, installed as flat twin and earth clipped direct (Reference Method C), with an ambient temperature of 30 degrees C and no grouping.
Ca = 1.0 (30 °C is the reference temperature), Cg = 1.0 (not grouped) and the device is an MCB, so Cf does not apply. Table 4D5, Reference Method C, gives 2.5 mm² twin and earth 27 A. That is below the 32 A device, but a ring is not sized like a radial. Regulation 433.1.204 deems a ring of 2.5 mm² copper on a 30 A or 32 A device to meet Regulation 433.1.1 when the cable's current-carrying capacity is at least 20 A. 27 A clears 20 A, so 2.5/1.5 mm² twin and earth is correct.
A three-phase submain to a workshop 45 m away, on an 80 A device with a 75 A design current, in multicore SWA clipped to a wall in a single layer beside two other circuits, at 35 °C. Ca = 0.94 (Table 4B1, 70 °C thermoplastic) and Cg = 0.79 (Table 4C1, item 2, three circuits). It ≥ 80 ÷ (0.94 × 0.79) = 107.7 A. Table 4D4A, Reference Method C, three- or four-core: 25 mm² is 102 A (too small), 35 mm² is 125 A. Volt drop from Table 4D4B: 1.1 mV/A/m × 75 A × 45 m ÷ 1000 = 3.7 V, 0.9% of 400 V, well inside 5%. 35 mm² SWA.
Where the earth fault check changes the answer: a 32 A Type B radial, 30 m of twin and earth clipped direct, on a TN-S supply with Ze 0.80 Ω. Current and volt drop give 4 mm² (37 A, 4.6% drop). But 4/1.5 mm² has (R1 + R2) of 16.71 mΩ/m (On-Site Guide Table I1) × 30 m × 1.20 = 0.60 Ω, so Zs = 1.40 Ω against a 1.37 Ω limit (Table 41.3). It fails. 6/2.5 mm² gives 10.49 × 30 × 1.20 = 0.38 Ω and Zs = 1.18 Ω, which passes, and its 2.5 mm² CPC is well above the 0.54 mm² the adiabatic equation asks for. Enter Ze in the calculator and it makes this step for you.
The calculator above works all of this through: the cable size and the Appendix 4 table it came from, volt drop in volts and per cent, and, when you enter Ze, Zs against Tables 41.2 to 41.4 and the CPC against the adiabatic equation.
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Free BS 7671 cable sizing calculator: enter load, length and installation method to get the cable size, volt drop and current rating. Works on any phone.
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