Cable sizing under BS 7671:2018+A4:2026 comes down to two checks. First, the cable must carry the load without overheating: the design current Ib must not exceed the device rating In, and In must not exceed the installed current-carrying capacity of the cable Iz, after applying correction factors for ambient temperature, grouping, thermal insulation and the type of fuse. Second, the voltage at the load must stay within the limits in Appendix 4: 3 per cent for lighting and 5 per cent for other circuits, which at 230 V is 6.9 V and 11.5 V. Table 4D5 gives the ratings for flat twin and earth by reference method, and the same table gives the mV/A/m figures for the volt drop sum.
- The overload rule is Ib ≤ In ≤ Iz: start from the load, pick the device, then find a cable whose installed rating covers the device.
- Iz = It × Ca × Cg × Ci × Cf, or equivalently It ≥ In ÷ (Ca × Cg × Ci × Cf).
- The reference method matters: 6 mm² twin and earth is 47 A clipped direct and 23.5 A buried in a stud wall's insulation.
- Volt drop = mV/A/m × Ib × length ÷ 1000, limited to 6.9 V for lighting and 11.5 V for other circuits.
- On the EIC the answer goes in the csa and reference method columns; on an EICR the inspector checks that device, cable and method still make sense together.
What do Ib, In and Iz mean?
Ib is the design current the circuit is intended to carry in normal service. In is the rated current of the protective device. Iz is the current-carrying capacity of the cable as installed, and It is the tabulated capacity from Appendix 4 before correction.
The rule is Ib ≤ In ≤ Iz. The load must not exceed the device, or it trips in normal use; the device must not exceed the cable, or an overload passes until the insulation is damaged. A companion requirement, that the current causing effective operation of the device must not exceed 1.45 times Iz, is met automatically by MCBs to BS EN 60898 when In ≤ Iz but not by BS 3036 rewireable fuses, which is why Cf exists.
Design current for a fixed load is power divided by voltage: an 8.5 kW shower at 230 V is 37 A. For socket circuits the load is taken as the device rating, since the designer has no control over what is plugged in.
What are the correction factors?
Correction factors reduce the tabulated rating to reflect the installation, and they multiply together.
| Factor | Accounts for | Source in BS 7671 | Typical values |
|---|---|---|---|
| Ca | Ambient temperature other than 30 °C | Table 4B1 | 1.00 at 30 °C; 0.94 at 35 °C; 0.87 at 40 °C for 70 °C thermoplastic |
| Cg | Grouping with other loaded circuits | Table 4C1 | 0.80 for 2 circuits bunched, 0.70 for 3; 0.85 and 0.79 for a single layer clipped to a wall |
| Ci | Thermal insulation in contact with the cable over part of its length | Table 52.2 | 0.88 at 50 mm; 0.78 at 100 mm; 0.63 at 200 mm; 0.51 at 400 mm; 0.50 at 500 mm or more |
| Cf | Rewireable fuse to BS 3036 as the protective device | Chapter 43 | 0.725 |
The working formula is It ≥ In ÷ (Ca × Cg × Ci × Cf). Look up It for the reference method and choose the first csa that meets it. Ci in Table 52.2 covers short lengths in contact with insulation; where flat cable runs in insulation for longer, methods 100 to 103 in Table 4D5 already build the effect in and Ci is not applied twice.
Grouping is the factor most often ignored on domestic work. Six twin and earth cables bunched through one hole in a joist are grouped for that length, and if they are all loaded at once the factor applies. BS 7671 lets the designer disregard grouping for cables that will not be loaded above 30 per cent of their grouped rating, but that is a judgement, not a default.
What are the reference methods?
The reference method describes how the cable is installed, which determines how well it sheds heat. Appendix 4 Table 4A2 lists them. For twin and earth the relevant ones are A (enclosed in conduit in a thermally insulated wall), B (enclosed in conduit or trunking on a wall), C (clipped direct or embedded in plaster), 100 (above a plasterboard ceiling, covered by thermal insulation not exceeding 100 mm), 101 (above a plasterboard ceiling, covered by thermal insulation exceeding 100 mm), 102 (in an insulated stud wall, touching the inner wall surface) and 103 (in an insulated stud wall, not touching the inner wall surface). Methods D to G cover buried cables and cables in free air and are used with the tables for armoured and single-core cables rather than 4D5.
Table 4D5: twin and earth ratings by method
Single-phase current ratings in amperes for 70 °C thermoplastic flat cable with protective conductor at 30 °C ambient, from Table 4D5 of BS 7671, with the volt drop figure in mV per amp per metre.
| csa (mm²) | 100 | 101 | 102 | 103 | A | B | C | mV/A/m |
|---|---|---|---|---|---|---|---|---|
| 1.0 | 13 | 10.5 | 13 | 8 | 11.5 | 13 | 16 | 44 |
| 1.5 | 16 | 13 | 16 | 10 | 14.5 | 16.5 | 20 | 29 |
| 2.5 | 21 | 17 | 21 | 13.5 | 20 | 23 | 27 | 18 |
| 4 | 27 | 22 | 27 | 17.5 | 26 | 30 | 37 | 11 |
| 6 | 34 | 27 | 35 | 23.5 | 32 | 38 | 47 | 7.3 |
| 10 | 45 | 36 | 47 | 32 | 44 | 52 | 64 | 4.4 |
| 16 | 57 | 47 | 63 | 42.5 | 57 | 69 | 85 | 2.8 |
Two things stand out. A 2.5 mm² cable is 27 A clipped direct but 13.5 A buried in a stud wall's insulation, the commonest reason a 32 A radial turns out to be undersized. And 6 mm² drops from 47 A to 23.5 A across the same change, which is why a shower cable that was fine when installed can be over-fused after a loft is insulated. The reduced cpc in twin and earth is verified separately by confirming Zs is low enough for the device to disconnect in time; the maximum Zs values guide covers that check.
How is volt drop calculated?
Voltage drop is the voltage lost in the cable between the origin and the load; too much dims lamps, slows motors and takes equipment outside its rated voltage. Appendix 4 of BS 7671 limits it, for an installation on a public low-voltage supply, to 3 per cent for lighting (6.9 V at 230 V) and 5 per cent for other uses (11.5 V), from the origin to the furthest point of the circuit. The calculation uses the mV/A/m figure from the cable table:
Volt drop (V) = (mV/A/m × Ib × L) ÷ 1000
where L is the one-way length in metres; the tabulated figure already accounts for the return conductor. For a ring final circuit with distributed load, the conventional On-Site Guide approach treats the load as concentrated at the mid-point, so each leg carries half the current over half the length and the result is the single-cable figure divided by four. Long runs to outbuildings and EV chargers are where volt drop, rather than current, usually decides the cable size.
Worked example 1: a 32 A ring final circuit
A kitchen ring on a 32 A Type B MCB, 2.5 mm² twin and earth, total ring length 60 m, clipped direct and, for a few metres, above the plasterboard ceiling under 100 mm of loft insulation.
Current rating. For a ring final circuit BS 7671 requires Iz of the 2.5 mm² conductors to be not less than 20 A, because current shares between the two legs. The worst method on the route is 100, rated 21 A, with no grouping or ambient correction. Pass. Through a fully insulated stud wall (method 103, 13.5 A) the ring would not comply.
Volt drop. Load at the mid-point: (18 × 32 × 60) ÷ 4 ÷ 1000 = 8.64 V, under 11.5 V. Pass. A 32 A ring much beyond 80 m starts to fail on volt drop while still passing on current. The ring final circuit testing guide covers proving the ring is intact once installed.
Worked example 2: an 8.5 kW shower on 6 mm² or 10 mm²
An 8.5 kW electric shower on a dedicated radial, 15 m from the consumer unit, up a stud wall and across a loft with 270 mm of insulation.
Design current. Ib = 8500 ÷ 230 = 37 A. The device is a 40 A Type B MCB, so In = 40 A and Ib ≤ In. (A 9.5 kW shower is 41.3 A and needs a 45 or 50 A device with the cable sized to match.)
Option 1: 6 mm². Under 270 mm of loft insulation the cable is method 101, rated 27 A; even under 100 mm or less (method 100) it is only 34 A. Neither reaches 40 A. Fail.
Option 2: 10 mm². Under 270 mm of insulation, method 101 gives 36 A, still below 40 A. The cable only passes if it is kept out of the insulation: clipped to the side of the joists above the insulation (treated as method C, 64 A) or covered by no more than 100 mm (method 100, 45 A). In the stud wall it must touch the inner face (method 102, 47 A) rather than sit within the insulation (method 103, 32 A). Pass, provided the installer controls the route; if it cannot be controlled, the next size up is needed.
Volt drop. 10 mm²: (4.4 × 37 × 15) ÷ 1000 = 2.4 V; 6 mm² would have been 4.1 V. Both well under 11.5 V.
Showers are run in 10 mm² in a modern house because of the loft insulation, not the 40 A MCB, and the route through that insulation has to be managed even at 10 mm². The bathroom brings further requirements under Section 701, including 30 mA RCD protection.
How does cable sizing appear on the EIC?
The Electrical Installation Certificate records the design on the schedule of circuit details: for each circuit, the reference method, the csa of the live conductors and of the cpc, and the protective device's type, rating and maximum permitted Zs. The designer is signing to say those columns together satisfy Ib ≤ In ≤ Iz and the volt drop limit. There is no column for the calculation, so keep a design record, a line per circuit showing Ib, In, method, correction factors and Iz, which is what a scheme assessor asks for when they pick a circuit at random. The schedule of test results then proves the installed circuit matches: R1+R2 in proportion to length and csa, and Zs within the maximum for the device.
What does an inspector check on an EICR?
On a periodic inspection the design is rarely available, so the inspector works backwards from the device rating, the cable marking, the route as far as it can be traced and the connected load, asking whether the device can still be relied upon to protect the cable.
| Finding | Typical code | Notes |
|---|---|---|
| 32 A MCB on a 2.5 mm² radial serving several sockets | C2 or C3 depending on load and method | Overload possible; C2 where the circuit supplies a kitchen or heaters |
| 6 mm² shower cable buried in loft insulation on a 40 or 45 A device | C2 | Cable rating well below device; sustained load |
| 2.5 mm² ring in a fully insulated stud wall | C3, or C2 if signs of heating | Iz below 20 A; check terminations for discolouration |
| Reduced cpc with Zs above maximum for the device | C2 | Disconnection time not achieved |
| Long run with lamps visibly dim under load | C3 | Not dangerous; improvement recommended |
Where the route cannot be traced, record a limitation rather than guessing; where the finding depends on the load, an FI is often more honest than a C2 until the load is confirmed.
How Certio helps
Certio produces the EICR, EIC, Minor Works certificate and PAT records on the BS 7671 model forms and turns them into branded PDFs. Photograph the consumer unit and the circuit schedule drafts itself; point the phone camera at a Megger MFT-X1 and the reading goes straight into the schedule of test results, or dictate it while your hands stay on the probes. Suggested observation wording with the regulation reference is there for you to check and confirm. The judgement stays yours; the typing goes. Starter is free for 7 days, then £5 a month.