Cable Sizing and Volt Drop: Ib, In, Iz and Appendix 4 Explained

Cable sizing under BS 7671:2018 comes down to two checks, and most domestic mistakes are failures of the second. First, the cable must carry the load without overheating: Regulation 433.1.1 requires the design current Ib not to exceed the device rating In, and In not to exceed the installed current-carrying capacity of the cable Iz after correction for ambient temperature, grouping, thermal insulation and fuse type. Second, the voltage at the load must stay within Appendix 4 Table 4Ab: 3 per cent for lighting and 5 per cent for everything else, which at 230 V is 6.9 V and 11.5 V. The single most important number to remember is that 6 mm² twin and earth is 47 A clipped direct and 23.5 A buried in the insulation of a stud wall.

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, or current setting, 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 any correction is applied.

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 for long enough to damage the insulation. There is a companion requirement in 433.1.1 that the current causing effective operation of the device, I2, must not exceed 1.45 × Iz. MCBs to BS EN 60898 and RCBOs to BS EN 61009 have I2 = 1.45 In, so that condition is met automatically when In ≤ Iz. A BS 3036 rewireable fuse has I2 = 2 In, and the only way to satisfy 1.45 × Iz is to make Iz at least In ÷ 0.725, which is where the factor Cf comes from (Regulation 433.1.202).

Design current for a fixed load is power divided by voltage: an 8.5 kW shower at 230 V is 37 A, a 3 kW immersion is 13 A, a 7.4 kW EV charger is 32 A. For socket circuits the design current is taken as the device rating, because the designer has no control over what gets plugged in. For a cooker circuit, diversity from the On-Site Guide is applied first; the cooker circuit and diversity guide at Cooker Circuit Design and Diversity: Sizing It Right works through that.

What are the correction factors?

Correction factors reduce the tabulated rating to reflect how the cable is actually installed, and they multiply together.

The working formula is It ≥ In ÷ (Ca × Cg × Ci × Cf). Look up It in the right table for the reference method and choose the first csa that meets it.

Two points trip people up. Table 52.2 is for a cable that touches thermal insulation over a short length, up to 0.5 m; 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 again on top. And grouping is the factor most often ignored on domestic work. Six twin and earth cables bunched through one notch in a joist are grouped for that length; if they could all be loaded at once, Table 4C1 applies. Appendix 4 lets the designer disregard a cable that will not carry more than 30 per cent of its grouped rating, which is a judgement to record, not a default to assume.

Ambient temperature is rarely a factor indoors in the UK, but a loft in summer or a plant room with a boiler in it can sit well above 30 °C, and Table 4B1 is designed to be used, not admired.

What are the reference methods?

The reference method describes how the cable is installed, which determines how well it sheds heat. Table 4A2 in Appendix 4 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, cable touching the inner plasterboard surface) and 103 (in an insulated stud wall, cable not touching the inner surface). Methods D (buried), E, F and G (free air) are used with the tables for armoured and single-core cables rather than 4D5.

Regulation 523.9 gives the back-stop: a cable totally surrounded by thermal insulation for 0.5 m or more is rated as if it were installed with a factor of 0.5, which is what method 103 effectively does.

Table 4D5: twin and earth ratings by method

Single-phase current ratings in amperes for 70 °C thermoplastic flat cable with protective conductor, 30 °C ambient, from Table 4D5 of BS 7671:2018, with the volt drop figure in millivolts per amp per metre.

Two things stand out. A 2.5 mm² cable is 27 A clipped direct but 13.5 A in a stud wall's insulation, which is 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 the day a loft is insulated. Always verify figures against your own copy of the current edition before signing a design.

How is volt drop calculated?

Voltage drop is the voltage lost in the cable between the origin and the load. Too much dims lamps, makes motors run hot and takes equipment outside its rated voltage. Regulation 525 points to Appendix 4 Table 4Ab, which limits it, for an installation supplied from a public low-voltage network, to 3 per cent for lighting (6.9 V at 230 V) and 5 per cent for other uses (11.5 V), measured from the origin of the installation to the furthest point of the circuit. Where the installation has its own private supply, the limits are 6 and 8 per cent.

The calculation uses the mV/A/m figure from the cable table:

Volt drop (V) = (mV/A/m × Ib × L) ÷ 1,000

L is the one-way length in metres; the tabulated figure already accounts for the return conductor. For three-phase circuits use the three-phase column of the table and compare against 3 or 5 per cent of 400 V, which is 12 V and 20 V line-to-line. The three-phase testing guide at Three-Phase Testing Basics for Domestic Electricians covers what you then measure.

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.

Appendix 4 section 6.2 also allows a refinement for cables running well below their rating: the tabulated mV/A/m assumes the conductor is at 70 °C, so a lightly loaded cable has lower resistance and a smaller real drop. Most designers ignore it on domestic work because it only helps, but it is there when a long run is marginal.

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 except for a few metres above the plasterboard ceiling under 100 mm of loft insulation.

Current rating. Regulation 433.1.204 allows a 30 or 32 A device on a 2.5 mm² ring provided Iz of each leg is 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. Run through a fully insulated stud wall (method 103, 13.5 A), the same ring would not comply, and neither would it bunched with two other loaded circuits through the insulation (21 × 0.70 = 14.7 A).

Volt drop. Load at the mid-point: (18 × 32 × 60) ÷ 4 ÷ 1,000 = 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. Once installed, the ring final circuit testing guide at Ring Final Circuit Testing: r1, rn, r2 and R1+R2 Explained shows how r1, rn and r2 prove it is a ring and not two radials.

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 = 8,500 ÷ 230 = 37 A. The device is a 40 A Type B MCB or RCBO, 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 passes only 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 plasterboard face (method 102, 47 A) rather than sit within the insulation (method 103, 32 A). Pass, provided the installer controls the route. If the route cannot be controlled, 16 mm² (method 101, 47 A) is the honest answer.

Volt drop. 10 mm²: (4.4 × 37 × 15) ÷ 1,000 = 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 because of the 40 A MCB, and the route through that insulation still has to be managed. The bathroom brings further requirements under Section 701, including 30 mA RCD protection, covered at Bathroom Zones, IP Ratings and Bonding: Section 701 Explained, and the electric shower certificate guide at Electric Shower Installation Certificate: EIC or Minor Works? covers what paperwork the job needs.

Worked example 3: a 7.4 kW EV charger on SWA

A 7.4 kW single-phase charger on a detached garage wall, fed by two-core steel wire armoured cable buried direct in the garden, 40 m from the consumer unit, on a 32 A Type B device.

Design current. Ib = 7,400 ÷ 230 = 32 A, In = 32 A.

Current rating. Table 4D4A gives ratings for 70 °C thermoplastic multicore armoured cable. Two-core figures, from BS 7671:2018:

Method D ratings in the 18th Edition assume a soil thermal resistivity of 2.5 K·m/W, which is deliberately pessimistic for typical UK ground; the earlier assumption of 1.2 K·m/W gave higher buried ratings and is why older designs show smaller buried cables. 6 mm² at 47 A covers 32 A with room for a grouping factor. Pass.

Volt drop. 6 mm²: (7.3 × 32 × 40) ÷ 1,000 = 9.3 V. Pass, but only just, and a charger draws its full 32 A for hours at a time. At 49 m the same cable hits 11.5 V. 10 mm²: (4.4 × 32 × 40) ÷ 1,000 = 5.6 V, comfortable, and the cable will also carry a future 32 A socket in the garage without a re-pull. Most designers go to 10 mm² beyond about 35 m.

Amendment 4:2026 adds new installation methods 72 and 73 and a new buried reference method D2 to Tables 4A2A and 4D4A, per SELECT's summary of the amendment, so a buried run designed to A4 should be checked against the A4 table rather than the figures above. The EV guide at EV Charger Installation Certificate: Section 722 and the EIC covers the Section 722 requirements for the charger itself, and the outbuilding guide at Outbuilding and Garden Supply: SWA, TT or PME Earth? covers the TT versus PME decision, which often decides whether the armour can be the cpc.

How do I check the cpc and fault current?

Overload is one duty of the cable; withstanding a fault until the device clears it is the other. Regulation 434.5.2 requires that the fault current does not raise the conductor above its limit temperature, checked with the adiabatic equation:

S = √(I²t) ÷ k

where S is the minimum csa in mm², I²t is the energy let through by the device (from the manufacturer's data, or I² × t using the disconnection time from the device curve) and k is the conductor constant from Tables 54.2 to 54.6.

For the line conductor, a circuit that satisfies Ib ≤ In ≤ Iz with an MCB will normally satisfy 434.5.2 as well, because the MCB's let-through at the prospective fault current is small. The check that matters on twin and earth is the reduced cpc. A 2.5 mm² cable has a 1.5 mm² cpc, a 6 mm² has 2.5 mm², a 10 mm² has 4 mm². Table 54.7 would require a cpc equal to the line conductor up to 16 mm², so flat cable never meets it and the cpc must be verified by calculation. With k = 115 (Table 54.3, copper cpc incorporated in a 70 °C thermoplastic cable) and a Type B 32 A MCB clearing in 0.1 s at, say, 1,000 A fault current, S = √(1,000² × 0.1) ÷ 115 = 2.75 mm², which is why a 2.5 mm² cable with a 1.5 mm² cpc on a 32 A device relies on either a faster disconnection (higher fault current, magnetic trip at 0.01 s gives S = 0.87 mm²) or the manufacturer's I²t data rather than the 0.1 s convention. In practice the design is verified by confirming Zs at the far end is below the maximum for the device, which guarantees the magnetic trip operates; the maximum Zs guide at Maximum Zs Values BS 7671: Tables, 80% Rule and Ze tabulates those limits and the high Zs troubleshooting guide at High Zs Readings: Causes, Checks and How to Fix Them covers what to do when they are not met.

Prospective fault current at the origin also has to be recorded and compared with the breaking capacity of the devices, 6 kA for most domestic MCBs; see What is prospective fault current (PFC)? Explained.

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, the type and rating of the protective device, and the maximum permitted Zs. The designer signs 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, reference method, Ca, Cg, Ci, It, Iz and volt drop. That is exactly what a scheme assessor asks for when they pick a circuit at random, and the QS sign-off guide at Qualified Supervisor (QS) Sign-Off Explained explains why the qualified supervisor is the one who has to defend it. The schedule of test results then proves the installed circuit matches the design: R1+R2 in proportion to length and csa, and Zs within the maximum for the device. The column-by-column guide at How to Fill In a Schedule of Test Results, Column by Column shows where each figure goes, and the EIC guide at Electrical Installation Certificate (EIC) Explained covers the rest of the form.

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.

Where the finding depends on the load, an FI is often more honest than a C2 until the load is confirmed. The observation wording guide at How to Write EICR Observations That Stand Up and the ten most common EICR fails at EICR Fails: The Ten Most Common C1 and C2 Observations both show how to phrase these so a landlord or the next electrician can act on them, and the loft and cellar guide at Loft and Cellar Wiring: EICR Observations and Codes deals with the insulation problem where it is usually found.

Common cable sizing mistakes

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 £4.99 a month.

Written by the Certio team. Last checked 29 September 2026.