BS 7671 Cable Sizing Calculator — 18th Edition
Cable size, protective device and volt drop to BS 7671, worked through the way the standard does it — reference method first, then the four correction factors, then Ib ≤ In ≤ Iz.
The circuit
How and where it is installed
It = In / (Ca × Cg × Ci)
= 32 / (1.00 × 1.00 × 1.00)
= 32 / 1.000 = 32.0 A
4 mm² in method C is 36.0 A
Iz = 36.0 × 1.000 = 36.0 A
Ib 28 A ≤ In 32 A ≤ Iz 36.0 A — the rule the whole method exists to satisfy.
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BS 7671 Cable Sizing Calculator — 18th Edition · BS 7671 18th Ed · Appendix 4 · EnergyCalcHQ · energycalchq.com
Preliminary calculation. The figures behind it are representative values for the stated conditions, not a substitute for the current edition of the standard or the manufacturer's published data. Verify before issuing for construction. Not a substitute for a qualified engineer or a protection study.
For page numbers, keep Headers and footers ticked under More settings in the print dialog.
Ib ≤ In ≤ Iz, and everything else is detail
One inequality carries the whole method. The design current must not exceed the device rating, and the device rating must not exceed what the cable can actually carry where it is installed.
Ib ≤ In ≤ IzIb is the load. In is the protective device, chosen at or above Ib from the standard ratings. Iz is the cable's tabulated capacity after the correction factors — and that last word is where designs go wrong.
BS 7671 works the inequality backwards. Rather than derating a cable and checking it, you divide the device rating up by the factors to get the tabulated capacity you need, then look for a cable that has it:
It ≥ In / (Ca × Cg × Ci × Cf)This is the opposite arithmetic to the NEC method, which multiplies an ampacity down. The two reach similar places, and mixing them produces answers that are wrong in ways nobody notices.
Reference method comes first
Before any factor is applied, the standard asks how the cable is installed — and the answer moves the capacity more than anything else on the page. A 2.5 mm² twin-and-earth is 18.5 A in conduit in an insulated wall and 27 A clipped direct. Same cable, same standard, forty per cent apart.
- A — in conduit in an insulated wall. The harshest common method; the heat has nowhere to go.
- B — in conduit or trunking on a wall. The usual commercial case.
- C — clipped direct. Open to air on one side.
- E — free air, multicore. Tray or ladder, spaced off the wall.
- F — free air, single-core touching.
If a cable can be clipped instead of buried in a wall, that choice is usually worth a size. It is the cheapest derating you will ever recover.
The four factors
They multiply together, and the combined figure is often harsher than people expect — three of them at 0.8 is already 0.51.
- Ca — ambient. Tables assume 30 °C. A 40 °C plant room costs 13 %, a 50 °C one costs 29 %.
- Cg — grouping. Two circuits bunched take 0.8, six take 0.57. Count what actually shares the enclosure over its length, not what is in the drawing.
- Ci — thermal insulation. Regulation 523.9, and the one that catches domestic work. A cable through 100 mm of loft insulation keeps 81 % of its capacity; totally surrounded, half. A 30 A circuit that needs 4 mm² clipped needs 16 mm² buried in insulation — try it in the calculator.
- Cf — BS 3036 rewireable fuse. An extra 0.725, because those fuses run hot and blow imprecisely. Rare in new work, common in anything not rewired since the eighties, and easy to forget when you meet one.
Volt drop
Regulation 525.202: 3 % for lighting, 5 % for everything else, of the nominal voltage. Lighting gets the tighter figure because of what happens either side of the limit — a lamp visibly dims where a motor merely runs a little warmer.
ΔU = mV/A/m × Ib × L / 1000On long runs it is usually volt drop, not capacity, that sets the size. The calculator says which of the two governed, and gives the longest run the chosen cable supports at your design current — often more useful than the drop figure itself, because it tells you how much slack you have.
Capacities by reference method
70 °C thermoplastic insulated, copper. Blank cells are combinations the standard does not tabulate.
| mm² | A | B | C | E | F | mV/A/m 1φ | 3φ |
|---|---|---|---|---|---|---|---|
| 1 | 11 | 13 | 15 | 17 | — | 44 | 38 |
| 1.5 | 14 | 16.5 | 19.5 | 22 | — | 29 | 25 |
| 2.5 | 18.5 | 23 | 27 | 30 | — | 18 | 15 |
| 4 | 25 | 30 | 36 | 40 | — | 11 | 9.5 |
| 6 | 32 | 38 | 46 | 51 | — | 7.3 | 6.4 |
| 10 | 43 | 52 | 63 | 70 | — | 4.4 | 3.8 |
| 16 | 57 | 69 | 85 | 94 | — | 2.8 | 2.4 |
| 25 | 75 | 90 | 112 | 119 | 131 | 1.75 | 1.5 |
| 35 | 92 | 111 | 138 | 148 | 162 | 1.25 | 1.1 |
| 50 | 110 | 133 | 168 | 180 | 196 | 0.93 | 0.8 |
| 70 | 139 | 168 | 213 | 232 | 251 | 0.63 | 0.55 |
| 95 | 167 | 201 | 258 | 282 | 304 | 0.47 | 0.41 |
| 120 | 192 | 232 | 299 | 328 | 352 | 0.38 | 0.33 |
| 150 | 219 | 258 | 344 | 379 | 406 | 0.3 | 0.26 |
| 185 | 248 | 294 | 392 | 434 | 463 | 0.25 | 0.21 |
| 240 | 291 | 344 | 461 | 514 | 546 | 0.19 | 0.165 |
| 300 | 334 | 394 | 530 | 593 | 629 | 0.155 | 0.136 |
| 400 | — | — | 634 | 715 | 754 | 0.125 | 0.11 |
What this does not do
- Earth fault loop impedance. Zs must be low enough for the device to disconnect in time — Regulation 411.4. A cable that passes capacity and volt drop can still fail this, and it is a separate calculation.
- The adiabatic check on the CPC. Covered by the earth conductor calculator.
- Ring final circuits. A ring is not two radials and is not sized this way.
- Diversity. Ib is what you supply, not the connected load.
Which calculator to use
This one for the UK, Ireland and jurisdictions that adopt BS 7671. The NEC calculator for the United States and Canada, where the arithmetic runs the other way and the terminal temperature limit governs. The IS 732 calculator for India, which follows the same IEC parent as BS 7671 but with its own tables and 415 V defaults.
Questions people ask
- Why does BS 7671 divide by the correction factors instead of derating the cable?
- Because it sizes from the protective device, not from the load. You take the device rating In and divide it up by the factors to get the tabulated capacity you need — It greater than or equal to In divided by Ca times Cg times Ci times Cf — then find a cable that has it. The NEC does the opposite arithmetic, multiplying a tabulated ampacity down. Both reach similar sizes; mixing the two produces answers that are wrong in ways nobody notices.
- What cable does a 32 A circuit need?
- It depends entirely on the reference method, and the spread is a full size. With no correction factors applied, 4 mm² ends up at 36 A clipped direct (method C) and carries the 32 A device comfortably. The same 4 mm² in conduit in an insulated wall (method A) is only 25 A, so that circuit needs 6 mm². Decide how the cable is installed before you look at any table.
- Does a cable in loft insulation really need to be that much bigger?
- Yes, and it is the factor that catches domestic work. Regulation 523.9: a cable through 100 mm of insulation keeps 81 per cent of its capacity, and one totally surrounded keeps about half. A 30 A circuit that needs 4 mm² clipped needs 16 mm² buried in insulation — two and a half sizes, from nothing but the insulation around it.
- What is the 0.725 factor, and when do I need it?
- It is Cf, and it applies only where the circuit is protected by a BS 3036 rewireable fuse. Those fuses run hot and blow imprecisely, so the standard asks for an extra 27.5 per cent of cable capacity to cover it. You will not meet one in new work, but you will meet it in anything not rewired since the eighties, and it is easy to forget exactly when it matters most.
- Is the volt drop limit 3 per cent or 5 per cent?
- Regulation 525.202: 3 per cent for lighting, 5 per cent for everything else, measured as a percentage of the nominal voltage. Lighting gets the tighter figure because of what the reader notices — a lamp visibly dims where a motor merely runs a little warmer. The limit covers the whole path from the origin of the installation, so a submain that has already used 2 per cent leaves the final circuit 3 per cent of a 5 per cent allowance.