How cable voltage drop works

When to use this: sizing a sub-main to an outbuilding, running power to a machine at the far end of a workshop, wiring a 12 V solar or automotive system, or working out why the motor at the end of a long run keeps tripping on undervoltage.

Every conductor has resistance. Push current through it and some of the supply voltage is consumed by the cable itself rather than delivered to the load. The formula is just Ohm’s law applied to the cable:

ΔV = 2 × L × I × R′  (DC and single-phase)
ΔV = √3 × L × I × R′  (balanced three-phase)

L is the one-way run in metres, I the design current in amps, and R′ the resistance of one conductor per metre at its operating temperature. The factor of 2 is there because the current has to come back — the loop is twice the cable length.

Three-phase uses √3 because in a balanced system the three phase currents are 120° apart and largely cancel in the return path. The practical result: the same cable carries the same current with about 13% less drop on three-phase than single-phase.

The BS 7671 method: Appendix 4 and mV/A/m

If you work to BS 7671 you almost certainly do not evaluate ΔV = 2LIR′ on site. You look up a number. Appendix 4 tabulates voltage drop for every cable type and conductor size as a single figure in (mV/A/m) — millivolts of drop, per ampere of design current, per metre of run — and you apply it like this:

ΔV = (mV/A/m) × Ib × L ÷ 1000

That one number has already absorbed the conductor resistance, the operating temperature and the go-and-return path. It is the same calculation as the formula above; the table just does the arithmetic for you. A 2.5 mm² two-core 70 °C thermoplastic copper cable is 18 mV/A/m, so 20 A over 30 m is 18 × 20 × 30 ÷ 1000 = 10.8 V.

The calculator shows 17.7 rather than 18 for that cable, and 10.6 V rather than 10.8 V. That is not a disagreement — Appendix 4 rounds its figures to two significant figures, and 17.7 is what the underlying IEC 60228 conductor resistance gives. Use the table value if you are writing it on a certificate; the difference is well inside the tolerance the conductor is manufactured to.

With BS 7671 selected, this calculator reports the mV/A/m figure alongside the volts, so you can check it directly against the Appendix 4 table for your cable type rather than taking our word for the answer. The figures it produces reproduce the published table to within 2% for every size up to 16 mm².

Above 16 mm², the table has a second column — and we do not model it

For 25 mm² and larger, Appendix 4 splits the tabulated figure into a resistive component r and a reactive component x, and the value you must design to is z = √(r² + x²). Reactance depends on the cable’s construction and conductor spacing; it is not derivable from conductor resistance, and the tabulated values are BSI’s.

This calculator models r only. Above 16 mm² it says so on screen rather than quietly returning a number that is too small — which is the failure mode that matters, because a sub-main is exactly where the reactive term stops being negligible. For those sizes, take z from the table for your cable type and multiply it out yourself.

The limits

None of these is a hard legal limit — BS 7671 Appendix 4 is informative, and the NEC figures are informational notes. In practice they are what an inspector expects and what equipment manufacturers assume, so treat them as binding unless you have a documented reason not to.

The BS 7671 private supply row catches people out. If the installation is fed from a generator, a private transformer or an off-grid system rather than the DNO’s network, the permitted drop roughly doubles — because the voltage at the origin is under your control rather than sitting somewhere inside a statutory tolerance you cannot see.

Lighting gets the tighter limit because incandescent output falls off sharply with voltage, and because flicker from voltage sag on other circuits is most visible in lighting.

Why low-voltage systems suffer most

The limit is a percentage, and that is what makes low-voltage DC so unforgiving. Consider the same 1 V of drop:

It compounds, too. For the same power, a 12 V system carries roughly 19 times the current of a 230 V one, and drop is proportional to current. This is why a 12 V solar installation uses conductors that look absurd for the wattage involved, and why the industry keeps moving to 48 V wherever it can.

Temperature matters more than people expect

Copper resistance rises about 0.393% per kelvin. The 20 °C figure in most tables is a reference condition, not an operating one:

A design that lands at exactly 5% at 20 °C is really at 6% once the cable warms up. This is not a refinement you can skip: BS 7671 Appendix 4 and IEC 60364 Annex G both tabulate at the insulation’s rated temperature, so a calculation done at 20 °C is not comparable with the limit it is being checked against. That is why this page asks for the insulation type rather than a temperature — 70 °C for thermoplastic, 90 °C for thermosetting — and keeps the free-text temperature for the specific case of a cable you know is lightly loaded.

Worked examples

Garage sub-main: 20 A over 30 m in 4 mm²

4 mm² 70 °C thermoplastic copper = 11.0 mV/A/m
ΔV = 11.0 × 20 × 30 ÷ 1000 = 6.62 V
6.62 / 230 = 2.88%

Inside the 5% limit with room to spare. Note the cable is also dissipating 20 × 6.62 = 132 W as heat when fully loaded — over a year of continuous use that is about 1,160 kWh, which is usually a stronger argument for the larger cable than the drop limit is.

The same run in 2.5 mm²

2.5 mm² = 17.7 mV/A/m
ΔV = 17.7 × 20 × 30 ÷ 1000 = 10.6 V = 4.63%

2.5 mm² is rated for 20 A in free air, so it passes on current-carrying capacity, and it scrapes inside a 5% limit. But it fails a 3% lighting limit outright, and it wastes 60% more energy as heat. This is the case where current rating and voltage drop give different answers — and voltage drop is the one that should decide.

It is also the example worth checking a nominal-area calculator against. Divide resistivity by 2.5 mm² and you get 9.9 V, or 4.3% — comfortably inside 5%, and wrong. The regulation works from the IEC 60228 conductor resistance, and by that measure the same cable is at 4.63%.

12 V DC over 10 m at 10 A in 2.5 mm²

ΔV = 17.7 × 10 × 10 ÷ 1000 = 1.77 V = 14.8%

Badly over limit despite a short run and a cable size that would be generous at 230 V. To hit 5% you would need 10 mm² — four times the copper for a tenth of the power. That is the price of low voltage.

SWA and armoured cable

Steel wire armoured is the default for UK sub-mains — a run to a garage, a workshop, an outbuilding, anything buried or exposed. Three things about it matter for voltage drop:

• It is normally XLPE, not PVC. Multicore SWA is usually thermosetting insulated, so the conductor is rated to 90 °C rather than 70 °C. That is about 7% more resistance at full load than the same size in twin & earth, and it means you should select 90 °C thermosetting above rather than leaving the default.
• The armour does not carry load current. The steel wire armour is normally the circuit protective conductor. It carries fault current, not load current, so it does not appear in the voltage drop calculation at all — only the phase and neutral conductors do.
• Size it for both jobs. The armour has to be adequate as a CPC in its own right. Sizing the phase conductors for voltage drop does not automatically make the armour adequate for the earth fault loop impedance, and that is a separate check.

One trap worth naming: single-core armoured cable in an AC circuit. Steel armour around a single core sits in an alternating magnetic field that does not cancel, and the induced eddy currents cause real heating and loss. That is why single-core armoured cables use aluminium wire armour rather than steel. It does not change the voltage drop arithmetic, but it does change which cable you are allowed to buy.

The sizes SWA is typically used in — 16 mm² and up — are also where the reactive component of the Appendix 4 figure starts to matter. Read the note above about z = √(r² + x²) before relying on this page for a large sub-main.

Copper or aluminium

Aluminium has about 61% the conductivity of copper, so it needs roughly 1.6× the cross-section for the same drop. Against that, it is far cheaper and much lighter, which is why service entrances and large feeders are routinely aluminium.

The real difficulties are mechanical rather than electrical. Aluminium creeps under sustained pressure, so terminations loosen over time unless they are torque-rated and listed for aluminium. It also forms an insulating oxide layer almost instantly on exposure to air, so joints normally need an anti-oxidant compound. Neither issue is a reason to avoid it — they are reasons to terminate it properly.

Reducing voltage drop

• Larger cross-section. Drop is inversely proportional to area, so doubling the cross-section halves the drop. Usually the first and simplest fix.
• Shorter run. Drop is directly proportional to length. Moving a distribution board closer to the load is often cheaper than upsizing a long cable.
• Higher voltage. This helps twice — the current falls for the same power, and the percentage is taken against a larger number. Going from 12 V to 24 V cuts the percentage drop by a factor of four for the same load.
• Split the load. Two circuits each carrying half the current have half the drop each, for the same total copper.

Size for both current and drop

Cable sizing has two independent constraints, and you take whichever demands more copper:

• Current-carrying capacity protects the cable. Exceed it and the insulation overheats and eventually fails. This is a safety limit.
• Voltage drop protects the load. Exceed it and equipment runs undervoltage — motors draw more current and run hot, electronics reset, lighting dims. This is a performance limit.

On short runs the current rating almost always decides. Past roughly 20–30 m, voltage drop takes over and frequently demands a cable one or two sizes larger than the current rating alone would suggest.

How this calculator is checked

Technical reviewer

Salamot Hok

Electrician · 10+ years of installation work in Bangladesh and the wider South Asian region

He reads the result the way an installer would: are the defaults values people actually meet, does the warning fire where you would stop and think, and is the answer something you could buy and fit? The code figures themselves come from the published standards cited below, not from him — that boundary is set out on his profile.

• The maths lives in a pure function with its own test suite, asserted against worked examples from published references and standards. A calculator does not ship until those tests pass.
• 5 sources cited by name and linked, so any figure on the page can be traced back to the document it came from.
• Last reviewed 19 August 2026. Review dates are advanced only when the page is actually re-read, never to look fresh.
• Unusable input returns no answer. Where the inputs do not describe a real design, the calculator says so and withholds the number rather than printing a plausible-looking wrong one.

The full process is written up in the methodology and editorial policy. Results are engineering guidance, not a code sign-off — see the disclaimer. If a result looks wrong, tell us; corrections are answered before anything else.