Wire Size Calculator

A = 2 × L × I × ρ ÷ ΔV, where ΔV = V × maxDropPercent ÷ 100 and copper's ρ is 0.0172414 Ω·mm²/m at 20 °C

L

The one-way run length, in metres or feet — the distance from the supply to the load, not the total length of copper in the circuit. This is the field that goes wrong without looking wrong: a 30 metre run contains 60 metres of conductor, because the current goes out and comes back, and the leading 2 in the expression is that fact written down. Enter 60 instead of 30 and every figure on the page doubles, and the answer still looks like nothing more than a longish cable

I

The current the load draws, in amperes. Both conductors carry it, so it multiplies the resistance of the whole loop rather than half of it. The area needed is directly proportional to the current, which is why the same run has to be thicker for a heavier load — and why the current is usually worked out first, from the power, by the watts to amps calculator

ΔV

The voltage drop you are prepared to accept, in volts, which is the percentage you enter multiplied by the supply voltage. It sits in the denominator, so a tighter limit demands more copper: halving the permitted drop doubles the area needed. On a 230 V supply, 3 per cent is 6.9 V; on a 12 V supply the same 3 per cent is only 0.36 V, which is why low-voltage runs need conductors that look absurdly large next to their current

ρ

The resistivity of the conductor: annealed copper at 20 °C, 0.0172414 ohm square millimetres per metre, which is the value corresponding to 100 per cent conductivity on the International Annealed Copper Standard. Aluminium is about 1.64 times higher, so an aluminium conductor of the same size drops about sixty per cent more volts. Temperature is not modelled, and a hot conductor has a higher resistance than the figure in a table, so what this page returns is a lower bound on the drop

2

The two conductors in the loop, out and back. It is not a safety margin and not a correction factor: leaving it out halves the answer, and the result still looks entirely reasonable. Single-phase and direct-current circuits both need it; a balanced three-phase circuit is a different piece of arithmetic with the square root of three in it, which is why the page does not offer a supply-type field and asks a three-phase reader to divide the permitted drop before entering it

Use it whenever a run has to be installed and the conductor can still be chosen: sizing a new circuit, extending one, working out what a low-voltage run needs, or checking whether a cable already on the shelf is thick enough before it gets pulled in. It is also the reverse of the voltage drop calculator — that page takes a conductor size you already have and reports the drop, this one takes the drop you are willing to live with and reports the conductor, and the same circuit entered on both should tell a consistent story. The value to decide before typing is the permitted drop, because it is the criterion rather than a measurement, and it is the one input with no physical answer: 3 per cent is a convention, not a law of nature

1. The defaults: 230 V, 20 A, 30 m one way, 3 per cent

   1. The permitted drop is 3 per cent of 230 V, so 6.9 V
   2. Area needed: 2 × 30 × 20 × 0.0172414 / 6.9 = 2.9985 mm²
   3. The smallest gauge at least that large is AWG 12, which is 3.3088 mm²
   4. The drop on the gauge you would actually buy: 2 × 30 × 20 × 0.0172414 / 3.3088 = 6.25 V, or 2.72 per cent

   This is the same circuit the voltage drop calculator opens with, and the pair is worth reading together: there, a 2.5 mm² conductor is already installed and drops 3.60 per cent, just past the recommended band; here, the answer is AWG 12 at 2.72 per cent, inside it. Two and a half square millimetres is fractionally too thin for this run and AWG 12 is fractionally enough, which is the whole reason a page that picks the conductor needs to exist next to a page that judges one. The fourth and fifth rows are the self-check: the drop on the gauge actually recommended has to come out at or under the limit you entered, and here it does.

2. A 12 V run: 2 A over 3 metres, 3 per cent

   1. The permitted drop is 3 per cent of 12 V, so 0.36 V
   2. Area needed: 2 × 3 × 2 × 0.0172414 / 0.36 = 0.5747 mm²
   3. The smallest gauge at least that large is AWG 18, at 0.823 mm²
   4. The drop on AWG 18: 2 × 3 × 2 × 0.0172414 / 0.823 = 0.25 V, or 2.09 per cent

   Two amps is a trivial current and the answer is still AWG 18, which is thicker than most people would guess for a two-amp load. The reason is the supply, not the load: 3 per cent of 12 V is 0.36 V, so the run has an absolute budget of a third of a volt. The same 3 per cent on a 230 V supply is 6.9 V, nineteen times more room for the same conductor to work in, which is why low-voltage installations use cable that looks oversized next to their current rating.

3. A long lighting circuit: 230 V, 6 A, 40 m one way

   1. The permitted drop is 6.9 V, the same as the default case
   2. Area needed: 2 × 40 × 6 × 0.0172414 / 6.9 = 1.1994 mm²
   3. The smallest gauge at least that large is AWG 16, at 1.3087 mm²
   4. The drop on AWG 16: 2 × 40 × 6 × 0.0172414 / 1.3087 = 6.32 V, or 2.75 per cent

   A third of the current of the default case and a longer run, and the answer is still a substantial conductor: length more than makes up for the smaller load. This is the case that surprises people who size cable by current alone, and it is the reason lighting circuits in large buildings end up thicker than the lamps would suggest. The arithmetic is the same multiplication as ever — 40 metres at 6 amps is 240 ampere-metres, against 600 ampere-metres for the default, so the area needed falls by more than half, but not to anything small.

4. A 24 V distribution run: 10 A over 4 metres

   1. The permitted drop is 3 per cent of 24 V, so 0.72 V
   2. Area needed: 2 × 4 × 10 × 0.0172414 / 0.72 = 1.9157 mm²
   3. The smallest gauge at least that large is AWG 14, at 2.0809 mm²
   4. The drop on AWG 14: 2 × 4 × 10 × 0.0172414 / 2.0809 = 0.66 V, or 2.76 per cent

   Set this beside the 12 V case and the pattern is plain: five times the current and only a third longer, yet the two answers are two gauge steps apart, while the same 10 A on a 230 V supply would be far thinner still. What the recommendation tracks is not the current but the current times the length divided by the voltage — 10 A over 4 metres at 24 V is a tighter proposition than 2 A over 3 metres at 12 V, and AWG 14 against AWG 18 is what that difference costs.

5. No current at all: the series floor rather than an error

   1. With no current the area needed is 2 × 30 × 0 × 0.0172414 / 6.9 = 0 mm²
   2. The smallest gauge that is at least zero is the thinnest in the series, AWG 20
   3. AWG 20 is 0.5176 mm², so the recommended area is not zero even though the need is
   4. With no current there is no drop: 0 V and 0 per cent

   Zero current is a legitimate input rather than a mistake — a control circuit before anything is connected, or a run sized ahead of its load — and the page answers it instead of refusing. The answer is the floor of the series, AWG 20, and that is a statement about the page rather than about electrical practice: thinner conductors exist, and small signal wiring is chosen for other reasons entirely. Nothing about the arithmetic picks a minimum for you, which is why the practical floor is discussed in the limitations instead.

The page sizes for voltage drop only. It does not check ampacity, the current a conductor may carry continuously, and that limit depends on the insulation's temperature rating, how many conductors share a raceway, the ambient temperature, whether the run is buried or in free air and whether it is wrapped in thermal insulation — none of which are inputs here. The consequence is concrete: on the reference table, AWG 0 comes out at 356.7 A by this page's arithmetic, while the same 1/0 copper conductor in a raceway with 75 °C insulation is ordinarily limited to around 150 A. A real selection uses the thicker of the two answers, and for small sizes the ampacity answer usually wins. Only two-wire single-phase and direct-current circuits are covered, because the expression carries the factor of two for the out and return conductors; on a balanced three-phase circuit the drop is smaller by the square root of three, so a three-phase reader should divide the permitted percentage before entering it. The conductor is copper at 20 °C, and aluminium is about 1.64 times more resistive, so an aluminium run of the same size drops roughly sixty per cent more. The recommendation is the thinnest standard gauge that satisfies your drop limit and nothing else — mechanical strength, the minimum cross-section some national codes impose on lighting circuits regardless of drop, and any allowance for a future heavier load are all outside it.

Does this check the cable's ampacity as well?

No, and it is the most important thing to know about the page. Ampacity is the current a conductor may carry continuously, and it depends on the insulation's temperature rating, how many conductors share the raceway, the ambient temperature and the installation method — none of which are inputs here. What this page returns is the smallest conductor that satisfies your voltage drop limit. The reference table shows how far apart the two criteria can be: AWG 0 comes out at 356.7 A on this page's arithmetic, while a 1/0 copper conductor in a raceway with 75 °C insulation is ordinarily limited to around 150 A. Choose the thicker of the two answers, and for small sizes expect the ampacity answer to win.

How is this different from the voltage drop calculator?

They are the same expression read in two directions. The voltage drop calculator takes a conductor size you already have and tells you what it drops; this one takes the drop you are willing to accept and tells you what conductor to buy. Put the same circuit through both and they should agree, and the defaults on the two pages are deliberately that same circuit: 2.5 mm² gives 3.60 per cent on the other page, just outside the recommended band, while AWG 12 gives 2.72 per cent here, inside it.

Is the length the one-way distance or the total?

The one-way distance, from the supply to the load. The expression carries a factor of two because the current has to travel out along one conductor and back along the other, so the loop contains twice the length of copper you would measure along the route. Entering the round-trip figure instead doubles every result, and because the answer is still a plausible gauge number rather than an obvious error, nothing on the page will tell you that you did it.

Can I use it for a three-phase circuit?

Not directly. On a balanced three-phase circuit the drop is smaller by the square root of three, so the same conductor performs better than this page says. The page has no supply-type field on purpose: unlike the watts to amps calculator, where missing the square root of three corrupts the answer the reader is looking for, here it only scales the criterion, and a three-phase reader can divide the permitted percentage by 1.732 before entering it and get the right conductor.

Why is the answer a gauge number rather than an area?

Because that is what you can buy. The area the run needs is a continuous figure — 2.9985 mm² in the default case — and no one stocks it. Conductors come in a series of standard sizes, so the page reports both: the area you need and the area of the smallest standard gauge that is at least that large, which here is AWG 12 at 3.3088 mm². The difference between the two numbers is the margin you get for free, and it is also the thing to look at when deciding whether to move up another step.

What happens when the run needs more copper than AWG 0?

The page refuses rather than recommending the thickest gauge it has, which is deliberate: answering AWG 0 to a run that needs 89.96 mm² would suggest a conductor that is nowhere near adequate, and that answer would look usable. The thin end is treated the other way round — a run that needs less than AWG 20 simply gets AWG 20, because the thinnest gauge in the series is already generous and nothing is gained by naming a size the page does not cover. One end can round down safely; the other cannot.

What percentage should I enter?

Three per cent of the supply is the customary figure for a branch circuit, and five per cent is sometimes used for a run that includes the feeder. Those are conventions rather than physics — nothing in the arithmetic knows about them — and what actually sets the limit is the equipment at the far end: a motor with a wide voltage tolerance will run on a larger drop than a piece of electronics, and some standards set their own figures. Enter the criterion you are working to, because this field is the one input with no physical answer behind it.

• American wire gauge — how the gauge system is defined, why dropping three gauge numbers doubles the area, and the areas the recommendations here are drawn from — Wikipedia
• Voltage drop — the two-conductor drop expression this page inverts, and the 3 per cent and 5 per cent figures used as design guidance — Wikipedia
• Ampacity — why the current a conductor may carry depends on insulation, ambient temperature and installation method, and therefore why this page does not compute it — Wikipedia