How do I calculate voltage drop on a wire run?

Multiply the current by the conductor’s resistance over the run, and by two for a single-phase circuit (the current travels out and back) or by √3 for three phase. The result is the volts lost along the way. Divide that by the supply voltage for the percentage drop, and subtract it from the source to see the voltage that actually arrives.

Where does the resistance value come from?

By default the tool derives it from physics: R = ρ × length ÷ area, using the published resistivity of copper or aluminium at about 20 °C and the conductor size you enter — in mm² or as an AWG number, which it turns into mm² from the gauge definition. That is a cool, solid conductor carrying DC. A conductor resistance out of a code book is a WARMER one, usually quoted at an operating temperature, and it is stranded, so it reads higher — metal gains about 0.4% of its resistance per °C, so a conductor at 75 °C is roughly a fifth more resistive than the same one at 20 °C, and stranding adds a little more on top. The derived figure is therefore the OPTIMISTIC end. If you have an ohms-per-kilometre from your own book or the cable maker, switch to the advanced option and type it — that is the number your inspector is working from.

Why is my answer lower than the one from the code book?

Because the default path derives the resistance from bulk resistivity at about 20 °C for a solid conductor, and a code table gives you a stranded conductor at its operating temperature, with reactance included. All three push the same way: the book figure is higher, so the drop it gives is higher. On a long run at a big size the gap is worth having — copper picks up roughly 0.4% of its resistance per °C, and reactance starts to matter once the conductor is large. Take the Ω/km from your own source, switch to the advanced option and type it, and the tool does the rest of the arithmetic on your number instead of its own.

What is an acceptable voltage drop?

That is set by your code and the authority having jurisdiction, and it differs by circuit type and by where the run sits in the system — so this page does not suggest a number. Type the limit your job specifies into the "What to do with this number" box and the tool compares your run to it, then lets you re-run at the next size up.

Why does the AWG option exist on a metric tool?

Because a lot of conductor is bought by gauge number and a first-year thinks "#12", not "3.31 mm²". American Wire Gauge is a geometric definition — diameter in inches = 0.005 × 92^((36 − n)/39) — so the tool computes the area from the number you type rather than looking it up. The derived mm² is shown beside the field so the two ways of naming a size line up in your head.

Why multiply the length by two for single phase?

Current has to return, so a single-phase run is really twice the one-way distance of conductor carrying current. The tool takes the one-way run length you enter and applies the ×2 itself. Three-phase circuits use the √3 factor instead, from the geometry of the three line voltages.

What does "next size up" actually do?

It lowers the gauge number by one — the next size in the AWG series — and re-runs the drop, or in mm² mode it grows the area by the same ratio one gauge step would (about 1.26 times). Whether that size is stocked, allowed or big enough for the load is not something this page knows; it is showing you how much the drop moves per step so you can see what a size change buys.

Voltage drop calculator

Corriente

Largo del tramo (solo ida)

Del tablero a la carga, una sola vez. La herramienta agrega el regreso.

Tensión de la fuente

Fase

Resistencia

Calibre del conductor

Metal

— % de caída

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Ver el procedimiento — con sus números

What to do with this number

A drop percentage means nothing on its own — it is compared to a limit. That limit comes from your job: the plans, your code, the authority. Type it here and the tool does the comparing; it will not suggest one.

El límite de caída que exige su obra

Before you type — how to measure

Run length is one way, in feet or metres (the switch above), along the path the cable actually takes — up the wall, across the ceiling, down to the box — not the straight line on the drawing. The tool doubles it for single phase; if you double it too, the drop comes out twice too big.
Current is the load current, the amps that will actually flow, not the breaker size. A 20 A breaker feeding a 12 A load drops at 12 A.
Size is the conductor, not the cable. For a two-conductor cable, type the size of one conductor once.
If you are typing an Ω/km (or, in feet mode, Ω per 1,000 ft — the km figure × 0.3048), make sure it is per one conductor at the temperature your source states. A loop resistance is already doubled.
The derived figure is a cool conductor — bulk resistivity at about 20 °C, solid, DC, no reactance. Your code book's conductor resistance is warmer and stranded, so it gives a bigger drop. If the run is close to the limit, type that figure in instead of deriving one.

This is arithmetic on the current, length and resistance you provide. Derive the resistance and you get it from bulk resistivity at about 20 °C for a solid conductor, direct current, no reactance — the optimistic end. A conductor resistance from a code book is warmer, stranded and, at larger sizes, carries reactance too, so it gives a higher drop; type that figure into the advanced option when the run is anywhere near the limit. This page does not decide the limit your job must meet — that is set by your code and your authority having jurisdiction. Confirm the conductor and the acceptable drop before you rely on it.

Enter the load current, the one-way run length (feet or metres) and the supply voltage, then the conductor size — in mm² or as an AWG number — and whether it is copper or aluminium. The tool derives the resistance from the metal's resistivity and the area, doubles the run for single phase (or applies √3 for three), and reports the volts lost, the percentage, and the voltage that reaches the far end. The derived resistance is the metal at about 20 °C, solid and on DC; if you have an ohms-per-kilometre (or ohms-per-1,000-ft) figure of your own — warm, stranded, out of your book — the advanced option takes it instead, and that is the one to use near a limit.

Worked examples

All four derive the resistance the way the tool does by default — bulk resistivity at about 20 °C, solid conductor, DC, reactance ignored. Work the same runs from a warm stranded conductor resistance out of your code book and every percentage here comes out higher; on the aluminium service, big enough that reactance counts, higher again. They show the arithmetic, not the margin you have.

The defaults in feet. 20 A, single phase, 100 ft (30.5 m) one way, 12 AWG copper on a 120 V circuit. The gauge definition makes 12 AWG 3.31 mm², so 5.21 Ω/km — 1.59 Ω per 1,000 ft. Over 100 ft that is 1.588 × 100 ÷ 1000 = 0.159 Ω, and out-and-back at 20 A the loss is 2 × 20 × 0.159 = 6.35 V — a 5.29% drop, leaving 113.6 V at the load.

The defaults in metres. 20 A, single phase, 50 m (164 ft) one way, 4 mm² copper on a 240 V supply. Copper's resistivity is 1.724 × 10⁻⁸ Ω·m, so one conductor is 1.724 × 10⁻⁸ × 10⁹ ÷ 4 = 4.31 Ω/km. Over 50 m that is 4.31 × 50 ÷ 1000 = 0.2155 Ω, and out-and-back at 20 A the loss is 2 × 20 × 0.2155 = 8.62 V — a 3.59% drop, leaving 231.4 V at the load. One gauge step larger (× 1.26 area, 5.04 mm²) would bring it to about 2.85%.

A 120 V branch. 15 A on a 30 m (98 ft) run of 12 AWG copper. The gauge definition makes 12 AWG 3.31 mm², so 5.21 Ω/km. Over 30 m: 0.156 Ω; times 2 × 15 A: 4.69 V, which on 120 V is 3.91%. Two gauge steps up, at 10 AWG (5.26 mm², 3.28 Ω/km), the same run drops 2.95 V, 2.46% — a bigger conductor, a smaller loss, in exactly the ratio of the areas.

A three-phase motor feeder. 30 A per line, 208 V three phase, 75 m (246 ft) of 6 AWG copper (13.3 mm², 1.30 Ω/km). Over 75 m: 0.0972 Ω. Three phase uses √3 in place of the 2: 1.732 × 30 × 0.0972 = 5.05 V, a 2.43% drop, 202.9 V at the motor. At 4 AWG (21.2 mm²) it would be 3.18 V, 1.53%.

An aluminium service. 100 A, single phase, 40 m (131 ft) of 2/0 aluminium. 2/0 is gauge number −1, which the definition makes 67.4 mm²; aluminium's resistivity is 2.826 × 10⁻⁸ Ω·m, so 0.419 Ω/km. Over 40 m: 0.0168 Ω; 2 × 100 × 0.0168 = 3.35 V, 1.40% on 240 V. At 3/0 (85.0 mm²) it is 2.66 V, 1.11%. Whether any of these is acceptable is the limit box's job, with your number in it.

Practice — estimate the drop

Five random runs. For each, work the volts lost in your head or on a scrap of paper before you type it — within 10% counts. The runs come in feet or metres to match the switch at the top. The working appears after each answer so you can see where a miss came from. Like the tool's default, the drill derives its resistance at about 20 °C: it is practice at the arithmetic, not a figure to take to a job.

Quick referencefor the arithmetic

Conductor resistance: Ω/km × one-way metres ÷ 1,000 — 5.21 Ω/km over 30 m is 0.156 Ω per conductor
In feet: Ω/1000 ft × one-way feet ÷ 1,000 — 1.59 Ω/1000 ft over 100 ft is 0.159 Ω · Ω/1000 ft = Ω/km × 0.3048
Monofásico: drop = 2 × amps × that resistance — out and back
Trifásico: drop = 1.732 × amps × that resistance — line to line
Percent: drop ÷ supply volts × 100 · the same volts are a bigger percentage at 120 than at 240
Halving the drop: half the run, half the current, or twice the copper — each halves it; three gauge numbers doubles the copper
Feet to metres: × 0.3048 · 100 ft = 30.5 m
The limit: is your job's — typed in the field above, never supplied here

Mistakes to catchbefore you trust the number

Typing the round-trip length: the run field is one way, and the tool doubles it — a doubled input doubles the drop.
Using the source voltage of the panel when the load is on a sub-panel already down by a few percent; the limit you typed is usually judged from the service.
Entering the resistance per 1,000 ft into a field set to Ω/km, or the reverse — a factor of about three in the drop.

What an inspector looks forthe electrical inspector

On this part of the job

Conductor size against the load and the length of the run. Drop is usually a performance requirement rather than a hard rule, but an undersized run will be questioned and a branch circuit that cannot hold voltage at the end gets rejected on its own merits.
That the conductor material and temperature rating you calculated with are what is actually installed, and that the termination rating has not quietly limited the whole run.
Where the design relied on upsizing for drop, that every part of it was upsized — including the neutral and the bonding conductor where the rules require it to follow.
The route length as installed, which is nearly always longer than the length someone scaled off the plan.

Have this ready

The load calculation and the run lengths you worked to.
Your own code book, for the conductor properties and the acceptable drop.
Cable markings visible at both ends of the run.

Every figure an inspector measures against comes from the code your jurisdiction has adopted, the stamped drawing, or the manufacturer's printed instructions — so none of them are printed here. Look the number up in your own book, type it into the tool, and the arithmetic is yours.

La fórmula

drop = k × I × (R/km × length ÷ 1000) · k = 2 (1φ) or √3 (3φ) · R/km = ρ × 10⁹ ÷ mm²

in feet: drop = k × I × (R/1000 ft × feet ÷ 1000) · R/1000 ft = R/km × 0.3048

Resistance over the run is the ohms-per-km times the length in kilometres. Multiply by the current and the phase factor for the volts lost. When derived from size, the resistance is R = ρ × length ÷ area, with ρ the resistivity of the metal — a measured property, not a tabulated code value. AWG feeds the same line: the gauge number becomes a diameter (0.005 × 92^((36 − n)/39) inches), the diameter becomes an area, and the area goes into ρ ÷ A. Working in feet is the same line with the resistance per 1,000 ft and the run in feet — a kilometre is 3,281 ft, so the per-1,000-ft figure is the per-km figure × 0.3048.

Preguntas

How do I calculate voltage drop on a wire run?: Multiply the current by the conductor’s resistance over the run, and by two for a single-phase circuit (the current travels out and back) or by √3 for three phase. The result is the volts lost along the way. Divide that by the supply voltage for the percentage drop, and subtract it from the source to see the voltage that actually arrives.
Where does the resistance value come from?: By default the tool derives it from physics: R = ρ × length ÷ area, using the published resistivity of copper or aluminium at about 20 °C and the conductor size you enter — in mm² or as an AWG number, which it turns into mm² from the gauge definition. That is a cool, solid conductor carrying DC. A conductor resistance out of a code book is a WARMER one, usually quoted at an operating temperature, and it is stranded, so it reads higher — metal gains about 0.4% of its resistance per °C, so a conductor at 75 °C is roughly a fifth more resistive than the same one at 20 °C, and stranding adds a little more on top. The derived figure is therefore the OPTIMISTIC end. If you have an ohms-per-kilometre from your own book or the cable maker, switch to the advanced option and type it — that is the number your inspector is working from.
Why is my answer lower than the one from the code book?: Because the default path derives the resistance from bulk resistivity at about 20 °C for a solid conductor, and a code table gives you a stranded conductor at its operating temperature, with reactance included. All three push the same way: the book figure is higher, so the drop it gives is higher. On a long run at a big size the gap is worth having — copper picks up roughly 0.4% of its resistance per °C, and reactance starts to matter once the conductor is large. Take the Ω/km from your own source, switch to the advanced option and type it, and the tool does the rest of the arithmetic on your number instead of its own.
What is an acceptable voltage drop?: That is set by your code and the authority having jurisdiction, and it differs by circuit type and by where the run sits in the system — so this page does not suggest a number. Type the limit your job specifies into the "What to do with this number" box and the tool compares your run to it, then lets you re-run at the next size up.
Why does the AWG option exist on a metric tool?: Because a lot of conductor is bought by gauge number and a first-year thinks "#12", not "3.31 mm²". American Wire Gauge is a geometric definition — diameter in inches = 0.005 × 92^((36 − n)/39) — so the tool computes the area from the number you type rather than looking it up. The derived mm² is shown beside the field so the two ways of naming a size line up in your head.
Why multiply the length by two for single phase?: Current has to return, so a single-phase run is really twice the one-way distance of conductor carrying current. The tool takes the one-way run length you enter and applies the ×2 itself. Three-phase circuits use the √3 factor instead, from the geometry of the three line voltages.
What does "next size up" actually do?: It lowers the gauge number by one — the next size in the AWG series — and re-runs the drop, or in mm² mode it grows the area by the same ratio one gauge step would (about 1.26 times). Whether that size is stocked, allowed or big enough for the load is not something this page knows; it is showing you how much the drop moves per step so you can see what a size change buys.

Herramientas relacionadas

Going deeper? The arithmetic is here; the why — what the ×2 and the √3 mean, how a drop shows up at a motor, how to work it without a calculator — is taught on sparkytools, a separate site built for drills rather than answers.

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