Voltage Drop Calculator

A conductor is a resistor. Push current through it and some of the supply voltage is spent getting to the far end rather than arriving there. How much depends on three things: how much current is flowing, how far it has to travel, and how much metal there is to travel through.

The industry-standard form of the calculation works in circular mils, a unit invented for exactly this purpose — the area of a circle one thousandth of an inch across, defined so that area in circular mils is simply the diameter in mils squared, with no π anywhere:

Vdrop = (2 × K × I × D) ÷ CM for single phase, andVdrop = (1.732 × K × I × D) ÷ CM for a balanced three-phase load. The percentage is then that figure divided by the system voltage and multiplied by a hundred.

D is the one-way distance. Panel to load, measured along the route the cable takes. The factor of 2 in the single-phase version is the return conductor — current goes out on one leg and comes back on the other, so the loss happens over twice the one-way length. This is worth being pedantic about because calculators differ: some ask for total conductor length instead, and the two conventions produce answers exactly a factor of two apart. Three phase uses 1.732 rather than 2 for a different reason entirely. A balanced three-phase load has essentially no neutral current, so there is no return leg to account for; the √3 is the relationship between line and phase quantities.

K is a resistivity constant, about 12.9 for copper and 21.2 for aluminium, in ohm-circular-mils per foot. It is a convention rather than a measurement: it folds a reference temperature of roughly 75 °C and an allowance for stranding into one number, and published values run from 12.0 to 12.9 for copper depending on which assumptions a given reference makes. The higher figure used here is the conservative one — it predicts slightly more drop than a lightly-loaded conductor in a cool space will actually show.

What this leaves out, honestly. AC impedance is not just resistance. A conductor carrying alternating current also has inductive reactance, and the true drop is the vector sum of a resistive and a reactive component at the load's power factor. The formula above ignores reactance entirely. That is the standard simplification, and it is a good one: reactance is roughly constant per foot regardless of conductor size while resistance falls as the conductor gets fatter, so on the small conductors and modest runs that make up almost all branch-circuit work the reactive term disappears into the rounding. It stops being negligible on large conductors — roughly 4/0 and up — and in steel raceway, where the magnetic path amplifies it. There, NEC Chapter 9 Table 9 publishes real effective-impedance values per raceway type and power factor, and it is the right reference. This calculator says so when your inputs land in that territory rather than quietly being a percent or two optimistic.

A 10 AWG copper conductor feeding a 50 A load, 100 ft from a 240 V single-phase panel. Every number below comes from the same module the calculator above runs.

Step one, the conductor area. 10 AWG is10,380 circular mils. That figure is geometry, not code — AWG is a fixed geometric progression, so it does not change between editions.

Step two, the volts. Single phase, so the multiplier is 2: (2 × 12.9 × 50 × 100) ÷ 10,380 = 129,000 ÷ 10,380 =12.43 V.

Step three, the percentage. 12.43 ÷ 240 × 100 =5.18%. The load sees227.57 V rather than 240 V.

Step four, the comparison. 5.18% is past the 3% branch-circuit figure and past the 5% combined figure as well. Neither is enforceable code text in most installations, so this is not automatically a violation — but it is well outside normal design practice, and the run would have to stop at about 58 ft to hold 3% at this load. To make 100 ft work you would need17,917 cmil, which means stepping up to 6 AWG.

And a fifth step the arithmetic does not tell you. 50 A on 10 AWG is not a legal circuit in the first place. Table 310.16 gives 10 AWG copper 35 A in the 75 °C column and 40 A even at 90 °C, and 240.4(D) caps its overcurrent device at 30 A regardless. The drop figure above is perfectly correct and completely beside the point: this is a diagnostic tool, and it will tell you the drop on any combination you hand it, including one that should never be installed. That is why it warns rather than refuses, and why the ampacity question belongs to a different page.

This is the table to reach for when the question is "how far can I run this", which is most of the time. Every row is the longest one-way run that stays inside 3% for that conductor, in copper, at the load in the second column.

About that load column. It is the 60 °C ampacity from Table 310.16, which is what NM cable and most residential terminations are held to — so 14 AWG pairs with 15 A, 12 AWG with 20 A, 10 AWG with 30 A and 8 AWG with 40 A. Those are the pairings people mean when they ask about twelve-gauge on a 20 amp circuit. If your actual load is lighter, the distance stretches in exact proportion: half the current goes twice as far, and the numbers below are the pessimistic end of the range rather than a wall.

Two patterns are worth taking away. The 240 V column is close to double the 120 V column throughout, and the 208 V three-phase column lands almost exactly on the 240 V single-phase one — which is why moving a long run to a higher voltage is so often cheaper than moving it to a bigger conductor.

The second is subtler and more useful: distance grows far more slowly than the conductor does. Going from 14 AWG to 4/0 multiplies the cross-section by fifty and the run length by only four, from 76 ft to 303 ft, because each step up the ladder is paired with a load that has grown almost as fast — 15 A becomes 195 A. Upsizing a conductor while the load stays put does buy distance in direct proportion to area; upsizing because the load grew buys you almost nothing. That is the trap behind "just run a bigger wire" on a circuit whose load is growing at the same time.

For aluminium, multiply every distance by roughly 0.61 — the ratio of the two K values — or just run the calculator, which does it exactly.