How this voltage drop calculator works
Enter the current, the voltage, the one-way distance and the conductor size, and it gives you the drop in volts and as a percentage, the voltage left at the load, and the two ways out if it fails: how far that conductor could go, and what size would pass at the distance you actually have. It also prices something most calculators leave out, which is the power being burned in the wire itself.
One thing to be completely clear about before anything else: this page sizes for voltage drop only. Ampacity is a separate calculation, it is a hard requirement rather than a recommendation, and it is the one that stops fires. A conductor has to satisfy both, and the right answer is whichever of the two gives the larger wire.
The formula
three phase: VD = 1.732 × K × I × L ÷ CM
K = 12.9 for copper, 21.2 for aluminum
maximum length = limit × V × CM ÷ (factor × K × I)
K is the conductor's resistivity in ohm-circular mils per foot at 75 °C, I is the load current, L is the one-way distance in feet, and CM is the conductor's cross-section in circular mils. The leading 2 on a single-phase run is there because the current goes out on one conductor and comes back on another, so the wire actually in the circuit is twice the distance you measured. Three phase uses 1.732, which is the square root of 3, because in a balanced load the return currents partly cancel and each conductor carries less of the burden.
Worked example
An ordinary 20 A circuit on 12 AWG copper, 120 V, 100 feet out. VD = 2 × 12.9 × 20 × 100 ÷ 6,530 = 7.90 V, which is 6.58%. The load sees 112.1 V instead of 120.
That is more than double the usual 3% recommendation, on a circuit that is entirely legal from an ampacity point of view. 12 AWG is the correct size to carry 20 A. It is simply the wrong size to carry 20 A a hundred feet.
The two answers: that conductor stays inside 3% out to 45.6 feet, and passing at 100 feet needs 8 AWG, which is two standard sizes up. And the run is currently burning 158 W in the wire, continuously, whenever the load is on. That is a small space heater installed inside your wall by accident.
The 3% rule is not a rule, and almost every source says otherwise
Search for voltage drop and you will be told, confidently and repeatedly, that the NEC requires a maximum of 3% on a branch circuit and 5% overall. It does not. Those figures appear in an Informational Note to 210.19(A), and NEC 90.5(C) states in as many words that informational notes are explanatory material and are not enforceable as requirements of this Code. An inspector working from the NEC alone has no voltage drop to fail you on.
That is the correction, and here is the part that stops it being an excuse. A meaningful number of jurisdictions adopt voltage drop limits by local amendment, California and New York City among the better-known ones, and in those places it is enforceable and you will be held to it. Separately, ASHRAE 90.1 requires a voltage drop analysis for commercial work in jurisdictions on recent energy codes, so on a commercial permit it can arrive through the energy code rather than the electrical one.
The honest summary is therefore not "it is code" and not "ignore it", but this: 3% is excellent engineering practice that is sometimes also the law where you are standing, and the only way to know which applies is to ask your AHJ. Design to it regardless, because the reasons behind it are real.
What voltage drop actually costs
The 3% figure exists because equipment behaves badly on low voltage, and the way it behaves badly depends on what it is. Resistive loads like heaters and incandescent lamps simply do less: power falls with the square of voltage, so a 5% drop costs about 10% of the heat or light. Motors are worse, because a motor delivering a fixed mechanical load compensates for low voltage by drawing more current, which increases the drop, which raises the current again. They run hotter, start harder, and die sooner. Electronics with switching supplies mostly do not care until they abruptly do, holding their output steady right up to a dropout voltage and then falling over.
And there is the number in every result on this page: the drop times the current is power turned into heat inside the wall. On the worked example that is 158 W running continuously. At 15 cents a kilowatt hour and eight hours a day it is about $70 a year, spent on warming a stud bay. Upsizing that run from 12 to 8 AWG cuts the loss to roughly 60 W. Over the life of a building, conductor upsizing for voltage drop is one of the few electrical decisions that genuinely pays for itself, which is exactly why the energy codes started asking about it.
Where this approximation stops being exact
The formula above is the one the NEC's own examples use, and it is a simplification worth understanding rather than trusting blindly. It uses direct-current resistance and ignores reactance, which is fine for small conductors and short runs and increasingly optimistic for large ones: above about 1/0, and on long runs with power-factor-heavy loads, the true drop is higher than this predicts. NEC Chapter 9 Table 9 gives the fuller treatment with separate resistance and reactance figures, and any serious commercial design uses it.
It also assumes a conductor temperature of 75 °C. A cool conductor has lower resistance and drops less; one running hot in a bundled conduit drops more, and the difference between 25 °C and 75 °C is around 20%. And it assumes a steady current, so anything with a large inrush, a motor starting or a welder striking, sees a much deeper momentary sag than the steady-state number here. For a house circuit, a feeder to a shop, or a run to a well pump, this approximation is what everybody uses and is accurate enough. For a 400 A three-phase feeder across a plant, get Table 9 out.