What the formula is doing
Voltage drop is Ohm's law applied to the wire instead of the load. The conductor has resistance, current through resistance produces a voltage difference, and that difference is subtracted from what arrives at the far end. The version used here is the long-standing circular-mil form:
Volts dropped = (length multiplier × K × amps × one-way feet) ÷ circular mils
The length multiplier is 2 for single-phase and DC, because the current goes out on one conductor and back on another and both of them drop voltage. For three-phase it is the square root of three, about 1.732, which falls out of the phase relationship between the legs rather than from any physical difference in the wire.
K is the resistivity of the metal expressed in ohm-circular-mils per foot. The figures used here are 12.9 for copper and 21.2 for aluminium, which are the values conventionally quoted for a conductor operating warm, around 75 °C. A cold conductor has lower resistance, so a run measured on a chilly morning with a light load will do slightly better than this predicts. That is the right direction to be wrong in.
Circular mils are just a way of writing conductor area that makes the arithmetic clean: a 12 AWG conductor is 6,530 circular mils, a 4/0 is 211,600, and a 250 kcmil is 250,000 by definition. Doubling the area halves the drop, which is why the fix for a long run is always a bigger conductor rather than a cleverer route.
Where 3 percent and 5 percent came from
You will see 3 percent on a branch circuit and 5 percent total from the service to the far end quoted everywhere, often in a tone that suggests they are hard limits. They are not. In the model code these figures live in informational notes rather than in enforceable requirements, and the point of them is efficiency and equipment behaviour rather than safety. A run at 6 percent is not dangerous, it is wasteful and it will make some equipment behave badly.
What does change behaviour:
| Drop | What you notice |
|---|---|
| Under 3% | Nothing. Everything works as the nameplate says. |
| 3-5% | Incandescent lamps dim slightly, resistive heaters run measurably below rating, long motor runs get warmer. |
| 5-10% | Motors draw more current to make the same torque and run hot. Compressors struggle to start. Saws bog in cuts they should walk through. |
| Over 10% | Motor starting becomes unreliable, contactors chatter, electronics with marginal supplies reset. The conductor is dissipating real power along its length. |
The asymmetric case is a motor. A motor is a roughly constant-power device: give it less voltage and it pulls more current to deliver the same mechanical output, which increases the drop, which lowers the voltage further. That feedback is why long runs to well pumps, compressors and shop machinery are the ones that fail, while long runs to a heater just quietly under-perform.
Watts, amps and power factor
If you only know the load in watts, the conversion to amps depends on the system. Single-phase, amps equals watts divided by volts divided by power factor. Three-phase, amps equals watts divided by the square root of three, divided by volts, divided by power factor. Resistive loads — heaters, incandescent lamps, elements — have a power factor of 1 and the division does nothing. Motors typically run somewhere between 0.8 and 0.9 at full load and considerably worse lightly loaded.
One caution on the arithmetic: this calculator uses the simple resistive method, which ignores conductor reactance. On small conductors and ordinary building circuits reactance is a minor term. On large conductors carrying reactive loads it is not, and a proper calculation uses the resistance and reactance figures for the specific conductor in the specific raceway type. If you are in that territory, the tables that carry those figures are the ones to use.
The distinction this page will not blur
Voltage drop is arithmetic. It is the same in every jurisdiction, it does not depend on which code edition your county adopted, and the number above is as correct as the inputs you gave it. Picking a conductor is a different activity entirely. The current a conductor may carry depends on its insulation type, the temperature rating of the terminals at both ends, the ambient temperature, how many current-carrying conductors share the raceway or cable, whether it is in sunlight on a rooftop, whether it is buried, and which edition of the code your Authority Having Jurisdiction adopted along with whatever amendments came with it. Two of those factors can easily halve a table figure.
So when this page tells you that a larger conductor would bring the drop under your target, read it as one input among several. These figures are planning arithmetic, not an electrical design. In most jurisdictions permanent wiring needs a permit and a licensed electrician, and the ampacity table in the code edition your Authority Having Jurisdiction has adopted is what governs the installation. Do not use a number from this page to size a conductor without a qualified person reviewing the whole circuit.
Questions people ask
Is a 3 percent voltage drop actually required?
No, and being precise about this matters. In the widely referenced model code the 3 percent branch-circuit and 5 percent total figures appear as recommendations in informational material, not as enforceable installation requirements, and they exist for efficiency and equipment performance rather than for safety. Some jurisdictions have amended them into local requirements, some specifications and utility rules impose tighter numbers, and some plan reviewers will ask about it on long feeders. Your Authority Having Jurisdiction is the one who decides what applies to your job. Treat 3 percent as a good design target that keeps equipment happy, and ask locally if you need to know whether it is enforceable where you are.
Why does the calculator double my distance?
Because current has to come back. On a single-phase circuit the current flows out along one conductor and returns along another, and both of them have resistance and both of them drop voltage. Entering the one-way distance and letting the tool double it is less error-prone than asking you to double it yourself, which is a mistake people make in both directions. On a three-phase circuit the multiplier is the square root of three rather than two, which comes out of the 120-degree phase relationship between the legs, not from a shorter wire path. DC is treated like single-phase: out and back, so a multiplier of two.
Does aluminium really need a bigger size?
Yes, roughly two sizes for the same drop. Aluminium has about 1.64 times the resistivity of copper for the same cross-section, so at equal size it drops about 64 percent more voltage and it also carries less current for the same insulation temperature. Going up two AWG sizes roughly restores both. Aluminium is still commonly used for service entrances and large feeders because at those sizes the material cost difference is substantial and the weight difference makes the conductor easier to pull. It is used far less for branch circuits, and small aluminium branch conductors from a particular era have a poor reputation at terminations. Where aluminium is used, the terminations at both ends need to be rated for it and prepared in the way the connector manufacturer specifies.
Can I fix a long run by using a higher voltage instead of a bigger wire?
Often, and it is usually the cheaper fix. Voltage drop in volts is roughly proportional to current, and delivering the same power at twice the voltage halves the current. Because the drop is then also measured against a supply that is twice as large, the percentage drop falls by a factor of about four. That is why a detached shop at the end of a long run gets a 240 volt feeder rather than a heroic 120 volt one, and why 480 volt distribution exists in industrial buildings. Whether that option is available to you depends on the service, the equipment at the far end and what the panel can accommodate, which is a conversation to have with an electrician before you buy anything.