Voltage drop
Voltage drop on a copper run from the wire size, length, current and phase — with the percent drop and a check against the 3% branch-circuit target, so a long run doesn’t starve the load.
Conductor voltage drop & wire sizing · copper or aluminum, 1φ & 3φ
Vd = k·I·R·L/1000 (k = √3 3φ, 2 1φ) · aim ≤ 3%
What this gives you
This is the voltage lost pushing current down a copper run and back — the difference between the voltage at the panel and the voltage that actually reaches the load. Enter the current, the one-way length, the wire size, the source voltage and the phase, and it returns the drop in volts, the drop as a percent of the source, and the voltage left at the load. For three-phase the drop is Vd = √3 × I × R × L ÷ 1000; for single-phase swap the √3 for a factor of 2. R is the conductor's resistance in ohms per 1000 feet, L is the one-way length in feet, and the phase factor already accounts for the return path, so you enter the run length once.
Why 3 percent matters
The NEC doesn't mandate a voltage-drop limit, but it recommends no more than 3 percent on a branch circuit and 5 percent total across feeder plus branch. The reason is practical: a motor fed low voltage draws more current to make the same power, runs hotter, and loses starting torque; contactor coils chatter; heaters and lighting fall off fast because output scales with voltage squared. When the calculator flags a run over 3 percent, the fix is almost always to step up one or two wire sizes — bigger copper means lower R and a proportionally smaller drop.
Field note — this is a DC-resistance estimate
The formula uses conductor DC resistance only. That's accurate for small conductors and typical runs, but on large conductors (roughly 1/0 and up) carrying high current, AC reactance adds to the drop and this figure will read a little low — reach for an impedance-based table for those. Resistance also climbs with temperature, so a hot conductor drops slightly more than the 75 °C book value used here. And always enter the one-way length: the √3 or ×2 factor already covers the return conductor, so doubling the length yourself double-counts it.
Worked example
A 20 A load on 100 feet of #10 copper at 240 V three-phase: with R = 1.24 Ω/1000 ft, the drop is 1.732 × 20 × 1.24 × 100 ÷ 1000 = 4.29 V, which is 1.79 percent and leaves 235.7 V at the load — comfortably inside the 3 percent target. Drop the same load onto #12 at 120 V single-phase and the picture flips: 2 × 20 × 1.98 × 100 ÷ 1000 = 7.92 V, or 6.6 percent — well over the limit, and a clear call to upsize the wire.
Copper or aluminum — and solving for size
Switch the material to aluminum and the calculator swaps in aluminum's resistance, which runs about 60 percent higher than copper of the same gauge — #10 aluminum is 2.03 Ω/1000 ft against copper's 1.24, so the same run that dropped 1.79 percent on copper drops 2.93 percent on aluminum. That is why an aluminum feeder is typically run a size or two larger than the copper you would have picked.
Flip the mode to Solve for wire size and the problem runs backward: give it the current, length, source voltage, phase, material and a target percent drop, and it walks the AWG table from smallest to largest and returns the smallest conductor whose drop still lands at or under your target. Ask for 20 A over 100 feet at 240 V three-phase in copper under 3 percent and it lands on #12 (6.86 V, 2.86 percent) — #14 would have blown past at 4.53 percent. It is the fast way to size a run without stepping through the table by hand.