Calculate the voltage drop across a wire run from the wire gauge, current, one-way length and system voltage, for both single-phase and three-phase circuits.
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This calculator uses the standard formula: Voltage Drop = (2 × K × I × L) ÷ CM for single-phase circuits, or (√3 × K × I × L) ÷ CM for three-phase circuits, where K is a resistivity constant for the wire material (12.9 for copper, 21.2 for aluminum), I is current in amps, L is the one-way wire length in feet, and CM is the wire's cross-sectional area in circular mils, looked up from standard AWG wire gauge tables.
Excessive voltage drop wastes energy as heat and can cause equipment to run inefficiently or fail to operate correctly. The National Electrical Code recommends keeping voltage drop under 3% for branch circuits and under 5% for the total of feeder and branch circuit combined, though it isn't a hard code requirement in most cases.
A single-phase 120V circuit carrying 20 amps over a 100-foot copper 12 AWG run: voltage drop = (2 × 12.9 × 20 × 100) ÷ 6,530 circular mils = 7.90V, a percent drop of 7.90 ÷ 120 × 100 = 6.58%, leaving 112.10V at the load. That 6.58% exceeds the commonly recommended 3% limit for branch circuits, stepping up to a thicker 10 AWG wire (10,380 circular mils) would roughly cut the voltage drop in proportion to the increase in circular mils.
Entering the round-trip distance instead of the one-way length, this calculator's formula already accounts for the return path internally (via the phase factor of 2 for single-phase), entering the full there-and-back distance effectively double-counts it.
Mixing up copper and aluminum, aluminum has a notably higher resistivity constant (21.2 vs 12.9), using the wrong material setting understates voltage drop on an aluminum run by more than half.
Ignoring a voltage drop percentage over 3% because the circuit still technically works, excessive voltage drop isn't usually an immediate failure, it shows up as wasted energy, dimmer lighting, or equipment running inefficiently over time.
In a single-phase circuit, current flows out and back through two conductors, doubling the effective resistance path. Three-phase circuits are more efficient due to their phase relationships, so the standard formula uses √3 (about 1.732) instead of 2.
Use a larger wire gauge (a lower AWG number means a thicker wire with more circular mils), or shorten the run if possible, both reduce voltage drop for the same current.
No, this is a planning estimate only. Always have electrical work verified by a licensed electrician and checked against your local electrical code.
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