Voltage Drop Calculator
Long wire runs lose voltage to the resistance of the conductor. Choose copper or aluminum, the AWG size, the one-way length, the load current, the phase, and the supply voltage, and this calculator returns the voltage drop in both volts and percent — flagged against the NEC 3% guideline.
Why voltage drops over a wire
Every conductor has some resistance, and pushing current through that resistance uses up a little voltage — the longer the run and the higher the current, the more is lost before it reaches the load. Excessive drop makes lights dim, heaters underperform, and motors run hot. Four things drive it: the conductor material (copper beats aluminum), the wire size (fatter is better), the run length, and the current. The usual fix for a long run is simply a larger conductor.
How it’s calculated
The calculator looks up the conductor’s DC resistance per 1000 ft from standard AWG tables, scaled for aluminum (about 1.6× copper’s resistivity). For a single-phase or DC circuit it uses Vdrop = 2 × I × R × L, where L is the one-way length and the factor of 2 accounts for the return conductor. For three-phase it uses Vdrop = √3 × I × R × L. Percent drop divides by the source voltage; the NEC check compares that to 3%.
Uses DC resistance values, which suit typical branch circuits. It ignores conductor reactance and temperature correction, so very long AC runs or high-frequency loads may differ slightly. NEC figures are recommendations, not hard limits.
Copper resistance by gauge
| Wire | Ω / 1000 ft (Cu) | Ω / 1000 ft (Al) |
|---|---|---|
| 14 AWG | 2.525 | 4.14 |
| 12 AWG | 1.588 | 2.60 |
| 10 AWG | 0.999 | 1.64 |
| 8 AWG | 0.628 | 1.03 |
| 6 AWG | 0.395 | 0.648 |
| 4 AWG | 0.249 | 0.408 |
| 2 AWG | 0.156 | 0.256 |
| 1/0 AWG | 0.098 | 0.161 |
DC resistance at 20°C; aluminum ≈ 1.6× copper. Source: standard AWG conductor tables.
Worked example
A 120 V single-phase circuit uses 12 AWG copper to feed a 15 A load 50 ft away. At 1.588 Ω per 1000 ft, 50 ft of conductor is 0.0794 Ω. Voltage drop = 2 × 15 × 0.0794 = 2.38 V, which is 1.98% — comfortably under the NEC 3% guideline, leaving about 117.6 V at the load.
Common mistakes
- Entering the round-trip length — use the one-way run; the formula already doubles it.
- Using copper resistance for an aluminum conductor, understating the drop.
- Forgetting that 3-phase uses √3, not 2, in the formula.
- Treating the 3% figure as a hard code limit rather than an NEC recommendation.
Where it is used
- Sizing wire for long runs to sheds, garages, and well pumps.
- Planning EV charger and subpanel feeders.
- Low-voltage lighting and solar wiring, where drop is a big deal.
- Troubleshooting dim lights or underperforming equipment.
Frequently asked questions
How do you calculate voltage drop?
For a single-phase or DC circuit, voltage drop = 2 × I × R × L, where I is current, R is the wire’s resistance per unit length, and L is the one-way run (the 2 accounts for the return path). For three-phase, use √3 × I × R × L instead of 2.
What is an acceptable voltage drop?
The National Electrical Code recommends keeping voltage drop under 3% on a branch circuit, and under 5% for the combined feeder and branch. These are recommendations in NEC informational notes (210.19 and 215.2), not hard mandates, but exceeding them can cause dim lights and struggling motors.
Does copper or aluminum drop more voltage?
Aluminum has higher resistance than copper for the same gauge, so it drops more voltage over the same length and current. That is why aluminum runs are often upsized one or two gauges compared with copper to hit the same voltage-drop target.
How does wire length affect voltage drop?
Voltage drop is proportional to length: doubling the run doubles the drop. Long runs to outbuildings, well pumps, or EV chargers are where voltage drop matters most, and are usually fixed by choosing a larger conductor.
Why multiply the length by two?
Current has to travel out to the load and back, so the total conductor length carrying current is twice the one-way distance. That is why the single-phase formula uses 2 × I × R × L with L as the one-way run.