Already have a wire size in mind? This runs the drop math on it: volts lost over the run, the percentage, what actually arrives at the load, and whether it clears the 3% recommendation.
The formula behind the number
Every conductor is a resistor. The classic field formula multiplies the round-trip length by current and a material constant, divided by the wire's cross-section in circular mils:
VD = (2 × K × I × L) ÷ CM where K = 12.9 (Cu) or 21.2 (Al)
The NEC's guideline (it's a recommendation, not a hard rule) is 3% max on a branch circuit and 5% total including the feeder. Motors, compressors, and pumps are the loads that punish you for ignoring it — low voltage means high current, and high current means heat.
Worked example
#12 copper feeding a 20 A load 150 ft away on 120 V: VD = (2 × 12.9 × 20 × 150) ÷ 6,530 = 11.9 V — a 9.9% drop. Massive fail. Bumping to #6 copper cuts it to about 2.9 V (2.5%) and passes. This is why "the breaker fits" is not the same as "the wire is right."
Reading your result
Under 3%: run it. Between 3–5%: acceptable for non-critical loads like lighting, marginal for motors. Over 5%: expect dim lights, hot motor windings, and nuisance trips — size up until it passes.
Frequently asked questions
Is 3% voltage drop a code requirement?
In the NEC it appears as a recommendation (informational note), not an enforceable rule for most circuits — but some local codes and many equipment manufacturers make it binding. Treat 3% as the design target either way.
Why does 12V wiring have such huge voltage drop problems?
Because the percentage is calculated against the system voltage. Losing 1.2 volts is 1% on a 120V circuit but a brutal 10% on a 12V circuit. That’s why trailer, solar, and automotive runs use very thick wire for small loads.
Does voltage drop waste electricity?
Yes — the lost volts become heat in the wire, which you pay for. On a long, heavily loaded, undersized run it’s a small but permanent tax on every hour the load operates, on top of the performance problems.