The Second Law You Meet on Every Job Site
Ohm's law governs the load; the same law governs the wire feeding it. Every conductor has resistance — a 12 AWG copper wire measures 1.588 Ω per 1,000 ft — and the current flowing through it produces a voltage drop exactly like a resistor in the circuit. The subtle part is the round trip: the current leaves the panel on the hot wire and returns on the neutral, so a 100 ft run is 200 ft of copper. That doubles the drop, which is why the number in the table above is 0.3176 Ω, not 0.1588 Ω.
volts dropped = amps × conductor ohms
Vdrop = I × R, with R = 2 × length × resistance-per-unit-length
Compare Vdrop to 3% of the supply voltage — the NEC branch-circuit ceiling.
Real Runs, Real Drops (12 AWG Copper)
| current | round-trip ohms | volts dropped | % of 120 V | Verdict |
|---|---|---|---|---|
| 15 A | 0.3176 Ω (100 ft) | 4.76 V | 4.0% | Over 3% — upsize to 10 AWG or shorten the run. |
| 15 A | 0.1588 Ω (50 ft) | 2.38 V | 2.0% | Within the 3% rule. |
| 12 A | 0.3176 Ω (100 ft) | 3.81 V | 3.2% | Borderline — the usual argument for 10 AWG on 100 ft circuits. |
| 20 A | 0.1268 Ω (100 ft, 10 AWG) | 2.54 V | 2.1% | 10 AWG carries 20 A over 100 ft comfortably. |
| 5 A | 0.3176 Ω (100 ft) | 1.59 V | 1.3% | Lighting circuit — fine, but the same run at 12 V would be 13%. |
The last row is the trap that catches solar and marine installers: the identical wire and current that lose 1.3% on a 120 V circuit lose 13% on a 12 V circuit, because the drop in volts is the same but the supply is ten times smaller. That is why 12 V systems get short, fat wire runs and 48 V e-bike and solar systems exist at all — quadrupling the voltage quarters the current and the percentage drop together. Every row above is I × R; the only variable that changes is which voltage you measure the percentage against.
Engineering Context
Voltage drop is Ohm's law applied to the wire, so its roots are the same family as ohms to amps and volts to amps. The conductor resistance comes from the wire's gauge — AWG to mm² translates the US gauge to the metric cross-section, and mm² to AWG back. The current in the circuit is set by the load: watts to amps at your supply voltage. And when the drop eats too much of a battery system, the fix is usually rethinking the voltage — see amp hours to watt hours and watt hours to amp hours for the pack side of that trade. All paths lead to the electric hub.
More: AWG to mm² · W to A · Ω to A · Electric Hub
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Frequently Asked Questions
How do I calculate voltage drop?
Multiply the current by the total conductor resistance: Vdrop = I × R. For a 15 A load on 100 ft of 12 AWG copper (0.3176 Ω round-trip), the drop is 15 × 0.3176 = 4.76 V. Count the round trip — outbound and return — because the current travels the wire twice.
What is the 3% voltage drop rule?
NEC guidance keeps branch-circuit voltage drop to no more than 3% at the farthest outlet, and 5% total including feeders. For a 120 V circuit that is 3.6 V of drop; for a 240 V circuit, 7.2 V. Above that, motors overheat, electronics brown out, and lighting dims. The rule is why long runs need a wire size up from what ampacity alone would allow.
Why does voltage drop matter more on 12 V systems?
Because the same percentage costs far fewer volts. A 4% drop on 120 V is 4.8 V — usually harmless. A 4% drop on a 12 V trolling-motor or lighting circuit is 0.48 V, and a 12 V system is often already near the edge of its operating range. This is why 12 V wiring uses short heavy runs: wire resistance matters ten times more at one-tenth the voltage.