The formula, in the forms you will actually use
Single-phase, resistance only (conductor resistance in ohms per 1000 ft, length in ft):
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VD = 2 × I × L × R / 1000
The 2 is the round trip: current goes out on one conductor and back on the other.
Three-phase, resistance only:
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VD = √3 × I × L × R / 1000
With reactance — the form used for feeders above roughly 1/0 AWG, where X is no longer negligible:
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VD = 2 × I × L × (R·cosφ + X·sinφ) / 1000 (single-phase)
VD = √3 × I × L × (R·cosφ + X·sinφ) / 1000 (three-phase)
Metric, the IEC tabulated form (mV per amp per metre from the cable tables):
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VD = (mV/A/m) × I × L / 1000 (L in metres)
Circular-mil sizing — solves for conductor size instead of drop:
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CM = 2 × K × I × L / VD K ≈ 12.9 for copper, 21.2 for aluminium
And always: %VD = VD / V_nominal × 100. Percentage is against the nominal voltage of the system the load sees, not against 120 V by habit.
Limits: where 3% and 5% come from
| Circuit | NEC guidance | IEC 60364-5-52 Annex G |
|---|---|---|
| Branch circuit | 3% maximum | 3% for lighting |
| Feeder | 3% | 5% for other uses |
| Feeder + branch combined | 5% | 5% (public distribution) |
| Private installations | — | 6% / 8% may be used by agreement |
| Status | Informational notes, not mandatory rules | Informational annex |
Two things get lost in translation. First, NEC 210.19(A) Informational Note No. 4 and 215.2(A)(1) Informational Note No. 2 are guidance, not enforceable requirements — but designers, specifiers and clients routinely make them contractual, and a 7% drop on a 120 V circuit will be rejected on review regardless of what the code says. Second, IEC Annex G allows wider limits for private installations; that is permission, not a target. Check the current code edition and the project specification, in that order.
Worked example 1: 240 V single-phase branch
A 100 A load, 200 ft from the panel, 2 AWG copper, power factor 0.85.
From the conductor tables: R = 0.194 Ω/1000 ft, X = 0.060 Ω/1000 ft. With cosφ = 0.85, sinφ = 0.527.
- R·cosφ = 0.194 × 0.85 = 0.1649
- X·sinφ = 0.060 × 0.527 = 0.0316
- Sum = 0.1965 Ω/1000 ft
VD = 2 × 100 × 200 × 0.1965 / 1000 = 40,000 × 0.1965 / 1000 = 7.86 V
%VD = 7.86 / 240 × 100 = 3.28% — over the 3% branch guidance.
At unity power factor the same conductor gives 2 × 100 × 200 × 0.194 / 1000 = 7.76 V, or 3.23%. Reactance added only 0.1 V here; at this size it is nearly irrelevant, which is why simple calculators ignore it and usually get away with it.
To get under 3% (7.2 V), move up one size to 1/0: R = 0.122, X = 0.050.
- 0.122 × 0.85 = 0.1037; 0.050 × 0.527 = 0.0264; sum = 0.1301
- VD = 40,000 × 0.1301 / 1000 = 5.20 V, or 5.20 / 240 = 2.17%
One size, 1.1 percentage points. That is the usual shape of the trade: cable cost versus a number nobody sees until the lights flicker.
Worked example 2: 400 V three-phase feeder (IEC)
An 80 A feeder, 45 m, 35 mm² copper, three-phase, from the IEC mV/A/m tables (approximately 1.10 mV/A/m for this size at typical power factor):
VD = 1.10 × 80 × 45 / 1000 = 3,960 / 1000 = 3.96 V
%VD = 3.96 / 400 × 100 = 0.99% — comfortably inside the 5% limit, which for 400 V is 20 V.
Now take a smaller circuit on the same 400 V system: 4 mm², roughly 9.5 mV/A/m, carrying 32 A — the tabulated capacity of 4 mm² at method C in PVC (42 A in XLPE). Over 80 m:
VD = 9.5 × 32 × 80 / 1000 = 24,320 / 1000 = 24.3 V, or 24.3 / 400 = 6.08%
Ampacity says 4 mm² is adequate — 32 A sits inside its thermal rating. Voltage drop says it is not, on this run. The same 80 m on 35 mm² (1.10 mV/A/m) gives 1.10 × 32 × 80 / 1000 = 2.82 V, or 0.70%. That is the classic long-run mistake: the conductor is chosen on amps and the circuit fails on volts.
If you would rather not run these by hand: try the calculation in the demo project.
Quick reference: 240 V, 100 A, 200 ft, unity PF
| Conductor | R, Ω/1000 ft | VD, V | %VD |
|---|---|---|---|
| 12 AWG | 1.93 | 77.2 | 32.2% |
| 10 AWG | 1.21 | 48.4 | 20.2% |
| 8 AWG | 0.764 | 30.6 | 12.7% |
| 6 AWG | 0.491 | 19.6 | 8.2% |
| 4 AWG | 0.308 | 12.3 | 5.1% |
| 2 AWG | 0.194 | 7.8 | 3.2% |
| 1/0 AWG | 0.122 | 4.9 | 2.0% |
Read it as a sensitivity table, not a recommendation: drop is inversely proportional to area, so halving it needs roughly twice the copper (12 AWG to 8 AWG is 2.5× the area for 2.5× less drop), and past 1/0 the AC resistance falls slower than the area grows.
Where voltage drop calculators go wrong
- Forgetting the round trip. Using one-way length halves the answer. The most common single error, and the hardest to spot because the result still looks plausible.
- Resistance at the wrong temperature. Tabulated values are at 75 °C for NEC Chapter 9 Table 8. Using 20 °C resistance underestimates drop by roughly 18%.
- Ignoring power factor on large feeders. Below 1/0 AWG it barely matters; at 500 kcmil with PF 0.8 it changes the answer materially.
- Parallel conductors. Two runs of 250 kcmil are not "250 kcmil". Resistance halves, and the reactance term needs the paralleled equivalent.
- Aluminium treated as copper. K = 21.2 versus 12.9 in the circular-mil method; a 64% difference in required area.
- Applying 3% to feeder plus branch separately and calling it 3% total. If both are at 3%, the load sees 6%.
- DC and AC mixed up. DC resistance with no skin effect is not the AC value at 60 Hz for large conductors.
The cable size calculator handles the ampacity side of the same decision; both results feed the panel schedule. Voltage drop and ampacity are independent checks and a conductor must pass both.
Motor starting and other dynamic cases
Running current is the easy case. Starting current is typically 5–8 times full-load current for a direct-on-line induction motor, and the drop during the start is what decides whether the motor actually accelerates or the contactor drops out.
A 55 A motor with 6× starting current (330 A), 35 mm² feeder, 40 m:
VD = 1.10 × 330 × 40 / 1000 = 14.5 V, or 3.6% of 400 V — usually acceptable.
At 100 m: 1.10 × 330 × 100 / 1000 = 36.3 V, or 9.1%. That is where starters, soft starters and oversized feeders enter the conversation. A common design target is no more than about 15% at the motor terminals during starting, but the manufacturer's data and the applicable standard govern; verify both rather than trusting a rule of thumb.
Next step
If you want voltage drop to run on every circuit instead of one at a time, open the demo project: it includes a plan, a panel and the calculation set, and the drop percentages update as you change conductor sizes. Export formats and the full calculation list are in the docs; plan limits are on the pricing page.
Sources
- NFPA 70 (NEC) — code development and current edition: https://www.nfpa.org/codes-and-standards/nfpa-70-standard-development/70
- NFPA free access to codes and standards: https://www.nfpa.org/codes-and-standards/free-access
- IEC 60364-5-52:2009 — wiring systems, including Annex G voltage drop: https://webstore.iec.ch/en/publication/1878
- IEC 60364-5-53:2001 — selection and erection of switchgear: https://webstore.iec.ch/en/publication/1880
- Southwire voltage drop calculator (independent cross-check): https://www.southwire.com/calculator-vdrop
Values are from the editions current at the time of writing. Verify against the current edition of the applicable code and any local amendments before issuing a design. This is technical guidance, not a compliance opinion.
In the NEC it is guidance in informational notes, not an enforceable rule in most editions. In IEC 60364-5-52 Annex G it is also informational. Specifications, utility interconnection rules and client standards routinely make it mandatory. Check the current edition and the project documents.
Because the circuit is a loop: current travels to the load and back. Three-phase uses √3 instead, because the return path is shared between phases and the geometry differs.
It matters in proportion to the reactance term. For small conductors on short runs the effect is negligible; for large feeders and long runs it can move the result by several percent.
The circular-mil method is quick, conservative, and ignores reactance and power factor. The resistance/reactance method is more accurate and is what serious calculators use. For a first pass, circular mils; for the issued design, R and X.
Resistance divides by the number of sets, reactance roughly divides too, and each set still carries its own ampacity. Two sets of 3/0 are not equivalent to one 500 kcmil for every purpose — check both ampacity and drop.
The nominal voltage at the load terminals: 120 V for a line-to-neutral branch, 240 V or 208 V for line-to-line, 400 V for a three-phase IEC system. Using the wrong base is a common source of percentages that look fine and are not.
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