How to use the voltage drop calculator
- Circuit: single-phase or DC (two current-carrying wires) or three-phase (three line conductors).
- Conductor and size: copper or aluminum, and the size in AWG/kcmil (North America) or mm² (IEC sizes used almost everywhere else).
- One-way length: the distance from the panel or battery to the load. The calculator doubles it for the return wire in single-phase and DC circuits.
- Load current and supply voltage. If you know watts rather than amps, convert them first with the watts to amps calculator.
- Allowed drop: the limit you design to, usually 3% or 5%.
You get the drop in volts and percent, the voltage left at the load, the smallest wire size that stays within your limit, the longest run the chosen size allows and the power lost as heat in the cable. The table shows every standard size side by side, so it works as a wire size calculator too.
The voltage drop formula
A cable is a resistor. Its drop is Ohm’s law, V = I × R, applied to the conductor length that carries the current:
- Single-phase and DC: Vd = 2 × L × R × I — current flows out and back, so twice the one-way length.
- Three-phase (balanced): Vd = √3 × L × R × I — the drop between lines; √3 ≈ 1.732.
L is the one-way length, R the conductor resistance per unit length and I the load current. Many textbooks, especially in Europe, write the same formula with the conductivity κ and the cross-section A: ΔU = 2 × L × I × cos φ ÷ (κ × A), with κ ≈ 56 m/(Ω·mm²) for copper and 35 for aluminum at 20 °C. The power factor cos φ in that form applies the drop to the real part of the current; leave it at 1 for the more conservative result.
Worked example: 12 AWG copper, 100 feet, 16 A at 120 V
NEC Chapter 9, Table 8 lists stranded 12 AWG uncoated copper at 1.98 Ω per 1,000 ft at 75 °C. The drop is 2 × 100 ft × 1.98 Ω/1,000 ft × 16 A = 6.34 V, which is 5.28% of 120 V — the load sees only 113.7 V. 10 AWG (1.24 Ω/1,000 ft) still gives 3.31%, so to stay under 3% this run needs 8 AWG (2.07%). The same 16 A at 240 V loses the same 6.34 V, but that is only 2.64% — and a 240 V load of the same power draws half the current, so its drop is a quarter: 1.32%.
Metric example: 2.5 mm² copper, 20 m, 16 A, 230 V. IEC 60228 gives 7.41 Ω/km at 20 °C, so the drop is 2 × 20 m × 7.41 Ω/km × 16 A = 4.74 V (2.06%). At 70 °C — the limit of PVC insulation at full load — the resistance is 20% higher, 8.87 Ω/km, and the drop rises to 5.67 V (2.47%).
Recommended voltage drop limits
| Rule | Recommendation |
|---|---|
| NEC 210.19(A), informational note | 3% on a branch circuit; 5% in total for feeder plus branch circuit |
| NEC 215.2(A), informational note | 3% on a feeder |
| IEC 60364-5-52, Annex G (supply from the public low-voltage grid) | 3% for lighting, 5% for other uses, from the origin of the installation |
| Low-voltage DC (12 V and 24 V systems) | often 3% for critical loads such as navigation lights, up to 10% for non-critical loads (boat and RV wiring guides) |
The NEC notes are recommendations, not requirements; local codes and utilities may set their own limits.
Resistance values and temperature
For AWG and kcmil sizes the calculator uses the DC resistances of NEC Chapter 9, Table 8 (stranded, uncoated copper and aluminum at 75 °C). For metric sizes it uses the maximum resistances of IEC 60228 at 20 °C. Copper’s resistance rises about 0.39% per °C (aluminum 0.40%), so both are corrected to the conductor temperature you enter: 75 °C matches the NEC tables, 70 °C is the maximum for PVC-insulated cable, and 20 °C gives the textbook value for a cold cable.
The calculation ignores the cable’s reactance. That is accurate for small conductors, but from roughly 1/0 AWG or 50 mm² upward, with AC and a low power factor, the real drop is somewhat higher — use NEC Chapter 9, Table 9 or the manufacturer’s impedance data there.
What this calculator does not check
Voltage drop is only one test a cable must pass. The wire must also carry the current without overheating — its ampacity, which depends on insulation, installation method, ambient temperature and grouping (NEC 310.16, IEC 60364-5-52, DIN VDE 0298-4) — and it must be protected by a suitable fuse or breaker, with a short enough disconnection time in a fault. Use the result to compare options, and have the installation designed and checked by a qualified electrician under your local code.
Frequently asked questions
How do I calculate voltage drop?
Multiply the one-way length by 2 (single-phase or DC) or by 1.732 (three-phase), by the wire’s resistance per unit length and by the current. 100 ft of 12 AWG copper (1.98 Ω per 1,000 ft) at 16 A drops 2 × 100 × 1.98 ÷ 1,000 × 16 = 6.34 V.
What is an acceptable voltage drop?
The NEC recommends no more than 3% on a branch circuit and 5% in total; IEC 60364 suggests 3% for lighting and 5% for other loads. On 120 V, 3% is 3.6 V; on 230 V it is 6.9 V.
What size wire do I need for 100 feet?
It depends on the current and voltage. At 120 V and 16 A, 100 ft one-way needs 8 AWG copper to stay under 3%, while 12 AWG gives 5.3%. At 20 A, 8 AWG still works (2.6%); at 30 A it takes 6 AWG. Enter your numbers and read “Smallest size within your limit”.
Is the length one-way or round trip?
One-way: the distance from the panel to the load. The single-phase formula already multiplies by 2 for the return conductor.
Does aluminum wire have more voltage drop?
Yes. Aluminum has about 61% of the conductivity of copper, so a cable of the same size has about 1.6 times the resistance and drop. Aluminum is usually installed one or two sizes larger to compensate.
Last updated: October 19, 2026