Voltage Drop Calculator

Choose the circuit type, wire size and material, then enter the current, the one-way distance and the source voltage.

For three-phase, the line-to-line voltage.
Copper's resistance rises about 0.39 % per °C above 20 °C.
Distance from the source to the load, not the total wire length.
A design target you choose; the page only compares the result with it.

Voltage drop

4.956 V

Voltage drop

4.13%

Voltage at the load

115.04 V

Against your limit

Exceeds the 3% limit

Resistance of one conductor

0.1239 Ω

Conductor resistance per 1,000 ft

1.23904 Ω

Resistivity K at the conductor temperature

12.865 Ω·cmil/ft

How it works

Current flowing through a wire loses voltage in the wire's resistance, V = I × R, so less reaches the load. In a two-wire circuit (single-phase AC or DC) the current goes out on one conductor and back on the other, so the drop is twice that of one conductor of the one-way length.

In a balanced three-phase circuit the return currents cancel in the neutral, and the line-to-line drop is √3 (≈ 1.732) times the drop in one conductor rather than twice it. Enter the line-to-line voltage for the percentage.

The conductor's resistance per foot comes from its cross-section and the standard resistivity of annealed copper or EC-H19 aluminum at 20 °C, raised to the conductor temperature you enter with the published temperature coefficient, plus 2 % for a stranded conductor. The product ρ × (cross-section in cmil per foot) is the K value that voltage-drop tables quote; it is shown so you can compare.

Formula

K (Ω·cmil/ft) = ρ₂₀ × (1 + α₂₀ × (T − 20)) × k    (k = 1.02 stranded, 1 solid)
Single-phase or DC:  Vd = 2 × K × I × L ÷ cmil
Three-phase:         Vd = √3 × K × I × L ÷ cmil
Vd % = Vd ÷ V × 100;  voltage at load = V − Vd
ρ₂₀: copper 10.371, aluminum 17.002 Ω·cmil/ft;  α₂₀: 0.00393, 0.00403 /°C

Example

A 20 A load 100 ft from a 120 V single-phase source on stranded 10 AWG copper at 75 °C: K = 10.371 × (1 + 0.00393 × 55) × 1.02 = 12.865 Ω·cmil/ft, and 10 AWG is 10,383 cmil, so Vd = 2 × 12.865 × 20 × 100 ÷ 10,383 = 4.956 V, or 4.13 %, leaving 115.04 V at the load. That exceeds a 3 % limit.

With solid 10 AWG instead, one conductor is 1.2147 Ω per 1,000 ft (NBS Handbook 100, Table 6, lists 1.215 Ω at 75 °C), so 200 ft of conductor drops 20 × 0.24295 = 4.859 V.

Assumptions and limitations

  • DC resistance only. On AC the drop also depends on the conductors' reactance (which grows with conductor size and in steel conduit) and on the load's power factor; neither is included.
  • Resistivity and temperature coefficient are the published standard values for annealed copper (NBS Handbook 100, 1966) and EC-H19 aluminum (NBS Handbook 109, 1972); stranded conductors are taken as 2 % above solid. A particular cable's resistance may differ by a few percent.
  • Temperature correction is linear with the 20 °C coefficient; Handbook 100 tabulates copper by this rule only from 0 to 200 °C (Table 6) and Handbook 109 aluminum from 0 to 100 °C, so temperatures above 200 °C (copper) or 100 °C (aluminum) are refused and results below 0 °C are extrapolations. The conductor temperature is an input and is assumed uniform along the run. The default 75 °C is the temperature at which the widely quoted K values of 12.9 (copper) and 21.2 (aluminum) apply, not a prediction of how hot this circuit runs.
  • Three-phase results assume a balanced load and give the line-to-line drop. The current is taken as constant; in reality a resistive load draws slightly less current as its voltage falls.
  • The drop limit is a value you enter for comparison; this page applies no code limit of its own.
  • This is an estimate, not an NEC compliance determination. Conductor size, ampacity and overcurrent protection are decided under the applicable code by a licensed electrician and the authority having jurisdiction.

Frequently asked questions

Why is K 12.9 for copper in some voltage-drop tables?

K is the resistivity in Ω·cmil/ft. Annealed copper is 10.371 at 20 °C; at 75 °C it is 10.371 × 1.216 = 12.61 for a solid wire, and with the 2 % stranding allowance 12.87, which rounds to 12.9. The same steps give 21.19 for stranded EC-H19 aluminum, quoted as 21.2.

Does three-phase use 2 or 1.732?

For a balanced three-phase load the line-to-line drop is √3 ≈ 1.732 times the drop in one conductor. The factor 2 applies to two-wire single-phase and DC circuits, where the current returns on the second conductor.