Wire Resistance Calculator
Pick an AWG or kcmil size or enter a cross-section, choose the material and conductor temperature, and enter the length of wire.
Resistance
158.8219 mΩ
Resistance per 1,000 ft
1.58822 Ω
Resistance per km
5.21069 Ω
Cross-section
3.3088 mm²
Cross-section
6,530 cmil
Resistivity at the conductor temperature
1.7241 µΩ·cm
How it works
A conductor's resistance is its resistivity times its length divided by its cross-sectional area, R = ρ × L ÷ A. Doubling the length doubles the resistance; doubling the area halves it.
For an AWG size the area comes from the gage's definition: 4/0 is 0.4600 in across and 36 AWG is 0.0050 in, with 38 sizes between them in geometric progression, so every size is 0.005 × 92^((36 − n)/39) in in diameter. Sizes above 4/0 are named by their area in thousands of circular mils (kcmil): 250 kcmil is 250,000 cmil. The resistivities are the published standard values: 1.7241 µΩ·cm for annealed copper (the International Annealed Copper Standard) and 2.8264 µΩ·cm for EC-H19 (1350) aluminum, both at 20 °C.
Metals conduct less well as they warm. The calculator adjusts the 20 °C resistivity with the material's temperature coefficient, ρ_T = ρ₂₀ × (1 + α × (T − 20)), the rule the National Bureau of Standards wire tables use. A stranded conductor is about 2 % more resistive than a solid one of the same area, because its outer strands spiral and so are longer than the cable.
Formula
d (in) = 0.005 × 92^((36 − n) ÷ 39) (AWG n; 1/0 = 0, 2/0 = −1, 3/0 = −2, 4/0 = −3) A (cmil) = (1000 × d)², or kcmil × 1000; A (m²) = cmil × π/4 × (25.4 × 10⁻⁶ m)² ρ_T = ρ₂₀ × (1 + α₂₀ × (T − 20 °C)) R = ρ_T × k × L ÷ A (k = 1 solid, 1.02 stranded)
Example
100 ft of solid 12 AWG copper at 20 °C: the wire is 0.0808 in across (6,529.9 cmil, 3.3088 mm²), so R = 1.7241 × 10⁻⁸ Ω·m × 30.48 m ÷ 3.3088 × 10⁻⁶ m² = 0.1588 Ω (158.8 mΩ), or 1.5882 Ω per 1,000 ft. NBS Handbook 100, Table 5 lists 12 AWG copper at 1.59 Ω per 1,000 ft at 20 °C.
1,000 ft of solid 10 AWG copper at 75 °C is 1.2147 Ω; Handbook 100, Table 6 lists 1.215 Ω. The same wire in EC-H19 aluminum is 2.0004 Ω; Handbook 109, Table 14 lists 2.000 Ω.
Assumptions and limitations
- DC resistance only. The extra AC resistance from skin and proximity effects, and a cable's reactance, are not computed.
- Resistivities and coefficients are the published standard values for annealed copper (NBS Handbook 100, Copper Wire Tables, 1966) and EC-H19 aluminum (NBS Handbook 109, Aluminum Wire Tables, 1972, Table 1). A particular wire's alloy, temper, purity or coating changes them; hard-drawn copper is about 2.5 % more resistive (Handbook 100, Table 5, note 2). Other materials can be entered as Custom.
- AWG areas are computed from the gage's definition. The handbooks print diameters rounded to 0.1 mil, so their tabulated resistances for the finest sizes differ slightly from the exact values shown here.
- Stranded conductors are taken as 2 % above solid, the ASTM figure Handbook 100 uses for sizes up to 2,000,000 cmil; actual stranding factors vary by class and manufacturer by a few tenths of a percent.
- Temperature correction is linear and uses the coefficient at 20 °C, as the handbooks' tables do. Handbook 100 tabulates copper by this rule only from 0 to 200 °C (Table 6) and Handbook 109 aluminum from 0 to 100 °C; temperatures above 200 °C (copper) or 100 °C (aluminum) are refused, and below 0 °C the result is an extrapolation. Custom materials have no upper limit. The conductor temperature is an input; this page does not estimate how hot a wire runs under load.
- This is an estimate of a physical property, not an NEC compliance determination or a substitute for the conductor manufacturer's data. Conductor sizing is decided under the applicable code by a licensed electrician and the authority having jurisdiction.
Frequently asked questions
How much more resistive is aluminum wire than copper of the same gauge?
At 20 °C, EC-H19 aluminum's resistivity is 2.8264 µΩ·cm against copper's 1.7241, so the same size is 1.64 times as resistive; 10 AWG is 0.9988 Ω per 1,000 ft in copper and 1.637 Ω in aluminum in the NBS wire tables.
Why does going up three gauge sizes halve the resistance?
Each AWG step multiplies the diameter by 92^(1/39) ≈ 1.1229, so three steps multiply the area by 1.1229⁶ ≈ 2.005. Twice the area is half the resistance, which is why 10 AWG (0.9988 Ω per 1,000 ft) is close to half of 13 AWG (2.00 Ω).
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