Wire Resistance Table: Ohms per 1,000 ft by AWG

DC resistance in ohms per 1,000 feet for every AWG from 14 to 4/0, copper and aluminum side by side, with circular mils — NEC Chapter 9 Table 8 values.

SizeCircular MilsCopper Ω / 1,000 ftAluminum Ω / 1,000 ft
14 AWG4,1073.075.05
12 AWG6,5301.933.17
10 AWG10,3801.211.99
8 AWG16,5100.7641.26
6 AWG26,2400.4910.807
4 AWG41,7400.3080.506
3 AWG52,6200.2450.403
2 AWG66,3600.1940.319
1 AWG83,6900.1540.253
1/0 AWG105,6000.1220.200
2/0 AWG133,1000.09670.159
3/0 AWG167,8000.07660.126
4/0 AWG211,6000.06080.0999

How to read ohms per 1,000 feet

The figures above are the conductor's DC resistance for a 1,000-foot length, which is how the NEC and wire manufacturers publish them. To get the resistance of a real run, scale by the length: R = (Ω per 1,000 ft) × feet ÷ 1,000. So 250 ft of 4/0 aluminum, at 0.0999 Ω per 1,000 ft, is 0.0999 × 250 ÷ 1,000 = 0.0250 ohm one-way. A circuit needs a conductor out and a conductor back, so double that for the round-trip value you use to work out voltage drop.

Reading down a column shows the pattern worth remembering: every three steps down in gauge number roughly doubles the cross-section and halves the resistance. Reading across a row shows the metals compared at the same size — aluminum is about 1.64 times the copper figure throughout, which is the whole reason an aluminum feeder is normally specified one or two sizes larger than the copper equivalent for the same load. The circular-mils column is the cross-sectional area that drives both numbers, and it is the quantity the voltage-drop formula wants as CM.

Where the numbers come from: the K factor

The resistivity constant K is the single number that lets a simple voltage-drop formula stand in for a full physics calculation. It is expressed in circular-mil-ohms per foot, and it bundles a conductor material's intrinsic resistivity into a form that pairs directly with wire sizes measured in circular mils. Copper's K is about 12.9 and aluminum's about 21.2 at a typical operating temperature near 75 °C — and that ratio, roughly 1.6 to 1, is exactly why aluminum needs a larger conductor than copper for the same run. The aluminum column above is derived from the copper column by exactly that ratio, so the chart and the calculators can never drift apart.

K appears in both formulas the wire pillar depends on. Voltage drop for a single-phase run is (2 × K × I × L) / CM, and a conductor's resistance is (K × L) / CM, where I is current in amps, L is one-way length in feet and CM is the cross-section in circular mils. Because resistance rises slightly with temperature, K is quoted at a reference temperature; a conductor running hot drops a little more than the nominal figure, which is one more reason to keep a design margin rather than sizing to the exact limit.

These constants feed the Voltage Drop and Wire Resistance calculators. They are stable engineering values, not a live feed, so the tools built on them need no maintenance and always return the same answer for the same inputs.

Sources: NEC Chapter 9 Table 8 (conductor resistance), NEC 310.16 (ampacity), NEC Chapter 9 Tables 1/4/5 (conduit areas & THHN), ASTM B258 (AWG geometry), NREL (peak sun hours) · All sources