Electrical Wire Sizing Calculator
Determines the minimum wire cross-sectional area needed to keep voltage drop within safe limits for a given load current and run length. Use it when wiring outlets, motors, or solar systems to avoid overheating and energy loss.
Last updated: September 2026
Formula below · 2 sources (asme.org, Wikipedia) · Updated Sep 2026
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About this calculator
Wire resistance causes a voltage drop proportional to current and conductor length. The conductor area needed to keep the drop within a limit is: CM = (2 × K × I × L) / (V × VD), where CM is the area in circular mils, K is the conductor resistivity in ohm-circular mils per foot (12.9 for copper and 21.2 for aluminum at the 75 °C operating temperature used in NEC voltage-drop calculations), I is the load current, L is the one-way distance in feet, V is the system voltage and VD is the allowed drop as a fraction. The factor 2 accounts for the out-and-back path; for a three-phase circuit the multiplier is √3 (1.732), so using 2 errs on the safe side. Choose a standard conductor whose area (NEC Chapter 9 Table 8: 12 AWG 6,530 cmil, 10 AWG 10,380, 8 AWG 16,510, 6 AWG 26,240, 4 AWG 41,740) is at least this value. This is the voltage-drop requirement only: the conductor must ALSO meet the ampacity of NEC Table 310.16 for the load (with 125% for continuous loads) and the NEC 240.4(D) limits, whichever requires the larger wire. NEC recommends limiting voltage drop to 3% for branch circuits and 5% for feeder plus branch combined.
How to use
Suppose you're running a 20 A circuit at 120 V over 75 feet of copper wire, with a maximum 3% voltage drop (0.03). Allowable drop = 120 × 0.03 = 3.6 V. Copper K = 12.9. Area = (2 × 12.9 × 20 × 75) / (120 × 0.03) = 38,700 / 3.6 = 10,750 circular mils. 10 AWG copper (10,380 cmil) is slightly too small, so 8 AWG (16,510 cmil) is the minimum for voltage drop; it also exceeds the 20 A ampacity requirement. The defaults (20 A, 100 ft, 120 V, 3%, copper) need 14,333 cmil, also 8 AWG.
Frequently asked questions
What happens if I use a wire gauge that is too small for the circuit current?
Undersized wire has higher resistance, which causes excessive voltage drop and heat generation. Over time this can degrade insulation, trip breakers, or—in severe cases—start an electrical fire. The NEC mandates minimum wire sizes specifically to prevent these hazards. Always select a gauge whose ampacity exceeds your load current and whose resistance keeps voltage drop within acceptable limits.
How does one-way distance affect wire sizing for long cable runs?
Because current must travel out to the load and return through the neutral or ground conductor, the total resistive path is twice the one-way distance. This means doubling the run length doubles the voltage drop for the same wire size and load. Long runs to outbuildings, motors, or EV chargers often require upgrading one or two AWG sizes beyond what a short run would need, even if the load current is modest.
Why do copper and aluminum wires need different sizes for the same current?
Aluminum has roughly 61% of the conductivity of copper, so it needs about 1.6 times the cross-section for the same resistance (K = 21.2 vs 12.9 here), typically two AWG sizes larger: where 10 AWG copper works, 8 AWG aluminum is usually needed, and the aluminum ampacity column of NEC Table 310.16 must also be checked. Aluminum is lighter and cheaper per foot, making it common for large feeders and service entrances, but connections must use aluminum-rated terminals (and anti-oxidant compound where required) to prevent corrosion and overheating.