Wire Size Calculator
Find the minimum wire cross-sectional area in circular mils needed to keep voltage drop within code limits for any DC or single-phase AC circuit. Use this when wiring long runs to panels, subpanels, motors, or outdoor equipment.
Last updated: September 2026
Formula below · 2 sources (ieee.org, Wikipedia) · Updated Sep 2026
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About this calculator
Wire size is set by two separate limits, and the calculator returns the smallest standard copper conductor (14 AWG to 750 kcmil) that meets both. Voltage drop: the required area is CM = (2 × I × d × 12.9) / (V × VD% / 100), where 12.9 ohm-cmil/ft is the copper K-factor at 75 °C, the factor 2 accounts for the round trip and d is the one-way distance in feet; the calculator checks each size with its NEC Chapter 9 Table 8 resistance (Vdrop = 2 × I × d × R / 1,000). The NEC recommends 3% or less for branch circuits and 5% or less for feeder plus branch combined (informational notes to 210.19 and 215.2). Ampacity: from NEC Table 310.16, using the 60 °C column for required ampacities of 100 A or less (the 110.14(C)(1)(a) default) and the 75 °C column above that, with the 240.4(D) limits (14 AWG 15 A, 12 AWG 20 A, 10 AWG 30 A); continuous loads are multiplied by 1.25 (210.19(A)(1)). Sizes 1/0 AWG and larger are shown as text. Assumes copper, 30 °C ambient, no more than three current-carrying conductors together and a single-phase two-wire run (a three-phase run with the same inputs drops less, so the result stays on the safe side). Aluminum conductors need larger sizes (K ≈ 21.2).
How to use
Example: You need to wire a 20 A, 120 V circuit to an outbuilding 75 ft away, with a maximum 3% voltage drop. Step 1 — Enter Current: 20 A. Step 2 — Enter Distance: 75 ft. Step 3 — Select 120 V and 3%. Step 4 — Required area: (2 × 20 × 75 × 12.9) / (120 × 0.03) = 38,700 / 3.6 = 10,750 circular mils. Step 5 — 10 AWG copper (10,380 CM) is slightly under: its drop is 2 × 20 × 75 × 1.24 / 1,000 = 3.72 V, more than 3.6 V. 8 AWG (16,510 CM) drops 2.33 V. Result: 8 AWG. The default 20 A over 100 ft at 120 V also returns 8 AWG; at 240 V it returns 10 AWG.
Frequently asked questions
What percentage of voltage drop is acceptable for residential wiring circuits?
The NEC recommends, but does not mandate, a maximum of 3% voltage drop for individual branch circuits and 5% total for the combined feeder and branch circuit. Exceeding 3% on a branch circuit causes lights to dim noticeably, motors to run hotter and less efficiently, and sensitive electronics to malfunction. For critical loads like medical equipment or data centers, designers often target 1–2% to provide additional margin. Voltage drop limits are among the most commonly overlooked NEC recommendations in DIY and light-commercial work.
How does wire run distance affect the required wire gauge?
Wire resistance increases linearly with length, so doubling the run distance doubles the voltage drop for the same wire size. To maintain the same percentage voltage drop over twice the distance, you must reduce resistance by half, which means approximately doubling the wire's cross-sectional area — roughly two AWG sizes larger. This is why a 14 AWG wire is perfectly adequate for a 15 A outlet 10 ft from the panel but completely inadequate for the same outlet 150 ft away. Always calculate voltage drop for long runs rather than relying on ampacity tables alone.
Why does the wire size formula use a factor of 2 for the distance?
The factor of 2 accounts for the fact that current travels the full length of the circuit twice — once through the hot (line) conductor to the load, and once back through the neutral (return) conductor to the source. Both conductors contribute resistance, so the effective resistive length is twice the one-way physical distance. In three-phase circuits, the factor changes because the return path is shared among three phases, which is why three-phase voltage drop formulas use √3 instead of 2.