Wire Gauge & Ampacity Calculator
Determine the minimum wire cross-sectional area (in mm²) needed to safely carry a given current over a set distance without exceeding your allowed voltage drop. Essential for wiring solar systems, EV chargers, and long cable runs.
Last updated: May 2026
About this calculator
Voltage drop in a conductor is governed by Ohm's Law applied to the wire's own resistance: V_drop = I × R_wire, where R_wire = ρ × L / A (resistivity ρ, length L, cross-sectional area A). For a two-conductor run the current travels out and back, so the total length is doubled: V_drop = 2 × I × ρ × L / A. Rearranging for the minimum area gives A = 2 × I × ρ × L / V_drop, with the run distance converted from feet to metres (× 0.3048) and the allowable drop equal to voltage × (voltageDrop% / 100). The calculator deliberately uses resistivity at 75 °C conductor operating temperature — copper 2.14 × 10⁻⁸ Ω·m and aluminium 3.52 × 10⁻⁸ Ω·m, equivalent to the NEC K-factors of 12.9 and 21.2 Ω·cmil/ft — rather than the lower 20 °C handbook values, because a loaded conductor runs hot and sizing it on cold-wire resistance would undersize it by roughly 20%. The computed area is rounded up, never down. Two safety notes: always step up to the next standard conductor size (metric mm² or AWG) at or above the result, and check the NEC ampacity table as a separate constraint — voltage drop is only one of the two limits, and a 20 A branch circuit needs at least 12 AWG copper no matter how short the run is.
How to use
Scenario: a 120 V branch circuit carrying 20 A over a 100-foot run in copper, with the NEC-recommended 3% maximum voltage drop. Enter current = 20, distance = 100, and select voltage = 120, voltageDrop = 3, wireType = copper. Step 1 — allowable drop: 120 × 0.03 = 3.6 V. Step 2 — run length in metres: 100 × 0.3048 = 30.48 m. Step 3 — minimum area: A = 2 × 20 × 2.14 × 10⁻⁸ × 30.48 / 3.6 = 7.247 × 10⁻⁶ m². The calculator returns 7.25 mm² (rounded up) — step up to the next standard size, 8 AWG (8.37 mm²), which also satisfies the ampacity requirement for 20 A. Switching wireType to aluminum raises the result to 11.93 mm², calling for 6 AWG aluminium (13.3 mm²); running the same load at voltage = 240 instead halves the area to 3.63 mm², where 10 AWG (5.26 mm²) suffices.
Frequently asked questions
What is the difference between wire ampacity and voltage drop when sizing a cable?
Ampacity is the maximum continuous current a wire can carry without overheating, determined primarily by its cross-sectional area and insulation temperature rating. Voltage drop is a separate concern — even a wire with adequate ampacity can cause excessive voltage drop if the run is very long, reducing efficiency and potentially damaging equipment. Best practice is to calculate both: choose the wire size that satisfies whichever constraint is more demanding. For runs longer than about 50 feet at low voltages, voltage drop is almost always the limiting factor.
Why does a longer wire run require a larger gauge wire for the same current?
Resistance increases proportionally with conductor length (R = ρ × L / A), so a longer wire drops more voltage for the same current. To keep the voltage drop within an acceptable percentage, the cross-sectional area must increase to reduce resistance. This is why a 20 A circuit in a 200-foot run may need 4 AWG wire, while the same 20 A load just 10 feet away is safely served by 12 AWG. The trade-off is cost and weight of the larger conductor.
How much voltage drop is acceptable for residential or automotive wiring?
The NEC recommends a maximum of 3% voltage drop on any branch circuit and no more than 5% total from the service panel to the end device for residential wiring. Automotive and marine systems typically target 3% or less because 12 V systems are particularly sensitive — a 3% drop is only 0.36 V, but it can noticeably affect motor speed and LED brightness. For sensitive electronics or critical loads, aim for 2% or below. Always check the equipment manufacturer's voltage tolerance before finalising wire size.