Inputs and units
Enter one-way feet, operating amps and system voltage. Start with a candidate AWG size, then change the wire-size selector while leaving the load and length fixed. A smaller AWG number means a larger conductor until the sizes progress into 1/0 and 2/0.
How the calculation works
Drop (V) = 2 × length (ft) × current (A) × resistance (Ω/1,000 ft) / 1,000Use three separate checks. First, determine the conductor ampacity required for the actual installation and load. Second, check that the voltage reaching the equipment meets its requirement. Third, confirm that terminals, protection, connections and physical routing suit that conductor. Passing the voltage-drop check cannot compensate for inadequate ampacity, and a conductor that carries current safely may still lose too much voltage on a long run.
Worked example
A 50 ft one-way run carries 10 A at 24 V. The built-in reference gives 0.999 V drop for 10 AWG, or 4.16%. At 8 AWG the estimate is 0.628 V, or 2.62%. At 6 AWG it is 0.395 V, or 1.65%. This comparison shows the resistance tradeoff; it does not establish which cable construction or protective device the installation needs.
Assumptions and limits
The table covers only the included copper sizes and fixed reference resistance values. It does not select aluminium conductors, flexible cord, automotive cable ratings, underground arrangements or high-temperature installations. The 60°C ampacity display is not a substitute for the applicable ampacity table and adjustments. Use the cable manufacturer’s data and the rules that apply to the actual installation. Do not treat the absence of a warning as design approval.
References and calculation details
Page prepared September 18, 2026. Examples describe this calculator’s implementation; reference links do not imply independent certification.