How this instrument works
A conductor is not a perfect wire; it has resistance, and any current forced through that resistance sheds some of the supply voltage as heat before it reaches the load. Voltage drop is that loss, expressed either in volts or as a percentage of the 220 V supply. The factor of two in the sizing formula exists because current has to travel the length of the circuit and then back again through the return conductor — a 30 ft run means 60 ft of copper actually carrying current, not 30.
Circular mils are the wire industry's own area unit: the diameter in thousandths of an inch, squared, with no need for π. It survives in wiring practice because the annealed-copper resistivity constant, K = 12.9 ohm·circular-mil per foot, was tabulated against it long before metric units took over electrical work. Rearranging Ohm's law, R = K·L ⁄ CM, for the cross-section a target drop allows gives CM = 2·K·I·L ⁄ VD — more current, a longer run, or a tighter drop budget all demand more copper.
The result is a floor, not a recommendation: it is the smallest circular-mil area that keeps the drop inside budget, so the wire gauge chosen must round up to the next standard size, never down. The formula also assumes ordinary resistive behaviour at a typical wiring temperature — heavily inductive loads, very long high-current feeders, or conductors bundled hot inside conduit carry extra impedance this simple K-factor arithmetic does not capture, and those cases call for the fuller AC tables in a wiring handbook.
- Enter the Circuit current the load draws, in amps.
- Enter the One-way wire run length, ft — the distance from the panel to the load, not the round trip.
- Set the Allowed voltage drop, % — 3% is the usual branch-circuit budget; the Allowed voltage drop, V field converts it to volts automatically.
- Read Minimum wire size, circular mils, then round up to the nearest standard wire gauge with at least that area.
Worked example — 16 A on 220 V over a 30 ft run
Take a 220 V appliance circuit — the standard household and workshop supply across most of the world outside North America — drawing 16 A, with the panel 30 ft from the outlet and a 3% voltage-drop budget, a common branch-circuit target. The allowed drop in volts follows straight from the percentage: VD = 220 × 3 ÷ 100 = 6.6 V, the most the conductor is permitted to lose along that run before the load sees meaningfully less than 220 V.
Feeding those numbers into the sizing formula: CM = 2 × 12.9 × 16 × 30 ÷ 6.6 = 1,876.36 circular mils, the minimum copper cross-section that keeps the drop at or under 6.6 V. Standard 14 AWG wire carries about 4,110 circular mils and 16 AWG about 2,580 — both clear the requirement, so 16 AWG is the leaner choice on this run, while 18 AWG's roughly 1,620 circular mils would fall short. Run the identical circuit at 110 V instead of 220 V and the allowed drop halves to 3.3 V; because circular mils scales inversely with allowed drop, the wire would need to nearly double in area, to about 3,753 circular mils — a clean illustration of why doubling the supply voltage roughly halves the copper a given percentage-drop budget demands.
Questions
Why does the formula multiply the length by 2?
Because current flows out to the load through one conductor and back through another, so a 30 ft one-way run means 60 ft of resistive copper carries the current. Doubling the length inside the formula — CM = 2 × K × I × L ÷ VD — accounts for that return path; entering the round-trip distance instead would double-count it and oversize the wire.
What is the K value of 12.9 in the formula?
It is the approximate DC resistance, in ohms, of one foot of copper wire with a cross-sectional area of one circular mil, at the temperature electrical handbooks standardize on for building-wiring calculations. Aluminum's equivalent constant is about 21.2 — noticeably higher, which is why aluminum circuits need a larger cross-section than copper for the same current and drop.
Why use circular mils instead of square millimetres?
Circular mils are inherited from the American Wire Gauge system, where a conductor's area is the square of its diameter in thousandths of an inch — no π required, since every AWG table and the K constant already assume that unit. Converting to mm² is possible (1 circular mil ≈ 0.0005067 mm²), but the K-factor formula and standard wire charts both work natively in circular mils.
How do I turn the circular-mil result into a wire gauge?
Round up to the nearest standard AWG size whose rated circular-mil area is at least the calculated minimum — never down. In the worked example, 1,876 circular mils calls for 16 AWG (about 2,580 circular mils) or anything heavier; 18 AWG, at roughly 1,620 circular mils, falls short of the requirement.
Is a 3% voltage drop always the right target?
It is a common branch-circuit guideline, not a physical law — many wiring codes suggest 3% for branch circuits and up to 5% combined with the feeder, but motors, electronics, and long agricultural or industrial runs often call for tighter limits. Change the Allowed voltage drop, % field to match whatever standard or manufacturer spec applies to the installation.
Does this formula account for wire temperature or AC effects?
No — K = 12.9 assumes ordinary DC-like resistance at a typical wiring temperature and ignores skin effect, reactance, and power factor, which only matter on larger conductors or higher-frequency AC systems. For everyday branch circuits the resistive approximation is accurate; for large industrial feeders, consult full AC impedance tables instead.