Calculate a cross-section from a voltage-loss target
The section of a conductor affects its electrical resistance. For a fixed route and current, increasing that section reduces the voltage lost in the cable. This tool rearranges the resistive two-wire DC equation to find a mathematical minimum section for your chosen percentage loss. It reports a continuous area in square millimetres and circular mils. It does not select a standard cable, an AWG number, a breaker or an installation method. A commercially available conductor must satisfy additional requirements beyond this single calculation.
The inverse resistance equation
Allowed loss = supply voltage × percentage ÷ 100; section (mm²) = 2 × resistivity × one-way metres × current ÷ allowed lossResistivity is entered in ohm square millimetres per metre in both unit modes. Imperial route lengths are converted from feet to metres before the equation runs. The factor two represents equal outward and return conductors. The sample resistivity of 0.0237 follows the copper resistance factor used in the Schneider Electric installation guide. It is an example operating-condition value, not a universal constant for every copper cable. Replace it when manufacturer data or the design temperature requires another factor. A circular mil is the area of a circle with a diameter of one thousandth of an inch; it is not a square mil.
A thirty-metre circuit example
Take a 230 volt supply, a 20 amp steady load and a 30 metre one-way cable route. A 3 percent loss limit allows 230 × 0.03 = 6.9 volts. At a resistivity factor of 0.0237, the continuous minimum section is 2 × 0.0237 × 30 × 20 ÷ 6.9 = 4.122 square millimetres. That number answers the voltage-loss question only. A cable close to that area could still fail a thermal or protection requirement, so the result must not be read as an installation recommendation. Reducing the permitted loss increases the computed area in inverse proportion.
Limits of a section-only estimate
Actual resistance changes with temperature, conductor construction and material. Terminations and joints also add resistance that is absent from this ideal equation. AC circuits can require reactance and power-factor terms, and a motor may impose a much higher starting current than its steady running value. This page covers none of those effects. Separately verify current-carrying capacity, fault protection, mechanical requirements and applicable installation rules with a qualified designer. Percentage loss must be greater than zero and no more than 100. At zero load current the mathematical area is zero, which is an unloaded-circuit result, not permission to omit a conductor or choose a physically zero-sized cable.
Frequently asked questions
Does the output approve an installed wire size?
No. It is a voltage-drop-only continuous section. Ampacity, protective devices, fault conditions and installation rules must be verified separately.
Why is the route length doubled?
The current uses an outgoing and a returning conductor. Enter only the one-way distance; the equation includes the equal return route itself.
Can I obtain an AWG number here?
The tool reports section and circular mils without selecting a gauge. A standard size and its actual resistance must be checked against the full design.
Is resistivity the same as resistance per metre?
No. Resistivity includes cross-section in its units. Divide resistivity by a conductor area to obtain its ideal resistance per metre.
What if I enter zero permissible drop?
A finite resistive conductor cannot meet an exactly zero loss target under load. The calculator rejects a zero percentage rather than returning an infinite area.
Last reviewed: 2026-10-03 · How we calculate