Voltage drop in supply wiring

Current must return too

In a two-wire supply, current travels through both outgoing and return conductors. For equal-length copper wires with equal cross-section, R = ρ·2L/A. L here is the one-way distance between source and load. A three-metre separation therefore means six metres of current-carrying conductor in the loop.

The calculator approximates copper with ρ = 0.0175 Ω·mm²/m at 20 °C and a linear temperature coefficient of 0.00393/K. Higher temperature raises resistance. Material condition and exact alloy can differ; contact resistance in connectors, fuses and terminals is not included in the wire-only model.

Consider voltage and heat together

At 20 °C, a 3 m one-way length and 2.5 mm² cross-section give about 0.042 Ω loop resistance. At 10 A, this drops 0.42 V and dissipates 4.2 W in the wire. A 13.8 V source therefore produces 13.38 V at the radio in this model.

Voltage drop is proportional to current, while dissipation grows with current squared. Double current gives twice the voltage drop and four times the heating at unchanged resistance. More cross-section or less length reduces both. A voltage measurement without load can easily hide these problems.

The model is not complete cable sizing

Compare voltage at the radio under representative load with its documented range. A real installation also depends on current rating, installation method, temperature, connectors and protection. A low calculated voltage drop alone does not verify these points. If calculated drop exceeds source voltage, the assumed combination of current and resistance is not possible.