Every conductor has resistance, so every conductor keeps some of the voltage for itself. $V_{d} = \dfrac{2 \rho L I}{A}$, read aloud V-d equals two rho L I over A. $V_{d}$ is the voltage drop in volts; $\rho$ — rho — is the conductor's resistivity, 0.0172 Ω·mm²/m for copper; $L$ is the one-way run length in metres, panel to load; $I$ is the load current in amperes; and $A$ is the cross-section of one conductor in mm².

That leading 2 is the named trap of this lesson, and it is not a fudge factor: the current goes out along one conductor and comes back along another, and both of them drop volts. Measure the run one way, let the 2 do the return trip, and never do both.

A balanced three-phase run has no return conductor at all — the three currents return through each other — so the 2 becomes $\sqrt{3}$: $V_{d} = \dfrac{\sqrt{3} \, \rho L I}{A}$, with $I$ the line current and $V_{d}$ the line-to-line drop. Same cable, same load, about 13% less lost. That is one of the quiet reasons industry runs three-phase.

Codes are written in percent, not volts: $\%V_{d} = \dfrac{100 \, V_{d}}{V_{s}}$, where $V_{s}$ is the nominal supply voltage. The familiar allowances are 3% for a branch circuit and 5% for the whole path from the service to the load. Three volts is nothing on 400 V and ruinous on 24 V — which is exactly why the rule is a ratio.