Calculating voltage drop is straightforward for DC and single-phase AC: ( V_d = 2 \times I \times R \times L ) (where L is length in meters, R is resistance per meter). For three-phase AC, the formula becomes ( V_d = \sqrt{3} \times I \times (R \cos\phi + X \sin\phi) \times L ), where ( \phi ) is the power factor and X is the inductive reactance. This extra complexity is essential: long motor feeders with poor power factor suffer voltage drops far beyond simple resistance calculations.

[ S = \frac{\sqrt{I^2 \times t}}{k} ]

The is deceptively simple:

A motor running on low voltage will draw higher current (to maintain power), potentially overheating and failing prematurely. Discharge lighting can flicker or fail to strike. For these reasons, most standards limit voltage drop to 3–5% from the service point to the farthest outlet.

Where ( S ) is the minimum cross-sectional area (mm²), ( I ) is the fault current (A rms), ( t ) is the fault clearing time (seconds), and ( k ) is a constant dependent on the conductor material and insulation type (e.g., 115 for copper/PVC, 76 for aluminum/PVC). This equation ensures that the cable’s temperature rise during the fault stays below the insulation’s damage threshold (e.g., 160°C for PVC, 250°C for XLPE).