Example : 2.5 mm² PVC copper (30 A tabulated), ambient 45°C (k₁=0.79), 4 circuits (k₂=0.65) → effective = 30×0.79×0.65 = 15.4 A. For a 16 A load, this cable fails. Increase to 4 mm². Voltage drop reduces torque in motors, causes flicker in lights, and wastes energy.
$V_d = \frac2 \times L \times I \times (R \cos\phi + X \sin\phi)1000$ (L in meters, Vd in volts)
$V_d = \frac\sqrt3 \times L \times I \times (R \cos\phi + X \sin\phi)1000$
Example : 230 V single-phase, L=80 m, I=20 A, cosφ=0.85, 4 mm² Cu (R=4.6 Ω/km, X=0.09). Vd = [2×80×20×(0.0046×0.85 + 0.00009×0.526)] / 1000 = 12.8 V → 5.6% > 3%. Fail. Increase to 6 mm². During a short circuit, heat is generated faster than it can dissipate (adiabatic process). The cable must survive until protection clears the fault.
Drainage Nottingham