Example : Isc = 3 kA, t = 0.1 s (breaker trip), Cu/XLPE, k=143. $S_min = \sqrt(3000^2 × 0.1) / 143 = \sqrt900,000 / 143 = 948 / 143 = 6.6 mm²$. Minimum = 6.6 → choose 10 mm² (next standard size).
$V_d = \frac\sqrt3 \times L \times I \times (R \cos\phi + X \sin\phi)1000$ how to size a cable
| Copper, XLPE, 90°C, 30°C ambient, free air | 1.5 mm² → 24 A | 2.5 mm² → 32 A | 4 mm² → 42 A | | Aluminum, PVC, 70°C, buried | 16 mm² → 70 A | etc. | Example : Isc = 3 kA, t = 0
Cable sizing is not merely about matching a conductor to a load current. It is a multi-variable optimization problem that ensures safety, reliability, efficiency, and longevity of an electrical installation. An undersized cable causes overheating, voltage drops, energy losses, and fire hazards. An oversized cable wastes material, increases installation costs, and may create termination difficulties. $V_d = \frac\sqrt3 \times L \times I \times
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.
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.
$V_d = \frac2 \times L \times I \times (R \cos\phi + X \sin\phi)1000$ (L in meters, Vd in volts)