How To Calculate Cable Size -

: Select ( I_n = 100A ) (circuit breaker).

: ( I_b = \frac50000\sqrt3 \times 400 \times 0.85 \times 0.94 = \frac50000553.6 \approx 90.3 A ) how to calculate cable size

: The protective device (circuit breaker or fuse) must satisfy: [ I_b \leq I_n \leq I_z ] Where ( I_n ) = nominal rating of protective device, and ( I_z ) = cable’s corrected ampacity. Additionally, for overload protection: [ I_2 \leq 1.45 I_z ] (Where ( I_2 ) is the current ensuring operation of the protective device, typically 1.3–1.45 ( I_n ) for circuit breakers). 3. Voltage Drop Constraint Excessive voltage drop causes poor equipment performance, increased current, and reduced efficiency. Standards (e.g., IEC 60364, BS 7671, NEC) limit total voltage drop from supply to load to typically 3–5% for lighting and 5–8% for other loads. 3.1 Voltage Drop Formula (AC, single-phase and three-phase) For a single-phase circuit: [ V_d = 2 \times I_b \times (R \cos\phi + X \sin\phi) \times L ] : Select ( I_n = 100A ) (circuit breaker)

: Table for 70°C PVC copper, Method C: 50 mm² → 193A base. So select 50 mm². : Voltage drop. For 50 mm²

: Check earth fault loop impedance (not shown in detail here) – likely passes with 70 mm².

: Short-circuit withstand. ( k=115 ). ( S_min = \sqrt10000^2 \times 0.4/115 = \sqrt40\times 10^6/115 = 6324/115 = 55 mm^2 ). Required minimum 70 mm². Our 50 mm² fails. Therefore increase to 70 mm².

: Voltage drop. For 50 mm², AC resistance at 70°C ≈ 0.494 Ω/km, reactance ≈ 0.088 Ω/km. Three-phase drop: [ V_d = \sqrt3 \times 90.3 \times (0.494 \times 0.85 + 0.088 \times \sin(\cos^-10.85)) \times 0.120 ] sinϕ = 0.527. R term = 0.494×0.85 = 0.4199 X term = 0.088×0.527 = 0.0464 Sum = 0.4663 Ω/km per phase. ( V_d = 1.732 \times 90.3 \times 0.4663 \times 0.120 = 8.74V ) Percent = 8.74/400×100 = 2.18% < 3% OK.