: ( I_b = \frac50000\sqrt3 \times 400 \times 0.85 \times 0.94 = \frac50000553.6 \approx 90.3 A )
For a three-phase circuit (line-to-line): [ V_d = \sqrt3 \times I_b \times (R \cos\phi + X \sin\phi) \times L ] how to calculate cable size
: Select ( I_n = 100A ) (circuit breaker). : ( I_b = \frac50000\sqrt3 \times 400 \times 0
: 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. Three-phase drop: [ V_d = \sqrt3 \times 90
: 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 ]