: 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.

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

[ I_z = I_ref \times k_temp \times k_group \times k_soil \times k_depth \times k_therm ]

Total cost over ( N ) years at interest rate ( r ): [ PV_loss = E_loss \times \textelectricity price \times \frac(1+r)^N - 1r(1+r)^N ] Optimum conductor size is where incremental capital cost equals incremental loss savings. Following IEC/BS 7671 approach: Step 1: Determine Design Current ( I_b ) [ I_b = \fracP\sqrt3 \times V_L \times \cos\phi \ \text(three-phase) \quad \textor \quad \fracPV_P \times \cos\phi \ \text(single-phase) ] Add margin for future expansion (typically 20%). Step 2: Select Protective Device Rating ( I_n ) Choose standard rating (e.g., 16, 20, 25, 32, 40, 50, 63A) such that ( I_n \ge I_b ). Step 3: Determine Installation Method and Correction Factors From standards tables (e.g., IEC 60364-5-52 Table B.52.1 for installation methods: clipped direct, trunking, buried, etc.). Apply ( k_temp ), ( k_group ), ( k_soil ), etc. Step 4: Calculate Required Tabulated Ampacity ( I_tab ) [ I_tab \ge \fracI_nk_total ] Where ( k_total ) = product of all correction factors. Step 5: Select Cable Size from Manufacturer Table Pick smallest standard size (mm²) with base rating ≥ ( I_tab ). Step 6: Check Voltage Drop If drop exceeds limit, increase size until compliant. Step 7: Check Short-Circuit Withstand Compute ( S_min ) from adiabatic equation. If selected size < ( S_min ), increase size. Step 8: Check Earth Fault Loop Impedance (for protection against indirect contact) Ensure disconnection time < 0.4s (for final circuits) or 5s (distribution). This may require larger CPC (circuit protective conductor) or larger line conductor. 7. Worked Example System : 400V three-phase, 50 kW motor, ( \cos\phi = 0.85 ), efficiency 0.94, length 120m, PVC insulated copper cable, clipped direct to wall (Reference Method C), ambient 40°C, grouped with 3 other circuits (4 cables touching), prospective short-circuit 10 kA, permissible voltage drop 3% (12V), disconnection time 0.4 sec.

: ( I_b = \frac50000\sqrt3 \times 400 \times 0.85 \times 0.94 = \frac50000553.6 \approx 90.3 A )

Abstract The correct sizing of electrical cables is a critical task in power system design, directly impacting safety, efficiency, reliability, and economic viability. An undersized cable leads to overheating, insulation breakdown, voltage drops, energy losses, and fire hazards. An oversized cable results in unnecessary material costs, difficult installation, and reduced fault detection sensitivity. This paper provides a rigorous, step-by-step examination of the scientific and regulatory principles governing cable sizing. It explores the four fundamental determinants: current-carrying capacity (ampacity), voltage drop, short-circuit temperature rise, and economic optimization. The paper derives key formulas, explains correction factors for installation conditions, and presents worked examples based on international standards (IEC 60364 and BS 7671, with reference to NEC guidelines). 1. Introduction Cable sizing is not a simple lookup from a table; it is a multi-variable optimization problem. The primary goal is to select a conductor cross-sectional area (usually in mm² or AWG/kcmil) such that under all expected operating conditions, the cable’s temperature remains within the insulation’s rating, the voltage at the load remains within tolerance, and the cable can withstand fault currents without damage.

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

: 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 ]