Three-Phase Circuit Cable Sizing — Method
Three-phase circuits — 400 V (Europe / IEC), 415 V (UK / older Australia), 480 V (US industrial) — power most motors, sub-mains and commercial loads. Cable sizing follows the same Iz × Ca × Cg × Ci ≥ Ib procedure but uses the line current Ib = P / (√3 × V × cosφ × η). The neutral can be reduced for balanced loads ≥ 16 mm² (IEC 524.2.3) but must be full-size or larger for harmonic-rich loads (variable-speed drives, LED lighting, IT loads), per BS 7671 524.2.4.
Where:
- Ib — design current of the circuit (A), from the load calculation
- Ca — ambient temperature correction (1.00 at 30 °C reference)
- Cg — grouping / bunching factor (1.00 for a single circuit)
- Ci — thermal-insulation factor (1.00 if the cable is in free air; 0.50 if fully buried in insulation)
Then pick the smallest cable cross-section in IEC 60364-5-52 Table B.52.4, Method B1 (3-core in conduit on wall) whose tabulated ampacity Iz ≥ It.
Related cable sizing calculators
Other standard- and method-specific cable-sizing calculators in the same series — same procedure, different reference tables and defaults:
- Cable Sizing Calculator (universal) — the seed page covering all standards in one tool
- Single-Phase Cable Sizing Calculator — Single-Phase · 230 V / 240 V
- IEC 60364 Cable Sizing Calculator — IEC 60364-5-52 · International
- NEC Cable Sizing Calculator — NEC NFPA 70 · USA
- All Electrical Engineering Calculators →
Frequently Asked Questions
The design current uses Ib = P / (√3 × V_LL × cosφ × η) instead of Ib = P / (V × cosφ). Once Ib is computed, the cable sizing procedure (Iz × Ca × Cg × Ci ≥ Ib) is identical. Voltage drop is 1/√3 of single-phase for the same conductor size and current — so 3-phase cables can be smaller for the same kW transferred over the same distance.
For balanced linear loads (no significant 3rd harmonic) and line conductor ≥ 16 mm² Cu / 25 mm² Al: yes — the neutral can be ~50 % of the line per IEC 60364-5-52 524.3 / BS 7671 524.2.3. For loads with > 15 % 3rd-harmonic content (LED lighting, VSDs, computers, switchmode supplies), the neutral can carry up to 1.7× the line current and must be sized accordingly — often requiring a 'high-neutral' cable with 200 % neutral.
Vd = √3 × I × L × (R cosφ + X sinφ) / 1000 (V). For the same conductor and current, Vd is 1/√3 ≈ 0.577 of an equivalent single-phase circuit (which uses 2× the per-line drop). Convert to percentage: % Vd = Vd / V_LL × 100. Limit is 5 % per IEC / BS 7671 / 3 % per NEC for branch circuits.
Use the motor full-load current (FLA) from the nameplate (not the calculated kW/√3·V·PF·η — actual FLA accounts for slip and inefficiency). NEC requires conductors sized to 125 % of motor FLA (Article 430.22). IEC / BS 7671 require the cable to safely carry the FLA continuously and the starting current (typically 6–7× FLA) for the start duration without thermal damage — the adiabatic equation S² ≥ I²t / k² governs.
63 A 3-phase in conduit on a wall (Method B1) at 30 °C with 3 circuits grouped: It = 63 / 0.80 = 78.75 A. From IEC 60364-5-52 Table B.52.4 (XLPE Cu): 16 mm² = 85 A ≥ 78.75 A ✅. Select 16 mm² 4-core XLPE/PVC SWA cable, MCB rated 63 A. Check voltage drop separately for runs > 30 m.
For line conductors ≤ 16 mm² Cu: PE = line size (BS 7671 Table 54.7). For 16–35 mm² line: PE = 16 mm². For > 35 mm² line: PE = line/2 (minimum 16 mm²). NEC Table 250.122: 6 AWG PE for 100 A, 4 AWG for 200 A, 1/0 AWG for 400 A, 250 kcmil for 800 A. Always verify with the adiabatic equation for the prospective fault current and disconnection time.