Single-Phase Circuit Cable Sizing — Method
Single-phase circuits — 230 V (Europe / UK) or 120 V / 240 V split-phase (North America) — dominate domestic, small-commercial and lighting installations. Cable sizing is identical to three-phase except: only one live conductor carries the load (no √3 factor), the neutral always equals the live cross-section, and voltage drop is twice that of an equivalent 3-phase cable for the same current and length.
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 (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
- Three-Phase Cable Sizing Calculator — Three-Phase · 400 V / 415 V / 480 V
- BS 7671 Cable Sizing Calculator — BS 7671:2018+A2 · UK / NICEIC
- NEC Cable Sizing Calculator — NEC NFPA 70 · USA
- All Electrical Engineering Calculators →
Frequently Asked Questions
32 A single-phase in conduit on a wall (Method B1) at 30 °C with no grouping: tabulated current Iz must be ≥ 32 A. From IEC 60364-5-52 Table B.52.4 (XLPE Cu): 4 mm² = 36 A ≥ 32 A ✅. UK BS 7671 typically allows 4 mm² T&E for a 32 A radial in domestic installations; 6 mm² is required if the cable is buried in thermal insulation (Ci = 0.5 → 4 mm² @ 36 × 0.5 = 18 A < 32 A).
Yes for circuits up to 16 mm². BS 7671 524.2.2 / IEC 60364-5-52 524 require the neutral cross-section to equal the line cross-section in single-phase circuits ≤ 16 mm² Cu. Above 16 mm², the neutral can be reduced if the load is balanced, but in single-phase the neutral always carries the full line current — keep them equal.
Vd = 2 × I × L × (R cosφ + X sinφ) / 1000 (V). The factor of 2 accounts for both the line and neutral resistance (current goes out and returns). Example: 30 m of 4 mm² Cu at 32 A unity PF: Vd = 2 × 32 × 30 × 4.61 / 1000 = 8.85 V = 3.85 % of 230 V — borderline 5 % limit.
In single-phase, current flows down the line and back through the neutral — both conductors contribute resistance. In balanced three-phase, the three line currents cancel in the neutral (no return current), so only the line conductors contribute. For the same conductor size and load current per phase, three-phase Vd is √3 / 2 ≈ 0.866 of single-phase — but for the same total power transferred, three-phase needs only 1/√3 of the current per phase, so Vd is 1/3 of single-phase.
BS 7671 Table 54.7: CPC = 1.0 mm² for 1.5 mm² line, 1.5 mm² for 2.5 mm² line (or 2.5 mm² in the same flat T&E cable), 2.5 mm² for 4 mm² line, 4 mm² for 6 mm² line, 6 mm² for 10 mm² line. NEC Table 250.122 gives equipment-grounding conductor: 14 AWG for 15 A, 12 AWG for 20 A, 10 AWG for 60 A, 8 AWG for 100 A breaker. Always verify with the adiabatic equation S ≥ √(I²t) / k for fault-clearing.
Yes — BS 7671 Appendix 15 allows a 32 A ring final on 2.5 mm² T&E because the load current is shared between the two halves of the ring (max 16 A per leg). The Iz of 2.5 mm² Method 100 (cable in conduit in insulated wall, ring) is rated 27 A — but with two paths in parallel the total is 27 × 2 / √2 ≈ 38 A, satisfying the 32 A breaker. For a radial, 4 mm² is the minimum for 32 A.