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Calculations 2025-07-10 · 7 min

How to calculate cable cross-section: step-by-step guide

Detailed guide to cable cross-section calculation by heating and voltage drop with examples for copper and aluminum.

Author: GorkyCAD Team

Why correct cross-section matters

Cable cross-section is a critical parameter. Too small — cable overheats, insulation melts, fire risk. Too large — wasted money on copper, difficult installation. The engineer must find the exact value.

Calculation follows two criteria. The larger of the two is selected.

Criterion 1: Heating (continuous current rating)

Cables heat up when current flows. IEC 60364 / IS 732 specify maximum currents for each cross-section.

Example

Given: Icalc = 25 A, copper cable, 3 cores, in conduit.

Per IEC table: copper 2.5 mm² → 27 A (sufficient).

Check: Icalc (25 A) ≤ Imax (27 A) → 2.5 mm² passes heating criterion.

Criterion 2: Voltage drop

Per IEC, voltage drop from main panel to end consumer — max 5%.

Formula: ΔU% = (2 × L × Icalc × cos φ) / (γ × S × Unom) × 100%

Where:

  • L — cable length (m)
  • Icalc — calculated current (A)
  • γ — conductivity (copper ≈ 57 m/(Ω·mm²))
  • S — cross-section (mm²)
  • Unom — nominal voltage (230 V single-phase)

    Example

    L = 30 m, Icalc = 25 A, cos φ = 0.92, copper.

    At S = 2.5 mm²: ΔU% = 4.2% < 5% → passes.

    At S = 1.5 mm²: ΔU% = 7.0% > 5% → fails.

    Result: choose 2.5 mm² (passes both criteria).

    GorkyCAD Algorithm

    GorkyCAD performs this calculation automatically for every circuit segment:

    1. Determines Icalc from group total load

2. Selects cross-section by heating (IEC/IS 732/GOST tables)
  • 3. Calculates voltage drop along entire circuit
  • 4. Picks the larger of the two cross-sections
  • 5. Rounds to nearest standard: 1.5, 2.5, 4, 6, 10, 16, 25 mm²...

    You just need to review and approve.