Cable Cross-Section Calculator

Load
Supply system
Load given as
Cable
Insulation
Installation conditions
Grouped cables are
Length and voltage drop

The size (2.5 mm²) is governed by current-carrying capacity — the voltage drop is already within the limit.

Copper conductors, reference methods A1–C, B-curve circuit-breakers to EN 60898-1 only (C/D curves need a fault-loop disconnection check). This does not check short-circuit and earth-fault protection (disconnection time, fault-loop impedance), harmonics in the neutral, or local regulations — it does not replace a design by a qualified electrician.

Minimum conductor cross-section2.5mm²
Circuit-breaker rating I_n20A
Voltage drop at that size2.06%
Design current and protective device
Design current I_B16.30 A
Circuit-breaker rating I_n (I_B ≤ I_n)20 A
Conventional tripping current I₂ = 1.45 × I_n29.00 A
Correction factors
Temperature row used (Table B.52.14)30 °C
Ambient temperature factor k₁ (Table B.52.14)1.00
Grouping factor k₂ (Table B.52.17)1.00
Combined factor k₁ × k₂1.000
Required tabulated capacity I_n / (k₁ × k₂)20.0 A
Size by current-carrying capacity
Smallest size by capacity2.5 mm²
Tabulated capacity at that size27.0 A
Corrected capacity I_z (≥ I_n)27.00 A
1.45 × I_z (≥ I₂)39.15 A
Voltage-drop check (IEC 60364-5-52 Annex G)
Voltage-drop limit (Table G.52.1)5.00 %
Allowance left for this circuit5.00 %
Voltage drop at the capacity size2.06 %
Smallest size by voltage drop2.5 mm²
Voltage drop u4.73 V
Voltage drop Δu2.06 %
Selected size
Selected cross-section2.5 mm²
Corrected capacity I_z of the selected size27.00 A

About this calculator

This calculator picks the smallest copper conductor cross-section and the circuit-breaker rating for one final circuit — a socket circuit, a cooker, a heat pump, a lighting circuit — the way IEC 60364 (in Lithuania LST HD 60364) sets it out:

  1. the breaker must carry the load: I_B ≤ I_n;
  2. the cable must carry what the breaker lets through, after correcting the tabulated capacity for ambient temperature and grouping: I_n ≤ I_z;
  3. the voltage drop at the load must stay within the Annex G limit.

It shows the size each condition needs on its own, and says which one governs. On a short run the current-carrying capacity usually governs; on a long run the voltage drop does.

The calculator covers copper conductors only, installation methods A1, A2, B1, B2 and C, PVC or XLPE insulation, and B-curve miniature circuit-breakers to EN 60898-1 (6–125 A). It does not check short-circuit protection or automatic disconnection: a C- or D-curve breaker needs a higher fault current to trip instantly, so it must be confirmed by an electrician's fault-loop impedance and disconnection-time check — see the limits below.

Formula

Design current (from power, if the load is given in watts):

single-phase:  I_B = P / (230 × cos φ)
three-phase:   I_B = P / (√3 × 400 × cos φ)

Breaker rating: the smallest standard rating I_n with I_B ≤ I_n, from 6, 10, 13, 16, 20, 25, 32, 40, 50, 63, 80, 100, 125 A.

Current-carrying capacity (IEC 60364-5-52 Annex B, IEC 60364-4-43 cl. 433.1):

I_z = I_table × k₁ × k₂ ≥ I_n
I₂  = 1.45 × I_n ≤ 1.45 × I_z
  • I_table — Tables B.52.2 (PVC, 2 loaded conductors), B.52.3 (XLPE, 2), B.52.4 (PVC, 3), B.52.5 (XLPE, 3), copper, ambient 30 °C. A single-phase circuit has 2 loaded conductors, a balanced three-phase circuit 3.
  • k₁ — Table B.52.14, the highest ambient air temperature along the route in service (PVC: 0.87 at 40 °C; XLPE: 0.91 at 40 °C). Values below the 30 °C reference raise k₁ above 1 and must be justified — a winter or average temperature gives an undersized cable.
  • k₂ — Table B.52.17, grouping (item 1, bunched: 0.80 for 2 circuits, 0.70 for 3; item 2, single layer on a wall, method C only).
  • For an EN 60898-1 circuit-breaker the conventional tripping current is I₂ = 1.45 I_n, so the second condition holds whenever the first does.

Voltage drop (IEC 60364-5-52 Annex G):

u  = b × (ρ₁ × L / S × cos φ + λ × L × sin φ) × I_B
Δu = 100 × u / U₀
  • b = 2 single-phase, 1 three-phase; U₀ = 230 V (line to neutral);
  • ρ₁ = 0.0225 Ω·mm²/m — copper resistivity in service, 1.25 × its 20 °C value;
  • λ = 0.08 mΩ/m — conductor reactance when nothing more precise is known;
  • L — one-way length (m), S — cross-section (mm²).

The size is stepped up from the capacity-governed size until Δu is within the limit (Table G.52.1, installation supplied from the public network: 3 % lighting, 5 % other uses), less any drop already used upstream of the distribution board.

Worked example

3 kW socket circuit, single-phase, cos φ = 0.8, PVC cable clipped to a wall (method C), 30 °C, not grouped, 20 m:

I_B = 3000 / (230 × 0.8) = 16.30 A        → breaker I_n = 20 A
k₁ × k₂ = 1.00 × 1.00
B.52.2, method C: 1.5 mm² = 19.5 A < 20;  2.5 mm² = 27 A ≥ 20   → 2.5 mm²
I₂ = 1.45 × 20 = 29 A ≤ 1.45 × 27 = 39.2 A                        ✓

u  = 2 × (0.0225 × 20 / 2.5 × 0.8 + 0.00008 × 20 × 0.6) × 16.30
   = 2 × (0.1440 + 0.00096) × 16.30 = 4.73 V
Δu = 100 × 4.73 / 230 = 2.06 %  ≤ 5 %                             ✓

Result: 2.5 mm², 20 A breaker, governed by current-carrying capacity. Lengthen the same circuit to 60 m at a resistive 16 A load and the drop governs instead: 1.5 mm² would carry the current, but only 4 mm² keeps the drop within 5 % (4.70 %).

FAQ

Why is the breaker rated above the load current, and the cable above the breaker? That is the overload coordination rule of IEC 60364-4-43: I_B ≤ I_n ≤ I_z. The breaker must not trip in normal use, and the cable must survive anything the breaker lets through without tripping. A 16.3 A load therefore gets a 20 A breaker, and the cable is chosen for 20 A, not for 16.3 A.

Why does 2.5 mm² sometimes come out as 4 mm² for the same breaker? Because of the corrections. In a hot loft (40 °C, k₁ = 0.87) bunched with two other circuits (k₂ = 0.70), a cable keeps only 61 % of its tabulated capacity. Insulation around the cable (methods A1/A2) lowers the tabulated capacity itself.

Which installation method do I have? A1/A2 — in conduit inside a thermally insulated wall (a frame wall with mineral wool); B1/B2 — in conduit or trunking on a wall; C — sheathed cable clipped direct to a wall or laid in plaster or masonry (IEC 60364-5-52 Table A.52.3, items 57–58). If in doubt, choose the less favourable method.

Why 3 % and 5 %? IEC 60364-5-52 Table G.52.1 recommends, for an installation supplied directly from the public low-voltage network, at most 3 % for lighting and 5 % for other uses, measured from the origin of the installation. If the feeder to your distribution board already uses part of that, enter it as the upstream drop.

Why can't I choose aluminium? Its tabulated capacities could not be checked against a second independent reproduction of the standard, and this site does not publish a constant it could not verify. Aluminium may be added once it has been.

Assumptions and limits

  • Copper only; methods A1, A2, B1, B2, C only (no buried D1/D2, no free-air E/F); conductors 1.5–120 mm².
  • Protection by a B-curve EN 60898-1 circuit-breaker (I₂ = 1.45 I_n). C- and D-curve breakers satisfy the same overload rule, but their instantaneous tripping needs a larger fault current: an electrician must check the fault-loop impedance and disconnection time.
  • The ambient temperature is the highest one during operation; anything below 30 °C must be justified for the whole route. For a gG fuse, I₂ can be up to 1.6 I_n and the second overload condition must be checked separately.
  • The ambient temperature is rounded up to the next 5 °C row of Table B.52.14.
  • Grouping assumes all grouped circuits are similarly loaded.
  • The neutral of a three-phase circuit is assumed unloaded (balanced load, no significant third harmonic).
  • Not checked: short-circuit protection, automatic disconnection and fault-loop impedance, RCD selection, motor starting currents, and any local requirement stricter than IEC 60364. This is a pre-sizing tool and does not replace a design by a qualified electrician.
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