How do you select the cross-section of an electrical conductor?
The cross-section is chosen so that it simultaneously satisfies three conditions per N-SEP-E-002: long-term current-carrying capacity (I_B ≤ I_z after correction by factors k₁ and k₂), permissible voltage drop (ΔU% ≤ limit per PN-HD 60364-5-52), and the minimum mechanical cross-section. The most demanding condition governs — the calculator checks all three at once and shows which one is decisive.
What cross-section for sockets and lighting circuits?
In typical residential installations, 1.5 mm² Cu is used for lighting circuits (10 A protection) and 2.5 mm² Cu for general socket circuits (16 A). Dedicated circuits such as an electric cooker or water heater often need 4–6 mm². These are guideline values — for longer runs or higher loads the cross-section should be recalculated against voltage drop and current-carrying capacity.
What are the factors k₁ and k₂?
They are factors that correct a conductor's current-carrying capacity. k₁ accounts for an ambient temperature different from the reference (30 °C in air, 20 °C in the ground), and k₂ accounts for the mutual heating of conductors grouped together or sharing a route. The corrected capacity is I_z = I_z(tab) × k₁ × k₂; the more adjacent circuits and the higher the temperature, the lower the permissible capacity.
Copper or aluminium — which to choose?
Copper has higher capacity and conductivity, so a smaller cross-section suffices for the same current; it is the standard in residential installations. Aluminium is cheaper and lighter, and economical for larger cross-sections (≥ 16 mm²) in supply lines and service cables. For the same current, aluminium requires a larger cross-section and appropriate terminals.
Is the cross-section governed by current or by voltage drop?
It depends on the length of the run. For short circuits, current-carrying capacity (linked to protective-device selection) usually governs, while for long lines it is the voltage-drop condition that forces a larger cross-section than the current alone would. Both criteria must therefore be checked at once; the calculator indicates which one is decisive.
How do installation methods A1–F differ?
The reference methods per PN-HD 60364-5-52 describe how the conductor is routed, which affects heat dissipation and therefore current-carrying capacity: A1/A2 — in conduits in an insulated wall, B1/B2 — in trunking and cable channels, C — directly on a wall, E/F — on ladders and brackets in air, D — in the ground. Routing with better cooling (E, F, in the ground) allows a higher capacity for the same cross-section.
Can a 2.5 mm² cable be protected by a B16 breaker?
Usually yes. For Cu/PVC with 2 loaded conductors, the ampacity Iz of 2.5 mm² is: A1 19.5 A, A2 18.5 A, B1 24 A, B2 23 A, C 27 A — each ≥ 16 A, and I₂ = 1.45·16 = 23.2 A ≤ 1.45·Iz. The condition only fails after the grouping correction: with method A1 and 2 circuits in one conduit, Iz = 19.5·0.8 = 15.6 A < 16 A — then you need 4 mm² or a B13 breaker. In the calculator pick “Protective device In = 16 A” as the current basis, and in the protection calculator compute Iz from the installation method, number of grouped circuits and temperature.