Is cable cross-section chosen by current or by distance?
By both. The cross-section is calculated with two independent criteria: current carrying capacity determines the heating of the cable, while voltage drop determines the voltage at the device end. The larger of the two results applies. Over short distances current is usually the deciding factor, while over long distances at a low voltage such as 12 V it is almost always voltage drop.
The common mistake is to look at a table, pick the cross-section that suits the current and call the job done. That table only tells you at how many amps the cable works without overheating; it does not tell you how many volts are left at the device end.
Do the two calculations separately:
- Ampacity: figures of around 18 A for 1.5 mm² and 76 A for 16 mm² are typically given. These values are for free installation and normal ambient conditions; assume lower figures inside conduit, in a bundle or in a hot compartment.
- Voltage drop: A = 2 x 0.0175 x L x I / ΔU.
A comparison example: 20 A at 12 V. In terms of ampacity, assuming free installation and normal ambient conditions, 2.5 mm² is borderline and 4 mm² looks comfortable; inside conduit or in a hot compartment this assessment changes. But if the one-way distance is 5 m and the target is 3% (0.36 V), voltage drop calls for 9.7 mm². The result is 10 mm²; a 4 mm² cable will not burn but the device will not work correctly.
The reverse also happens: if 100 A is drawn over a 1 m battery connection, the voltage drop stays small and the cross-section is determined entirely by ampacity and mechanical strength.
The summary decision rule: do both calculations, take the larger result, and round up to a commercial cross-section.
The full topic: Installation, redundancy and field continuity