How is cable cross-section calculated in a 12V system?
At 12 V DC the cross-section is chosen against two criteria: current-carrying capacity and voltage drop; the larger result applies. For voltage drop, A = 2 x ρ x L x I / ΔU is used; for copper ρ = 0.0175 Ω·mm²/m. Example: at 20 A, 5 m one-way, with a target drop of 3% that is 0.36 V, A = 2 x 0.0175 x 5 x 20 / 0.36 = 9.7 mm², so 10 mm² in practice.
Think of the calculation as the answer to two separate questions. First, can the cable carry this current without overheating, and second, is there enough voltage left at the device end. You do both calculations and choose the larger cross-section.
The formula for the voltage drop side is this: A = 2 x ρ x L x I / ΔU. Here ρ is 0.0175 Ω·mm²/m for copper, L is the one-way distance, I is the current and ΔU is the drop you allow. If you target 3% on a critical circuit at 12 V, ΔU becomes 0.36 V.
Numerical example: a device drawing 20 A, 5 m one-way from the battery. A = 2 x 0.0175 x 5 x 20 / 0.36 = 9.7 mm². Rounding to commercial cross-sections you run 10 mm². Had the same device been 1 m away, the voltage drop would have been met even with 2 mm²; the cross-section would then have been set by the current capacity.
- The length is always entered as one way, the 2 in the formula already counts the return.
- The values in ampacity tables are typical values; in a bundle, inside conduit or in a hot engine bay the applicable limit is lower.
- If the ambient is 40 °C multiply the table value by about 0.87, and by 0.71 if it is 50 °C.
Compare the result of the calculation with the cable you have; if there is one step between them, prefer the larger one, because changing the cross-section after installation is the most expensive operation of all.
The full topic: Installation, redundancy and field continuity