Why is thinner cable enough for the same power in a 24V system?
At constant power, doubling the voltage halves the current. Since the voltage drop ΔU = 2 x L x I x ρ / A is directly proportional to current, halving the current also halves the drop at the same cross-section. On top of that the allowed margin in percentage terms grows: the 3% limit is 0.36 V at 12 V but 0.72 V at 24 V. The two effects combine to reduce the required cross-section to about a quarter.
Two independent gains stack up. First: if you draw the same power from twice the voltage, the current halves. A 1200 W load means 100 A at 12 V and 50 A at 24 V. Since the voltage drop is directly proportional to the current, this alone halves the drop.
Second: the accepted drop in absolute volts also doubles. The 3% rule gives 0.36 V at 12 V and 0.72 V at 24 V. So the drop produced is halved while the drop permitted is doubled.
Combining them in the cross-section formula A = 2 x ρ x L x I / ΔU; with I halved and ΔU doubled, A falls to about a quarter. A run needing 10 mm² at 12 V is generally solved with around 2.5 mm² at 24 V.
Practical consequences:
- On long-distance, high-current installations a 24 V system markedly reduces copper cost and cable weight.
- Heat loss also falls; the power dissipated in the cable is proportional to the square of the current, so halving the current cuts the loss to a quarter.
- On the other hand the choice of fuses, relays and connectors must be reassessed at 24 V in terms of arc quenching; a contact rated for 12 V may not break the same current safely at 24 V.
The full topic: Installation, redundancy and field continuity