What is the voltage drop formula and why is it multiplied by 2?
In a DC circuit the voltage drop is calculated with ΔU = 2 x L x I x ρ / A. ρ is 0.0175 Ω·mm²/m for copper, L is the one-way distance, I the current and A the cross-section. The factor of 2 in the formula comes from the current going out on the positive conductor and returning on the negative one: the circuit contains two cable lengths of copper and both produce a drop.
Current does not go somewhere and disappear; it has to return to the source. When you connect a 12 V device 5 m away, the circuit contains not 5 m but 10 m of copper. The 2 in the formula expresses exactly this.
Written out: ΔU = 2 x L x I x ρ / A. Since the resistance of the conductor is R = ρ x length / A, the total resistance is R = 0.0175 x (2L) / A and by Ohm's law the drop is ΔU = I x R. So the formula is not a separate rule but Ohm's law applied to the cable.
Example: 20 A, 5 m one way, 10 mm² cable. ΔU = 2 x 5 x 20 x 0.0175 / 10 = 0.35 V. At 12 V this comes to about 2.9%, that is, below the 3% limit accepted for a critical circuit.
Points to watch:
- The factor of 2 applies even if you use a chassis return; the chassis is also a conductor, and a painted, corroded chassis is a far worse conductor than a cable.
- The formula is for DC; with alternating current, skin effect and inductance come into play.
- The result depends on current and cross-section rather than on voltage, but as a percentage the effect at 12 V is twice what it is at 24 V.
The full topic: Installation, redundancy and field continuity