Cable voltage drop DC:
formula and worked example

Why long low-voltage runs lose volts

Every metre of copper has a small resistance, and a current through resistance always drops voltage. On a 230 V mains circuit a couple of volts hardly registers. On a 12 V battery or solar circuit, the same couple of volts is a large share of what you started with and appears as dim lights, slow motors or a charger that never finishes.

That is why DC cable sizing for low-voltage systems is mostly a voltage-drop problem, not a heating problem. The cable may carry the current without warming, and still waste more voltage than the load can spare.

• Current: the load current in amperes.
• One way length: the distance from source to load in metres, measured along the cable route.
• Resistance ohm per m: the resistance of a single conductor per metre, from the cable data.

The formula and the factor of two

The factor of 2 is there because the current must travel to the load and come back, so the loop is twice the run. The product 2 × length × resistance per metre is the total loop resistance, and current times that resistance is the drop by Ohm's law.

If the cable data gives resistance per kilometre, divide by 1,000 before using it. Resistance also rises with temperature, by roughly 0.4% per degree Celsius for copper, so a cable in a hot space is slightly worse than its catalogue figure.

Worked example: 10 A over 25 m

A 10 A load sits 25 m from its battery, connected by a twin cable of 0.003 Ω per metre per conductor, which is close to a 6 mm² copper conductor at room temperature.

1. 1Loop length: 2 × 25 = 50 m of conductor.
2. 2Loop resistance: 50 × 0.003 = 0.15 Ω.
3. 3Voltage drop: 10 × 0.15 = 1.5 V.
4. 4Power lost in the cable: 1.5 V × 10 A = 15 W.

On a 12 V system that 1.5 V is 12.5% of the supply, and the load only sees 10.5 V. On a 24 V system it is 6.25%, and on 48 V it is about 3%.

Judging whether the drop is acceptable

What counts as acceptable depends on the load and the standard you follow. Designers commonly aim for a few percent of the supply voltage, and tighter for sensitive electronics or batteries being charged, but the limit varies by application and regulation.

• Lighting and motors: low voltage reduces brightness and torque, and motors draw more current to compensate.
• Battery charging: a drop in the cable lowers the voltage that reaches the battery, so charge controllers may under-charge it.
• Electronics: some devices shut down below a minimum input voltage, long before the cable is overloaded.

The 15 W lost in this cable also heats it, which is modest here but grows with the square of current. Check the cable's current rating as a separate test from the voltage drop.

What shifts the drop most

The drop is proportional to current, so a 10% higher current gives a 10% higher drop: 1.5 V becomes 1.65 V. It is also proportional to length and to resistance per metre, so each variable can be traded against the others.

The practical levers are therefore: shorten the run, raise the cross-section, or raise the system voltage. Doubling the copper area halves the resistance and the drop. Moving from 12 V to 24 V halves the current for the same power, which quarters the cable loss, a far stronger effect than it looks.

Mistakes with DC cable drop

• Forgetting the return conductor and using the one-way length alone, which halves the drop.
• Measuring the run in a straight line when the cable route is longer.
• Using the cable's ampacity rating as the design limit, even when the drop is the real constraint.
• Using the resistance per metre of the whole twin cable as if it were per conductor, which doubles the loop twice.

A scenario: sizing for a 3% budget

A 12 V system with the same 10 A and 25 m run is allowed a 3% drop, which is 0.36 V. The loop resistance may be at most 0.36 ÷ 10 = 0.036 Ω, and with 50 m of conductor that is 0.00072 Ω per metre. The conductor therefore needs about four times the cross-section of the 0.003 Ω per metre cable, around 25 mm² of copper.

That is a heavy cable, so the alternative is worth weighing: move the source closer, or convert to 24 V so the same power takes half the current and the drop target is far easier to meet.