How wire sizing works
A wire that's too thin wastes energy as heat — the voltage drops along the run, so less of it reaches your load. On solar, battery and inverter DC runs this is almost always the deciding factor, so the goal is to choose the smallest gauge that keeps the drop under your target.
1. Voltage drop
As current flows down a wire and back, the wire's resistance burns off some of the voltage. The DC voltage drop is:
Voltage drop = (2 × K × I × L) ÷ circular mils
where K is the resistivity constant (12.9 for copper, 21.2 for aluminum), I is the current in amps, and L is the one-way length in feet. The 2 is the round trip — current goes out on one conductor and back on the other. Express it as a percentage of system voltage, and you size the wire so that stays under your target:
Voltage drop % = voltage drop ÷ system voltage × 100
2. Short, high-current runs
Voltage drop sizes the wire on almost every solar and battery run, because low voltage over distance makes the drop the binding constraint. The one place it doesn't tell the whole story is a very short, very high-current cable — a battery-to-inverter link, say — where the run is too short for drop to matter but the current is large. There, pick the wire by its own current rating: check the gauge stamped on the cable or its datasheet and make sure it comfortably handles your current. This tool sizes by voltage drop, so on those short, heavy runs treat the cable's rated capacity as the deciding number.
Why 12V is so unforgiving
Voltage drop is about the share of voltage lost, so the same loss hurts far more at low voltage:
- Losing 0.36 V on a 120 V circuit = 0.3% — negligible
- Losing 0.36 V on a 12 V system = 3% — your whole budget
Low-voltage DC also carries high current for the same power, which increases the drop further. That's why a short, fat cable is normal on 12 V battery systems, and why raising the system voltage to 24 V or 48 V lets you use dramatically smaller wire.
A worked example
A 12 V charge controller sits 20 ft from the battery and carries 30 A. You hoped to use 10 AWG:
- 10 AWG (10,380 cmil): drop = (2 × 12.9 × 30 × 20) ÷ 10,380 = 1.49 V = 12.4% — far over budget, and the wire runs hot
- 4 AWG: about 3.1% — just over a strict 3%
- 2 AWG: about 1.9% — comfortably within 3%
So that run wants 2–4 AWG, not 10. Getting this wrong is the difference between a battery that charges and one that never reaches full, with a warm cable to show for it.
Quick reference: copper circular mils
| AWG | Circular mils |
|---|---|
| 10 | 10,380 |
| 8 | 16,510 |
| 6 | 26,240 |
| 4 | 41,740 |
| 2 | 66,360 |
| 1/0 | 105,600 |
| 2/0 | 133,100 |
| 4/0 | 211,600 |