How voltage drop is calculated
A cable is a resistor: drop = 2 × length × current × material resistivity ÷ cross-section. The factor two is there because current travels to the load and back. For example, 20 A through 5 m of 4 mm² copper drops 0.88 V — 7.3% of a 12 V system, well over the 3% planning guideline, and 17.5 W turned into heat along the way. The same run in 10 mm² drops 2.9% and passes.
Drop is not the only limit
This calculator answers "will my load see enough voltage?" — it does not answer "is this cable safe?". Every cable also has an ampacity rating (how much current it can carry without overheating) that depends on insulation, bundling and ambient temperature, and every DC circuit needs a correctly sized fuse at the battery end. Check both against your cable's datasheet and local rules.
Why system voltage matters so much
| System | Current for 1,000 W | Drop in 5 m of 10 mm² |
|---|---|---|
| 12 V | 83 A | 1.46 V = 12.2% |
| 24 V | 42 A | 0.73 V = 3.0% |
| 48 V | 21 A | 0.36 V = 0.8% |
This is the practical reason bigger inverters use higher system voltages — see the Inverter Sizing calculator for the battery-side currents involved.
Methodology
The calculator UI, the cross-section table and the displayed formula all call the same calculation function. Resistivity values and the guideline come from the sources below.
- Copper resistivity ≈ 0.0175 Ω·mm²/m, aluminium ≈ 0.0286 Ω·mm²/m (20 °C) — Standard electrical engineering reference values (IEC 60228 conductor classes)
- ≤ 3% drop guideline for critical DC circuits — Common marine and off-grid wiring practice, e.g. ABYC guidance (3% critical / 10% non-critical)
- Ampacity must be checked separately — Cable manufacturer datasheets and local electrical codes
Assumptions last reviewed on 26 August 2026. If a source disagrees with your battery or equipment datasheet, trust the datasheet.