1. What voltage drop is and why it matters
Voltage drop is the loss of electrical potential that happens as current flows through a conductor. Copper and aluminium are not perfect conductors; their resistance converts a small part of the electrical energy into heat, leaving less voltage at the load.
Too much drop means motors run hot, lights dim, inverters shut down early and battery systems waste energy. That is why standards such as the NEC and IEC set maximum allowable voltage drop limits for installations.
2. Ohm's Law and conductor resistance
The fundamental relationship is Ohm's Law: V = I × R. If a cable has resistance R and carries current I, the voltage lost across it is I × R.
Conductor resistance depends on material, length and cross-sectional area. For copper at 20 °C, resistivity ρ is about 0.0172 Ω·mm²/m; for aluminium it is about 0.0282 Ω·mm²/m. Resistance R = ρ × L ÷ A, where L is the one-way length in metres and A is the conductor area in mm².
3. DC voltage drop formula
For DC circuits, multiply the current by the total conductor resistance of both outgoing and return paths: voltage drop = 2 × I × R, where R is the resistance of one conductor.
Percentage drop = (voltage drop ÷ supply voltage) × 100. This percentage is what most codes and equipment datasheets care about.
4. AC single-phase voltage drop
In single-phase AC, current travels out on the line conductor and back on the neutral, so the loop length is twice the one-way distance. The approximate formula is: voltage drop = 2 × I × L × (R' cos φ + X' sin φ), where R' and X' are resistance and reactance per unit length and φ is the phase angle between voltage and current.
For small installations with resistive loads (cos φ ≈ 1), this simplifies to voltage drop ≈ 2 × I × R, the same form as DC.
5. AC three-phase voltage drop
For balanced three-phase systems, the formula is: voltage drop = √3 × I × L × (R' cos φ + X' sin φ). The √3 factor comes from the geometry of three-phase voltages.
When reactance is small compared with resistance, a quick estimate is voltage drop ≈ 1.732 × I × R, where R is the total resistance of one conductor to the load.
6. Worked example: 12 V DC solar cable
A 12 V solar panel feeds a 10 A load through 10 m of 2.5 mm² copper cable. Copper resistivity gives R = 0.0172 × 10 ÷ 2.5 = 0.0688 Ω one way, so the loop resistance is 0.1376 Ω.
Voltage drop = 10 A × 0.1376 Ω = 1.38 V. Percentage drop = (1.38 ÷ 12) × 100 = 11.5%. That is too high for most 12 V systems; upsizing to 4 mm² would cut it to about 7.2%.
7. Worked example: 230 V AC single-phase run
A 230 V socket circuit carries 16 A over 25 m of 2.5 mm² copper. One-way resistance R = 0.0172 × 25 ÷ 2.5 = 0.172 Ω, so the loop resistance is 0.344 Ω.
Voltage drop = 16 A × 0.344 Ω = 5.5 V. Percentage drop = (5.5 ÷ 230) × 100 = 2.4%. This sits comfortably inside the common 3% limit for final circuits.
8. Acceptable voltage drop limits
NEC recommendations are roughly 3% for branch circuits and 5% overall from service point to load. IEC 60364 typically uses similar figures, though national annexes vary.
Sensitive electronics, inverter input circuits and long LED runs often need tighter limits. Always check the equipment datasheet; some devices stop working below 10.5 V on a 12 V nominal supply.
9. How to reduce voltage drop
The simplest fix is a larger conductor cross-section. Doubling the area roughly halves the resistance and the drop. Shortening the cable run or moving the source closer has the same effect.
Raising the system voltage also helps because the same absolute drop is a smaller percentage of a higher voltage. That is why solar strings run at hundreds of volts and utility grids use thousands of volts.
10. Calculate it instantly
Use the OneCalcApp Voltage Drop Calculator to check percentage drop for DC, single-phase or three-phase runs. It also suggests whether your cable size passes common code limits.