How to use
- Choose DC, single-phase or three-phase and the conductor material.
- Choose the units and enter the cross-section in mm² or pick the AWG gauge.
- Enter the one-way length of the cable — from the supply to the load, not the length of both wires.
- Enter the load current and the supply voltage (line-to-line for three-phase, for example 400 V).
Formulas
- resistance of one conductor: R = ρ × L / A
- DC and single-phase: drop = 2 × I × R, loss = I² × 2R
- three-phase: drop = √3 × I × R, loss = 3 × I² × R
- percent = drop / supply voltage × 100; voltage at the load = supply − drop
ρ = 0.017241 Ω·mm²/m for copper and 0.028264 Ω·mm²/m for aluminium at 20 °C. AWG sizes are turned into mm² with the AWG definition (12 AWG = 3.309 mm²).
Worked examples
Single-phase, 12 AWG copper, 50 ft one way, 10 A at 120 V: 50 ft = 15.24 m, R = 0.017241 × 15.24 / 3.309 = 0.0794 Ω. Drop = 2 × 10 × 0.0794 = 1.588 V, which is 1.32 %; the load gets 118.41 V and the cable turns 15.88 W into heat.
Single-phase, 2.5 mm² copper, 20 m, 10 A at 230 V: R = 0.1379 Ω, drop 2.759 V, or 1.2 %.
Three-phase, 16 mm² aluminium, 50 m, 32 A at 400 V: R = 0.028264 × 50 / 16 = 0.088325 Ω, drop = √3 × 32 × 0.088325 = 4.895 V, or 1.22 %.
Tips
- Doubling the length doubles the drop; doubling the cross-section halves it.
- Low-voltage circuits suffer most: the same 2 V that is under 1 % at 230 V is about 17 % at 12 V. Size 12 V and 24 V wiring by voltage drop, not only by current.
- Voltage drop is a separate check from current capacity: a wire can be thick enough not to overheat and still lose too many volts over a long run.