Voltage Drop Calculator
Find the voltage at the end of the line and the percentage voltage loss, taking into account how conductor resistance changes with temperature and the reactance of AC lines.
Formula used
Three-phase: ΔU = √3 × I × L × (R cos φ + X sin φ)
Voltage drop by cross-section (example line)
Calculator results for three-phase 400 V, 30 kW, cos φ 0.90, efficiency 0.95, 80 m one way, copper conductor at 70 °C, reactance 0.08 Ω/km.
| Cross-section (mm²) | Voltage drop (V) | Voltage drop (%) | Voltage at line end (V) |
|---|---|---|---|
| 10 | 13.47 | 3.37 | 386.53 |
| 16 | 8.51 | 2.13 | 391.49 |
| 25 | 5.53 | 1.38 | 394.47 |
| 35 | 4.02 | 1.01 | 395.98 |
| 50 | 2.89 | 0.72 | 397.11 |
| 70 | 2.13 | 0.53 | 397.87 |
| 95 | 1.64 | 0.41 | 398.36 |
Conductor resistance used in the calculation (Ω/km)
The tool calculates resistance from resistivity: copper 0.0175 and aluminum 0.0282 Ω·mm²/m at 20 °C, corrected for temperature. The maximum conductor resistance in IEC 60228 and the manufacturer’s AC resistance data are higher than these values (example: 25 mm² copper at 20 °C: 0.727 Ω/km); use manufacturer data for the final calculation.
| Cross-section (mm²) | Copper · 20 °C | Copper · 70 °C | Copper · 90 °C | Aluminum · 20 °C | Aluminum · 70 °C | Aluminum · 90 °C |
|---|---|---|---|---|---|---|
| 1.5 | 11.6667 | 13.9592 | 14.8762 | – | – | – |
| 2.5 | 7.0000 | 8.3755 | 8.9257 | – | – | – |
| 4 | 4.3750 | 5.2347 | 5.5786 | – | – | – |
| 6 | 2.9167 | 3.4898 | 3.7190 | – | – | – |
| 10 | 1.7500 | 2.0939 | 2.2314 | – | – | – |
| 16 | 1.0938 | 1.3087 | 1.3946 | 1.7625 | 2.1176 | 2.2597 |
| 25 | 0.7000 | 0.8376 | 0.8926 | 1.1280 | 1.3553 | 1.4462 |
| 35 | 0.5000 | 0.5983 | 0.6376 | 0.8057 | 0.9681 | 1.0330 |
| 50 | 0.3500 | 0.4188 | 0.4463 | 0.5640 | 0.6776 | 0.7231 |
| 70 | 0.2500 | 0.2991 | 0.3188 | 0.4029 | 0.4840 | 0.5165 |
| 95 | 0.1842 | 0.2204 | 0.2349 | 0.2968 | 0.3567 | 0.3806 |
| 120 | 0.1458 | 0.1745 | 0.1860 | 0.2350 | 0.2824 | 0.3013 |
| 150 | 0.1167 | 0.1396 | 0.1488 | 0.1880 | 0.2259 | 0.2410 |
| 185 | 0.0946 | 0.1132 | 0.1206 | 0.1524 | 0.1831 | 0.1954 |
| 240 | 0.0729 | 0.0872 | 0.0930 | 0.1175 | 0.1412 | 0.1506 |
Example: 30 kW three-phase load, 80 m, 25 mm² copper cable
- Current: I = 30,000 W / (√3 × 400 V × 0.90 × 0.95) = 50.64 A.
- Resistance: for 25 mm² copper at 70 °C, R = 0.8376 Ω/km; reactance X = 0.08 Ω/km.
- Voltage drop: with L = 0.08 km, ΔU = √3 × I × L × (R cos φ + X sin φ) = 5.53 V.
- Result: 1.38% drop, voltage at line end 394.47 V.
What is the voltage drop formula?
Single-phase uses ΔU = 2 × I × L × (R cos φ + X sin φ) and balanced three-phase uses ΔU = √3 × I × L × (R cos φ + X sin φ). I is the line current (A), L the one-way line length (km), R and X the conductor resistance and reactance (Ω/km). Percentage drop = ΔU / U × 100, where U is the supply voltage of the circuit (phase-to-neutral for single-phase, line-to-line for three-phase).
How is the length entered in a voltage drop calculation?
Length is entered one way. The single-phase formula applies a factor of 2 for the outgoing and return path; the balanced three-phase formula applies √3.
How should the voltage drop result be assessed?
Compare the volt and percentage results with the applicable project criteria, the tolerances of connected equipment and the cable manufacturer's AC resistance and reactance data.