ELECTRICAL / CALCULATOR
Cable Voltage Drop
Estimate steady-state single-phase or three-phase cable voltage drop and percentage.
Input Parameters
Edit the example values. Results update as you type.
Enter circuit voltage for single phase and line voltage for three phase.
Enter a decimal greater than 0 and no greater than 1.
Default estimate; replace with manufacturer data when available.
Enter reliable active power at the load end to calculate transmission efficiency.
Used to correct candidate cable ampacity; range: 10–60 °C.
Results
- Steady-State Voltage Drop
- 7.702V
- Relative Voltage Drop
- 2.027%
- Conductor Resistance
- 3.438Ω/km
- Conductor Loss
- 464.122W
- Transmission Efficiency
- Enter load-end active power
- Reference Cross-Section
- 6mm²
Formula
ΔU ≈ k × I × L × (R′ × cosφ + X′ × sinφ) · ΔU% = ΔU / U × 100% · Ploss = 3 × I² × R′ × LSubstitution
ΔU ≈ √3 × 30 × 50 × (0.003437943 × 0.85 + 0.00008 × √(1 − 0.85²)) = 7.702 VCalculation Steps
- Convert cross-section to 0.000006 m²; calculate R′ using the resistivity of Copper at 70 °C.
- Use voltage-drop factor √3; divide the drop by 380 V to obtain 2.027%.
- Sum losses across 3 current-carrying conductors: Ploss = 3 × 30² × 0.003437943 × 50 = 464.122 W.
- Candidates use the same conductor temperature and reactance to estimate drop and loss; check against 30 °C ambient, installation method B2 and a 3% drop budget.
Result Explanation
Steady-state approximation for lagging power factor. Efficiency includes conductor loss only. Candidate ampacities use a single-circuit PVC 70 °C reference table. The five cross-sections nearest the recommendation are shown; sizes without ampacity data are excluded.
| Conductor Cross-Section | Drop V | Drop % | Loss W | Ampacity Check | Voltage-Drop Check | Overall Status |
|---|---|---|---|---|---|---|
| 2.5 mm² | 18.331 | 4.824 | 1,113.894 | 19.5 A · Insufficient | Over Limit | Review required |
| 4 mm² | 11.498 | 3.026 | 696.184 | 26 A · Insufficient | Over Limit | Review required |
| 6 mm² | 7.702 | 2.027 | 464.122 | 34 A · Pass | Pass | Recommended Reference |
| 10 mm² | 4.665 | 1.228 | 278.473 | 46 A · Pass | Pass | Meets Requirements |
| 16 mm² | 2.957 | 0.778 | 174.046 | 61 A · Pass | Pass | Meets Requirements |
Continue to Component Selection
Formulas and FAQ
ΔU ≈ k × I × L × (R′ × cosφ + X′ × sinφ)Why is the length one-way?
The single-phase factor 2 includes the outgoing and return conductors. Balanced three-phase circuits use √3.