Calculators
Calculate voltage drop
This calculator works the other way round from the cross-section calculation. The cross-section is fixed and the question is what still arrives at the end of the cable. You need exactly this calculation for an existing installation or a later extension.
Result
- Voltage drop
- —%
- Voltage drop, absolute
- —V
- Voltage at the load
- —V
- Does the cross-section carry the current?
- —
Please fill all fields with valid numbers.
A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.
The grey values are an example. Simply enter your own.
Matching services
Calculating is the easy part.
A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.
Method and questions
Voltage drop
Voltage drop across a cable of known cross-section.
Discuss a projectWorked example
The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.
Inputs
- System
- Three-phase (3~)
- Operating current
- 32 A
- One-way cable length
- 45 m
- Conductor cross-section
- 6 mm²
- Nominal voltage
- 400 V
- Power factor cos φ
- 0.9
- Conductor material
- Copper (κ = 56 m/Ω·mm²)
Result
- Voltage drop
- 1.67 %
- Voltage drop, absolute
- 6.68 V
- Voltage at the load
- 393 V
- Does the cross-section carry the current?
- In free air, yes. Installation method, grouping and ambient temperature remain to be checked to IEC 60364-5-52. The cable cross-section calculator does that.
Formula and standards
Formula
- Three-phase Δu = (√3 · L · I · cos φ) / (κ · A)
- Single-phase Δu = (2 · L · I · cos φ) / (κ · A)
- DC Δu = (2 · L · I) / (κ · A)
- Δu% = Δu / U · 100
Assumptions and standards
- κ = 56 m/(Ω·mm²) for copper, 35 m/(Ω·mm²) for aluminium at 20 °C.
- Resistive component only, without inductive reactance.
- With heated conductors the resistance rises. At 70 °C the actual voltage drop is around 20 % above the value calculated here.
- The formula is a linear approximation for voltage drops well below the nominal voltage. Once the calculated drop reaches the nominal voltage the design is unusable anyway, the voltage at the load is then left blank.
Why does the measured value differ from the calculated one?
The formula assumes a conductor temperature of 20 °C. In operation the conductor is warmer and its resistance rises, by about a fifth at 70 °C. On top of that comes inductive reactance, which becomes noticeable at large cross-sections.
Does this apply to long motor cables on a drive?
Only to a limited extent. Inverter-fed cables add switching edges, leakage currents and screening effects. The calculation is usable for a rough sizing but not sufficient for the final decision.
Contact
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