Minimum cross-section according to voltage drop criterion. A = k·ρ·L·I·cosφ / ΔU It is found using (ΔU = allowable drop, V; 3F: k=√3, 1F: k=2). For example, 400 V, 63 A, 45 m, %3 target, copper, cosφ 0.9 for ≈ 6.5 mm² → standard 10 mm².
A = k·ρ·L·I·cosφ / ΔU (3F: k=√3, 1F: k=2)
Example calculation: 3-phase 400 V, 63 A, 45 m, %3 target, copper (ρ=0.0178), cosφ 0.9 → A = √3×0.0178×45×63×0.9/12 ≈ 6.5 mm² → standard 10 mm².
Minimum cable cross-section from target voltage drop.
Find the minimum cable cross-section required based on the allowed voltage drop percentage. This is the inverse of the voltage drop tool.
Calculated cross-section
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Standard cross-section
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Allowed ΔU
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Notes: A = k·ρ·L·I·cosφ / ΔU (3F: k=√3, 1F: k=2). It is an omics approach. (Reactance and temperature neglected). The selected cross-section is also current carrying capacity and short-circuit withstand This should also be verified in terms of other criteria — this tool only provides the voltage drop criterion.
How is it calculated?
- Select the system (1F/3F) and voltage.
- Enter the load current (A) and line length (m).
- Target %ΔU, conductor and cosφ, then press Calculate.
Frequently Asked Questions
How is cable cross-section selected based on voltage drop?
The allowable voltage drop (e.g., %3–5) is determined, and the minimum required cross-section is calculated using A = k·ρ·L·I·cosφ/ΔU and rounded to the next higher standard value.
Is voltage reduction alone sufficient?
No. The selected cross-section must also be verified in terms of current carrying capacity (ampacity) and short-circuit thermal withstand capability.
Which voltage drop limit is used?
General reference: ≤ %3 for lighting, ≤ %5 for other circuits (TS EN 60364).
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