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Single-Phase • 230V • Copper Cable • Current Input

Calculate Wire Cross-Section with Current in Amps

Calculate the required wire cross-section for a copper cable on a single-phase system based on current, cable length, cos φ and permissible voltage drop.

Calculation Values

Note: This calculation is a technically sound simplification typical of online calculators. For practical sizing, standards, installation method, temperature, grouping, protective devices, cable type and manufacturer specifications must also be checked.

Formula Used

For single-phase systems:

A = (2 × L × I × cos φ) / (κ × ΔU)

Here, A is the conductor cross-section in mm², L the one-way cable length in meters, I the current in amps, κ the conductivity of copper, defaulting to 58, and ΔU the permissible voltage drop in volts. The power factor cos φ is properly accounted for in the current and voltage drop calculation.

FAQ

Frequently Asked Questions about Calculate Wire Cross-Section for Single-Phase Copper Cable with Current in Amps

Briefly explained: formula, use case and key limits of the calculation.

Which formula does the calculator with amp input use? +

The calculator uses A = (2 × L × I × cos φ) / (κ × ΔU). The factor 2 accounts for the outgoing and return conductor on a single-phase system.

When should I use this calculator? +

This calculator is useful when the current in amps is already known and does not need to be derived from power first.

Which κ value is preset? +

For copper, κ = 58 is preset.

Is the current-carrying capacity checked automatically? +

No. The calculator looks at voltage drop. The permissible current-carrying capacity must be checked separately based on installation method, cable type and protective devices.

Practical guidance

Check a 230 V copper cable from known current

Known operating current and one-way length are used with factor 2 for the complete outgoing and return path.

Input values explained

Current
Enter the expected operating current, not a fuse rating. Motors and electronic power supplies may draw much higher starting or peak currents.
Cable length
Enter the one-way distance from supply point to load. The required return path is already represented by the selected formula.
Voltage
Enter the voltage between line and neutral. This is commonly 230 V in low-voltage systems.
Power factor cos φ
cos φ is the ratio of active to apparent power. It is close to 1 for resistive heating loads; use the data-sheet value for motors and transformers.
Permitted voltage drop
This percentage limits the calculated voltage loss. It is a design assumption and must suit the circuit and applicable requirements.
Conductivity κ
κ is the calculation value for the conductor material. Common approximations are 58 for copper and 37 for aluminium; heating increases actual resistance.

Example: 16 A over 25 m

For 16 A, 230 V, 25 m copper, cos φ 0.90 and 3%, the available voltage loss is 6.9 V.

A = (2 × 25 × 16 × 0.90) ÷ (58 × 6.9) ≈ 1.80 mm²

The calculator selects 2.5 mm². Thermal ampacity and protection still need independent verification.

Voltage drop is one design check

Long runs may be governed by voltage drop; short runs are often governed by ampacity. The stricter requirement wins.

Calculation limits

  • Contact resistance is excluded.
  • Installation method and fault-disconnection conditions can require a larger size.