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Single-Phase • 230V • Copper Cable

Calculate Wire Cross-Section for Single-Phase 230V

Calculate the required wire cross-section for a copper cable on a single-phase system based on power, 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:

I = P / (U × cos φ)

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

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

Which formula does the single-phase copper calculator use? +

The calculator uses I = P / (U × cos φ) for the current and A = (2 × L × I × cos φ) / (κ × ΔU) for the cross-section.

Why does the cross-section formula include a factor of 2? +

Single-phase circuits account for both the outgoing and return conductor. This is why the cable length enters the voltage drop calculation with a factor of 2.

Which κ value is used for copper? +

For copper, κ = 58 is the default. The value stays editable so other material or temperature values can be considered if needed.

Is the recommended cross-section automatically standards-compliant? +

No. The calculator provides a guideline based on voltage drop. Installation method, current-carrying capacity, protective devices and applicable standards must also be checked.

Practical guidance

Size a copper cable for a 230 V load

This page converts active power into single-phase current and then calculates the cross-section required by the selected voltage-drop limit.

Input values explained

Power in kilowatts
Enter the electrical active power in kW. One kW equals 1,000 W; motor output power and electrical input power are not necessarily the same.
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: 3.5 kW over 25 m

A 230 V load draws 3.5 kW at cos φ 0.90 through 25 m of copper with a 3% voltage-drop target.

I ≈ 16.91 A; A = (2 × 25 × 16.91 × 0.90) ÷ (58 × 6.9) ≈ 1.90 mm²

The next standard size is 2.5 mm². The circuit must still meet installation-method, protection and disconnection requirements.

Why outgoing and return paths matter

Factor 2 represents line and return conductors. Enter the one-way route length; do not double it manually.

Calculation limits

  • Loop impedance and thermal ampacity are not verified.
  • High-inrush and nonlinear loads require additional assessment.