Three-Phase • 400V • Copper Cable • Current Input

Calculate Wire Cross-Section with Current in Amps

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

Three-Phase • 400V • Copper Cable • Current Input

Calculate the required wire cross-section for a copper cable on a three-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 three-phase systems:

A = (√3 × 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 Three-Phase Copper Cable with Current in Amps

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

When does entering amps directly make sense? +

Direct current input is useful when the operating current is already known, for example from a nameplate, datasheet or an existing fuse.

Which formula is used? +

For three-phase with direct current input, the calculator uses A = (√3 × L × I × cos φ) / (κ × ΔU).

Which material value applies to copper? +

For copper, κ defaults to 58.

Why is cos φ still requested? +

cos φ affects the voltage drop for single- and three-phase loads. For purely resistive loads, cos φ = 1 can be used.

Practical guidance

Compare a three-phase copper cable using known current

The entered line current goes directly into the voltage-drop formula; no current is inferred from power.

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 line-to-line voltage between two phases. This is commonly 400 V in low-voltage systems; 230 V is not the correct input here.
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: 32 A over 40 m

For 32 A, 400 V three-phase, 40 m copper, cos φ 0.90 and 3% voltage drop, the minimum is calculated.

A = (√3 × 40 × 32 × 0.90) ÷ (58 × 12) ≈ 2.87 mm²

The next standard value is 4 mm². Depending on installation method, 32 A may require a larger conductor for thermal reasons.

Check ampacity separately

Meeting a voltage-drop target does not prove acceptable conductor temperature or protective-device operation.

Calculation limits

Starting current, harmonics and phase imbalance are not modelled.
Loaded-core count and installation correction factors are absent.