Cable • Amps • Voltage Drop

Calculate Maximum Current Over a Cable

Calculate the maximum current theoretically possible for a given voltage drop.

Cable • Amps • Voltage Drop

Calculate the maximum current theoretically possible for a given voltage drop.

Calculation Values

Conductor Cross-Section in mm²

Cable Length in m

Voltage in V

Permissible Voltage Drop in %

Power Factor cos φ

Conductivity κ

System type: 1 = single-phase, 3 = three-phase

Note: This calculator provides a technical estimate for guidance only. For electrical installations, standards, protective measures, installation method, ambient temperature, grouping, protective devices and manufacturer specifications must also be checked.

Formula Used

Imax = (ΔU × κ × A) / (factor × L × cos φ)

The factor is 2 for single-phase and √3 for three-phase.

Frequently Asked Questions

Is the calculated current also the fuse rating?

No. Fuses are selected based on cable protection, installation method and permissible current-carrying capacity.

Why does cable length affect the current?

Longer cables produce more voltage drop at the same current.

Why should voltage drop be limited?

Excessive voltage drop can impair device function, efficiency and safety.

Practical guidance

Calculate a voltage-drop current limit

The reversed formula finds the current at which the selected cable reaches the chosen voltage-drop target, not its thermal ampacity.

Input values explained

Conductor cross-section: Enter the cross-section of one loaded conductor in mm², not the sum of all cores in a cable.
Cable length: Enter the one-way distance from supply point to load. The required return path is already represented by the selected formula.
Voltage: Match the voltage to the selected electrical system. Single-phase calculations use line-to-neutral voltage, while three-phase calculations use voltage between two line conductors.
Permitted voltage drop: This percentage limits the calculated voltage loss. It is a design assumption and must suit the circuit and applicable requirements.
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.
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.
Electrical system: Choose single-phase AC for a 230 V circuit and three-phase for a balanced three-phase system. This selection changes the formula factor.

Example: 2.5 mm² over 25 m

At 230 V single-phase, copper, cos φ 0.90 and 3%, calculate the current limit.

I = (6.9 V × 58 × 2.5) ÷ (2 × 25 × 0.90) ≈ 22.23 A

22.23 A is a voltage-drop result only. Permitted thermal current may be lower.

Do not use it as a breaker rating

Protective-device selection must include cable ampacity, overload, fault current, disconnection time and load behaviour.

Calculation limits

Installation method, temperature and grouping are unknown.
Starting currents and nonlinear loads are excluded.