Cable • Loss • Power

Calculate Cable Loss

Calculate how much power is lost over a cable.

Cable • Loss • Power

Calculate how much power is lost over a cable.

Calculation Values

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

R = L / (κ × A)

Single-phase: P_loss = 2 × I² × R

Three-phase: P_loss = 3 × I² × R

The calculator determines the loss from the conductor resistance and the current.

Frequently Asked Questions

Why do cable losses occur?

Every cable has electrical resistance. When current flows, some of the energy is converted into heat.

Why does current matter so much?

Power loss increases quadratically with current. Doubling the current quadruples the loss.

How can cable losses be reduced?

Through larger cross-sections, shorter cables, a suitable voltage level and proper sizing.

Practical guidance

Interpret cable loss in watts and percent

Resistive loss rises with the square of current, so a modest current increase can produce much more heat.

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.
Conductor cross-section: Enter the cross-section of one loaded conductor in mm², not the sum of all cores in a cable.
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.
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.
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: 16 A on 2.5 mm²

A 25 m one-way copper run of 2.5 mm² has about 0.1724 Ω per conductor.

Ploss = 2 × 16² × 0.1724 Ω ≈ 88.3 W

At constant load the cable dissipates about 88 W, or roughly 0.44 kWh over five hours.

Loss changes with conductor temperature

Heating increases resistance and actual loss. Percentage loss must also be compared with the real active power transferred.

Calculation limits

Load profile and power factor are simplified.
Contact losses, conductor temperature and harmonics are excluded.