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Electrical • Fuse • Amps

Calculate Fuse Rating

Calculate the current draw from power and voltage and find a suitable standard fuse rating.

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

Single-phase: I = P / (U × cos φ)

Three-phase: I = P / (√3 × U × cos φ)

The fuse rating is estimated as the next common standard value above the calculated current draw.

Frequently Asked Questions

Can I select the fuse directly from this result?

No. Fuse selection depends on the cable, installation method, environment, protective device, tripping characteristic and applicable standards.

Why is the result rounded up?

Standard fuses come in fixed rated values. The calculator shows the next standard value above the current draw.

What matters for motors?

Motors can have high inrush currents. This makes the protection concept and tripping characteristic especially important.

Practical guidance

Estimate load current, then select protection properly

The calculator finds steady load current and displays the next standard value. This is a starting point, not a completed protective-device design.

Input values explained

Power in watts
Use the electrical active power drawn by the load. A nameplate, data sheet or power meter is more reliable than an estimate.
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.
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.
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: 3,500 W at 230 V

A resistive 3,500 W load operates at 230 V and cos φ 1.

I = 3,500 W ÷ 230 V ≈ 15.22 A; next listed value: 16 A

16 A is numerically plausible but not approval. Cable, installation method, inrush, trip curve and fault conditions must all match.

Design cable and protection together

Rating, characteristic and breaking capacity must coordinate with the conductor and prospective fault current.

Calculation limits

  • Selectivity, let-through energy and disconnection time are not checked.
  • Motors and power supplies may require a different trip characteristic.