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trójfazowy • Prąd • Amps

Oblicz prąd dla trójfazowego

Prąd w amperach z mocy, napięcia i współczynnika mocy cos φ.

Wartości obliczeń

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Zastosowany wzór

For balanced three-phase systems:

I = P / (√3 × U × cos φ)

I is the current in amps, P the real power in watts, U the line-to-line voltage in volts and cos φ the power factor.

Convert kW to Amps at 400V trójfazowy

To select the right cables, fuses and protective devices, the power of a device or system must be converted to amps. For a 400V three-phase network, the square root of 3 is used.

What does 1.732 mean?

This is the rounded square root of 3 (more precisely ≈ 1.73205). It is used in three-phase calculations because a three-phase system has three phases, each shifted by 120°.

1. Formula without power factor (purely resistive load)

I = P / (U × 1.732)

  • I = current in amps (A)
  • P = power in watts (W) or kW (1 kW = 1000 W)
  • U = three-phase voltage, usually 400 volts (V)
  • 1.732 = rounded square root of 3

Worked example:

P = 5 kW = 5000 W, U = 400 V

I = 5000 / (400 × 1.732)

I = 5000 / 692.8

I ≈ 7.22 A

The current is approximately 7.22 A.

2. Formula with power factor (cos φ)

I = P / (U × 1.732 × cos φ)

Many devices, especially motors and transformers, have a power factor (cos φ) below 1. This factor is typically between 0.7 and 1.0.

Worked example with cos φ = 0.8:

P = 5 kW = 5000 W, U = 400 V, cos φ = 0.8

I = 5000 / (400 × 1.732 × 0.8)

I = 5000 / 554.24

I ≈ 9.02 A

The current is approximately 9.02 A.

Why is the current higher with cos φ?

When cos φ is less than 1, not all of the electrical power is converted into useful power. Part of it flows back and forth as reactive power.

Quick Overview as a Table

Without cos φ (assuming cos φ = 1)

P (kW) Prąd (A) with I = P / (400 × 1.732)
5 kW5000 / 692.8 ≈ 7.22 A
10 kW10000 / 692.8 ≈ 14.45 A
15 kW15000 / 692.8 ≈ 21.67 A

With cos φ = 0.8

P (kW) Prąd (A) with I = P / (400 × 1.732 × 0.8)
5 kW5000 / 554.24 ≈ 9.02 A
10 kW10000 / 554.24 ≈ 18.05 A
15 kW15000 / 554.24 ≈ 27.07 A

Converting kilowatts (kW) to amps (A) is essential for planning electrical systems on a 400V three-phase network. For complex calculations or high power values, a qualified electrician should be consulted.

Practical guidance

Apply a three-phase current result to motors and equipment

First establish whether the stated kW is electrical input or mechanical shaft output before using the current result.

Input values explained

Moc w watach
Use the electrical active power drawn by the load. A nameplate, data sheet or power meter is more reliable than an estimate.
Napięcie
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.
Moc 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.

Example: 11 kW at cos φ 0.90

For 11 kW electrical active power and 400 V line voltage, the balanced line current is calculated.

I = 11,000 W ÷ (√3 × 400 V × 0.90) ≈ 17.64 A

Use a manufacturer's rated current when available, particularly if the 11 kW value is mechanical output.

Moc factor is not efficiency

Moc factor relates active and apparent power. Efficiency describes losses between electrical input and useful output.

Calculation limits

  • Starting method, variable-speed drives and inrush are excluded.
  • Phase imbalance requires measurements and a different assessment.