Three-Phase • Current • Amps
Calculate the current in amps from power, voltage and power factor cos φ.
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.
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.
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°.
I = P / (U × 1.732)
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.
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.
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.
Without cos φ (assuming cos φ = 1)
| P (kW) | Current (A) with I = P / (400 × 1.732) |
|---|---|
| 5 kW | 5000 / 692.8 ≈ 7.22 A |
| 10 kW | 10000 / 692.8 ≈ 14.45 A |
| 15 kW | 15000 / 692.8 ≈ 21.67 A |
With cos φ = 0.8
| P (kW) | Current (A) with I = P / (400 × 1.732 × 0.8) |
|---|---|
| 5 kW | 5000 / 554.24 ≈ 9.02 A |
| 10 kW | 10000 / 554.24 ≈ 18.05 A |
| 15 kW | 15000 / 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.
FAQ
Briefly explained: formula, use case and key limits of the calculation.
For balanced three-phase systems, I = P / (√3 × U × cos φ) applies.
√3 results from the relationship between line-to-line voltage and phase quantities in a three-phase system.
Usually the line-to-line voltage is used, for example 400 V in a typical three-phase network.
cos φ = 1 applies approximately to purely resistive loads. Motors and inductive loads are usually below that.
Practical guidance
First establish whether the stated kW is electrical input or mechanical shaft output before using the current result.
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.
Power factor relates active and apparent power. Efficiency describes losses between electrical input and useful output.
