Three-Phase Current and Power Factor: From kW to Feeder Current
A worked balanced three-phase example converting real power, line voltage, power factor, and efficiency into line current and apparent power.
Why this calculation matters
Three-phase equipment is often specified by real output power while feeders and transformers must carry apparent power and current. Power factor and efficiency therefore matter even when the useful kW is fixed.
This example uses a balanced 480 V three-phase load delivering 30 kW of useful output at 0.85 power factor and 92% efficiency.
What you will calculate
- Separate useful output power from electrical input power.
- Convert kW and power factor into apparent power.
- Calculate balanced three-phase line current.
- Understand how poor power factor increases current without increasing useful real power.
Given values
- Useful output power = 30 kW
- Line-to-line RMS voltage = 480 V
- Power factor = 0.85
- Efficiency = 92%
- Balanced sinusoidal three-phase system
Governing equations
Electrical input power
Pin = Pout/ηEfficiency loss means the source must supply more real power than the useful output.
Apparent power
S = Pin/PFFor sinusoidal balanced operation.
Three-phase line current
I = S/(√3 VLL)S in VA and line-to-line voltage in volts.
Reactive power
Q = √(S² − P²)Magnitude for the balanced sinusoidal case.
Worked solution
1. Correct for efficiency
The source must supply about 32.61 kW of real input power to deliver 30 kW at 92% efficiency.
Pin = 30/0.92 = 32.61 kW2. Convert to apparent power
At 0.85 power factor, the apparent power rises to about 38.36 kVA. This is the quantity that drives RMS current and equipment VA loading.
S = 32.61/0.85 = 38.36 kVA3. Calculate line current
Using the balanced three-phase relation gives about 46.1 A line current at 480 V.
I = 38360/(√3×480) ≈ 46.1 A4. Interpret power factor
If power factor were improved while real power and voltage stayed the same, line current would fall. That can reduce conductor I²R loss and free capacity in upstream equipment, although the correct correction strategy depends on the load and harmonic environment.
Engineering interpretation
A 30 kW useful load at 92% efficiency and 0.85 PF requires about 32.6 kW real input, 38.4 kVA apparent power, and 46.1 A at 480 V three phase.
Feeder and protection selection still requires the applicable load classification, duty, code factors, harmonics, starting/inrush behavior, conductor installation, and equipment ratings.
Sanity checks
- At PF = 1 with the same real input power, apparent power should equal real power and line current should be lower.
- Efficiency below 100% must make input real power greater than output real power.
- For fixed kVA, doubling line voltage should halve line current.
- Do not apply the √3 equation to a single-phase circuit.
Common mistakes
- Using output mechanical kW as if it were electrical input kW.
- Leaving efficiency and power factor both out of the current calculation.
- Using phase-to-neutral voltage in an equation written for line-to-line voltage.
- Assuming nominal current captures motor starting or nonlinear-load harmonic current.
References and model boundaries
- Balanced three-phase real/apparent/reactive power relationships from standard AC circuit and power-system references.
- Motor and nonlinear-load feeder design requires equipment-specific and code-specific current rules beyond the balanced sinusoidal calculation.
For safety-critical, regulated, production, or otherwise consequential work, independently verify the result using the governing standard, current manufacturer data, and qualified engineering review. See the site methodology and engineering disclaimer.