Alternating single-phase
cosφ = P / (I * U₀ * η)
Alternating two-phase
cosφ = P / (I * U * η)
Alternating three-phase
cosφ = P / (√3 * I * U * η)
Where：
η = Efficiency(%)
P = Power (W)
U₀ = Voltage [L - N] (V)
U = Voltage [L - L] (V)
I = Current(A)
cosφ = Power factor
Calculate actual motor power factor (cosφ) from output power (kW), voltage (V), current (A), and efficiency. For engineers assessing NEMA-compliant motors under real load conditions.
Supports single-phase, two-phase, and three-phase AC systems.
Since:
Pout = Pin × η
Pin = U × I × cosφ (for three-phase: √3 × U × I × cosφ)
Therefore:
| System Type | Formula |
|---|---|
| Alternating single-phase | cosφ = P / (I × U × η) |
| Alternating two-phase | cosφ = P / (I × U × η) |
| Alternating three-phase | cosφ = P / (√3 × I × U × η) |
Note: All formulas assume balanced systems. P is output mechanical power in watts (W), η is efficiency as decimal (e.g., 85% → 0.85).
P (Power): Output mechanical power (shaft power) in watts (W). Enter this value in the “Power” input field.
U (Voltage): Line-to-line voltage in volts (V). Enter this value in the “Voltage” input field.
I (Current): Line current in amperes (A). Enter this value in the “Current” input field.
η (Efficiency): Motor efficiency as a decimal (e.g., 85% → 0.85). Enter this value as a percentage in the “Efficiency” input field; the calculator automatically converts it to decimal form.
Example 1: Three-phase industrial pump motor
Nameplate rating: 7.5 kW, PF = 0.85
Field measurement: U = 400 V, I = 12 A, P = 5.8 kW, η = 90% → 0.9
→ cosφ = 5800 / (√3 × 12 × 400 × 0.9) ≈ 0.70
→ Actual PF ≈ 0.70 vs rated 0.85 → drop of 18%
→ Insight: At partial load, PF decreases — may increase demand charges and losses.
Example 2: Single-phase HVAC compressor
Nameplate rating: 1.1 kW, PF = 0.90
Field measurement: U = 230 V, I = 6 A, P = 850 W, η = 80% → 0.8
→ cosφ = 850 / (230 × 6 × 0.8) ≈ 0.62
→ Actual PF ≈ 0.62 vs rated 0.90 → significant drop
→ Insight: Low PF may trigger utility penalties in commercial buildings.
Maintenance engineers monitoring motor health
Energy auditors assessing power quality
Facility managers reducing demand charges
Electrical technicians troubleshooting low PF issues
Measure under stable, representative operating conditions
Use a true RMS power analyzer — standard multimeters cannot measure real power accurately
Power factor cannot exceed 1.0; values above indicate measurement error
PF varies with load: always compare against expected performance at that load level
Industrial facilities with PF < 0.9 may face utility surcharges
≥0.95: Excellent (modern high-efficiency motors)
0.85–0.95: Good (typical at full load)
0.70–0.85: Acceptable (partial load)
<0.70: Poor — consider power factor correction
No. Voltage and current alone give apparent power (kVA), but you need real power (kW) — measured with a wattmeter or power analyzer — to calculate true PF.
Not directly, but it increases current draw for the same output, leading to higher winding temperatures, reduced lifespan, and larger cable/transformer requirements.
Nameplate PF is specified at full load and ideal conditions. In practice, motors often run at partial load, where PF drops significantly—sometimes below 0.7.
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