Sparky's Journal

📅 March 27, 2026

Power Factor (PF) 

Power factor (PF) tells you how effectively AC power is being converted into useful work. It’s one of the most important concepts for understanding system efficiency, conductor sizing, and overall electrical performance.

Power Triangle

Power factor comes from the relationship between three types of power:

• kW (Real Power) → Does actual work (heat, motion, light)

• kVAR (Reactive Power) → Supports magnetic fields (motors, transformers)

• kVA (Apparent Power) → Total power the system must supply

Key formulas:

PF = kW ÷ kVA
kVA² = kW² + kVAR²
PF = cos(θ)

Where θ (theta) is the phase angle between voltage and current.

What Power Factor Really Means

PF = 1.0 (100%) → Perfect efficiency (all power does work)
PF < 1.0 → Some power is reactive and does not perform useful work

Lower power factor means:

• Higher current draw

• Increased heat losses

• Larger conductors and equipment required

Lagging vs Leading Power Factor

Lagging PF (most common):

• Inductive loads

• Motors

• Transformers

Leading PF:

• Capacitive systems

• Capacitor banks

• Lightly loaded long cable runs

Single-Phase Formulas (1φ)

kVA = (V × A) ÷ 1000
kW = (V × A × PF) ÷ 1000
PF = kW ÷ kVA

Example:

240 V, 50 A, PF = 0.80

kVA = (240 × 50) ÷ 1000 = 12.0 kVA
kW = 12.0 × 0.80 = 9.6 kW

Three-Phase Formulas (3φ)

(Use line-to-line voltage)

kVA = (1.732 × V × A) ÷ 1000
kW = (1.732 × V × A × PF) ÷ 1000
PF = kW ÷ kVA

Why 1.732 (√3)?

1.732 = √3

This comes from the 120° phase separation in three-phase systems and is used to relate line values to phase values.

Three-Phase Example

480 V, 50 A, PF = 0.80

1.732 × 480 = 831.36
831.36 × 50 = 41,568 VA = 41.6 kVA
kW = 41.6 × 0.80 = 33.3 kW

Rearranging Formulas 

From PF relationship:

kW = kVA × PF
kVA = kW ÷ PF

Solve for current:

Single-phase:
A = (kVA × 1000) ÷ V

Three-phase:
A = (kVA × 1000) ÷ (1.732 × V)

Field Insight

Low power factor increases system demand without increasing useful output.

This results in:

• Higher current

• More system losses

• Larger required equipment

Common solution:
Power factor correction using capacitor banks

Quick Reference

PF ≈ 0.8 → Typical motor load
PF ≈ 0.9–0.95 → Efficient system
PF ≈ 1.0 → Ideal (rare in real systems)

• 1.732 × 480 = 831.36

• 831.36 × 50 = 41,568 VA = 41.6 kVA

• 41.6 × 0.80 = 33.3 kW

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