Formula & Calculator
Motor Full-Load Current (Single Phase)
Calculates a single-phase motor's rated full-load current from its power, voltage, power factor, and efficiency.
Interpretation
Motor full‑load current (single phase): I = P / (V × cosθ × η), where η is efficiency.
This gives the line current at rated load, accounting for power factor and losses.
Example: P=1kW, V=230V, cosθ=0.8, η=0.85 → I = 1000 / (230 × 0.8 × 0.85) = 1000 / 156.4 ≈ 6.39A.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| I | Full-load current | A |
| P | Output power (mechanical) | W |
| V | Voltage | V |
| cos θ | Power factor | dimensionless |
| η | Motor efficiency | dimensionless (0 to 1) |
What it means
For a single‑phase motor, the full‑load current I is given by I = P / (V × cosθ × η), where P is the mechanical output power (kW), V is the voltage, cosθ is the power factor, and η is the efficiency. This accounts for the motor’s power factor and losses, giving the actual line current drawn from the supply. This current is used for sizing cables, fuses, and starters. Example: A 1kW, 230V motor with power factor 0.8 and efficiency 85% draws I = 1000 / (230 × 0.8 × 0.85) = 1000 / 156.4 ≈ 6.39A. The nameplate current is typically provided by the manufacturer.
Worked example
Single‑Phase Motor FLC – Practical Example
Real‑World| Parameter | Value |
|---|---|
| P | 1492 W |
| V | 230 V |
| cos θ | 0.85 |
| η | 0.85 |
| Formula | I = P / (V × cos θ × η) |
Common mistakes
Watch for unit mismatches (W vs kW, single- vs three-phase) and remember to include power factor or efficiency where the formula requires it.Applications
Motor full‑load current (single phase): I = P/(V×cosθ×η) calculates the current drawn by a motor at rated load. Engineers use it to size conductors, to select overload protection, and to assess starting current. This is essential for motor circuit design.
- Single‑phase motor circuit design
- Conductor and protection sizing
- Motor performance and efficiency assessment
- Starting current and voltage drop analysis
- Educational understanding of motor current
Frequently Asked Questions
The full‑load current is I = P / (V × cos θ × η), where P is the output power (W), V is the voltage, cos θ is the power factor, and η is the efficiency.
PF: 0.6‑0.9 (larger motors have higher PF). Efficiency: 70‑85% for small motors, up to 90% for larger ones.
P = 2 × 746 = 1492 W. I = 1492 / (230 × 0.85 × 0.85) = 8.98 A.
Common errors: 1) using the output power instead of input, 2) forgetting the power factor or efficiency, 3) using the wrong voltage, 4) applying the three‑phase formula.
At start, the motor draws 5‑7 times full‑load current due to low impedance. This must be considered for protection.
Sizing conductors, breakers, and starters for single‑phase motors.
The nameplate gives the full‑load current; it is often used directly instead of calculating.
At lower voltage, the motor draws more current to maintain the same power, potentially causing overheating.