1. What is Full Load Current?

Full Load Current (FLC) is the maximum current a motor draws when operating at its rated power under full load conditions. It's crucial for selecting proper wire sizes, circuit breakers, and overload protection devices.

2. How Does the Calculator Work?

The calculator uses the three-phase motor current formula:

Where:

• \( I \) — Full load current (amps)
• \( P \) — Motor power rating (watts)
• \( V \) — Line voltage (volts)
• \( PF \) — Power factor (0-1, typically 0.85)
• \( \eta \) — Motor efficiency (0-1, typically 0.85-0.95)
• \( \sqrt{3} \) — Constant for three-phase systems (≈1.732)

Explanation: The formula accounts for both the power factor (ratio of real power to apparent power) and motor efficiency (ratio of mechanical output to electrical input).

3. Importance of Full Load Current

Details: Knowing the full load current is essential for proper motor protection, preventing overheating, and ensuring electrical system safety. It helps in selecting:

• Correct wire gauge to prevent voltage drop and overheating
• Proper circuit breaker size for overload protection
• Appropriate starter and contactor ratings

4. Using the Calculator

Tips:

• Enter motor power rating in watts (convert horsepower if needed: 1 HP ≈ 746 watts)
• Use line-to-line voltage for three-phase systems
• Typical power factor for motors ranges from 0.8 to 0.95
• Modern motors typically have efficiencies between 85-95%

5. Frequently Asked Questions (FAQ)

Q1: What's the difference between FLC and locked rotor current?
A: FLC is the steady-state current at full load, while locked rotor current (starting current) can be 5-7 times higher during motor startup.

Q2: How does voltage affect full load current?
A: Current is inversely proportional to voltage. A 10% voltage drop results in approximately 10% higher current draw.

Q3: What's a typical power factor for motors?
A: Induction motors typically have power factors between 0.8-0.9 at full load, lower at partial loads.

Q4: How do I account for single-phase motors?
A: For single-phase, use \( I = P / (V \times PF \times \eta) \) (remove the √3 factor).

Q5: Why is efficiency important in this calculation?
A: Efficiency accounts for power losses in the motor. A 90% efficient motor draws more current than a 95% efficient one for the same output power.