AC induction motors are widely used in industrial and commercial applications due to their robustness, reliability, and efficiency. However, it’s important to properly size the electrical circuit and components for a motor to ensure it operates safely and efficiently. One important factor to consider is the full load current of the motor, which is the amount of current the motor will draw when it is operating at its rated load. Here we will introduce a calculator tool that allows you to quickly and easily estimate the full load current of an AC induction motor based on its power rating, voltage, power factor, and efficiency. This tool can be a useful resource for engineers and technicians working with AC induction motors.

## Calculator:

# AC Induction Motor Full Load Current Calculator

## The full load current of an AC induction motor is calculated using the following formula:

Full Load Current (FLC) = (Power Rating (kW) * 1000) / (Efficiency * Power Factor * Voltage) for single phase induction motor

and

Full Load Current (FLC) = (Power Rating (kW) * 1000) / (1.73 * Efficiency * Power Factor * Voltage) for three phase induction motor

Where:

- kW is the motor’s power rating in kilovolt-amperes.
- Voltage is the voltage of the electrical system the motor will be connected to.
- 1.73 is the square root of 3, which is the ratio of the RMS value of the sinusoidal voltage to the peak value.

To calculate the full load current of a motor with a power rating of 10 KW, a voltage of 240V, a power factor of 0.85, and an efficiency of 75%, you can use the following formula:

Full Load Current (FLC) = (Power Rating (kW) * 1000) / (Efficiency * Power Factor * Voltage)

Plugging in the values given, we get:

FLC = (10 * 1000) / (0.75 * 0.85 * 240) = 65.36 A

This is the full load current of the motor under these conditions. It’s important to note that this is just an estimate, and the actual full load current of the motor may be slightly different. It’s always a good idea to verify the full load current of a motor before selecting a suitable electrical circuit and component ratings.