on . Posted in Rotating Equipment

Motor load amps, abbreviated as MLA, is the amount of current that an electric motor draws when operating under a specific load.  It is used for understanding and sizing electrical systems, as it helps determine the electrical capacity and requirements for a motor.  The load amps can vary based on factors such as the motor's power rating, efficiency, mechanical load, and the voltage supplied to it.  When selecting a motor or designing an electrical system, it's important to consider the motor load amps to ensure that the electrical components, such as wires, fuses, and circuit breakers, can handle the current without exceeding their rated capacities.

Motor load amps are typically measured using an ammeter in the circuit, and they represent the actual current consumed by the motor during normal operation.  This value is distinct from the Full Load Amps (FLA) or Rated Load Amps (RLA), which are the maximum current values a motor is designed to handle under full load conditions.  The motor load amps can be lower than the full load amps when the motor operates at less than full load.  Understanding motor load amps is used for maintaining the reliability and efficiency of electrical systems, preventing overloading, and ensuring that the motor and associated components operate within their specified limits.

## Single Phase Motor Full Load Amps formula

$$\large{ I = \frac{ P_{out} \; 1000 }{ V \; PF } }$$    (Single Phase Motor Full Load Amps)

$$\large{ P_{out} = \frac{ I \; V \; PF }{ 1000 } }$$

$$\large{ V = \frac{ P_{out} \; 1000 }{ I \; PF } }$$

$$\large{ PF = \frac{ P_{out} \; 1000 }{ I \; V } }$$

Symbol English Metric
$$\large{ I }$$ = current $$\large{A}$$ $$\large{\frac{C}{s}}$$
$$\large{ P_{out} }$$ = output power $$\large{W}$$ $$\large{\frac{kg-m^2}{s^3}}$$
$$\large{ V }$$ = voltage $$\large{V}$$ $$\large{\frac{kg-m^2}{s^3-A}}$$
$$\large{ PF }$$ = Power Factor $$\large{dimensionless}$$

## Three Phase Motor Full Load Amps formula

$$\large{ I = \frac{ P_{out} \; 1000 }{ 1.732 \; V \; PF } }$$     (Three Phase Motor Full Load Amps)

$$\large{ P_{out} = \frac{ I \; 1.732 \; V \; PF }{ 1000 } }$$

$$\large{ V = \frac{ P_{out} \; 1000 }{ I \; 1.732 \; PF } }$$

$$\large{ PF = \frac{ P_{out} \; 1000 }{ I \; 1.732 \; V } }$$

Symbol English Metric
$$\large{ I }$$ = current $$\large{A}$$ $$\large{\frac{C}{s}}$$
$$\large{ P_{out} }$$ = output power $$\large{W}$$ $$\large{\frac{kg-m^2}{s^3}}$$
$$\large{ V }$$ = voltage $$\large{V}$$ $$\large{\frac{kg-m^2}{s^3-A}}$$
$$\large{ PF }$$ = Power Factor $$\large{dimensionless}$$ Tags: Motor