For a 3-horsepower motor, what is the minimum conductor size allowed in a 120/208-V 3-ph system?

For a 3-horsepower motor operating in a 120/208-V three-phase system, it is important to determine the appropriate conductor size to ensure safe and efficient operation. The National Electrical Code (NEC) provides guidelines on sizing conductors based on the current that a motor will draw during operation.

To find the minimum conductor size, first, we calculate the full-load current for the motor. Using the formula for three-phase motors:

Full-load current (in amps) = (Horsepower x 746) / (Voltage x √3 x Efficiency x Power Factor)

Assuming an efficiency of around 90% and a power factor of 0.8, we can simplify this calculation. For a 3-horsepower motor:

Full-load current = (3 x 746) / (208 x √3 x 0.9 x 0.8)

This calculation results in a full-load current of approximately 9.6 amps.

Based on the NEC guidelines, conductors must be sized to handle this current safely, considering factors such as temperature ratings and the conditions of use. The minimum conductor size typically required for this current rating is 14 AWG copper wire when using standard derating factors.

While larger conductor sizes