A 25 hp motor has a nameplate SF = 1.10 and FLA = 60. Without damage, what is the maximum current in amps this motor can pull for a short period of time at rated voltage?

To determine the maximum current a motor can pull for a short period of time without damage, you need to consider both its Full Load Amps (FLA) and its Service Factor (SF). In this case, the motor has a nameplate SF of 1.10 and FLA of 60 amps.

The maximum current that the motor can safely draw is calculated by multiplying the FLA by the SF. This is because the service factor indicates how much additional load the motor can handle above its rated capacity for a limited time without suffering damage.

By multiplying the FLA (60 amps) by the service factor (1.10), you obtain the following:

Maximum current = FLA × SF

Maximum current = 60 amps × 1.10 = 66 amps

This calculation shows that the motor can safely handle a short-term current of up to 66 amps. Therefore, the correct answer indicates the maximum short-term current capability of this motor under specified conditions, aligning with the concepts of motor performance and safety standards.