What is the largest circuit breaker allowed to protect a 25 HP three-phase, 480-volt induction motor?

To determine the largest circuit breaker allowed to protect a 25 HP three-phase, 480-volt induction motor, you can start by calculating the full-load current (FLC) of the motor using the standard equations provided in the National Electrical Code (NEC). For a three-phase motor, the formula used to find the full-load current is:

[

FLC = \frac{HP \times 746}{\sqrt{3} \times V \times PF}

]

For a 25 HP motor, assuming a power factor of around 0.9 (which is typical for induction motors):

[

FLC = \frac{25 \times 746}{\sqrt{3} \times 480 \times 0.9} \approx 30.3 \text{ amperes}

]

According to NEC guidelines, for motor protection, you generally use a circuit breaker that can handle 125% of the full-load current for short-circuit protection.

To find the appropriate circuit breaker size:

[

\text{Breaker Rating} = 125% \times FLC = 1.25 \times 30.3 \approx 37.875 \text{ amperes}

\