What is the minimum standard size inverse-time circuit breaker for a 480-volt, 3-phase circuit with a 50-kilowatt continuous load?

To determine the minimum standard size inverse-time circuit breaker for a 480-volt, 3-phase circuit with a 50-kilowatt continuous load, you begin with calculating the full-load current (IL) using the following formula for 3-phase circuits:

[

IL = \frac{P}{\sqrt{3} \times V \times PF}

]

Where:

• ( P ) is the power in watts (50 kW = 50,000 watts)

• ( V ) is the voltage (480 volts)

• ( PF ) is the power factor (assumed to be 1 for continuous loads unless specified otherwise)

Substituting the given values into the formula:

[

IL = \frac{50000}{\sqrt{3} \times 480 \times 1}

]

[

IL \approx \frac{50000}{831.76} \approx 60.0 \text{ Amps}

]

Now, voltage drops for continuous loads typically require applying a 125% rule for the breaker size to ensure that the circuit can handle overload conditions without nuisance tripping. The calculation for the circuit breaker size thus becomes:

[

Breaker Size = IL