What is the required circuit ampacity of a noncontinuous, single-phase load of 8.5kW at 277V?

• A. 35.24A
• B. 30.69A
• C. 28.15A
• D. 27.50A

Answer: (B)

To determine the required circuit ampacity for a noncontinuous, single-phase load of 8.5 kW at 277V, we start by using the formula to calculate the current (I) based on the power (P) and voltage (V):

[ I = \frac{P}{V} ]

Here, P is the power in watts, which can be calculated by converting kilowatts (kW) to watts:

[ P = 8.5 , \text{kW} \times 1000 , \text{W/kW} = 8500 , \text{W} ]

Next, we substitute this value of power and the voltage into the current formula:

[ I = \frac{8500 , \text{W}}{277 , \text{V}} ]

Calculating this gives:

[ I = 30.69 , \text{A} ]

Since the load is classified as noncontinuous, we can use this value directly as the required circuit ampacity. Noncontinuous loads are those that do not operate continuously for three hours or more, meaning their maximum current can be used directly in circuit design without derating it for continuous operation.