For a 240-volt, single-phase circuit supplying a continuous load of 5 kilowatts and a non-continuous load of 6 kilowatts, what is the minimum standard size inverse-time circuit breaker required?

To determine the minimum standard size inverse-time circuit breaker required for the given loads, you'll need to calculate the total load and then apply the NEC (National Electrical Code) guidelines.

First, convert the continuous and non-continuous loads from kilowatts to amperes using the formula:

[

I = \frac{P}{V}

]

where (I) is the current in amperes, (P) is the power in watts, and (V) is the voltage in volts. For this circuit, the total continuous load is 5 kW (or 5000 watts), and the non-continuous load is 6 kW (or 6000 watts).

Calculating the continuous load:

[

I_{\text{continuous}} = \frac{5000}{240} \approx 20.83 \text{ A}

]

Calculating the non-continuous load:

[

I_{\text{non-continuous}} = \frac{6000}{240} = 25 \text{ A}

]

According to NEC guidelines, for a continuous load, circuit breakers must be rated at 125% of the continuous load. So, the continuous load portion for the