What is the required circuit ampacity for a noncontinuous, three-phase load of 6kW at 208V?

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Multiple Choice

What is the required circuit ampacity for a noncontinuous, three-phase load of 6kW at 208V?

Explanation:
To determine the required circuit ampacity for a noncontinuous, three-phase load of 6 kW at 208 V, we first need to calculate the current using the formula for three-phase power. The formula for calculating the current in a three-phase system is: \[ I = \frac{P}{\sqrt{3} \times V \times PF} \] In this case: - \( P \) is the power in watts (6000 W), - \( V \) is the line-to-line voltage (208 V), - \( PF \) is the power factor. For this calculation, if not specified, it's typically assumed to be 1 for purely resistive loads. By plugging in the values: \[ I = \frac{6000}{\sqrt{3} \times 208} \] Calculating \( \sqrt{3} \) gives approximately 1.732. So, the calculation becomes: \[ I = \frac{6000}{1.732 \times 208} \approx \frac{6000}{360.736} \approx 16.65A \] This result of approximately 16.65 amps is the required circuit ampacity for this three-phase load of

To determine the required circuit ampacity for a noncontinuous, three-phase load of 6 kW at 208 V, we first need to calculate the current using the formula for three-phase power.

The formula for calculating the current in a three-phase system is:

[ I = \frac{P}{\sqrt{3} \times V \times PF} ]

In this case:

  • ( P ) is the power in watts (6000 W),

  • ( V ) is the line-to-line voltage (208 V),

  • ( PF ) is the power factor. For this calculation, if not specified, it's typically assumed to be 1 for purely resistive loads.

By plugging in the values:

[ I = \frac{6000}{\sqrt{3} \times 208} ]

Calculating ( \sqrt{3} ) gives approximately 1.732.

So, the calculation becomes:

[ I = \frac{6000}{1.732 \times 208} \approx \frac{6000}{360.736} \approx 16.65A ]

This result of approximately 16.65 amps is the required circuit ampacity for this three-phase load of

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