If a 3Ø 240-volt load is supplied by a 100 KW generator and the load's nameplate rating is 59.2 kVA, what is the current in amps?

Study for the Independent Electrical Contractors (IEC) Year 2 Part 3 Test. Use flashcards and multiple choice questions with hints and explanations to prepare confidently. Get exam-ready now!

To find the current in amps for a three-phase (3Ø) load, you can use the formula:

[

I = \frac{P}{\sqrt{3} \times V}

]

Where:

  • ( I ) is the current in amps,

  • ( P ) is the power in watts (or kVA, converted to watts),

  • ( V ) is the line-to-line voltage.

Given that the load's nameplate rating is 59.2 kVA, this value needs to be converted to watts:

[

59.2 \text{ kVA} = 59,200 \text{ VA}

]

Next, use the voltage which is provided as 240 volts:

Now, substituting the values into the formula:

[

I = \frac{59,200}{\sqrt{3} \times 240}

]

Calculating ( \sqrt{3} ) gives approximately 1.732, so:

[

I = \frac{59,200}{1.732 \times 240}

]

Calculating the denominator:

[

1.732 \times 240 = 415.68

]

Now divide:

[

I = \

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