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?

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?

• A. 166 amps
• B. 142 amps
• C. 48.6 amps
• D. 82.7 amps

Answer: (B)

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 = \

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 = \