What is the maximum line current a 500 KVA 480-volt three-phase delta secondary transformer can deliver without exceeding its rating?

What is the maximum line current a 500 KVA 480-volt three-phase delta secondary transformer can deliver without exceeding its rating?

• A. 480 amps
• B. 601 amps
• C. 375 amps
• D. 720 amps

Answer: (B)

To determine the maximum line current that a 500 KVA, 480-volt three-phase delta secondary transformer can deliver, we need to use the formula for calculating current in a three-phase system. The formula to find the current (I) for a three-phase transformer is: \[ I = \frac{P}{\sqrt{3} \times V} \] Where: - \( P \) is the power in watts (in this case, 500 KVA, which is equal to 500,000 VA), - \( V \) is the line-to-line voltage (480 volts), - \( \sqrt{3} \) is approximately 1.732, representing the conversion in a three-phase system. Substituting the values into the formula: \[ I = \frac{500,000}{\sqrt{3} \times 480} \] Calculating it step by step: 1. Calculate \( \sqrt{3} \times 480 \): \[ \sqrt{3} \approx 1.732 \quad \text{thus,} \quad 1.732 \times 480 \approx 829.44 \] 2. Now, divide

To determine the maximum line current that a 500 KVA, 480-volt three-phase delta secondary transformer can deliver, we need to use the formula for calculating current in a three-phase system. The formula to find the current (I) for a three-phase transformer is:

[

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

]

Where:

• ( P ) is the power in watts (in this case, 500 KVA, which is equal to 500,000 VA),

• ( V ) is the line-to-line voltage (480 volts),

• ( \sqrt{3} ) is approximately 1.732, representing the conversion in a three-phase system.

Substituting the values into the formula:

[

I = \frac{500,000}{\sqrt{3} \times 480}

]

Calculating it step by step:

1. Calculate ( \sqrt{3} \times 480 ):

[

\sqrt{3} \approx 1.732 \quad \text{thus,} \quad 1.732 \times 480 \approx 829.44

]

2. Now, divide