Oklahoma Electrical Practice Exam

A transformer rated at 75 kva and operating at 460v, 3 phase with an efficiency of 90% requires what ampacity for the primary wire?

1. 90.5 amps

2. 104.6 amps

3. 115.7 amps

4. 125.3 amps

The correct answer is: 104.6 amps

To determine the required ampacity for the primary wire of a three-phase transformer, you can use the following formula, which takes into account the transformer’s power rating, voltage, and efficiency: \[ I = \frac{P}{\sqrt{3} \times V \times \text{Efficiency}} \] Where: - \(I\) is the current in amperes, - \(P\) is the apparent power in volt-amperes (VA), which for a transformer rated at 75 kVA means 75,000 VA, - \(V\) is the line-to-line voltage (in this case, 460 V), - Efficiency is expressed as a decimal (90% efficiency would be 0.90). Plugging in the values: 1. Convert kVA to VA: \[ P = 75 \, \text{kVA} = 75,000 \, \text{VA} \] 2. Substitute into the formula: \[ I = \frac{75,000}{\sqrt{3} \times 460 \times 0.90} \] 3. Calculate \(\sqrt{3}\) (approximately