CSLB C-10 Electrical ContractorElectrical Theory and CalculationsHard
A 480V, 3-phase, 60 Hz transformer is connected Delta on the primary side and Wye on the secondary side. If the primary line current is 10 amps, what is the approximate phase current in the primary windings?
- A3.33 A
- B5.77 A
- C10 A
- D17.32 A
Show answer & explanationAnswer & explanation
Correct answer: B. 5.77 A
In a Delta-connected system, the line current (IL) is √3 times the phase current (Iph). Therefore, Iph = IL / √3. Given IL = 10A, Iph = 10A / 1.732 ≈ 5.77A.
Why the other options are wrong
- A. This is a significant underestimation, likely a calculation error.
- C. This would be true for a Wye connection, where line current equals phase current.
- D. This would be line current if phase current was 10A, or a calculation error.
Delta Connection Currents
In a Delta (mesh) connection, the line current is √3 times the phase current (IL = √3 × Iph), and the line voltage equals the phase voltage (VLL = Vph).
- Commonly used for high power transmission and distribution.
- Provides three-phase power without a neutral.
- Line currents are displaced by 30° from phase currents.
Memory trick: Delta Current: Line is Root-3 times Phase.