A single-phase transformer is rated 10 kVA, 1000/500 V, 60 Hz. What is the maximum current on the 1000 V side at full apparent power (neglecting losses)?

Introduction / Context:
Transformer nameplates specify apparent power (kVA) and rated voltages on each winding. At full load, the maximum current on a given side is determined by I = S / V on that side. This question checks quick conversion from nameplate data to rated current, a routine design and sizing task.

Given Data / Assumptions:

• Rated apparent power S = 10 kVA = 10,000 VA.
• Primary (source) side voltage considered: 1000 V.
• Ideal transformer assumptions (ignore losses and power factor for rating current).

Concept / Approach:
For a single-phase transformer, I_rated (on a given winding) = S_rated / V_rated (of that winding). The kVA rating applies to both sides; voltage differs, so the current differs correspondingly to keep S the same (S = V * I).

Step-by-Step Solution:
Compute current on 1000 V side: I1 = S / V1I1 = 10,000 VA / 1000 V = 10 ATherefore, the maximum rated current on the 1000 V side is 10 A.

Verification / Alternative check:
On the 500 V side, rated current would be I2 = 10,000 / 500 = 20 A. The ampere-turns balance through the turns ratio, consistent with ideal transformer behavior.

Why Other Options Are Wrong:

• 2 A, 5 A, 7.5 A: Too small for 10 kVA at 1000 V.
• 20 A: That is the rated current on the 500 V side, not the 1000 V side.

Common Pitfalls:

• Confusing primary and secondary sides; always divide the kVA by the side voltage in question.
• Mixing up kW and kVA; rated current is based on kVA regardless of power factor.

Final Answer:
10 A