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GATE 2024 IN Q.61: solution and answer

GATE 2024 · INQ.61 · 2 marks · NAT

A 50 kVA transformer has an efficiency of 95% at full load and unity power factor. Assume the core losses are negligible. The efficiency of the transformer at 75% of the full load and 0.8 power factor is ________% (rounded off to one decimal place).

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Answer and solution

ANSWER

The efficiency of the transformer at 75% of full load and 0.8 power factor is 95.3%.

Official answer: 95.2 to 95.4Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

  1. Because core losses are negligible, the total loss at any load equals the copper loss alone.

  2. At full load and unity power factor: rated output = 50 kW, and eta_fl = 50/(50+Pcu,fl) = 0.95, giving Pcu,fl = 50*(1/0.95 - 1) = 2.6316 kW.

  3. Copper loss varies as the square of the per-unit load. At 75% load: Pcu = (0.75)^2 * 2.6316 = 1.4803 kW.

  4. Core loss is negligible at every load, so total loss at 75% load is also 1.4803 kW.

  5. Output at 75% load and 0.8 power factor: Pout = 0.75 50 0.8 = 30 kW.

  6. Efficiency: eta = Pout/(Pout+Pcu) = 30/(30+1.4803) = 30/31.4803 = 0.9530.

  7. So the efficiency is 95.3%.

KEY IDEA

With core loss negligible, all loss at any load is copper loss, which scales as the square of the per-unit load. Full-load unity-pf efficiency fixes the full-load copper loss; scale it to 75% load, recompute output at the new load fraction and power factor, and take the ratio.

WATCH OUT FOR

Forgetting that copper loss scales with the square of the load fraction (using 0.75x instead of 0.75 squared), or re-deriving core loss instead of taking it as negligible as stated.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 14 Transformers questions, with notes

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