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A 500 kVA transformer has an efficiency of 95% at full load and also at 60% of full load, both at upf. The efficiency η of the transformer at \(\frac{3}{4}th\) full load will be nearly
1. 98%
2. 95%
3. 92%
4. 87%

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Correct Answer - Option 2 : 95%

kVA rating of the transformer, S = 500 kVA

Efficiency, η= 95% at full load

Power factor, cos ɸ = 1

Let x = loading of transformer

For full load, x = 1

Efficiency, \(\eta = \frac{{x\; \times S\; \times cos\Phi }}{{\left( {x \times S\; \times cos\Phi } \right) + \left( {{P_i}} \right) + \left( {{x^2}{P_{cu}}} \right)}}\; \times 100\;\% \)

\(\Rightarrow 0.95 = \frac{{500}}{{500 + {P_i} + {P_{cu}}}}\)

⇒ Pi + Pcu = 26.3157      …. (1)

η at 60% of full load is 0.95, x = 0.6, cos ɸ = 1.

\(\Rightarrow 0.95 = \frac{{300}}{{300\; + \;{P_i}\; + \;0.36{P_{cu}}}}\)

⇒ Pi + 0.36Pcu = 15.789        …. (2)

By solving the above equations (1) and (2), we get

⇒ Pcu = 16.44 kW

And Pi = 9.86 kW (Iron loss at full load)

The efficiency η of the transformer at (3/4th) full load is,

\(\eta = \frac{{0.75\; \times \;500\; \times \;1}}{{\left( {0.75\; \times \;500\; \times \;1} \right)\; + \;9.86\; +\; [{{\left( {0.75} \right)}^2}\; \times \;16.44]}} = 0.95150 = 95.15\;\% \) 

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