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