The efficiency of a transformer at a given load and p.f. is expressed by the relation

$$ \eta=\frac{V_{2} I_{2} \cos \phi_{2}}{V_{2} I_{2} \cos \phi_{2}+P_{1}+I_{2}^{2} R_{e}}=\frac{V_{2} \cos \phi_{2}}{V_{2} \cos \phi_{2}+P_{1} / I_{2}+I_{2} R_{e}} $$ The terminal voluge $V_{2}$ is approximately constant. Thus for a given p.f., efficiency depends upen the lnad current $I_{2}$. In expression $(i)$, the numerator is constant and the efficiency will be maximum if dencminator is minimum. Thus the maximum condition is obtained by differentiating the quantity i the denominator w.r.t. the variables $I_{2}$ and equating that to zero i.e. $\begin{aligned} \frac{d}{d I_{2}}\left(V_{2} \cos \phi_{2}+\frac{P_{f}}{I_{2}}+I_{2} R_{e}\right) &=0 \text { or } 0-\frac{P_{t}}{I_{2}^{2}}+R_{e}=0 \\ I_{2}^{2} R_{e} &=P_{i} \\ \text { Copper losses } &=\text { Iron losses } \end{aligned}$ Thas, the efficiency of a transforater will be maximum when copper (or variable) losses are equed to iron (or constant) losses. $$ \eta_{\text {mar }}=\frac{V_{2} l_{2} \cos \phi_{2}}{V_{2} I_{2} \cos \phi_{2}+2 P_{2}} $$ |since $P_{c}=P$ | From cquation (ii), the value of cutput current $I_{2}$ at which the efficiency of the transformer will $\mathrm{x}$ max mum is given by : $$ I_{2}=\sqrt{\frac{P_{1}}{R_{e s}}} $$ If $x$ is the fraction of full load $\mathrm{kNA}$ at which the efficiency of the transformer is maximum. Then, copper losses $=x^{2} P_{e}$ (where $P_{c}$ is the full load Cu losses) L.on losses $=P_{1}$ For maximum efficiency. $\quad x^{2} P_{c}=P_{i} ; x=\sqrt{\frac{P_{1}}{P_{c}}}$ $\therefore$ Ourput kVA correspondingt ta maximum efficiency $=x \times$ full load $\mathrm{kVA}=$ full load $\mathrm{kVA} \times \sqrt{\frac{P_{\mathrm{i}}}{P_{C}}}$ $=$ full load $\mathrm{kVA} \times \sqrt{\frac{\text { iron losses }}{\text { copper lesses af full load }}}$