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The full-load iron losses and copper losses can be determined by open-circuit test and short-circuit test respectively. Therefore,

From the open circuit test,

$$\mathrm{Full \:load\: iron \:loss \:= \:P_{i} Watt}$$

From the short-circuit test,

$$\mathrm{Full \:load\: copper \:loss = P_{cu} Watt}$$

$$\mathrm{\therefore\:Total\: full\: load \:losses\: = P_{i} + P_{cu} Watt}$$

Therefore, the efficiency of transformer at full-load is

$$\mathrm{\eta_{fl} =\frac{VI_{fi} × cos\varphi2}{(VI_{fi} × cos \varphi2) + P_{i} + P_{cu}}… (2)}$$

Now, for fraction of load, i.e., if the load is x &tI_{m}es; Full load, then,

$$\mathrm{Corresponding\: total \:losses = P_{i} + x^2P_{cu}}$$

The iron loss (P_{i}) remains constant at all loads.

Thus, the efficiency corresponding to fraction of loading is given by,

$$\mathrm{\eta_{x} =\frac{(𝑥 × VI_{fi}) × cos\varphi2}{(𝑥 × VI_{fi} × cos\varphi2) + P_{i} + 𝑥^2P_{cu}}… (3)}$$

As the transformer is a static device, hence does not have rotational losses (such as windage and frictional losses). Therefore, a transformer can have efficiency as high as 99 %.

The output power of the transformer is given by,

$$\mathrm{P_{out} = V_{2}I_{2} cos \varphi_{2}}$$

If we assumed that the transformer is referred to the secondary side, then,

$$\mathrm{Total \:resistance \:of \:transformer\: referred\: to \:secondary \:= \:R_{02} \Omega}$$

Hence, the total Cu-loss in the transformer is,

$$\mathrm{P_{cu} = I_{2}^{2}R_{02}}$$

And, the total losses in transformer is,

$$\mathrm{Total\:losses = P_{i} + P_{cu} = P_{i} + I_{2}^{2}R_{02}}$$

Therefore, the efficiency is given by,

$$\mathrm{\eta =\frac{V_{2}I_{2} cos \varphi2}{(V_{2}I_{2} cos \varphi2) + P_{i} + I_{2}^{2}R_{02}}}$$

$$\mathrm{⇒\eta =\frac{V_{2} cos\varphi2}{(V_{2} cos\varphi2) + (\frac{P_{i}}{I_{2}}) + I_{2}R_{02}}… (4)}$$

Practically, the output voltage (V_{2}) of transformer is almost constant. Thus, for a load of given power factor, the efficiency of a transformer depends upon the load current (I_{2}). Hence, for a given load, the numerator of eqn. (4) is constant so, for the efficiency to be maximum, the denominator should be minimum, i.e.,

$$\mathrm{\frac{𝑑}{𝑑I_{2}}[(V_{2} cos\varphi2) + (\frac{P_{i}}{I_{2}}) + I_{2}R_{02}] = 0}$$

$$\mathrm{⇒ 0-\frac{P_{i}}{I_{2}^{2}} + R_{02} = 0}$$

$$\mathrm{⇒ P_{i} = I_{2}^{2}R_{02} … \:(5)}$$

$$\mathrm{⇒ Constant\:Losses = Variable\:Losses}$$

Thus, the efficiency of transformer will be maximum when the constant losses (or iron losses) being equal to variable losses (or copper losses).

Now, the load current corresponding to maximum efficiency is given by,

$$\mathrm{I_{2max} = \sqrt{\frac{P_{i}}{R_{02}}}\:… (6)}$$

Also,

$$\mathrm{I_{2𝑚𝑎𝑥}^{2} =\frac{I_{2fi}^{2} P_{i}}{I_{2fi}^{2} R_{02}}}$$

$$\mathrm{⇒ I_{2𝑚𝑎𝑥} = I_{2𝑓𝑙} × \sqrt{\frac{P_{i}}{P_{cu𝐹.𝐿.}}}\:… (7)}$$

$$\mathrm{\therefore \:Current\: at\: maximum \:efficiency \:= Rated \:load \:Current × \sqrt{\frac{Iron \:losses}{Full − load \:cu \:losses}}}$$

Now, multiplying eqn. (7) by V_{2} on the both sides, we get,

$$\mathrm{V_{2}I_{2𝑚𝑎𝑥} = V_{2}I_{2𝑓𝑙}× \sqrt{\frac{P_{i}}{P_{cu𝐹.𝐿.}}}\:… (8)}$$

$$\mathrm{\therefore \:kVA \:at \:maximum \:efficiency = Full \:load \:kVA × \sqrt{\frac{Iron \:losses}{Full − load cu losses}}}$$

The equation (8), gives the values of output kVA corresponding to the maximum efficiency. Also, it may be noted that the value of kVA at maximum efficiency is independent of the load power factor.

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