In fact, efficiency becomes maximum when copper losses Ri^2 equals iron losses.

Here you'll find the explanation.

Yours sincerely,

JF Antoine 

Ok but please tell me technically what happens when I^2R equals to iron losses.

while designing a machine the constant lossess are taken into consideration,so that the constant loss is kept minimum during designing itself. Therefore when copper loss is equal to constant loss,the copper loss is also minimum and hence efficiency is high

Dear Dinesh

Our aim should be to reduce the losses to the minimum extent to make efficiency maximum, on the other hand we say that efficiency is maximum when Cu losses becomes equal to Core losses ? 

The efficiency at full load is given by,

  efficiency=Output Power/(output power+iron loss+copper loss)

so efficiency is inversely proportional to losses. Therefore when iron loss is equal to copper loss (ie. iron loss is kept minimum during designing itself), copper loss is also minimum and hence loss is minimum, efficiency will be maximum since efficiency is inversely proportional to loss

Example: if output power=100KW  Powerfactor=1, Iron loss =1000W then

Efficiency=100000*1/((100000*1)+1000+1000)

Efficiency=98.04%

It is possible that Copper Losses are lower than Iron losses ? if yes than why efficiency is not higher in that case. Because Iron losses are constant--> Min and if Copper losses goes to lower than iron losses than efficiency should be higher in this case.