Question:- A 500/100 V, two-winding transformer is rated at 5 kVA. 

1. The transformer have maximum efficiency when the output of the transformer is 3 kVA.

2.. The transformer draws a current of 3 A, and the power is 100 W when a 100-V supply is applied on the low-voltage side with the high-voltage side open-circuited.

Find the  efficiency of the transformer at 0.8 PF lagging. 

Solution: we know the relation

KVA of transformer at maximum efficiency = Rated KVA * sqrt (pc/psc)

                                                                    3 = 5*sqrt (pc/psc)

              transformer draws a current of 3 A, and the power is 100 W when a 100-V supply is impressed on the low-voltage winding with the high-voltage winding open-circuit.

 so                                           core loss pc = 100 watt

hence                                 full load copper loss psc = 277.78

                                                                    Efficiency =  1-[(loses)/(input)

                                                                  

                                                                  Efficiency  =1 -[377.78/5000*0.8]

  

                                                             

                                                                        Efficiency =90.55%