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. ...
