Transformers - Study Mode
[#16] The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be
Correct Answer
(D) 400 W
Explanation
Solution: Copper loss in a transformer is due to the current flowing through the windings' resistance. It's also known as I 2 R loss (where I is current and R is resistance). Importantly, copper loss varies with the square of the load current . Let's denote the full-load current as I FL . The copper loss at full load (P cu_FL ) is proportional to (I FL ) 2 . P cu_FL = k * (I FL ) 2 = 1600 W (where k is a constant related to the winding resistance). At half-load, the current is I FL / 2. The copper loss at half-load (P cu_HL ) is proportional to (I FL / 2) 2 . P cu_HL = k * (I FL / 2) 2 = k * (I FL ) 2 / 4. Since k * (I FL ) 2 = 1600 W, then P cu_HL = 1600 W / 4 = 400 W. Therefore, the copper loss at half-load is 400 W.
[#17] The size of a transformer core will depend on
Correct Answer
(D) (A) and (B) both
[#18] A transformer cannot raise or lower the voltage of a D.C. supply because
Correct Answer
(C) Faraday's laws of electromagnetic induction are not valid since the rate of change of flux is zero
[#19] Which of the following loss in a transformer is zero even at full load ?
Correct Answer
(B) Friction loss
[#20] Part of the transformer which is most subject to damage from overheating is
Correct Answer
(C) winding insulation