Q6
The full-load copper loss of a transformer is 1600 W. At half-load, the copper loss will be
A.
6400 W
B.
1600 W
C.
800 W
D.
400 W
AnswerAnswer: Option D
Solution
Answer: Option D
Solution:
Copper loss in a transformer is due to the current flowing through the windings' resistance. It's also known as I2R 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 IFL.
The copper loss at full load (Pcu_FL) is proportional to (IFL)2.
Pcu_FL = k * (IFL)2 = 1600 W (where k is a constant related to the winding resistance).
At half-load, the current is IFL / 2.
The copper loss at half-load (Pcu_HL) is proportional to (IFL / 2)2.
Pcu_HL = k * (IFL / 2)2 = k * (IFL)2 / 4.
Since k * (IFL)2 = 1600 W, then Pcu_HL = 1600 W / 4 = 400 W.
Therefore, the copper loss at half-load is 400 W.