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
Answer: Option D
Solution (By Examveda Team)
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.
This Question Belongs to Electrical Engineering >> Transformers
how
Cu loss directly proportional to I^2. So if I become half loss will be reduced to one-fourth.
copper loss is proportional to KVA^2 or I^2 so for half load its is = (1/2)^2*1600 = 400 W
In the open circuit test of a 260/130 V single phase transformer, obtained data were: primary current 0.4 A, Power input 25 W. If the resistance of the primary winding is 0.5 ohm. Determine the core loss. Answer pleas?
for full load, copper loss = I^2 * R = 1600
For half load, I = I/2
therefore, half load copper loss = (I/2)^2 * R
= (1/4)*(I)^2 * R
= (1/4) * 1600
= 400
We know that,
Pcu= x^2* Pcu(full load)
Pcu= (1/2)^2 *1600w
= 400w