Question

At full load, the copper loss in a transformer is 600 watts. At half load, the copper load will be

Answer 1

Answer:

The copper losses of transformer at X times the full load is proportional to X^2 times the losses at full load. Hence at half the full load, the copper losses will be (0.5)^2 * 1600 = 400 Watts.

Answer 2

Given,

At full load, the copper loss in a transformer is 600 watts.

To Find,

At half load, the copper loss will be =?

Solution,

We can solve the question as follows:

It is given that at full load, the copper loss in a transformer is 600 watts. We have to find the copper loss at half load.

$Copper\: loss\: at\: full\: load = 600\: watts$

Copper losses in a transformer occur due to the resistance of the wire of the primary and secondary windings and the current that flows through them.

The formula for finding the copper loss at any given load is given as:

$P_{cu} = x^{2} P_{cf}$

Where,

$P_{cu} = Copper\: loss\: at\: any\: load$

$P_{cf} = Copper\: loss\: at\: full\: load$

$x = The\: given\: fraction\: of\: load$

In the given question,

$P_{cf} = 600\: watts$

$x = 1/2\:(half\: load)$

Substituting the values in the given formula,

$P_{cu} = (\frac{1}{2}) ^{2} *600$

      $= \frac{600}{4}$

      $= 150\: watts$

Hence, the copper loss at half load will be 150 watts.