# Power Factor OCT of Transformer

1. Dec 18, 2011

### jegues

1. The problem statement, all variables and given/known data

See statement below.

2. Relevant equations

3. The attempt at a solution

I've been reading through the notes for my course and in the section where we are determining the parameters for transformers.

For the OCT they give,

240V, 1.066A, 136.6W

From this they calculate Rc and Xm which I understand how to do but underneath in bold it says,

"Calculate the power factor of the transformer during the OCT and comment."

Wouldn't the PF of the transformer just be,

$$PF = \frac{136.6W}{240V \cdot 1.066A} = 0.533$$

Is there something to take away from this? It wouldn't be put in bold in my notes for no reason.

What am I to conlcude from this?

2. Dec 18, 2011

### Staff: Mentor

Just guessing, but perhaps it's what you write as the comment that is valued, rather than the easy calculation? It would be true that this pf has practically no bearing on the pf under significant load.

I suppose the ideal OC pf would be 0. That would mean zero real power losses when the device is connected to the mains but completely unloaded.

3. Dec 19, 2011

### rude man

What does "OCT" mean? Open secondary?

4. Dec 19, 2011

### Staff: Mentor

open circuit test

5. Dec 19, 2011

### rude man

I see nothing wrong with what you concluded

Maybe your instructor likes bold type ... ?