Hi. If you lose one of the phases on the Delta side what is your phase voltage on the secondary side? Or how do you work it out?

Thanks

 

Any feedback at all would be useful. Someone told me that one phase would be full voltage and the other two at half voltage. Don't see why that would be?

 

Patience, patience. You are getting anxious because no one has responded in 2.5 hours? It took 20 days before my first post received a response. In my time zone you posted at 3:30am and your second post was at 6am.

If you've been a member for over four years you should know that it can take a day or two for someone to respond. You should also know that you are expected to show your attempts at solving your own homework. Please do so. That will then form the basis for a discussion in which we can try to steer you in a direction in which you can discover the answer.

 

Try this.

Assume that the delta connected side of the transformer is connected to the source, and the wye connected side is connected to the load.

There are in fact three single phase sources connected to the delta side of the transformer. These sources are normally connected A-B, B-C, and C-A.
When the delta side is properly energised, three single phase voltages are developed on the wye side of the transformer. These are a-n, b-n, and c-n.

When connection A-B is properly energised, Voltage a-n is present on the load side of the transformer.

Proceed from this point and tell us what you can deduce.

 

Well by the looks of it a-n will be at the expected voltage as the delta coil A-B will be at the full voltage as both phases are present, but b-n and c-n should be half the expected voltage because if one phase on the delta side is missing then the voltage seen across the B-C and A-C coils is halved as the full voltage is seen between B-A and the C tap can be perceived as a mid-point tap between that voltage resulting in half the full delta voltage for the B-C coil and half for the A-C coil.

Sound about right?