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## Main Question or Discussion Point

I have a few confusions related to the delta configuration of a transformer (single-phase transformers used to make a 3-phase transformer):

1) In a closed delta, the windings are connected such that H2 end of one winding is connected to H1 of the other winding. What will happen if one of the windings is reversed?

The flux in that winding should be opposite and thus the induced emf should also have opposite polarity..but that isn't possible! because induced emf should be equal to voltage applied across the winding. Also, what would be the effect of one reversed winding on the other windings?

2) I've been thinking (and searching) about this for days but haven't reached anywhere:

How do you calculate the total power output of an open delta transformer? I don't understand the explanation given in various texts and on websites:

http://yourelectrichome.blogspot.com/2011/05/open-delta-or-v-v-connection-of-3-phase.html

The √3 E

But why are we using this formula for open delta?

And btw, in the conductor attached to the point where the two windings are joined together, would the line current be √3I

1) In a closed delta, the windings are connected such that H2 end of one winding is connected to H1 of the other winding. What will happen if one of the windings is reversed?

The flux in that winding should be opposite and thus the induced emf should also have opposite polarity..but that isn't possible! because induced emf should be equal to voltage applied across the winding. Also, what would be the effect of one reversed winding on the other windings?

2) I've been thinking (and searching) about this for days but haven't reached anywhere:

How do you calculate the total power output of an open delta transformer? I don't understand the explanation given in various texts and on websites:

http://yourelectrichome.blogspot.com/2011/05/open-delta-or-v-v-connection-of-3-phase.html

The √3 E

_{L}I_{L}formula was for closed delta because in a closed delta, we calculate the power delivered by one winding as E_{L}*I_{phase}which becomes E_{L}*(I_{L}/√3) since phase current=I_{L}/√3, and then multiply by three.But why are we using this formula for open delta?

And btw, in the conductor attached to the point where the two windings are joined together, would the line current be √3I

_{phase}?