Gee.... when I said brief, I was hoping for something a little more informative than what I could surmise. :slaphead:
If you do not care to elaborate just a little bit more, perhaps provide a link to some documentation.
OK. This has been discussed in other threads, but I can't find a good link right now.
Look at it this way:
In a wye-delta transformer each L-L voltage in the delta corresponds, with a factor of the turns ratio, to a single L-N voltage in the wye. Lets call the wye voltage that corresponds to V
AB on the delta side V
an.
Because the three L-L voltage vectors (phasors) form a closed triangle their sum must obey a simple vector equation:
VAB +
VBC +
VCA =
0 (where a bold character represents a vector.)
That means that on the wye side
Van +
Vbn +
Vcn must also =
0.
Since the actual wye side voltages are somewhat independent (taps, primary imbalance, IR drops from unbalanced load, etc.) this equation will NOT generally be satisfied. The result is that no-load current will flow in the delta windings and wye windings until the individual voltage drops allow both equations to be satisfied. Depending on the magnitude of the voltage imbalance this may result in neutral current which is far greater than the design full load current of the transformer. If large enough it will trip the OCPD. If smaller, it will just overheat the transformer or the unprotected neutral wire.
If you leave the star point unconnected it will simply float (with no need for current) to a point that allows the second equation to be satisfied. And there will be no neutral current regardless of how balanced or unbalanced the delta load may be.
If that does not get you thinking on the right track, let me know and I will try to answer more questions.