# NCEES 535



## colo_elec (Apr 15, 2010)

Ok, so the answer for this problem requires dividing the transformer power by 3, not root 3. It's 3 phase, wye connected. Why do we divide power by 3 to find output per phase in order to find current?


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## phowardtx (Apr 15, 2010)

colo_elec said:


> Ok, so the answer for this problem requires dividing the transformer power by 3, not root 3. It's 3 phase, wye connected. Why do we divide power by 3 to find output per phase in order to find current?


In this problem we are asked to solve per-phase and in order to do that we have to convert the Voltage to per-phase. That's where the root 3 comes in... |Vp|=|VL|/√3. In 3-phase systems line-line voltage is always assumed given unless specified otherwise. Once the voltages are in per-phase quantities you can treat it as 3 separate systems where the voltage and current magnitudes are identical and the power is 1/3 of the total system. Thus the 3-phase power equation using per-phase quantities is |S|=3*|Vp|*|Ip|

*using per phase voltage:*

|S|=3*|Vp|*|Ip|

*using line-line voltage:*

|S|=√3*|VL|*|IL|

*Wye-systems*

|Vp|=|VL|/√3

|Ip|=|IL|

*Δ-systems*

|Vp|=|VL|

|Ip|=|IL|/√3

PS: The power equations are identical for both wye and Δ systems, the one you use depends on which quantities (line or phase) are most convenient to work with.


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## BamaBino (May 24, 2011)

The 180 MVA rating is for the autotransformer and not the rating for the original isolation transformer that was used to form the autotransformer, right?


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## BamaBino (May 26, 2011)

The answer 16.5 MVA is also the rating of transformer (for each phase) before it was connected as an autotransformer, right?


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