# NCEES practice exam #11



## Gman (Mar 22, 2021)

Could someone please explain to me why the solution is calculating Van phase voltage using Wye connected formula? Isn't VAB=12.5KV the phase voltage for the delta connection?? Much appreciated


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## DLD PE (Mar 22, 2021)

Please click on the "search" tool in the upper right. You will find the answer since this problem (and pretty much all the NCEES practice exam problems) has been discussed here at some point in the past.

Also, NCEES changed the numbering, so for example #11, search "NCEES 111" and for the afternoon problems, search "NCEES 511."


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## akyip (Mar 22, 2021)

It is typical to use a per-phase line-to-neutral circuit analysis when dealing with 3-phase circuit analysis. When you have a wye load or source, the wye per-phase values are the line-to-neutral values, so there is no need to convert.

But when you have a delta load or source, the delta per-phase values are line-to-line values. You always want to use line-to-neutral values when doing per-phase line-to-neutral circuit analysis. This is why the delta load's phase-to-neutral voltage is used.


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## Zach Stone P.E. (Mar 22, 2021)

Gman said:


> Could someone please explain to me why the solution is calculating Van phase voltage using Wye connected formula?



The solution is using a single-phase equivalent circuit model to solve the problem, but the solution does not show the single-phase equivalent circuit so it is easy to miss it if you are not already familiar with applying it.

In order to use the single-phase equivalent circuit, you need to use one phase of a wye connected source, providing power to one phase of a wye connected load, through the line impedance.

If the source or load are delta connected, you'll need to convert them to their wye equivalents first, before using one phase of each in the single-phase equivalent circuit.



> Isn't VAB=12.5KV the phase voltage for the delta connection?? Much appreciated



Yes, the phase voltage across the A-phase of the delta connected load is VAB=12.5KV. However, you cannot use this value directly in the single-phase equivalent circuit because it is the phase voltage across a delta connection.


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## kris7o2 (Jun 16, 2021)

Hi, 

Is this formula correct

Since Vab = Van + VaA 

however, shouldn't it be Vab = Vna + VaA 

where Van = - Vna (or does reversing little a with the neutral matter?)


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## kris7o2 (Jun 26, 2021)

Also, is Zach is there any type of visualization for the single-phase equivalent circuit?


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## Zach Stone P.E. (Jun 27, 2021)

kris7o2 said:


> Also, is Zach is there any type of visualization for the single-phase equivalent circuit?


Sure. This is directly from our live class on circuit analysis:


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## kris7o2 (Jun 27, 2021)

Thanks much Zach.

I am still a little confused though. Is the little a and big A on the right side ( mean shouldn't the big A be on the left of the load, and the little a be on the right-hand side of the load)? I'm also confused on the right-hand side of the circuit because the problem shows a delta load, and the illustration provided shows little a, load, and a little n (neutral). Are we supposed to convert the delta load to a wye load so that we have the ability to have the little n (neutral) after the load?


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## akyip (Jun 28, 2021)

kris7o2 said:


> Thanks much Zach.
> 
> I am still a little confused though. Is the little a and big A on the right side ( mean shouldn't the big A be on the left of the load, and the little a be on the right-hand side of the load)? I'm also confused on the right-hand side of the circuit because the problem shows a delta load, and the illustration provided shows little a, load, and a little n (neutral). Are we supposed to convert the delta load to a wye load so that we have the ability to have the little n (neutral) after the load?


kris702:

I think Zach's diagram was just an example and does not necessarily reflect the designations given in problem 111. In any case, an equivalent 1-phase or per-phase circuit analysis is done by using the line-to-neutral versions of the sources and loads, regardless of how the sources and loads are actually connected.

This means that if the sources/loads are already connected in Y, you do not need to do any conversion since the Y connection is already line-to-neutral. But if the sources/loads are connected in delta, you need to use the Y equivalent by converting line-to-line voltages to line-to-neutral and delta impedances to Y impedances.


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