# NCEES Power Sample Exam #108



## Dodgeviper1017

A 3-phase, 4-wire, neutral-grounded, wye-connected utility line has a phase to phase voltage of 13.2 kv. A complex load of (200+j100) kVA is connected between phase A and neutral. An identical load is connected between Phase B and neutral. The neutral current is most nearly:

(a) 0

(b) 9.8

(c) 16.9

(d) 29.3

I know this has been on many threads, but I am not understanding a different part of the problem than the others.I solved the problem like this to start S = sqrt(3)VI then I = S/sqrt(3)V which gave me 9.8 or B. This is not correct, so I looked in the back and realized I should be phase to phase and not phase to neutral. So I divided voltage by sqrt(3) to get phase to phase and used S=sqrt(3)VI again but for some reason I still didnt get the answer they gave this time I got 16.9 or C. Why are they not using the sqrt(3) in the equations regardless if they are using it to get phase to phase voltage already? Also beyond this when do I know not to use sqrt(3) on a 3 phase question?


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## TWJ PE

This is an unbalanced load - right?

So Ia+Ib+In=0

They give you phase to phase - you need phase to neutral (so divide by sqrt(3)) - Vp

Ia=Sp/Vp; Ib=Sp/Vp

200+j100/7.62 @ 0 degrees + 200+j100/7.62 @ 120 degrees.

That should be close to the answer..


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## Dodgeviper1017

Thanks for the response, that clears it up a bit. However, I am still confused as to why you use Ia = Sp/Vp instead of Ia = Sp/((sqrt(3)*Vp)? I guess because Vp-p/sqrt(3) = Vp. In essence, what I am trying to say is shouldn't sqrt(3) be in the equation regardless of what is done with voltage as it is 3 phase?


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## TWJ PE

All depends on what they are asking for... sqrt(3) is a tricky little devil. Complex Imaginary has a good video on it if it's giving you trouble.



In regards to the problem... in short, no. You've already accounted for the sqrt(3) by converting phase-to-phase to phase-neutral.


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## Dodgeviper1017

New concern on this why is there no 40 degree phase shift when converting voltage and why didbthey not divide the MVA by 3. So its S/3/(V/sqrt (3)) to give Ia and Ib then Ia+Ib = In. So (74.53 angle 26.56/(7.62 angle -30))+(74.53 angle 26.56/(7.62 angle -150) this gives 9.78 which when multiplied by 3 gives 29.3 the answer. Is this an ok way to solve? Should I use the complex conjugate?


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## Dodgeviper1017

30 degree phase shift I mean.


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## Dodgeviper1017

So I got why the load isnt divided by 3 it is because it is hooked on a single phase. Therefore only the voltage should be converted to make simple s=VI work. However I still dont understand no 30 degree phase shift and how its known to use complex conjugate other than it says load is complex. Is this an ok assumption?


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## Dodgeviper1017

Just did it with 30 degree phase shift and complex conjugate. I got 29.5 however I still wanna know if I approached it correctly. Also should I do the conjugate of 26.56 also or just the voltage angle.pangle.p


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## AAGR2013

We have to forget 30° phase shift between line and phase voltage in wye, and conjugate of the phasor

(200+j100)/7.62∠0° + (200+j100)/7.62∠120°

=224∠26.6°/7.62∠0° + 224∠26.6°/7.62∠120°

=29.4∠26.6° + 29.4∠-93.4°

=29.4∠-33.2°


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## Panchito

Dodgeviper1017 said:


> However I still dont understand no 30 degree phase shift and how its known to use complex conjugate other than it says load is complex. Is this an ok assumption?


Since the problem does not give you any 0 degree reference angle, you can pick any quantity as your 0 degree reference, for example, Van. Then, your voltages would be:

Van = 7.62 @ 0 deg
Vbn = 7.62 @ -120 deg

or you could have picked Vbn as our reference angle. In that case, you would have the following:

Van = 7.62 @ 120 deg
Vbn = 7.62 @ 0 deg

Both options above will give you the same result, because the problem is asking only for the MAGNITUDE of the neutral current. The ANGLE of the neutral current will be different, because it will be relative to the quantity you picked as your 0 deg reference. But that does not matter because the problem is not asking to find the angle.

If the problem had asked for the magnitude and ANGLE of the neutral current, then they would have to give a reference 0 deg angle. For example, Vab. In that case, you would need to take into account the 30deg phase shift when doing the LL to LN conversion.

And regarding your second question, you use complex conjugate because the complex power (S) is the product of the voltage and the complex conjugate of the current. S = V.I*

Hope this helps.


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## nukem2k5

Panchito said:


> If the problem had asked for the magnitude and ANGLE of the neutral current, then they would have to give a reference 0 deg angle. For example, Vab. In that case, you would need to take into account the 30deg phase shift when doing the LL to LN conversion.


This is something that's still a bit unclear to me.  I know, for sure, that there's a 30 degree phase shift through a XFMR, depending on the primary and secondary winding connections.  

However, if we're working a problem involving complex numbers (and the angles are relevant to the problem)- then we need to add/subtract 30 degrees when simply converting to/from a line voltage and a phase voltage (e.g. to go from a "3-phase" to "per phase" value)?  Is that correct?  Most of the time this isn't required, but is that because the angle isn't relevant to the problem?  Or is there something unique that I'm missing wherein we apply the phase shift when converting from a line to phase value (or vice versa) only in certain situations?

If so, converting from V_LL to V_LN in ABC rotation, we would subtract 30deg, right?  Likewise, if we have a delta-connected load, and need to artificially create a "phase to neutral" voltage (there's a practice problem where this is done), then we would shift by -30 as well?  The way I determine whether to add or subtract the 30 degrees is simply by drawing the phasor diagram and comparing the phasors.


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## Armus

The reason why you don't use sqrt(3) in the equation I=S/V is because the given S is the per phase value.

Ordinarily when given S it is S (3 phase) which would be divided by 3. This is why we use V(l-l) instead of V-phase, because V(l-l) is V-phase * sqrt(3). so in using V(l-l) and adding sqrt(3) in the denominator, we are in effect, dividing by 3. My first power teacher called the three-phase values "Conversation bases" because everyone used them in communication but had to change to single phase for most calculations.


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## Opisthenar PE

Hello all,

I'm still a little confused on why the solution did not use the conjugate of the complex power as we are solving for the current. Any other information would be great.

Thank you.


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## EngrinSF

TWJ PE said:


> This is an unbalanced load - right?
> 
> So Ia+Ib+In=0
> 
> They give you phase to phase - you need phase to neutral (so divide by sqrt(3)) - Vp
> 
> Ia=Sp/Vp; Ib=Sp/Vp
> 
> 200+j100/7.62 @ 0 degrees + 200+j100/7.62 @ 120 degrees.
> 
> That should be close to the answer..


Why do we not consider the negative sign for the In value in the final answer ?


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## EngrinSF

Also, when I do node analysis should it not be In= Ia + Ib?


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## supra33202

1) Could you confirm my calculation (attachment) is correct?

2) I understand that the question doesn't ask about the angle. But if I set Van at angle 0 deg, should the In (neutral current) at angle -86.57 deg?

Notes:

a) I also think that current conjugate should be considered (for single phase, S = VI*), and I use -120 deg instead of 120 deg (abc/positive sequence -&gt; counterclockwise).


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## rg1

supra33202 said:


> 1) Could you confirm my calculation (attachment) is correct?
> 
> 2) I understand that the question doesn't ask about the angle. But if I set Van at angle 0 deg, should the In (neutral current) at angle -86.57 deg?
> 
> Notes:
> 
> a) I also think that current conjugate should be considered (for single phase, S = VI*), and I use -120 deg instead of 120 deg (abc/positive sequence -&gt; counterclockwise).
> 
> View attachment 10915


Your answer is the most right. From the KVA load given, it is clear that the load in inductive ( Both parts positive) and from your answer the phase current lags Voltage; matches perfect. You should not have doubt now.


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## Stephen2awesome

Is the problem Leading or Lagging?


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## rg1

RadioBox said:


> The problem reads.
> 
> A 3-phase, 4-wire, neutral-grounded, wye-connected utility line has a phase to phase voltage of 13.2 kv. A complex load of (200+j100) kVA is connected between phase A and neutral. An identical load is connected between Phase B and neutral. The neutral current is most nearly:
> 
> (a) 0
> 
> (b) 9.8
> 
> (c) 16.9
> 
> (d) 29.3


The problem is a lagging load, but how does it matter here. You have been given a load and Voltage. Rest is maths.  Yes if the solution has taken Vbn with +120 degrees, it is wrong as per convention.  Conventionally phase sequence is ABC. May be because the question asks only magnitude part which remains unaffected by talking any two phases. Taking Van &lt;0 and Vbn&lt;-120 is more correct. I give one more spin to the solution for better appreciation. Find out the impedance of the load from KVA and V and then find the currents in what ever phases you connect this impedance. Add them to find the neutral current.


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