# NCEES Sample Exam #110 - sqrt(3) question



## wiliki (Oct 15, 2019)

So I'm a bit lost here. 

In terms of magnitude, why was S=VI used here versus S=sqrt(3)*V_line*V_line, since this is a 3-phase problem? Why is the sqrt(3) eliminated here?


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## DLD PE (Oct 15, 2019)

Because in a wye system, the phase current and line current are equal.  Since they give you the phase voltage, you can use the power equation to solve for the phase current, and the magnitude is the same for the line current.

The problem does not specifically say it's a Wye system, but they tell you it's three phase, four wires, so that tells you there is a neutral, hence a wye and not a delta.


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## wiliki (Oct 15, 2019)

MEtoEE said:


> Because in a wye system, the phase current and line current are equal.  Since they give you the phase voltage, you can use the power equation to solve for the phase current, and the magnitude is the same for the line current.
> 
> The problem does not specifically say it's a Wye system, but they tell you it's three phase, four wires, so that tells you there is a neutral, hence a wye and not a delta.


But here, they gave the "phase-to-phase" voltage, and not the phase voltage. Isn't the phase voltage the same as the line-to-neutral voltage? 

It's my understanding that for 3PH circuits, the sqrt(3) should be used with the line-to-line or phase-to-phase values?


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## Zach Stone P.E. (Oct 16, 2019)

wiliki said:


> But here, they gave the "phase-to-phase" voltage, and not the phase voltage. Isn't the phase voltage the same as the line-to-neutral voltage?
> 
> It's my understanding that for 3PH circuits, the sqrt(3) should be used with the line-to-line or phase-to-phase values?


Hi @wiliki Even though it is a three-phase source, it is a single-phase circuit drawing current and power due to just one single-phase load connected across the B to C phase. If the load is connected across B to C (instead of B to neutral, or C to neutral), then the total phase to phase voltage is applied across the load (VBC). 

The formula being used is the single-phase apparent power formula:

|S1ø| = |Vp|•|Ip|

Single-phase apparent power formula since it is a single-phase load drawing single-phase power. 

|Vp| = 13.2kV since that is the voltage applied directly across the load (VBC)

IIp| = The single-phase current drawn by the single-phase load. 

There is no power factor term or angle in this formula, it uses magnitudes only. You could use the power angle from power factor to include the angle of Ip, when the angle of Vp is zero, but that is not called for by the question.

"Phase to phase" means the same thing as "line to line" or just "Line voltage".


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## DLD PE (Oct 16, 2019)

Zach, thanks for clarifying.  My explanation wasn't correct.


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## wiliki (Oct 16, 2019)

Zach Stone said:


> Hi @wiliki Even though it is a three-phase load, it is a single-phase circuit drawing current and power due to just one single-phase load connected across the B to C phase. If the load is connected across B to C (instead of B to neutral, or C to neutral), then the total phase to phase voltage is applied across the load (VBC).
> 
> The formula being used is the single-phase apparent power formula:
> 
> ...


Thanks so much Zach for taking the time to write this. And thanks MEtoEE for the efforts. Cheers!


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