# NCEES #512



## knight1fox3 (Oct 20, 2010)

Looking at the solution for this problem, doesn't the impedance for the cable need to be adjusted to account for a pf other than 0.85 (0.90 in the problem statement)? This one tripped me up when I went through the sample exam the first time. I thought note 2 of Table 9 would apply here.


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## cableguy (Oct 20, 2010)

No, power factor is only considered when you're doing voltage drop calculations, and then you can use the {Vdrop=I * (Rcos0+Xsin0) * Length} formula. This is a straight "what is the impedance of this length of wire" question.

They intentionally flooded the question with too much information.


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## knight1fox3 (Oct 20, 2010)

cableguy said:


> No, power factor is only considered when you're doing voltage drop calculations, and then you can use the {Vdrop=I * (Rcos0+Xsin0) * Length} formula. This is a straight "what is the impedance of this length of wire" question.
> They intentionally flooded the question with too much information.


Ok I understand, thanks. I did notice there was way more information in the problem than what necessary.


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## cruzy (Oct 23, 2010)

It seems like they are dividing by 2 because of the 2 sets of conductors in conduits. So does that mean there are 3 conductors per conduit (3 phase motor)?? And if there are two sets why does that make you divide by two?? Is that because you treat it like a parallel set of impedances like in a normal circuit (where if the two parallel impedances are equal, the equivalent is found by just dividing one of the impedances by 2? or z1*z2/(z1+z2)??


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## cruzy (Oct 23, 2010)

It seems like they are dividing by 2 because of the 2 sets of conductors in conduits. So does that mean there are 3 conductors per conduit (3 phase motor)?? And if there are two sets why does that make you divide by two?? Is that because you treat it like a parallel set of impedances like in a normal circuit (where if the two parallel impedances are equal, the equivalent is found by just dividing one of the impedances by 2? or z1*z2/(z1+z2)??


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## nmh0408 (Oct 23, 2010)

cruzy said:


> It seems like they are dividing by 2 because of the 2 sets of conductors in conduits. So does that mean there are 3 conductors per conduit (3 phase motor)?? And if there are two sets why does that make you divide by two?? Is that because you treat it like a parallel set of impedances like in a normal circuit (where if the two parallel impedances are equal, the equivalent is found by just dividing one of the impedances by 2? or z1*z2/(z1+z2)??



Correct. When calculating the impedance for two parallel sets of conductors, you deal with it as two impedance's (resistances) in parallel. In this case two identical impedancse, therefore the resulting impedance is halved.


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