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knight1fox3

knight1fox3

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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.

 
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.

 
Last edited by a moderator:
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.
Click to expand...
Ok I understand, thanks. I did notice there was way more information in the problem than what necessary.

 
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)??

 
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)??

 
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)??
Click to expand...

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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