Sparky Bill PE
Senior, Master, Professional, Licensed Electrical
Does anyone else get the same answer they do? I get closer to D. Why aren't they finding the "Z effective Impedance" instead of just using R and X?
CI exam test 2 question 12, voltage drop
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which Z effective impedance are you using? The table 9 value at 0.85pf? Or did you somehow modify that Z value to 0.75pf?
Also, are you using the same equation as the solution: "V1=V-IZ" where I is the complex current value? If so, the answer does not come out closer to D even if you use the Z effective value from the table, its closer to 590.9A.
Also, are you using the same equation as the solution: "V1=V-IZ" where I is the complex current value? If so, the answer does not come out closer to D even if you use the Z effective value from the table, its closer to 590.9A.
Sparky Bill PE
Senior, Master, Professional, Licensed Electrical
I used X = .04, and R = .038, then did .038cos(41.4)+.04sin(41.4) = .055. Then (.055*500)/1000 = .027. I*Z = 6.3 V, 600v -6.3 .
Sparky Bill PE
Senior, Master, Professional, Licensed Electrical
All of their voltage drop solutions they just multiple R and X times the distance divided by 1000. They never mention Z or convert. I don't know why they aren't doing it, and am I supposed to be doing it? Every voltage drop problem I solve rather it be single phase or 3 phase involves a Z.
Most of the practice exam problems that I've done involving NEC Chapter 9 Table 9 use the R and X values per unit length, instead of the effective Z at pf = 0.85.
I believe at one point there was kind of a debate between whether to use the R and X values or whether to use the effective Z value and convert it. But most problem solutions that I have seen involving Chapter 9 Table 9 use the individual R and X values instead the effective Z values at pf = 0.85, for higher accuracy.
I believe at one point there was kind of a debate between whether to use the R and X values or whether to use the effective Z value and convert it. But most problem solutions that I have seen involving Chapter 9 Table 9 use the individual R and X values instead the effective Z values at pf = 0.85, for higher accuracy.
Also, one more thing about the effective Z value. The table values listed for Z eff are specifically to be used ONLY for pf = 0.85. These listed table values cannot be listed for another pf value.
But, if you look closely at the notes under the end of Chapter 9 Table 9, you can use an alternate Z eff (per unit length) that can be calculated using:
Z eff = R cos(theta) + X sin(theta),
Where theta = arc cos(PF)
Hope this helps.
But, if you look closely at the notes under the end of Chapter 9 Table 9, you can use an alternate Z eff (per unit length) that can be calculated using:
Z eff = R cos(theta) + X sin(theta),
Where theta = arc cos(PF)
Hope this helps.
Sparky Bill PE
Senior, Master, Professional, Licensed Electrical
I'm not talking about .85 effective Z values. I'm talking about using the formula of Rcostheta+Xsintheta to obtain a Z. Then use your lengths to get the Z effective. I don't see how that path is ever the "wrong" path and when I "can't" use it.
I just saw an explanation a few days ago from Zach's Oct 2020 live class that basically this Z eff value using R cos(theta) + X sin(theta) is just an approximation. The individual R and X values from the table for each conductor are more accurate than the Z eff value, from what I understand.
Maybe if Zach Stone sees this post, he can elaborate more on this. This is the best I can explain it...
I see and agree with your method not being wrong, until someone is able to explain why. Maybe email Complex Imaginary with this question for clarity
