Voltage Drop Calc Method

[allenmn85]

Hello, I am taking the MN Master Electrician Exam tomorrow and am wondering about the voltage drop calculation method.

The exam prep book I have studied uses the Vd=2kId/CMIL method where k=12.9 ohms for copper and getting the CMIL value from Table 8. I am used to the method of using Table 9 to find ohm/k-ft for whatever the specific application is (and then doing the normal Vd=I*R calc).

Both methods should be valid, however using my method has lead me to choose the incorrect answer on the sample exam. Does anyone have an insight into which method should be used? Thanks for your help.

 

[Dennis Alwon]

Does the VD formula give the correct answer? I have always used that but I never took an exam that asked for a VD calculation.

 

[allenmn85]

The Vd=2kId/CMIL method results in a "nuts on" specific number to choose A,B,C or D on the practice exam. (BTW, this is Mike Holt's 2011 Master Exam.)

The method I like to use gives a similar number, but off slightly. So then I am forced to choose the next closest answer from the options. This has led me to choose the "incorrect" choice many times.

I haven't taken the exam before, so I don't even know for sure if this will be on it.

Also, as a side note, I have found 2-3 errors in his practice exam. Would it be worth it to try to send an email pointing these outs?

 

[Dennis Alwon]

I am sure Mike would love to hear about errors in the book. I would email them to him without a second thought.

It seems the calc method would work. You do realize that the only time that we must consider VD is for fire pumps however it is good practice to wire with VD in mind. I am unfamiliar with using the book for VD so I cannot help there. Maybe someone else will help out.

 

[marti smith]

Somewhere in the book is a link or number that is solely for the purpose of obtaining errata and giving feedback. Could the questions be for aluminum conductor? K= 17.

 

[bob]

This info came from Mike's web site

VD = 2 x K x Q x I x D/CM - Single Phase

VD = 1.732 x K x Q x I x D/CM - Three Phase

“VD” = Volts Dropped: The voltage drop of the circuit conductors as expressed in volts.

“K” = Direct Current Constant: This is a constant that represents the direct current resistance for a one thousand circular mils conductor that is one thousand feet long, at an operating temperature of 75? C. The direct current constant value to be used for copper is 12.9 ohms and 21.2 ohms is used for aluminum conductors. The “K” constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0. “Q” = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

K = 12.9 is used with the DC figures in table 8. You said you were using the formula VD = I x R with R from table 9. This may be why you are getting different values.

If you take the formula VD = 1.732 x K x Q x I x D/CM and re-arrange it to VD = 1.732 x I x (K x Q x D/CM) you can see that R = (K x Q x D/CM).

 

[allenmn85]

No, I am basing the k=12.9 ohms directly from the answer key.

I am doing both methods correctly, but get a consistent 8-10% deviation between the two results. I have sent an email to 'customerservice@mikeholt.com' bringing up this issue as well as the errors I found. If I get a prompt response, I will paste it here.

 

[allenmn85]

This info came from Mike's web site
The ?K? constant is suitable for alternating current circuits, where the conductors do not exceed No. 1/0. ?Q? = Alternating Current Adjustment Factor: Alternating current circuits No. 2/0 and larger must be adjusted for the effects of self-induction (skin effect). The "Q" adjustment factor is determined by dividing alternating current resistance as listed in NEC Chapter 9, Table 9, by the direct current resistance as listed in Chapter 9, Table 8.

This is why I prefer not to use the "k" method... I would have to introduce an adjustment factor for *some* conditions...

Why not just use the Vd=2LRI/1000 method and reference the same Table 9 every time, even for small conductors?

 

[bob]

This is why I prefer not to use the "k" method... I would have to introduce an adjustment factor for *some* conditions...

Why not just use the Vd=2LRI/1000 method and reference the same Table 9 every time, even for small conductors?

You can do that. The reason Mike refers to the 1/0 conductor is because the adjustment factor is negligible for smaller conductors.

 

[George Stolz]

This is why I prefer not to use the "k" method... I would have to introduce an adjustment factor for *some* conditions...

Why not just use the Vd=2LRI/1000 method and reference the same Table 9 every time, even for small conductors?

I think the main reason is, it's simpler to reference Table 8 and forget about it. One number, one column to keep track of.

Personally, I have added VD=2RID to my exam prep in addition to Mike's use of VD=2KID/CM. For me, 2RID is easier to understand and delivers virtually identical results. I do the same problems side by side during the class so people can use whatever they are most comfortable with, and to demonstrate that they are comparable methods.

VD=IR is okay, if you have all the information you need. Most voltage drop questions are the longer variety, though, I think.

Email corrections to corrections@mikeholt.com .

 

[allenmn85]

Just FYI in case anyone still cares:

I took the exam yesterday and there was one question on voltage drop. Luckily, the question addressed this very issue and said either method can be used. The possible answers had fairly large margins between them (>1V). This made both methods result in the same choice, after rounding. *whew*

 

[RonEl]

UPPER vs. lower case

UPPER vs. lower case

I'm very much a beginner on this stuff. Trying to learn enough NEC for a Master Plan Examiner exam ... it gets really tough to follow, understand and learn when it appears there is indiscriminate use of upper and lower case letters in equations and discussions.
Does "VD = Vd"?
Is "K" the same as "k"?
For many equations used in the building/construction trade, especially for structural calculations, there is a difference. Whether or not that is the case here, it would be beneficial (especially for us beginners) to apply a degree of consistancy in these discussions.

 

[ggunn]

I'm very much a beginner on this stuff. Trying to learn enough NEC for a Master Plan Examiner exam ... it gets really tough to follow, understand and learn when it appears there is indiscriminate use of upper and lower case letters in equations and discussions.
Does "VD = Vd"?
Is "K" the same as "k"?
For many equations used in the building/construction trade, especially for structural calculations, there is a difference. Whether or not that is the case here, it would be beneficial (especially for us beginners) to apply a degree of consistancy in these discussions.

Sometimes the difference is pretty significant. "M" means multiply by a million, while "m" means divide by a thousand. That's nine orders of magnitude difference between 1 MW and 1mW, or a factor of a billion. No partial credit. ;^)

 

[Strife]

On another side not, I'm pretty sure there were 2-3 errors in my master test.
Guess that's why I only got 93:-(

Also, as a side note, I have found 2-3 errors in his practice exam. Would it be worth it to try to send an email pointing these outs?