ECN Forum Forums Technical Discussion Electrical Theory and Applications Voltage Drop question

I'm doing the Mike Holt series of workbooks for basic electrical theory (I also have a test tonight at school), and I'd just like to check my work to make sure I'm "gettin' it" if you know what I mean.

Question #32:

What is the voltage drop of two 12 AWG conductors, each 100 feet in length, supplying a 2-wire, 16A Load? The resistance of 12 AWG conductors is 2? per 1,000 feet.

So, Evd = I X R

I = 16

R = 2? per 1000'/ 1000 = 0.002

.002 x 200' (round trip) = 0.4

E = I (16) X R (0.4)

Evd = 6.4v

----------------------------

Correct?

I don’t think I have ever seen a VD calculated like that.

This is how I was taught:

VD= (2*L*I*K)/CM

K = 12.9 for CU and 21 for Al
CM for conductors on Table 8 NEC 2006 (pp70-635)

2 x 100 x 16 x 12.9/6530 = 6.32V

Note 3ph is a little different.

The Mike Holt book IS a litle bit different than the way I've been taught in school. I was taught in school by the example you provided. I have to assume that 6.4v and 6.32v is the same thing. The question had multiple choices and 6.4 volts was one of them.

And thank you for chiming in. I can't thank this forum enough for all that I have learned here the past 2 years.

I got a voltage drop calculation book from
Tom Henry back in 1989 the formula is still basically the same.

Ken

Shock is using Ohms law to caulate the VD.
VD = Amps x Resistance.

Ito
Your formula is the same only in a different format. The resistance of a conductor is
R per ft =(K/A) where A = cross-sectional area in cmil. K varies with temperature and ranges from 10.4 at 20C to 12.9 at 65C.
Lets look at the VD=(2*L*I*K)/CM . R = 2xLxK/CM. The formula becomes
VD = amps x R.

[This message has been edited by Bob (edited 12-07-2006).]

[This message has been edited by Bob (edited 12-07-2006).]

On the SBCCI Master's Test K was give at 12.9

The only time copper resistance came up, I had to calculate it using K...go figure.

Also keep in mind it NEC specifically notes this is the DC resistnace at 75°C.

R = (K*L)/CM

[This message has been edited by ITO (edited 12-07-2006).]

Hi to all of you.

I make so few posts on this site, but Mike Holts’ computation formulae and opinions on voltage drop compels me to this comment, so please bear with me as this is only a matter of physics and math.

At the point of your test, just use the formula that your instructor or test wants you to use and nothing more as it is impossible to argue with “city hall” and win. If you are really interested in this subject, then after regurgitating their formula for the purpose of the test you might want to look into these comments.

Voltage loss is sensitive to ambient temperature, conductor and insulation types, distance, conduit type, ambient conditions, if buried then depth and soil conditions, power factor, harmonics, and a lot more. A brief Google search on Neher-McGrath will shed some light on all the factors and conditions that are associated with the R variable in V =IR (which is correct but your formula is far too simple to be of any value). I don’t think that a complete understanding of these formulae is necessary at this point, but an understanding of their existence and concept of use will be very useful.

The IEEE has developed standards that are globally respected for voltage drop computations. Understanding and using the IEEE standard 141 exact formula will produce and voltage drop value that will be the same as a site measurement. This is the only formula that can boost this level of accuracy.

Volts is an electrical design software program that uses Neher-McGrath and IEEE formulae in its voltage drop computations. It can be found and downloaded from this website’s bookstore and used freely for 10-days. Try entering a number of different devices under different environmental conditions and you will see the impact on the voltage drop.