# Voltage Drop revisited

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#### litesleeper

##### New member
This question is related to the Voltage Drop discussion already listed dated April 20, 2003. Several of the responders to the question used the figure 12.9 for a "K" value of copper. Is this number listed somewhere in the NEC and if so, where? If not, how is it calculated? Thanks.

#### bphgravity

##### Senior Member
Re: Voltage Drop revisited

"K" is not a value that is listed in the NEC. The NEC considers VD to be a design issue and not a concern for safety. "K" is the resistance of a circular mil-foot of copper. The resistance of this circular mil-foot is approx. 12.9 at 75 degrees C. To find exact "K", you must use the values of resistance given in Table 8 of the NEC. The equation is: R x CM / 100, R being the resistance at 1000 feet and cm being the circular mil of the size conductor used. 12.9 is good average value that will work well for most VD calculations. Other values that are sometimes used are based on other temperatures or other varying factors.

#### Ed MacLaren

##### Senior Member
Re: Voltage Drop revisited

As already stated, the actual specific resistance of copper varies with the alloy, hardness, temperature, etc.

Pure copper at 20 deg C - 10.09 ohms/mil-foot
Annealed drawn magnet wire - 10.4 ohms/mil-foot
Hard drawn building wire - 10.8 ohms/mil-foot

Ed

#### charlie

##### Senior Member
Re: Voltage Drop revisited

I thought you might be interested in the method I promote at IPL.

THE CONCEPT

The concept of voltage drop is very basic. Regardless of the formula that you use, it is a modification of E = IR. The formula that is given in the National Electrical Code Handbook is:

Vd = 2RLI / 1000

</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">Vd = voltage drop (in volts)</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">R = resistance of the conductor per 1000 feet</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">I = amperes</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">2 = both ways, going out to the load and back to the source</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">L = length of the run</font>
• <font size="2" face="Verdana, Helvetica, sans-serif">1000 = gets rid of the 1000 feet on the conductor resistance</font>
<font size="2" face="Verdana, Helvetica, sans-serif">
In order to change the formula to 3 phase, you must multiply by the Sin of 60 degrees, i.e. Sq. Rt. 3/2. This changes the formula to:

Vd = Sq.Rt.3RLI/1000

The concept and formulae above are based on direct current, not alternating current. If calculations must be accurate, reactance must be accounted for as well as the power factor of the load and the resistance of the circuit to alternating current. The question is, how important is the accuracy of your calculations?

When doing load calculations, you are making a ?best guess? as to the expected load and as to the future power factor. As long as the method used is not too cumbersome and as accurate as your needs, your method is fine. The above formula fits this criteria and may be used for most of the voltage drop calculations that the IPL project engineer will need to perform. Unless the run involves a significant amount of primary, open wire secondary, or metallic conduit, reactance will present no problem. Converting kW to kVA will take care of the problem of power factor since voltage drop is based on kVA amperes.

Remember this was written for PL engineers and we are guessing at the actual demand expected for a (or several) service.

#### wocolt

##### Member
Re: Voltage Drop revisited

to me 12.9 seems a tad high. From table 9, for example it lists No.14 solid at 3.07 ohms/1000',working backwards, 3.07 x CM/L(1000') = 12.618. Which would be a value for K.
According to Mike Holt, Q can be used when concerned wiht AC calculations, usually in circuits of 2/0 and larger. where Q is the AC resistance as listed in table 9.

Wm.Colt

#### wocolt

##### Member
Re: Voltage Drop revisited

Work has been slow in my area so I have been playing around with the VD formula.
I came up with a little bit different approach.
It uses largers numbers than say 3.07Ohms/1000'.
This works especially well when trying to determine in old construction how far the branch circuit is from the panel.
The old way is R/1000 x D x 2 x A = VD
or 3.07/1000 x 100 x 2 x 15 = 9.21 volts, however, not a new or novel approach but different
take the inverse of the 1/3.07/1000 = 325.73 which is approximately 326
Take the formula for voltage drop and Mhos/feet, or 326/200 = 1.628 then current is 15 amps/1.628 =
9.21 volts. (Today, I guess 1/R is called Siemens)
I havent got to 3-phase yet.
So back to determining how far the branch circuit is from the panel. You need two measurements the voltage at the receptacle and the load on line. Or I/V or 15/9.21 = 1.61Mhos
326/1.61 = 200 feet or the cable length is 100 feet from the panel at that particular point. I think this can work for old construction using a sur-test tester, it gives a quick ball-park of what you are dealing with.

Just a thought since we are on the voltage drop topic.

WmColt

#### bennie

##### Esteemed Member
Re: Voltage Drop revisited

Don't forget the voltage drop on the rest of the circuit when calculating loss.

[ April 30, 2003, 11:37 AM: Message edited by: bennie ]

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