1. rk1
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Aug 2007
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## resistance calculation

i have a study guide with a question i do not understand.

question: what is the total resisitance in a 2-wire circuit, 30 feet in length, utilizing NO. 10 AWG THWN?

the equation they say to use is R=KxL/CM; R=10.4x60/10380=.060ohms.

I don't understand were the 10.4 comes from. The question only explains were the leght comes from and the circular mils.
:confused:

2. The 10.4 is related to the physical properties of the conductor material. There is a different value to use, if the wire is aluminum. But I don't have that number handy.

3. rk1
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Aug 2007
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thank you for your reply. i have seen the table in the code book. i just don't see that number. the formula I was show to find exact-K does not come up with the 10.4. i'm just missing a step in the equation or i'm not reading the table right.

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You are correct 10.4 is not the value of K for Cu. the values for K are approx. 12.9 for Cu and 22.1 for Al. The actual calculation as I learned it is K = (Rdc x CirM)/1000.

5. The K value is dependent on the conductor material (copper or aluminum) and the ambient temperature. The value of 12.9 is currently used and corresponds to the resistance of a 1,000 cmil conductor 1,000 ft long at 75 degree C temperature. 75 degrees C is used as an ambient for Chapter 9 Table 8. In years past, (long ago when I was an apprentice), the table was based on 60 degrees C and the K value was then 10.4

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Wait until you get to the C factor

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CAPS I believe the correct constant for Alum is 21.2, not 22.1, as well 12.9 is the copper constant. To prove this, look at Table 8, Ch 9, go to the 1000 kcmil line, you will see copper is 0.0129 (0.0129 x 1000 = 12.9), go under the aluminum column and you will see 0.0212 (0.0212 x 1000 = 21.2). 12.9 cu and 21.2 Al are your K values, however I do know there is some variations by user, but this is a provable method.

8. ## c factor?

What the heck is that??
Alot of smart people on this forum, Im just glad that I somehow passed my Masters without having to be all that smart, and know all that!:grin:

9. Join Date
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## History of conductor resistance in NEC

In the 1981 edition of the NEC, conductor resistance (dc) was shown at 25 degrees C and 77 degrees F in Table 8 of Chapter 9. The resistivity (K) factor (or value) for 1000 MCM (now kcmil) uncoated copper was 10.8. The K factor is found by multiplying the conductor’s resistance by the conductor’s circular mil area and then dividing by 1000. The K factor for 1000 kcmil aluminum was 17.7. Table 8 was revised in the 1984 edition of the Code. DC conductor resistance was shown at 75 degrees C and 167 degrees F. The approximate K factor for uncoated copper was (and still is) 12.9 and aluminum was 21.2.

Electricians taking a test based on the 1981 (or earlier) edition of the Code calculated voltage drop by using 10.8 for copper and 17.7 for aluminum. Electricians taking a test based on the 1984 (or later) edition of the Code calculated voltage drop by using 12.9 for copper and 21.2 for aluminum.

Don’t do the math on this one but during my first master’s electrical exam, I used 10.8 as the K factor for copper. We didn’t worry about finding the exact K for conductors. When I started teaching my electrician’s exam prep class, I taught the students to use 12.9 for copper and 21.2 for aluminum. I also taught them to find the exact K if the conductor size was given.

Back to the original question . . . since the question wants the resistance and the conductor is given; use the resistance values shown in Table 8. Since the conductor has THWN insulation, it matches the 75 degrees C (or 167 degrees F) in Table 8. On electrical exams, if the question does not specify copper or aluminum, use copper. Also, if the question does not state whether the conductor is coated or uncoated, use uncoated. There is one piece of information missing; is it a solid or stranded conductor? If the conductor is a solid conductor, the answer is 0.0726 ohms. First find the resistance for one foot of the conductor. Divide the conductor resistance by 1000 feet. (1.21 ohms divided by 1000 feet = 0.00121 ohms per foot) Next multiply the total length of the conductor by the ohms per foot. (60 feet x 0.00121 = 0.00726 ohms)

Charles

10. ## Reference

Try Electrician's Handbook of Formulas and examples, Mine is from wing publishing CO. PO Box 2707 Livonia, Michigan 48151, BUT, I have had it since 1968 and the pages have coffee on them. Excellent reference, Alot like Ugly's.

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