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#### georgemcfly

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First off, I apologize if I'm breaking protocol but I'm a newbie and can't post yet so I figured this might be a good place to post this question as a reply. I'm studying for my Masters Exam and just when I though I had it licked I was humbled. Just to be clear, I'm using my Ugly's book as a refrence for this formula but keep getting an alternate and incorrect answer! Appriciate the help in advance!

Question: The voltage drop on a #10 AWG solid THHN uncoated copper conductor on a 120v circuit is 85' long and connected to a 2HP single phase motor is ?

When I use the 2K x L x I / Cm= VD formula I come up with 5.06v (INCORRECT)

The testing software has it simply at VD= I x R which gives me the correct answer of 4.94v

This simple formula (not listed in Uglys) is working and the other isn't. WHat am I not seeing here?!

#### btharmy

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For "L" (length) are you using 85' or 170'?

#### georgemcfly

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Here is my math: 2 x 12.9(K)= 25.8 x 85' x 24 amps= 52632 / 10380= 5.07 amps

If I use 170' in the above equation I come up with 10.14 amps. (way off the mark)

My understanding was that the 2K part of the formula means you don't have to double the conductor length when working the equation?

#### georgemcfly

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If I do it the long way: 1.21/1000 x 170' x 24 amps= 4.9 (correct) however I'm still baffled as to what I'm doing wrong when following the classic formula laid out in Uglys.

#### Dennis Alwon

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VD= I x R----- nope

Voltage = I x R

#### Dennis Alwon

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Here is my math: 2 x 12.9(K)= 25.8 x 85' x 24 amps= 52632 / 10380= 5.07 amps

If I use 170' in the above equation I come up with 10.14 amps. (way off the mark)

My understanding was that the 2K part of the formula means you don't have to double the conductor length when working the equation?
That is correct the 2 before the K is doubling the length. K is a constant which is different for copper or aluminum.

#### don_resqcapt19

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VD= I x R----- nope

Voltage = I x R
The current times the resistance of the current path gives you the voltage drop for that circuit and the current squared times the resistance gives you the power loss for that circuit.

#### Jhellwig

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Isn't a vd calculation what you do when you are deciding to get a hooker?

The current times the resistance of the current path gives you the voltage drop for that circuit and the current squared times the resistance gives you the power loss for that circuit.
Tis correct. My guess is why the answers aren't comming out the same is because in one you use the constant for the material of the conductor. The other you would have to take the resistance per foot out of the code book to do it that way. The tables might not jive with the formula. Aren't the tables just a cold resistance?

#### don_resqcapt19

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...
Tis correct. My guess is why the answers aren't comming out the same is because in one you use the constant for the material of the conductor. The other you would have to take the resistance per foot out of the code book to do it that way. The tables might not jive with the formula. Aren't the tables just a cold resistance?
The constant used in the formula is based on a specific temperature. I am not sure what that temperature is, but expect it is lower than the 75°C temperature that is used for the resistance values in Tables 8 and 9 in Chapter 9.

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