Cannot find a way to answer this question. Voltage Drop Paralelled Conductors

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tjb660

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What is the Alternating-Current Ohms-to-Neutral Resistance for three 1/0 AWG aluminum conductors run in parallel for 1000 FT installed in a metal raceway.

I have literally been through my Mike Holt text book and a few times questions like this have come up during sections where I am using formulas like:

VD=(2*K*I*D)/CMil

I would use this formula but I do not have Amperage.

My multiple choice answers are as follows:
A. 0.067ohms
B. 0.114ohms
C. 0.231ohms
D. 0.413ohms

What am I missing here. very frustrating to be presented with questions that don't seem to have been covered in the sections we are reviewing.
 
Hint: voltage drop = current * resistance.

You need to know current to figure out voltage drop, but you are not asked for voltage drop. You are being asked the first step only, finding the resistance.

Jon
Thanks for the reply. I was aware I am just looking for resistance but am I looking at the wrong table?

Chapter 9, Table 9 does not have anything close to the four options I was presented with, and neither does Table 8.

I'm probably overlooking something extremely simple, but I have exhausted all of my own options and would really appreciate a simple "where is the answer" reply, it would seriously be the only way I would learn this at this point.
 
What is the Alternating-Current Ohms-to-Neutral Resistance for three 1/0 AWG aluminum conductors run in parallel for 1000 FT installed in a metal raceway.

I have literally been through my Mike Holt text book and a few times questions like this have come up during sections where I am using formulas like:

VD=(2*K*I*D)/CMil

I would use this formula but I do not have Amperage.

Resistance only depends on the geometry and identity of the conductive material. Rearrange your equation, so that only geometry and identity properties are on one side, and the electrical operating conditions are on the other side. Recall the definition of resistance and recognize where you find it within this equation. This is how you would solve it, if the equation you presented were valid for this problem. It turns out it isn't valid for this problem and I'll explain what else comes in to play that you need to consider.

The formula you gave is built for the round trip voltage drop, hence the 2 in the numerator because current travels the full round trip in either a DC circuit, or a single phase AC circuit. For 3-phase, 2 gets replaced with sqrt(3), but that is a topic for another post. In your case, you are interested in the line-to-neutral resistance which is only the one-way resistance of the circuit. The 2 or sqrt(3) constant in the numerator disappears completely, and is now 1.

You also are given the fact that there are three conductors in parallel. This means three conductors are joined together at both ends, so that the current is shared among all three of them. Think about how three identical resistances combine in parallel, and apply it to your calculation. For other problems like this, think about how N identical resistances combine in parallel, to generalize this problem.

On top of all of that, there is another factor that comes in to play for AC circuits, called the skin effect, which is why it is relevant that it is AC, and why it is relevant that it is in a metal raceway. It turns out that since your choices are so far apart, that you can ignore this, and approximate through the formula you presented, modified to account for the factors discussed above. You may get an answer that is slightly different than the given options, but it will be close enough to conclude which of your options is correct.

There are resistance per unit length values in Chapter 9 / Table 9, that account for the skin effect, and you will notice that effective AC resistance is no longer inversely proportional to KCMIL area as it is for simple DC resistance. It is only a first order effect that resistance as inversely proportional to KCMIL area. The skin effect comes in to play, the larger the wire is, and it starts to matter for sizes 1/0 and greater. Metal raceways and the fact that it is AC, also increase this effect. It also matters what metal the raceway is, as this is also a factor in the skin effect. It doesn't matter what plastic the raceway is, if it is non-metallic, but identity of the metal does matter. They neglected to specify the metal raceway's identity, but I'd recommend assuming the metal is steel, as this is the most common in practice, and most likely what the author would have in mind.
 
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Resistance only depends on the geometry and identity of the conductive material. Rearrange your equation, so that only geometry and identity properties are on one side, and the electrical operating conditions are on the other side. Recall the definition of resistance and recognize where you find it within this equation. This is how you would solve it, if the equation you presented were valid for this problem. It turns out it isn't valid for this problem and I'll explain what else comes in to play that you need to consider.

The formula you gave is built for the round trip voltage drop, hence the 2 in the numerator because current travels the full round trip in either a DC circuit, or a single phase AC circuit. For 3-phase, 2 gets replaced with sqrt(3), but that is a topic for another post. In your case, you are interested in the line-to-neutral resistance which is only the one-way resistance of the circuit. The 2 or sqrt(3) constant in the numerator disappears completely, and is now 1.

You also are given the fact that there are three conductors in parallel. This means three conductors are joined together at both ends, so that the current is shared among all three of them. Think about how three identical resistances combine in parallel, and apply it to your calculation. For other problems like this, think about how N identical resistances combine in parallel, to generalize this problem.

On top of all of that, there is another factor that comes in to play for AC circuits, called the skin effect, which is why it is relevant that it is AC, and why it is relevant that it is in a metal raceway. It turns out that since your choices are so far apart, that you can ignore this, and approximate through the formula you presented, modified to account for the factors discussed above. You may get an answer that is slightly different than the given options, but it will be close enough to conclude which of your options is correct.

There are resistance per unit length values in Chapter 9 / Table 9, that account for the skin effect, and you will notice that effective AC resistance is no longer inversely proportional to KCMIL area as it is for simple DC resistance. It is only a first order effect that resistance as inversely proportional to KCMIL area. The skin effect comes in to play, the larger the wire is, and it starts to matter for sizes 1/0 and greater. Metal raceways and the fact that it is AC, also increase this effect. It also matters for whether the metal raceway is aluminum or ferrous, as the ferromagnetism of steel raceways also impacts this. They neglected to specify what metal the raceway is, but I'd recommend assuming the metal is steel, as this is the most common in practice, and most likely what the author has in mind.
Again, thank you for the reply.

Would there be anyone else out there willing to solve the problem themselves and post the answer in a reply. We all learn differently, and while both of the replies thus far have been thorough and probably complete in their own way of explanation, it still does not clarify anything.

I understand that the equation I presented is not useful in this situation, It requires Current to be solved. I understand that it is also not valid for use, as I am not being asked for voltage drop. I understand that I am being asked for simple resistance

Am I being asked for the resistance of all three 1/0 AWG conductors combined? Chapter 9, Table 9 lists 1/0 AWG Aluminum in Steel raceway as having 0.20ohms over 1000 feet. Shall I simply divide this by 3 and get .67. I suppose so, if Amperage gets cut into thirds by running parallel so shouldn't the resistance.
 
211128-1353 EST

tbj660:

If you have 3 equal valued resistors in parallel and you know the value of that parallel combination, then what is the value of one of those resistors? Knowing that one value, then go to an aluminum wire resistance table and determine the closest size value for 1000 ft.

.

.
 
Thanks for the reply. I was aware I am just looking for resistance but am I looking at the wrong table?

Chapter 9, Table 9 does not have anything close to the four options I was presented with, and neither does Table 8.

I'm probably overlooking something extremely simple, but I have exhausted all of my own options and would really appreciate a simple "where is the answer" reply, it would seriously be the only way I would learn this at this point.

Okay, you are looking in the correct tables.

What you are missing is correcting for the number of conductors in parallel.

You know how to find the resistance for _one_ 1/0 conductor 1000 feet long in a conduit. Just look it up in the table. Now correct for having 3 in parallel.

Here is a link for resistors in parallel: http://physics.bu.edu/py106/notes/Circuits.html (scroll down to the parallel resistors section)

-Jon
 
Again, thank you for the reply.

Would there be anyone else out there willing to solve the problem themselves and post the answer in a reply. We all learn differently, and while both of the replies thus far have been thorough and probably complete in their own way of explanation, it still does not clarify anything.

I understand that the equation I presented is not useful in this situation, It requires Current to be solved. I understand that it is also not valid for use, as I am not being asked for voltage drop. I understand that I am being asked for simple resistance

Am I being asked for the resistance of all three 1/0 AWG conductors combined? Chapter 9, Table 9 lists 1/0 AWG Aluminum in Steel raceway as having 0.20ohms over 1000 feet. Shall I simply divide this by 3 and get .67. I suppose so, if Amperage gets cut into thirds by running parallel so shouldn't the resistance.

That is correct reasoning. If you follow the link I posted you will see the equation that confirms you are correct...except for a decimal point error in your maths. 0.2/3 = 0.067

-Jon
 
Again, thank you for the reply.

Would there be anyone else out there willing to solve the problem themselves and post the answer in a reply. We all learn differently, and while both of the replies thus far have been thorough and probably complete in their own way of explanation, it still does not clarify anything.

I understand that the equation I presented is not useful in this situation, It requires Current to be solved. I understand that it is also not valid for use, as I am not being asked for voltage drop. I understand that I am being asked for simple resistance

Am I being asked for the resistance of all three 1/0 AWG conductors combined? Chapter 9, Table 9 lists 1/0 AWG Aluminum in Steel raceway as having 0.20ohms over 1000 feet. Shall I simply divide this by 3 and get .67. I suppose so, if Amperage gets cut into thirds by running parallel so shouldn't the resistance.

If you were to use the formula you originally presented, you would modify it as follows:
Vd=(2*K*I*D)/cmil

Remove the 2 as discussed, because we are only interested in 1-way resistance:
Vd=(K*I*D)/cmil

Divide both sides by current (I). Now the right side exclusively depends on geometry and material identity. The operating conditions of Vd and I are grouped on the left.
Vd/I = (K*D)/cmil

Vd/I is resistance, replace with capital R1. This is the resistance of a single wire.
R1 = (K*D)/cmil

Given N wires in parallel, we divide resistance by N to find the equivalent resistance of the group:
R = (K*D)/(cmil*N)

For K=21.2 Ohm-cmil/ft for aluminum, D=1000 ft, N=3, and cmil = 106000 cmil for 1/0, you get:
R = 0.067 Ohms

We get the same answer, as we'd get if we looked up the value in Chapter 9/Table 9, and then divided by N. The skin effect is not significant enough to impact the answer by a noticeable amount. If given a length other than 1000 ft, you would have to multiply the Chapter 9 / Table 9 values by D/(1000 ft).
 
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