# 2nd year AC Theory Homework Question



## App.Electrician (Jun 2, 2009)

Two rows of florescent lamps are installed in an office with each row drawing 12.5a. The source volatge is 277v, line resistance is .5 ohms. The wire used has a constant (k) of 12.6 ohms/cmil.

What is the voltage at the load?

E=IR
E=12.5a x 0.5 ohms
E=6.25 v

277v
- 6.25v
---------
270.75 V Voltage drop percentage - 2.25%

My question to the forum is....Is this right??
I feel, in my gut, that its not. Feels like I'm missing a much needed bit of information, the length of the wire run. Also, the last bit of the question also says theres a constant of 12.5 ohms/cmil. What is this and should I apply it somewhere or is it just information to throw me off? If you guys can work it and let me know what equations to use and/if you used anything out of the NEC. Just started my 2nd year and I'm already intimidated. I don't want to goto class without homework done and the teach hasn't gone over ANY of this with us yet.

:001_huh:


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## RxScram (Aug 26, 2009)

It looks right to me, at least at a first glance. The 12.6 ohms per cmil is just a red herring, like you suspected. (You asked what it is... it is the resistance of the wire per circular mil foot. You can use this, with the length of the wire run and the gauge of the wire, to determine the resistance of the wire. Obviously, you weren't given enough information to do this, so it is useless information.)

You don't need to know the length of the wire, because the line resistance that you are given is for the entire length of the line.


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## App.Electrician (Jun 2, 2009)

my other concern is the next question: 

Is this value within the recommendation of the NEC? (Voltage Drop Percentage)

Per NEC 210.19 FPN No. 4 
Conductors for branch circuits sized to prevent a voltage drop exceeding 3% at the farthest outlet of power, heating, and lighting loads, or combination of such loads, and where the maximum total voltage drop on both feeders and branch circuits at the farthest outlet does not exceed 5%, provide reasonable efficiency of operation. 

In other words a branch circuit is not to exceed 5% voltage drop at the furthest receptacle, for reasons of operation, not safety, requirements.

So the answer is yes.

BUT the next question says:

How would you fix the problem?

Which is leading me back to the beginning. 

Aye carumba!


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## uber stein (Aug 20, 2010)

The 5% is for feeder and branch (total VD), its 3% "just" for branch. If its to much VD just increase wire size.


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## jwjrw (Jan 14, 2010)

uber stein said:


> The 5% is for feeder and branch (total VD), its 3% "just" for branch. If its to much VD just increase wire size.


 
And its just a fine print note....not required by the nec.:thumbsup:


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## JohnR (Apr 12, 2010)

I think you missed the *2 rows @ 12.5A EACH* gives 25 A total. Not 12.5
This using your calculation will give adrop of *12.5V or 264.5V at load
*


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## wildleg (Apr 12, 2009)

each light is a load, and the voltage will vary, being the least at the last light. That being said, the calculation you did is the only one that makes sense (as corrected above), being that distances were not given.


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## 10492 (Jan 4, 2010)

RxScram said:


> The 12.6 ohms per cmil is just a red herring, like you suspected.


What is the purpose of this red herring?

Did someone get paid, to make things harder than they really have to be?


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## RxScram (Aug 26, 2009)

That's what test writers do! They give you enough information to solve the problem, but often throw in another piece of information, related but irrelevant. It's supposed to help make sure you can pick out the important facts in a problem, or something.


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## 10492 (Jan 4, 2010)

RxScram said:


> That's what test writers do! They give you enough information to solve the problem, but often throw in another piece of information, related but irrelevant. It's supposed to help make sure you can pick out the important facts in a problem, or something.


I think it is to save thier jobs. 

There is no need to make things more complicated than they have to be.

More people would be willing go along with testing and licensing, if this kinda crap wasn't built into the requirements. IMO


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## Lensky (Feb 12, 2010)

If I is 12.5A the two rows are wired in series? So I think you are right.


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## App.Electrician (Jun 2, 2009)

Lensky said:


> If I is 12.5A the two rows are wired in series? So I think you are right.


parallel


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## Nildogg (Jul 29, 2010)

ahhhhhhhhhhhh crap


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## BuzzKill (Oct 27, 2008)

App.Electrician said:


> Two rows of florescent lamps are installed in an office with each row drawing 12.5a. The source volatge is 277v, line resistance is .5 ohms. The wire used has a constant (k) of 12.6 ohms/cmil.
> 
> What is the voltage at the load?
> 
> ...


 I'd crack your nuts but I'm from TN originally.


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## App.Electrician (Jun 2, 2009)

BuzzKill said:


> I'd crack your nuts but I'm from TN originally.


I've pretty much heard em all. 

I leave my feelings at the door.

:thumbsup:

Squeal like a pig, boy! WEEEEEEEE WEEEEEEEE! ha


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## App.Electrician (Jun 2, 2009)

Well its been almost 2 weeks and we still haven't gone over this problem.

When we do I'll post it here that way we can all point fingers!


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## App.Electrician (Jun 2, 2009)

JohnR said:


> I think you missed the *2 rows @ 12.5A EACH* gives 25 A total. Not 12.5
> This using your calculation will give adrop of *12.5V or 264.5V at load
> *


This was the right answer.

:thumbsup: Kudos!


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