is this right?

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Re: is this right?

All "k" means is:
The specific resistance (K) of a material is the resistance by a wire of this material which is one foot long with a diameter of one mill at a specified temp.

NEC quotes copper of 12.9 @ 75 degrees Celsius. That is a very liberal number. I use 11.1 because of years of collected data, and 11.1 is still liberal with the copper.
 
Re: is this right?

still confused .....on the test should i use for copper..2 x k x i x d /cm 2 x 12.9 x amps x distance divided by cir mills will this work ? this needs to be right for the test not for rocket science calculations any ideas ? this is what it says in the mike holt book the test is hard enough dont need to make this more difficult then it needs to be ..thanks ...
 
Re: is this right?

Physis:

This one has nothing to do with English. It has everything to do with the integrity of mathematical formulas involving physical quantities. The units on the left side of an equation MUST equal the units on the right side of the equation.

Here is the first link I found with some insight involving our discussion:

web page
 
Re: is this right?

well that link makes this whole thing look simple thats much better ... you know the old saying you ask someone what time it is then they tell you how to build a watch... useless info when all you wanted to know is the time

[ March 14, 2005, 02:40 PM: Message edited by: jwest ]
 
Re: is this right?

if you are taking a code test, then use the info from table 8 for resistance.

For example, if the question asks for the resistance of 685 feet of 3/0 copper then:

Table 8 gives 3/0 as having a resistance of .0766 ohms per 1000 feet

685 feet x .0766 ohms/1000 feet = .05247 ohms

(notice how the units of feet cancel and leave ohms)

Now if this was a voltage drop question, you would multiply the computed ohm value by the given current to find the answer.

Also, pay attention to whether they tell you the length of the circuit or the length of the conductors.... multplying the length by 2 is necessary when you must include the length of the wire to the load and then also the length of the wire back from the load.

Now, if you do not have access to a code book, you can still do the resistance calculation based on the "K" that we are talking about.

Understand that K is an approximation based on a number of things as mentioned above, and includes the "pureness" of the copper, its annealing (hard drawn, soft drawn) and temperature.

Of course, the ohms/1000 feet numbers in the code book are approximations also.
 
Re: is this right?

thats what i was talking about keep it simple [did you catch the part about the time and the watch ]thanks again im studying for the new jersey electric lience test .i only want to take it once got to be ready for it.
 
Re: is this right?

Jwest,

yes, that is the correct use of the equation. And 12.9 is pretty much a universally accepted value for the K of copper.

And it isn't as complicated as this thread is making it look.

If you want to demonstate the error you'll get using K = 12.9 use the formula I posted on the last page.
 
Re: is this right?

Crossman,

I'm not going to read all of this right now, maybe later, but like Dereck said, I've said and even you said, it's just the resistance of a mil foot at a given temperature. :)


Edit: I had to change Bob to Dereck, all you moderators look alike to me. :D

[ March 14, 2005, 03:41 PM: Message edited by: physis ]
 
Re: is this right?

I agree, and my little sideline really had no point in this thread. But it is important in the grand scheme of things. In physical formulas, the units will always work out. ;)
 
Re: is this right?

physis I used your idea and the mike h. book idea #12 copper 10amps 200 ft for voltage drop. yours 7.72 voltage drop --- mike h. book 7.90 voltage drop .18 difference in the real world thats nothing .. on the test dont know ?
 
Re: is this right?

Pretty much true.

But why go through essentially the same process to get a result that is known in advance to be wrong.

Except in your case, where that's how you'll be tested.

It's good that you're taking a close look at this. :)
 
Re: is this right?

real world vs book world its amazing ..i did it the book way 12.9 copper 10 amps on a 20 amp ckt up to 100 ft use #12 up to 150 ft use #10 up to 200 ft use #8 didnt need a book to tell me that learned that 15 years ago from the old timmer who taught me nothing beats real world experence but backing it up with books is good too thanks for keeping me interested in the math
 
Re: is this right?

I know just a little about theory and real life. And you know what, they actually match! Provided you understand and apply the theory correctly.

It's easy to say the book stuff is mumbo jumbo when you haven't actually used it along side real world stuff.

Use your meter and ohms law together and you'll find when something comes out differnt than you expect it to there will be a real reason for it.

Edit: I left the law out of ohm's law.

[ March 14, 2005, 08:34 PM: Message edited by: physis ]
 
Re: is this right?

man just when you think your a pretty good electrian you get put back into place..got alot still to learn well back to study [ is my brain supposed to hurt i think i smell smoke ]
 
Re: is this right?

I thought it would be nothin' to go from what I used to do to electrical. I can design circuit boards, wiring a building is nothing. Hah, I'll never know all the stuff there is to know in this field. Plus, I don't learn things as easily as I used to.
 
Re: is this right?

Sam,

Are you a CAD man? I have spent a few hours in front of the tube myself. Never did any PCBs though.

Rattus
 
Re: is this right?

If you mean computer aided I'm afraid that industry was in it's relative infantcy in my day. Although it existed it was out of reach finacially. I had to use the manual stuff, 2:1 artwok on the light table and the rub on stuff, dounut pads and what not. That would usually get shipped out to turn into boards. At times I've done the photographic process and developed and etched the boards for prototyping and testing, and hurry hurry jobs, didn't get plated through holes that way though.

We were just starting to try out some less expensive systems that were coming out in those days. It was kind of cool watching the computer do in an hour what took me a couple days. I'm sure it's microseconds now.
 
Re: is this right?

Hi All, I am a first year instructor as well.

Just for giggles I teach BOTH methods in my class. I too feel that the constant of 12.5, or 12.9, or whatever is just a bit too variable for my tastes. I usually use the table 8 out of the code book to find ohms per foot( or 1000 ft). For my money it seems a tiny bit more accurate to use. Besides, if you have the code book handy anyway, use it!

2(K)(I)(L)/CM usually WILL get you in the ball park no matter WHAT value is used for K (within reason).

Just my $0.02 worth.
 
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