User Tag List

Page 1 of 3 1 2 3 LastLast
Results 1 to 10 of 29

Thread: Is resistance transferable ?

  1. #1
    Join Date
    May 2019
    Location
    Tomball,tx
    Posts
    1
    Mentioned
    0 Post(s)
    Tagged
    0 Thread(s)

    Is resistance transferable ?

    I am having a debate with another master on a tricky subject . We have 1000 foot of conductor with a 15 amp load at the end. The D rating word meaning that the conductor need it would be too large to fit under a standard breaker . So my thoughts are that you could use a standard number 12 conductor out of the breaker and Tie in to the larger conductor through a gutter. He says you cannot do this because the resistance would burn up the smaller conductor . My thoughts are that resistance is only calculated through the length of wire and is only impartrd as current travels through the conductor . For example if you cross a stream of water From side to side that is only 2 foot wide there would be very little resistance. If you tried to cross the same stream perpendicular for a mile the resistance would be great . I know it is alternating current but . Would not the increased conductor size relieve the resistance as the current is traveling through that conductor and then the smaller conductor only carries their resistance for 2 feet ? To look at it another way resistance is only being applied to the section of wire that the current is passing through and the resistance does not pass to the smaller wire . In addition the larger conductor would relieve the resistance end it would not build beyond the smaller conductors capabilities ? Help please.

  2. #2
    Join Date
    Dec 2006
    Location
    Chapel Hill, NC
    Posts
    34,744
    Mentioned
    4 Post(s)
    Tagged
    0 Thread(s)
    You are correct. Resistance is going to be low thru the run if you upsize the conductor so adding a short piece of #12 at the end will not cause any issues if the load is 20 amps or less.

    Also look at 250.122(B) for sizing the egc
    They say I shot a man named Gray and took his wife to Italy
    She inherited a million bucks and when she died it came to me
    I can't help it if I'm lucky



  3. #3
    Join Date
    Dec 2007
    Location
    NE Nebraska
    Posts
    41,338
    Mentioned
    0 Post(s)
    Tagged
    0 Thread(s)
    Your two conductors are connected in series.

    Your short piece of #12 likely has less resistance than the long run of larger conductor and will have very little voltage drop across it in comparison to the long larger piece.
    I live for today, I'm just a day behind.

  4. #4
    Join Date
    Dec 2006
    Location
    Chapel Hill, NC
    Posts
    34,744
    Mentioned
    4 Post(s)
    Tagged
    0 Thread(s)
    Think of it this way-- do a VD calculation of the 15 amp load at 999 feet and then do the same calculation for the 1' of #12 wire. Add them together and the #12 will add virtually no VD to the answer.

    There are many online VD calculators online. Here is one https://www.calculator.net/voltage-drop-calculator.html
    They say I shot a man named Gray and took his wife to Italy
    She inherited a million bucks and when she died it came to me
    I can't help it if I'm lucky



  5. #5
    Join Date
    Jun 2003
    Location
    NE (9.06 miles @5.9 Degrees from Winged Horses)
    Posts
    11,263
    Mentioned
    0 Post(s)
    Tagged
    0 Thread(s)
    Change in conductor size is done all the time. The only time there is a problem is when the current exceeds that of the smaller. (PP workmanship not included.)
    Tom
    TBLO

  6. #6
    Join Date
    Mar 2016
    Location
    The Motor City, Michigan USA
    Posts
    1,218
    Mentioned
    0 Post(s)
    Tagged
    0 Thread(s)
    "Transferable" is the wrong word and might be causing confusion.

    What you have here is a series circuit with six elements: Source, load, two long & fat conductors, and two short & thin conductors.

    The total resistance can be calculated by calculating the resistance of each element, then adding them.
    The current can be calculated by dividing the supply voltage by the total resistance.
    (which is the same everywhere because there's only one current path)
    The voltage drop across each element can be calculated by multiplying the current by each individual resistance.
    The total voltage drop can be calculated by adding the voltage drop across the four individual conductors.

    If the load has a large reactive (inductive or capacitive) component, these calculations will be somewhat inaccurate, but probably close enough.

    The "thin" conductors have sufficient ampacity to carry the load current. Regardless of how long or short they are, or what else is connected to them, the breaker will open before they will overheat.
    Last edited by drcampbell; 05-16-19 at 10:29 AM.

  7. #7
    Join Date
    Feb 2003
    Location
    Seattle, WA
    Posts
    19,581
    Mentioned
    1 Post(s)
    Tagged
    0 Thread(s)
    Quote Originally Posted by Clint B View Post
    He says you cannot do this because the resistance would burn up the smaller conductor.
    With all due respect to you fellow master electrician, this is utter nonsense. It is not resistance that creates a risk of a wire burning up. It is the heat generated by current flowing through the wire. That heat is generated at a rate that depends on resistance and current. You have 15 amps of current flowing through a wire (#12 AWG) that is rated to handle 20 amps (more, actually, but we are required to protect it at 20 amps). There is no way your current will result in overheating the wire.

    Welcome to the forum.

    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

  8. #8
    Join Date
    May 2005
    Location
    Richmond, Virginia
    Posts
    25,507
    Mentioned
    1 Post(s)
    Tagged
    0 Thread(s)
    I had a slightly similar discussion here a few years ago, when I mentioned using my solenoid tester to find an open fuse by testing from line to load. Someone was aghast because he thought the load would force its full current through the tester and destroy it.

    I explained that, if testing line-to-line, which causes the greatest tester current, would not damage the tester, then testing in series with the load would not either. It dawned on him when I explained the difference between a voltage tester and a current tester.


    In this discussion, having a wire with lower resistance does not cause the line current to somehow rise above that of the load. It merely reduces the loss of current which would otherwise be caused by the increased total circuit resistance of using a smaller wire.
    Code references based on 2005 NEC
    Larry B. Fine
    Master Electrician
    Electrical Contractor
    Richmond, VA

  9. #9
    Join Date
    Apr 2008
    Location
    Ann Arbor, Michigan
    Posts
    7,807
    Mentioned
    1 Post(s)
    Tagged
    0 Thread(s)
    Your "another master" totally lacks an understand of electrical circuit theory and thermal theory. I would suggest he should not be a "master".

    Others have answered the question for you.

    Basically one wants to prevent overheating the insulation. For each incremental length of wire, for example 1" or 1 ft, there is an amount of power dissipated in the wire. Temperature rise in the wire and thus the insulation adjacent to the wire is proportional to the power dissipated in that increment of wire. So the resistance of said length times I^2 thru that wire segment determines the power dissipated in that unit length. For a given fixed current thru the wire the temperature rise will be less for a larger wire because the resistance of the wire decreases as the diameter is increased.

    As an approximation wire resistance changes by a factor of about 2 for a change of 3 wire sizes. #10 copper is about 1 ohm per 1000 ft. #12 about 1.5 ohms, and #14 about 2.5 ohms. #13 about 2 ohms. See Cirris
    https://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table

    At 20 A #12 copper dissipates about 400*1.59*1/1000 = 0.64 W/foot. Look what happens when you go to 30 A in that same wire, 1.43 W/foot. Note 1.5*1/5 = 2.25 and that is why power goes up so much.

    .

  10. #10
    Join Date
    Dec 2012
    Location
    Placerville, CA, USA
    Posts
    20,313
    Mentioned
    3 Post(s)
    Tagged
    0 Thread(s)
    The only thing of even potential concern is that you now have two terminations (at the breaker and at the joint) very close to each other on the same short length of wire. Localized heating which is in excess of what is expected and allowed (workmanship, design, etc.) at both of these points can result in a higher wire temperature even though the individual substandard terminations might not cause a problem by themselves.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •