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Electrical Resistivity Math Question

  1. Oct 22, 2014 #1
    I am curious why in the resistivity formula you divide the length by the area? R=P*(L/A). I cannot figure why you would not take the L*A to get volume?
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  3. Oct 22, 2014 #2


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    Volume has no bearing on current for a fixed resistance. That is, the amount of current that flows through an X ohm object is V/X regardless of the volume. Internally, if you want to look at the current flow through a cross section of the object, you take the area. The current per unit volume is the same. That is, the current through a 2-inch section (length) of the object is identical to the current through a 1-inch section of the object.
  4. Oct 22, 2014 #3
    I am not sure that i follow what you are saying. I am under the impression that a 2" long section of say, aluminum, would have a greater resistance than a 1" long section of the same diameter. My understanding is that the larger the diameter the less the resistance, but the longer the length the more the resistance. Is this correct? Like a pipe? I do not understand why the length is divided by the cross section area. Can you explain the mathematical reasoning for how this operation works?

  5. Oct 22, 2014 #4


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    Resistance in wiring is measured by resitance per foot.....or resistance per hundred feet....or resistance per one thousand feet.

    However, if I put an ohmeter on a 1" long section compared to a 2" inch section the 2 inch section will have twice the resistance, but is still the same resistance per foot.

    You are correct in saying if you have two pipes of equal length, but different diameters, yes, the bigger diameter has less resistance....and also important....the larger diameter can carry a much larger current. Goes back to V=IR.
  6. Oct 22, 2014 #5


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    I am saying exactly what I said:
    Current is like a bicycle chain. You can't have a different current in one part of a series circuit or element than in another part.

    You are right that for a given resistance per unit length, you'll have a different current if that amount is bigger but if the CURRENT is the same through an object, it doesn't matter if you measure the current through one inch or one mile ... as long as it's a series circuit, the current is the same. That means that the current in any volume of a series circuit, given a fixed amount of current in the circuit. Identical to the current in any other volume of that same series circuit.

    On the other hand, the amount of current per unit area, taken as a cross section of an object, depends on the volume. If you have 1 amp flowing through a 1mm diameter wire, you get a different current per unit area than if that same 1 amp is flowing through a 1-inch diameter wire.

    you are confusing resistance with current. You started off asking about measures of current and then got off on a tangent about resistance.
    Last edited: Oct 22, 2014
  7. Oct 22, 2014 #6


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    If you look in the code book for example and take a single conductor 500 MCM wire, it has a diameter of nearly an inch (.813 inch) and has a resitance rating of .0845 ohm/km or .0258 ohm/kFT.

    Should be noted that the rating is for 75 degrees Celcius.

    Change the termperature and the resistance changes. Hotter generally increases resistance and cooler generally reduces resistance.
  8. Oct 22, 2014 #7
    "you are confusing resistance with current. You started off asking about measures of current and then got off on a tangent about resistance."[/QUOTE]

    I appreciate your input, but i did not however go off or a tangent about current. I made no mention of current in any of my posts. I was simply, and without attitude by the way, trying to ask a question about the formula for the electrical resistance of a given material. I was not asking about KCL, i was asking for a mathematical representation of how the formula works, not that current in one node= current exiting the same node. Sorry to have not been clear on my question.
  9. Oct 22, 2014 #8
    Thanks, so the formula works mathematically because it is (the ratio of the frontal area to the length) * Resistance of the material? I guess what i am trying to get at is why does that formula of cross section area/ length work? Does the current only ravel in straight lines and not need to fill the whole volume of the material?
  10. Oct 22, 2014 #9
    Hello Jeff - it almost seems that in your second post you answer the question - considering the length and area separately you have got it - but when you think of them both you get off track:
    "I am under the impression that a 2" long section of say, aluminum, would have a greater resistance than a 1" long section of the same diameter. My understanding is that the larger the diameter the less the resistance, but the longer the length the more the resistance. Is this correct?' .. YES.

    Multiply by length - then longer = greater resistance

    divide by area - then greater area = less resistance
  11. Oct 22, 2014 #10


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    You are getting into electrons and phsyics and chemistry and so forth. Sometimes the electrons just fill the outer layer of the wire. Sometimes they fill it up...

    Look at it this way. Say you have a long wire and you measure it with an ohmeter and ohm meter reads 10 ohms over a super long length.

    If you connect 240 volts to it, you will literally have 24 amps running thru the wire according to V=IR. Put an amp meter on it and it will confirm it.
    How the electrons behave and how they dispurse thru the wire is always more of an obscure lesson in possible theory.

    Also keep in mind you would never just hook up a wire....it would be the resistance of the wire plus the resitance of the load. Either way, V=IR.
  12. Oct 22, 2014 #11


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    The "resistivity formula" relates to the intrinsic quantity Resistivity to the extrinsic quantity Resistance. It is the constant of proportionality, for the chosen units of resistance and length. Resistance is the ratio of PD to Current, so it tells you the Joules per Coulomb needed for a given Coulombs per second.
    The "mathematical Representation" is the formula. You seem to after a Physical interpretation, in fact. There is proportionally more energy required per unit charge (Potential Difference) to get any given charge through a longer conductor and proportionally less energy per unit charge as the cross sectional area increases. Hence the formula contains l/A (which is a better way of expressing the relationship).
    In arm waving terms, having the wire longer makes it harder and having it fatter makes it easier to pass a certain current. That's pretty intuitive, I think, and no surprise to anyone.
    Last edited: Oct 23, 2014
  13. Oct 22, 2014 #12


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    I suppose this is off tangent a bit, but It's also important to keep in mind that if you put 1 amp thru a 10 K resistor.......That resistor will now have 10,000 volts across it! More than enough to kill you. A wire as small as a paper clip could carry 1 amp. So even the smallest wires with only 1 amp can be deadly in the right circumstances.

    Now that I mention this, say you did grab that resistor. Would it be 10,000 volts/ 2,000 ohms (very possible resitance of body) for about 5 amps?
    Or would it max out at 1 amp? Hmmmm......Sophie?

    Now that I think about it in schematic form, you have a current source in series with two parallel loads.
    Do current division and you get .83 amps thru human and .16 amps thru 10K resistor.......which would equal roughly 1600 volts thru each load. Interesting how the voltage drops as soon as you "parallel" it. That makes sense....follows KVL and KCL.

    So yes, .83 amps will kill you in a heartbeat.
    Last edited: Oct 22, 2014
  14. Oct 22, 2014 #13
    Thanks everybody for your help. I think that it is slowly becoming more clear for me, even though i tend to make my questions unclear. I understand the concept better now. I was probably thinking too much into it. I was curious how the mathematical operation worked out, and i think that everybody has helped to guide me into the right direction so that i can think more clearly about the topic.

    the resistivity times the length of wire (because it has a harder time the longer it goes) then divided by the cross section area (because more can fit into a hypothetical section at a time).

    In some way like a funnel maybe?
  15. Oct 22, 2014 #14


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    You are right (obviously) and I apologize. MY brain took off on a tangent and I was not answering the question you wrote but a different one. That's the second time today for me on that kind of goof. I think I need to take my brain back to bed. o:)
  16. Oct 22, 2014 #15
    Thanks, its no problem. I appreciate you trying to help me out in the first place!
  17. Oct 22, 2014 #16
    Jeff, Don't forget to just work the units, This gets over looked at times.

    ρ is in ohms / length, Think about how that number was derived. This is where you will get your understanding.

    Topic gets more interesting with solid wire, stranded, or plated wire. Frequency is another twist to consider.
  18. Oct 22, 2014 #17


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    Right. The skin effect at high frequencies seems quite weird when you first encounter it (well, it did for me anyway).
  19. Oct 22, 2014 #18
    Yep, skin effect is great topic of discussion. Have you seen Litz Wire? Stranded copper wire with each wire coated. High current loads for high freq power transmission. Costly material, different bonding methods.

    I had some nickel plated copper wire I was dealing with a few years ago. Be good topic thread, don't want to De-rail this thread going off topic. This was a topic I choose for presentation in a metallurgy class.
  20. Oct 22, 2014 #19
    Skin effect is not weird when compared with weirdness of proximity effect.
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