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Ohmic resistace

  1. Apr 6, 2014 #1
    Where does ohmic resistance arise from, is it from the collision of the current with the electrons of the wire or is it from the electric field of the electrons of the wire's atoms that cause repulsion force against the flowing current ??
     
  2. jcsd
  3. Apr 6, 2014 #2

    UltrafastPED

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    Electron scattering in the metal - some of the scattering is from the metal grain boundaries, others are from impurities, still others are from phonons (quantized sound). In a "perfect" metal there are no collisions or scattering events. The "electron gas" effectively cancels the net positive field from the stationary metallic cores; thus resistance is not due to the local electric fields.

    The "simple" model of Drude (~1900) simply posits collisions - it gives a good feel, but the calculated values are inconsistent with experiment.

    For the Drude model see http://www.doitpoms.ac.uk/tlplib/thermal_electrical/drude.php
    or the Wikipedia article: http://en.wikipedia.org/wiki/Drude_model

    For the modern theory you will have to take a condensed matter physics course that studies metals.
     
  4. Apr 6, 2014 #3
    Why does increasing the length of the wire increases R, and decreases the intensity, I know that when the length increases the charges travel for a further distance which increases the resistance, but my point is when The length increases, the resistance per unit length is still the same why would that affect intensity. I think an analogy could simplify this. Thanks in advance
     
  5. Apr 6, 2014 #4
    If you increase the length of the wire without changing the voltage, the electric field decreases.
    So the electric force "pushing" electrons through each unit length decreases and so does the current.
     
  6. Apr 7, 2014 #5

    I don't clearly understand what causes the electric field in the circuit :confused:
     
  7. Apr 7, 2014 #6

    UltrafastPED

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    The electric field is the rate of change of the voltage over distance ... in fact the field is more fundamental (charge x field = force on that charge), but voltage is easier to work with for most practical problems.

    Note that if the resistance per unit length is constant then by Ohms law the voltage drop per unit of length will be a constant: d/dx [V(x)] = constant; but this constant is also equal to the field strength - hence the electric field is uniform throughout a single loop of wire connected to a constant voltage source.

    This was understood when Ohm first published his results - and as it implies that there is a constant force applied to the charges we should then see an increasing current -- for the current is just the number of charges per unit of time, and if the charges are being accelerated (a=F/m) we should see an ever increasing speed for the charges as they move along the wire, and hence an increase in the current.

    Thus scientists familiar with how currents actually work denigrated Ohm's work for several years as being "inconsistent with Newton's laws of motion".

    But Ohm's law does work - but it is not a simple "physical law" - instead it is a statistical effect due to the properties of electrons in metals.
     
  8. Apr 7, 2014 #7

    There is a continuous force acting on each coulomb Field x coulomb but the coulomb isn't continuously accelerated due to the resistance which causes deceleration, how didn't ohm suppose that since he is the one that find out about resistance and induced it in a mathematical formula V=RI
     
  9. Apr 7, 2014 #8
    Ohm found his law experimentally. Did not have to worry about microscopic mechanisms of electron scattering. Electrons have not been discovered at that time yet.

    In case this is what you are asking. I cannot understand most of your post.
     
    Last edited: Apr 7, 2014
  10. Apr 7, 2014 #9
    You posed earlier that we should expect an increasing acceleration in the motion of the electrons since a=F/m and the force is continuous on the electrons, I'm saying that this continuous acceleration doesn't happen due to the resistance
     
  11. Apr 7, 2014 #10
    I don't think I said such things. Maybe someone else?
     
  12. Apr 7, 2014 #11
    Why is the potential difference is the same in parallel resistors? For an example of two parallel resistors of same material and cross-sectional area but their lengths is 2L and L, how can I prove that the potential difference across them is the same in terms of : F x L / Q
     
  13. Apr 7, 2014 #12
    What is (FxL)/Q?
     
  14. Apr 7, 2014 #13

    Voltage :smile:
     
  15. Apr 7, 2014 #14
    So you are asking
    "how can I prove that the potential difference across them is the same in terms of: voltage"?

    Voltage and potential difference means the same thing so I don't think this is what you actually want to ask.
    What do you want to prove and what significance do you associate with that formula?

    By definition, parallel resistor have common terminals. The terminals in each pair have the same potential (they are connected together). So obviously the difference is the same for both resistors.
    There is nothing to prove. If they don't have the same potential difference across them they are not in parallel.
     
  16. Apr 8, 2014 #15
    I got the point that the potential difference across parallel resistors is the same because they are connected to two fixed points but on applying this idea to the equation V = F x L / Q
    If one resistor is 2R because its length is 2L and the the other is R and they are connected in parallel then the Voltages across both of them is 2V = F x 2L / Q
    And the other is V = F x L / Q
    For sure I'm mistaken, so I want to know why I am not correct ?
     
  17. Apr 8, 2014 #16
    What are the meanings of these parameters? What is F, what is Q?
    You assume that they are the same in both branches? How do you know this?
    You never explained here what is the matter with that formula and what do you want to achieve with it. Why did you bring this equation into the discussion? Is it relevant?
     
  18. Apr 8, 2014 #17
    F is force acting on charge and Q is quantity of charge,
    F / Q force acting on one coulomb then F/ Q is the electric field intensity since voltage = field intensity x distance
    Distance is the length of the wire, I'm saying that although parallel resistors have the same field intensity and different lengths, they consume equal voltages.
     
    Last edited: Apr 8, 2014
  19. Apr 8, 2014 #18
    They don't have the same field intensity. Or rather, they don't have to.
    They only have the same potential difference. The current (and current density) as well as drift speed of the carriers can be different.

    The integral of the electric field along the resistor is the potential difference.
    It depends on both the electric field and the path, including the lengths of the path.
    So both electric field and length of the path may be different for the two resistors.
     
  20. Apr 8, 2014 #19

    What makes the electric field intensity different ?
     
  21. Apr 9, 2014 #20
    I am not sure what you mean by "what makes".
    If you have a point charge, what makes the field to be different at different points around the charge?
     
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