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Divide by Zero

  1. Oct 8, 2007 #1
    Hi,

    I'm currently doing an A level course in Physics, so this thought struck me then.

    I was looking at a simple problem, when a thought struck me, if V=I/R and at absolute zero (0 kelvin) there is no resistance, what happens to the voltage?

    In other words, V=I/0 I know that you cannot divide by zero so I asked a staff member and he said at that temperature there is no Voltage at all, the current just keeps going round and round.

    So does this mean that 0=I/0? So is the answer dividing by zero = zero??

    Sorry if I have posted in the wrong forum, but I wasn't sure where it was supposed to go.

    Regards.
     
  2. jcsd
  3. Oct 8, 2007 #2
    well if you have no resistance you have no change in electrostatic potential energy between 2 points right?

    and voltage is the change in electrostatic PE between 2 points.

    so all points have the same resistance 0, and so no change in electrostatic PE there's 0 voltage. So yea the current just keeps going around and around like the staff member says.
     
  4. Oct 8, 2007 #3

    Hurkyl

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    You should always be suspicious when you are probing the edge cases like this. Textbooks don't always fully state all the necessary conditions for a law to hold valid, and students often miss it when the textbooks do state it. So, when you're considering unusual cases, you need to take extra care to make sure everything you're doing is applicable!

    (And, of course, you can't actually get to absolute zero)
     
  5. Oct 8, 2007 #4
    Okay, thanks both of you answered most of my question.
     
  6. Oct 8, 2007 #5
    If V= I/R you have discovered some new relationship.
     
  7. Oct 8, 2007 #6

    G01

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    Yes, Ohm's Law is:

    v=IR

    not

    V=I/R
     
  8. Oct 8, 2007 #7
    Yes, so the problem is that I=V/0, somebody can explain what happens to the electric current?

    It's right that it's impossible to get absolute zero (for the moment). But We can also get zero electric resistance if we use perfect superconductors in low temperatures.
     
  9. Oct 8, 2007 #8

    russ_watters

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    There is always going to be some resistance somewhere, even if it is in the power source itself.
     
  10. Oct 8, 2007 #9
    ps. as far as mathematically goes 1/0 is typically infinity

    but infinity can have many different meanings when trying to apply it to real life
     
  11. Oct 8, 2007 #10
    A superconducting loop across its Josephson junction has voltage drop given by the Josephson equation

    I=2eV/(hbar), where hbar is Planck's constant.

    Note that voltage is independent of resistance.

    Aside from utilizing a Josephson junction, measurement of superconductivity may be impossible (without disrupting the current).
     
  12. Oct 9, 2007 #11

    Hurkyl

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    No, it's typically undefined.
     
  13. Oct 9, 2007 #12

    f95toli

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    No, you can use an external dc-SQUID to read out the flux in the loop.
    That's how e.g. the flux is read out in most flux-qubits (in the original "Mooij-configuration"), the junctions in the qubit are never biased above Ic meaning there is no voltage that can be read out.
    Of course you are still "disturbing" the current somewhat due to the mutal inductance, but that effect is quite small if you get the design right.
    It is also possible to do the same thing with just a loop without junctions but then you are limited to whatever flux ju can inject using a bias field, i.e. you can make a ring which will catch a flux [tex] /Phi_0/2[/tex]); when the field is removed the flux does not go back to zero (this is yet another "trick" used to bias flux qubits).
     
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