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Ohm's Law and a circuit loop

  1. Aug 26, 2004 #1

    cepheid

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    Now, this question may be incredibly stupid, but it has been bothering me nonetheless. I figure you guys can set me straight really easily. Say you have a basic circuit loop as shown in the picture.

    Ohm's law:

    [tex] V = RI [/tex]

    Now, according to a textbook I'm reading right now, "By Ohm's law, there is no voltage across an ideal (i.e. zero-resistance) wire regardless of the current flowing through it."

    I'm just wondering what the meaning of that statement is. Yeah, sure, obviously:

    [tex] V = 0I = 0 [/tex]

    But when I look at that picture, here's how I see it: there is definitely a potential difference between points A and B, because they are at either end of the battery. But these points are also the two ends of the wire that makes up the loop! So how could there possibly not be a potential difference across the wire?! And if there weren't, why would there be any current at all? Aren't the electrons moving from a point of high potential to low potential, gaining KE along the way? So there must be a voltage across the wire.
     

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  3. Aug 26, 2004 #2
    Have you taken into account the internal resistance of the source?
     
  4. Aug 26, 2004 #3
    Ummm. Not everything is as cut and dried as you assume. Yes, ohms law always works, but the real life components we have do things you may not expect when used outside their specs. Your drawing is somewhat of an undefined or non-permitted schematic. Kinda like dividing by zero. The wire has a near zero resistance so an IDEAL voltage source will push a very large amount of current through it. Depending on how small the wire is and how much current the voltage source can supply it can get VERY hot. Wire heats up and it's resistance goes up. Resistance goes up and the current goes down. Things will sort of stabilize. But with the average 9 volt battery you will only get a hot battery. What happens is the battery's internal resistance ends up dropping all of the voltage and you see nothing on the output terminals. Do what you have drawn with a 9 volt battery and what will happen is what I described. Do it with a car battery and you will have a glowing red hot wire in a split second. DON'T do it, take my word. The car battery has a lower internal resistance so it will supply more current. Enough current to get the wire red hot. Make sense? If not, shoot more questions out.
     
  5. Aug 26, 2004 #4

    chroot

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    If you connect an ideal battery (which is a perfect voltage source with no series resistance) to an ideal wire (which is a perfect conductor and has zero resistance), you'll draw an infinite current. Obviously Ohm's law is singular there, and really no longer applies. In reality, you can't have a perfect battery whose voltage never sags under load. As BoulderHead says, every real battery has some small, finite resistance effectively built into it by the nature of the chemical reactions it uses and the conductors from which they are constructed.

    - Warren
     
  6. Aug 26, 2004 #5

    cepheid

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    Understood...

    Thanks for the insights. Really it's only this statement that doesn't make sense to me:

    How can there be a current in that case, let alone an infinite one?
     
  7. Aug 27, 2004 #6
    You are getting current and voltage confused. Current is a rate. It is a given number of electrons past a certain point in a given amount of time. Voltage is thought of as a type of pressure.
     
  8. Aug 27, 2004 #7

    cepheid

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    I know the difference between current and voltage. What I'm saying is, in order for there to be a current through a wire, whether ideal or not, doesn't there have to be a potential difference between the two ends of the wire (that connect to each terminal). Wouldn't that qualify as "a voltage across the wire"?
     
    Last edited: Aug 27, 2004
  9. Aug 27, 2004 #8

    Integral

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    The whole point of neglecting the voltage drop across a wire is to simplify circuit analysis. You are completely losing track of what you need to learn by obsessing on this point. When you leave the resistance out of a simple DC circuit you have nothing instructive to gain. Why not put a resistor into your circuit and get on with it.

    Why are you surprised at an apparent contradiction when you incorrectly use a simplifying factor. If the wire is the only thing in the circuit you cannot neglect the wires resistance. A voltage drop across a wire can only be neglected when it is SMALL IN COMPARISON to other voltage drops in the circuit. This is NOT the case when it is the ONLY voltage drop in the circuit.
     
  10. Aug 27, 2004 #9
    Averagesupernova,
    I like you explanation; as clear and simple as it is.
     
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