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Only voltage source and wire?

  1. Feb 18, 2010 #1
    Only voltage source and wire?(PLEASE SEE THIS!!)

    Suppose a voltage source of any value(eg:5V,10V etc) is connected with a wire and thats all about the circuit,nothing else (resistors,bulbs or any kind of loads) is connected with the circuit.Assume that the internal resistance of the wire is totallyzero ,then what is the voltage drop and current at point B?(see the attached figure)Would the current flow become theoretically infinity? and voltage drop is zero?
     

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  3. Feb 18, 2010 #2
    Yes, but you can only get a wire like that at area 51. It's hidden in the lab where they divide by zero.

    In case you were wondering, superconductors cannot carry infinite currents. They have limits where they break down and turn back into normal conductors.
     
  4. Feb 18, 2010 #3

    sophiecentaur

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    A voltage source (ideal) has zero resistance an an ideal wire also has zero resistance. That would mean infinite current. Neither of the components you propose actually exist so the thought experiment goes nowhere, I'm afraid.
    What you can say is that, if you connect a thick wire across a car battery you will either melt the wire of boil the battery. I was so impressed when I did it with a coat hanger wire and an old battery of my Dad's (years ago and pre- health and safety). The wire got to yellow heat and collapsed - be careful though, the wire can burn through you, the battery case and the floorboards!
     
  5. Feb 18, 2010 #4
    Another thing about this, Ohm's law is really only an abbreviated description of electromagnetic phenomena. It's a subset of physics that's only useful for charged particles moving through real conductors that have resistance. Maxwell's equations combined with quantum physics explains electricity in much more detail.

    You could, for example, make a particle accelerator that sends a charged particle beam through space. That could be considered as a current that goes through a no-resistance medium. You can even apply a voltage difference to two points along the particle beam's path. It doesn't create an infinite current according to Ohm's law. Instead, it speeds the particles up to some finite number. Ohm's isn't meant to describe everything.
     
    Last edited: Feb 18, 2010
  6. Feb 18, 2010 #5

    sophiecentaur

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    Okefenokee
    That's an interesting point and it's given me a happy time thinking around it. I was looking for a 'flaw' or a 'loophole' in what you have written (as we all do) and I think that the point is that the energy which would be dissipated, normally, in a conductor - due to interaction with the atoms - is, in fact, transferred to KE of the electrons in a beam. The same amount of energy is involved in both cases (QV) but it turns up in a different form. Your "no-resistance medium" will, in fact, exhibit the same thing as resistance once the electrons hit the target / collector / anode
    I'm also not quite sure that you have, in fact, a constant voltage source. The current has already been defined by whatever limited the availability of the electrons (thermionic cathode / electron gun). These are more like 'Current Sources' than Voltage sources, so the available current is actually independent of the accelerating voltage across the gap you refer to. The "some finite number" you refer to is, in fact, not a factor of the accelerating voltage. There are times when the voltage is so high that it can 'tear' electrons from a cathode but I think that, under those conditions, what you would really be seeing would relate to some sort of 'surface resistance (work function related) and the current would then relate to the applied voltage.
    Any more thoughts?
    Btw, I agree with your first para. It's motherhood.
     
  7. Feb 18, 2010 #6
    When I said that you can apply a voltage difference to two points on the beam what I had in mind was something like a ring of charge around the beam. One ring could be positive and another along the beam could be negative. That would induce an electric field across some distance. In other words, voltage.

    The result would give us everything in samieee's question: a voltage, a current, and a medium with no resistance. Ohm's law isn't adequate for this situation by itself. The same can be said for a superconductor too. You can have a voltage, current, and a no resistance medium. However, the current is limited by quantum mechanics stuff that I don't know much about.
     
    Last edited: Feb 18, 2010
  8. Feb 18, 2010 #7

    sophiecentaur

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    Yes.
    Of course, the applied PD isn't a 'voltage source' (zero source impedance) as in circuit theory.
    In the end, a big enough beam current would be affecting the PD that was being applied and you'd have to consider how much actual power was being supplied from the supply which is accelerating the charged particles.
    In fact that's a bit of a puzzle and I haven't sussed it out. The beam is getting Kinetic energy so how does this manifest itself to the power supply, which must be supplying this energy? Someone must be paying for it! How? Is it in the form of capacitative energy as the PD changes somehow? Every Coulomb that is accelerated by 1 volt must be getting a Joule of energy.
    Help.
     
  9. Feb 18, 2010 #8
    Oh boy, this is going to get messy. Faraday's law of induction (one of Maxwell's 4 equations) says that the path integral of the dot product of the electric field around the path is equal to the time rate change of the magnetic flux. Wow, that's a mouthful. Anyway, if there is no changing magnetic field then this integral must be 0. In other words, the voltage around a loop must be 0.

    If the particles in the beam come back around in a loop, whatever happens between the rings I described will be canceled out by the fields everywhere else around the rings. This also happens if the beam goes off to infinity. Let me illustrate this with an example. Suppose the beam is made of protons and they pass through a positive ring then a negative ring. As a proton approaches the positive ring, it will be repelled (slowed down by a factor of 1). Between the rings, the proton is repelled by the positive ring but attracted to the negative ring (sped up by a factor of 2). Past the negative ring, the proton is pulled back by the negative ring (slowed down by a factor of 1). The total deceleration will equal the total acceleration so the net effect on the energy of the beam was zero. The interesting point that I was making for the OP was what happened between the rings and nowhere else.

    Now, in for a penny, in for a pound. If we wanted to permanently add some energy to the beam, we could send the beam in pulses. The rings start out neutral. When a pack of protons enters the the first ring, we send the charge into the rings. Because the protons are closer to the positive side, we have to do some work the push the charge. When the pack gets to the point before the second ring. We have to do some work to pull the negative charge out because it's attracted to the protons. That's how you could deliver energy with a voltage source which emits no current. Make sense?
     
  10. Feb 18, 2010 #9

    sophiecentaur

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    Yes, the bigger picture makes sense; those basic rules are fair enough. Around the loop = zero is along K2 lines, too.
    But I have a problem in deciding on details like how the power supply would be aware of (i.e. could measure) the fact that it had imparted (or even was constantly imparting) some energy to the beam. I imagine it might end up either acquiring a net charge, if it were 'floating' or pass a current to Earth. Would it be 'charging by induction' perhaps?
    Sort that out for me!
     
  11. Feb 18, 2010 #10
    Re: Only voltage source and wire?(PLEASE SEE THIS!!)

    voltage sources have an internal resistance. that internal resistance limits the available current.

    i = Vs / Rs, where Vs is source voltage at open circuit, and Rs is the source internal resistance.
     
  12. Feb 18, 2010 #11
    When you charge the rings (by any method you like) you are sending energy into the electric field between the rings. When you neutralize it, you get that energy back. Read in paragraph 3 of my last post how I described what happens as you charge the rings for an incoming beam pulse. You have to send a little extra push when the protons are near the positive ring and you get back a little less when the protons are near the negative ring. As you run through the cycle charging/neutralizing, you impart energy to the beam. You could run this scenario in the opposite direction and drain energy from the beam too.

    Also, sorry for the overly complicated discussion. This simple answer was: There is no such thing as an ideal conductor.
     
  13. Feb 18, 2010 #12

    sophiecentaur

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    But a steady PD across two rings will also impart energy to a continuous beam of charges. How does the power supply that charged the two plates know that the energy has gone? No current flows into the beam. What changes about the power supply? I guess the potential of the two rings gradually increases or a current flows to or from Earth via the Power supply?

    I realise that the pulsing effect transfers energy- as in the 'bunching' that occurs in a Klystron or traveling wave amplifier.
     
  14. Feb 18, 2010 #13
    Nothing changes overall. I suppose the source would feel a force as a single proton passes by. It would be like holding a spring and feeling a ball bounce off of it. Ideally, no energy would be transferred (you would have to be massive so that the momentum transfer is negligible compared to your mass). The energy of the ball was converted into potential energy then back into kinetic energy. The ball would leave the spring just as fast as it came. If there was a steady stream of balls bouncing on the spring, it would feel like a constant force because you wouldn't be able to sense each individual proton as they come and go. It would feel exactly the same as if someone put a static line of charge through the ring or, in the case of this analogy, pressed down on the spring with a steady force.

    If you could time it and push up on the spring as a ball was just starting to go up, the ball would bounce up higher. you would give the ball a little extra energy.

    By the way, these are just analogies according to how I see it. They aren't perfect. They may not even be the most correct way to view these things. Maxwell himself said that analogies are no substitute for math. Also, I took a serious interest in Maxwell's equations when I was in school but I'm certainly no expert on particle accelerators (or Maxwell's equations for that matter).
     
    Last edited: Feb 18, 2010
  15. Feb 18, 2010 #14
    For the circuit in the original post, When the battery is connected, the current rate of increase is theoretically given by

    dI/dt = V/L

    where V is the battery voltage, and L is the inductance of the loop. Very roughly, the inductance is

    L ≈ 100·μ0·d Henrys

    where μ0 is the permeability of free space and d is the dimension one side of the loop (in meters). This is the EE equation for the magnetic energy mentioned by Okefenokee earlier.

    Bob S
     
  16. Feb 19, 2010 #15

    sophiecentaur

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    Okefenokee
    You don't seem to have taken my point (my problem / my worry) that, if Kinetic Energy is transferred to the constant stream of charge then somewhere, somehow, there is a force times distance, a Charge times PD, a change in temperature or something, to account for this energy transfer. Merely saying there is some reactive force is just not enough to account for an energy change; force is not work.
    I have to admit that I am looking for a Nuts and Bolts explanation here - to suit my engineering mind.
    The other possibility is there is no energy actually imparted to the beam. I think this is actually what happens and the following case shows why.
    I suggest that your statement, that the whole 'loop' needs to be considered, has to be applied here. Take a CRT, which is a nice closed electrical circuit with every electron accounted for on its way round the loop.
    Suspend two rings in positions, 1/3 and 2/3 of the distance between Cathode and Anode (I think we can subsume the initial bits of electron optics which produce the beam into the initial part of the electron beam)

    Allow the rings to acquire the potentials that they would 'naturally assume' if there were some leakage, i.e. 2/3 and 1/3 EHT. Connect a voltage supply between them of 1/3EHT and the electron beam will go through unaltered by the presence of the rings. To accelerate the electrons through the gap, you need to increase the PD between the rings to greater than 1/3EHT (more than their natural share). The total energy that the electrons will acuire in the CRT will be EHTeV. This implies that before and after the rings, the electrons will be accelerated by LESS than 1/3EHT, . So, when they arrive at the first ring they will be going slower than in the original situation and when they leave the second ring they will be going faster than before. But, the total energy they will have acquired over the whole journey from C to A will be exactly the same. The power supply on the two rings needs to supply no energy.

    This, I am sure, is just a pictorial re-statement of what you have already said but it makes me feel better. :approve:

    Edit - addition: And, of course, when the supply to the rings is turned on, the apparent capacity between the two rings is higher when there is a flow of electrons going past so the stored energy is greater.
     
    Last edited: Feb 19, 2010
  17. Feb 19, 2010 #16

    sophiecentaur

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    Proton soup
    "voltage sources" are ideal sources of PD with NO internal resistance. They are not real devices but idealised for the purposes of analysis.
     
  18. Feb 19, 2010 #17
    what does PD stand for ? anyway, it really depends on what you want to analyze, how accurate your analysis needs to be, etc.

    also, would 1/0 be equal to infinity, theoretically?
     
  19. Feb 20, 2010 #18

    sophiecentaur

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    PD is Potential Difference.
    To model a real source you specify an ideal voltage (emf) in series with a specific (internal) resistance.
    I learned that at School and that's what is always done, in my experience,
     
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