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Short Circuiting an Ideal Capacitor

  1. Nov 7, 2012 #1
    Hi all,

    My question is what happens if I short circuit a capacitor with fully ideal wires. It is obvious that it will become chargeless but where does its energy be used? Is it possible to explain this system theoretically or we must always consider a parasitic resistor in these kind of problems?

    Thanks in advance.
     
  2. jcsd
  3. Nov 7, 2012 #2
    As the capacitor discharges a magnetic field forms around the conductor.
    When the capacitor has completely discharged the current in the wire drops.
    As the current in the wire drops the magnetic field shrinks.
    As the magnetic field shrinks it induces a voltage in the wire which is in the same direction as the original voltage.
    This voltage results in a current as it pulls charges off of 1 capacitor plate and pushes them onto the other, this continues until the magnetic field is completely gone.
    The situation now is that once again you have a capacitor shorted by a wire, except that the polarity of the capacitors charge is reversed.
    The same thing now happens in reverse until the capacitor has resumed it's original polarity, it's charge however has been reduced because during the cycle the wire has acted an antenna and emitted energy as an electromagnetic wave.
     
  4. Nov 7, 2012 #3
    Very enlightening explanation, thank you for it. :)

    But current flows in the wire must be infinite since there is no resistance, right? Therefore the magnitude of electromagnetic wave also must be infinite. Doesn't it reduce the whole energy immediately?
     
  5. Nov 7, 2012 #4
    No, there is still the inductance of the loop formed by the capacitor and the wire.

    see

    http://en.wikipedia.org/wiki/LC_circuit
     
  6. Nov 7, 2012 #5
    No. As the current starts to flow the magnetic field starts to grow. A growing magnetic field is a moving magnetic field. As the magnetic field grows it induces a voltage in the wire that opposes the voltage that is creating it, thus the net voltage on the wire is substantially less then the voltage being applied by the capacitor. I suppose for a superconducting wire the net voltage must be zero. With 0 resistance and 0 voltage current flow must be calculated using something other than ohms law. This is where my knowledge of electricity breaks down. Perhaps someone more familiar with superconductors will step in.
     
  7. Nov 7, 2012 #6
    Thank you very much. I will search for it.
     
  8. Nov 7, 2012 #7
    Look in the article on wikipedia on LC circuits under "Time domain solution"
     
  9. Nov 7, 2012 #8

    sophiecentaur

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    So far this thread has assumed that just an L and a C are involved here and. left to themselves, they will oscillate merrily for ever. Unfortunately, there is another factor that you cannot get away from (with or without Superconductors). A circuit with varying currents in it will radiate power out into space. The equivalent of a resistance appears in the simple LC resonator so the oscillations will gradually die down as the 'radiation resistance' dissipates the energy in the system. The rate of decay depends upon the dimensions of the set up and the L and C values.
     
  10. Nov 7, 2012 #9

    collinsmark

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    Real-world capacitors and other components are commonly modeled using ideal components. For example a real-world capacitor can be modeled using an ideal capacitor, a couple ideal resistors, and an ideal inductor. (See http://www.rfcafe.com/references/electrical/capacitance.htm.) You can even add several more ideal components depending on how precise you wish to make your model. You can maybe even add an ideal antenna to the model, which would allow radiated energy.

    (Note that the modeled capacitor now has parasitic inductance and resistance. But the ideal capacitor that is only a part of the model does not. The ideal components of the model are still ideal.)

    You can then use your tools of lumped parameter, circuit theory (and maybe some EM theory); or a computer SPICE simulation, to model what would happen if the terminals of this modeled capacitor are shorted together.

    But that doesn't answer the original question at all.

    What would happen if you ideally shorted the ideal terminals of a charged, ideal capacitor (assuming ideal wires)? Answer: you can't. It's nonsense. It's like asking what would happen if an unstoppable force meets an immovable object. It doesn't make any sense. It's undefined. It's not allowed.
     
    Last edited: Nov 7, 2012
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