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Capacitor in a circuit

  1. May 31, 2008 #1
    Hi,

    When you charge a capacitor in an RC-circuit, the charge q in the capacitor is given by
    [tex]q(t)=\epsilon C(1-e^{-\frac{t}{RC})[/tex]
    I was wondering what happens when the resistor is absent, that is when the circuit contains only an emf and a capacitor. Also assume that the wires have zero resistance. Does the capacitor then charge and uncharge instantaneously? If true, why does it take time when a resistor is there? What is it about the resistor that it takes time to charge and uncharge a capacitor?

    Thank you.
     
  2. jcsd
  3. May 31, 2008 #2
    The superconductor is an inductor; the combination of a capacitor plus an inductor is an oscillator.

    [tex]\omega[/tex] = [tex]\sqrt{\frac{1}{LC}}[/tex]

    Understanding the relation between voltage and magnetism is important. You should be able to answer the following question:

    In an open circuit, attaching a large versus small plates to a charged piece of Styrofoam produces high versus low capacitance. Energy and voltage in the Styrofoam does not change.

    Creating a closed circuit allows magnetism to force the Styrofoam to comply with the voltage to energy ratio of the large versus small capacitor. How?

    U = CV[tex]^{2}[/tex]
     
  4. May 31, 2008 #3

    Redbelly98

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    Yes, if connected to an ideal voltage source, the capacitor would charge instantaneously. (We're also ignoring relativistic effects imposed by the finite speed of light :smile: )

    As for "what is it about a resistor ...", the simple answer is: it resists current.
     
  5. May 31, 2008 #4

    Redbelly98

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    I think bringing inductors into this is unnecessarily confusing things. It's a basic question about what happens with a capacitor in an idealized circuit.
     
  6. May 31, 2008 #5
    I wasn't saying that the wire is supposed to be a superconductor. The wire I had in mind is just an idealized wire that doesn't exist physically.

    Ok thank you that was the answer I was looking for. The fact that the resistor resists current is that because the charges collide with the atoms in the resistor?
    I don't know a lot about relativity but now you made me curious about what happens when you put a capacitor in a circuit with a very low resistance.
     
  7. May 31, 2008 #6

    Redbelly98

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    Yes.

    In a real circuit, the resistance in the wires will slow things down enough that relativity is not important. The equation you gave in port #1 will still work.

    In an ideal zero-resistance circuit, relativity will limit the time at which the capacitor charges. I believe the capacitor charge would go from 0 to full-charge in an instant, but that instant will be delayed until sometime after the voltage is switched on. The amount of delay depends on the length of wires and the propagation speed of electrical signals within the wire.

    Of course, this is a highly idealized thought experiment. We're ignoring resistance and inductance (which a real circuit would have), and assuming a perfect voltage source.
     
  8. May 31, 2008 #7
    IN A CIRCUIT CONTAINING A VOLTAGE SOURCE, RESISTOR AND A CAPACITOR:
    Volts across the wire (inductor) initially equals source volts, because capacitor and resistor volts equal zero. The rate source volts converts to magnetism is maximum.

    Magnetism is produced in the inductor and all voltage in the resistor and capacitor was converted from magnetism. The amount of magnetic energy converting to volts in the resistor must be subtracted from the amount of magnetic energy converting to voltage's energy in the capacitor.

    Shortly after creation of a closed circuit, resistor voltage nearly equals source voltage. A high resistor volts to capacitor volts ratio is not in compliance with a current x resistance calculation (Ohm's law). Part of the reason for non-compliance is that resistors usually require less voltage's energy to produce a given voltage than a capacitor requires.

    As capacitor volts increase, the volts available to the resistor decreases. When the capacitor is fully charged, capacitor and resistor volts comply with Ohm's law.
     
  9. May 31, 2008 #8
    Correction: in the previous post,
    "The amount of magnetic energy converting to volts in the resistor must be subtracted from the amount of magnetic energy converting to voltage's energy in the capacitor."
    should be changed to "During charging, resistor volts + capacitor volts nearly equal source volts."

    I was trying to convey that voltage applied to a capacitor within a closed circuit does not instantly charge the capacitor. Some rate determining processes exist:

    1) The rate voltage's energy converts to magnetism. This rate depends on the amount of volts applied to an inductor.
    2) As capacitor volts increase, the rate source volts convert to magnetism decreases.
    3) The amount of voltage's energy (or magnetic energy) required to raise capacitor volts to source volts depends on capacitance.

    I realize this explanation seems incongruent with an RC time constant (RC = R x C for R=0), but zero resistance is an extrapolation beyond the limits that the time constant equation applies to. Similarly, the force and energy resulting from the force between charged particles does not become infinite when the distance between them is zero.
     
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