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Capacitor voltage of series LC circuit (inductor and capacitor)

  1. Dec 24, 2012 #1
    1. The problem statement, all variables and given/known data
    I have a series LC circuit that consists of a 1V DC voltage supply, an inductor and a capacitor. The inductance of the inductor is 2mH and the capacitance of the capacitor is 1μF. I performed a transient analysis using a simulation software and I observed the trace of the capacitor voltage.

    I have uploaded a picture of my simulation.
    2. Relevant equations
    May I know why the magnitude of the capacitor voltage exceed that of the 1V?


    3. The attempt at a solution
    I know that at the resonant frequency (159Hz) the impedances of the capacitor and the inductor cancel out each other hence maximum voltage.....but this is a time domain analysis and i don't get why it keeps oscillating ( is it because this is a dc supply?)
     

    Attached Files:

  2. jcsd
  3. Dec 24, 2012 #2

    NascentOxygen

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    Staff: Mentor

    Hi anandan111, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    The transient dies out as its energy is dissipated in the circuit's resistance. How much resistance is in your circuit? http://img803.imageshack.us/img803/4666/holly1756.gif [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Dec 24, 2012 #3

    gneill

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    Hi anandan111; Welcome to Physics Forums.

    You're circuit contains two so-called "reactive" components; a capacitor and an inductor. Both are energy storage devices which store energy differently, and can trade energy back and forth (oscillate). One stores energy in an electric field while the other stores it in a magnetic field.

    When power is first applied to your circuit it is equivalent to "hitting" the LC circuit with a step input (simulation starts and DC supply goes from 0V to 1V). This kicks off the oscillation that you're seeing. Since this is a simulation with ideal components (no resistive losses) the oscillations will continue indefinitely. In real life with real components with imperfect conductors, they would soon die away as energy is given up to Ohmic heating.

    The amplitude of the oscillation on the capacitor is 1V (yielding a 2V peak-to-peak measurement):

    Vcap = (cos(ωt) - 1) V

    assuming that the reference node (negative lead of your "meter") is placed at the junction of the C and L, thus the zero reference for the oscillation is shifted by a "DC bias" established by the "step" of the step function.

    In the first instant, when the +1V step is applied, the capacitor cannot change its potential instantly (initially uncharged, thus 0V) so this drives the potential at the L-C junction to +1V with respect to the battery negative lead. Subsequent energy trading with the inductor pushes the potential there from +1V through zero to -1V.
     
  5. Dec 24, 2012 #4
     
    Last edited by a moderator: May 6, 2017
  6. Dec 24, 2012 #5

    gneill

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    The turning on of the voltage supply sets a new equilibrium point of reference for the oscillation of the capacitor voltage that's 1V above ground (assumed to be the negative lead of the voltage source).

    To make a mechanical analogy, suppose you had a mass hanging on a spring, but had the ability to switch gravity off and on. Starting with gravity off, the mass sits at rest at the end of the spring which is at its un-stretched, natural length. There is no motion, and we set the zero reference for position where the mass is, and take any downward displacement to be positive (just a choice of reference point and axis direction).

    Now you suddenly switch on gravity. The new equilibrium position for the mass would be below the old one; it's where the mass would come to rest if there were damping forces involved. Gravity will pull the mass down through this new equilibrium position (where it will have maximum velocity), until it finally slows and then turns around at a distance below the new equilibrium point equal to the distance of the equilibrium point from the original rest length of the spring. The mass will continue to oscillate around the new equilibrium point, reaching only as high as the old equilibrium position (zero displacement); the displacement will never go negative.

    So the equation for the displacement becomes A*(1 - cos(ωt)) where A is the amplitude of the oscillation, equal to the displacement of the new equilibrium position to the old equilibrium position).

    The same thing is happening with the LC circuit. If there were a bit of resistance in the circuit, the capacitor would "come to rest" with a potential difference of 1V across it. That's its "new equilibrium position" around which it oscillates.
    Yes, it's a sort of filter. Replace the DC voltage source with a variable frequency signal generator and you will see a peak in current when the signal generator matches the resonance frequency. With no resistance in the circuit the signal generator will continue to feed energy to the oscillations, in phase, so the voltages and currents will grow without bound! Away from this "selected" frequency the response is not so dramatic.
     
  7. Dec 25, 2012 #6
    THank you very much for your explaination! At first I didn't get it...then what I did was referring to your analogy of the spring-mass system, I drew a pictorial view of it and interpreted your explaination line by line until I fully understood everything. Thank you. That is what I needed. =D
     
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