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Inductance and V=Lxdi/dt?

  1. Sep 13, 2011 #1
    When an inductor is oscillated at it's resonant frequency it becomes a parallel LC circuit (Tank circuit). At this frequency the circuit has a theoretical infinite impedance.

    However, due to the resistance of the inductor energy is dissipated. The power supply provides a small current (equal to the energy dissipated in the tank circuit) to try and keep the tank circuit charged.

    If the circuit were suddenly switched off the parallel tank circuit would continue to oscillate (between L and C) for a time until all the energy was dissipated. But, wouldn't the inductor also resist the sudden change in current from the power supply and induce a voltage across it equal to V=Lx di/dt?
     
  2. jcsd
  3. Sep 13, 2011 #2

    vk6kro

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    A perfect inductor does not have a resonant frequency, but real ones have capacitance between the windings which does cause a type of resonance.

    If you apply a sinewave at this resonant frequency, the inductor will draw minimum current at that frequency. If you remove the sinewave, then the current will drop and the exact waveform across the inductor will depend on where in the sinewave cycle you removed the input.

    If you apply a DC voltage to the inductor and then remove it, the inductor will generate a large pulse of the same polarity as the DC voltage was before it was removed. There will then be a series of oscillating pulses at the resonant frequency of the inductor and its stray capacitance. These decrease in amplitude until they stop. This is called "ringing".
     
  4. Sep 13, 2011 #3
    So if it was pulsed dc (square wave) at the resonant frequency would the pulsing and ringing combine and cause amplitude modulation across the inductor?
     
  5. Sep 14, 2011 #4

    vk6kro

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    No, it doesn't look much like amplitude modulation.

    It looks like this:

    images?q=tbn:ANd9GcRwD3zmdx_xbF3kU1siadtZxwXDj3T6S5jxOHgm9JCU4VnjsthVQ3VC545W.jpg

    This is for an actual square wave, though. A pulsing DC square wave has most of the ringing on the falling edge where voltage is removed from the inductor. Peak voltages of hundreds of volts can be produced.
     
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