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Resonance-overvoltage & undervoltage

  1. Nov 5, 2011 #1
    So I was playing in NI, with RLC circuits, trying to understand resonance, and what causes it etc.

    What I did is following:

    I connected an RLC circuit to a function generator, where I can change my frequency.

    What I found, that at some frequencies capacitor voltage was above the voltage source's voltage.

    I couldn't explain that at that moment but then I realised I saw this somewhere.

    [PLAIN]http://pokit.org/get/3f3d916aa8f70d541b889efc3736b968.jpg [Broken]

    This overvoltage and undervoltage as I call it, I do not understand. Why does this happen at specific frequency both for capacitor and inductor? I cannot find the google solution for this as I don't know the name of this phenomena in English.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 6, 2011 #2
    When you send a sine wave through a resistor, the current and voltage are in phase. To get the resistance of resistors in series you simply add their resistance values. In an inductor the current lags the voltage and in a capacitor the current leads the voltage. So to find the reactance of a series LC circuit you add their reactances just like you added resistances for resistors, but the impedance of the L and C are imaginary rather than real (their vectors are orthogonal to resistance). So "resistance" means the real component of impedance and "reactance" means the imaginary (orthogonal) component, at a given frequency. Look at the wikipedia article on phasors. The impedance of an inductor is in the positive imaginary direction and the impedance of a capacitor is in the negative imaginary direction. In a series RLC ckt you add the impedances. If the reactance magnitudes of the L and C are equal, then their sum is zero (because one is in the positive imaginary direction and the other is in the negative imaginary direction) so you only have the real resistance left. If the resistance is fairly low you can have a lot of current. If a lot of current goes through an inductor or capacitor, then there will be a lot of voltage across it.
  4. Nov 6, 2011 #3
    You basically explained to me the voltage resonance, which I understand.

    Voltage resonance will occur at ω0. But still that doesn't explain those spikes at ω1 for capacitor and ω2 for inductor. I understand that at ω0, a lot of current will flow through the circuit and, phasors of L and C will cancel out.
  5. Nov 6, 2011 #4
    Because the reactive voltage VC,L = I * XC,L is the product of two terms that both are functions of frequency. What is the current, I, in the circuit?
  6. Nov 6, 2011 #5
    ooh. That would be the solution of the differential equation. I'll get back to you when I solve it.
  7. Nov 7, 2011 #6
    I guess you could solve the diff equation, but it's not necessary for this problem. If you're learning this because you want to build circuits as an EE, you're (imo) better off here if you stick with Ohm's Law, V = I * ZT, where ZT is complex and represents the total impedance of the circuit. Find ZT and you have your current.
  8. Nov 7, 2011 #7
    Makes sense. Thank you.
  9. Nov 7, 2011 #8

    jim hardy

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    Gold Member

    let's peak briefly of a series R L C circuit. Keeps it simple.
    at the frequency where Xl= Xc the impeddance of the circuit is purely resistive.
    that's because ohms in series add up and it's a vector addition.
    so the Xl and Xc add to zero because they're vectors pointing in opposite directions.

    so current through the circuit is i= volts/ohms and there's only those resistive ohms left (because the reactive ohms cancelled one another out)

    i = volts/R and that same i flows through both X's, Xl and Xc .

    Since X migh be numericlly larger than R,

    volts across X may be larger than R in fact even larger than applied voltage.

    The ratio X/R is called the "Q" of the circuit and is also the ratio of energy stored to energy dissipated per cycle.

    thinnk on that for a minute, it leads to interesting conclusions.
    a series R L C circuit has voltage gain proportional to its Q.

    there's many mechanical analogies and the equations are identical to those of harmonic motion.

    a parallel resonant circuit has of course no voltage gain it's parallel but it has current gain.

    These voltage and current gains are real and so resonsnces must be watched for and dealt with - just as in bridge design.

    i hope this simple word picture helps.

    old jim
  10. Nov 7, 2011 #9
    yes yes yes I get it!! Didn't look at that way, regarding values of Xl Xc. Thank you very much mr. Jim !
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