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Resonance

Discussion in 'Electronic Basics' started by Deniz, Nov 20, 2004.

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  1. Deniz

    Deniz Guest

    -->------L--
    I | |
    C R
    | |
    = =

    Circuit seen above (series RL and a parallel C) is supposed to
    resonate somehow. There is a current source driving the circuit (since
    I thought that a voltage source driving the circuit would be useless).

    Overall impedance (Zo) is calculated as: 1/(R/(w^2*L^2 +
    R^2)+j*w*(R^2*C - L + w^2*L^2*C)/(w*L^2 + R^2))

    Now in order to maximize abs(Zo) we let imaginary part to be 0, by
    letting wo = [1/LC - (R/L)^2]^0.5 (here i assume that abs(Zo) is
    maximum when imaginary part is 0). Since Zo is maximized, I*Zo
    (voltage across the capacitor (Vc)) will reach the maximum, and at wo
    the circuit is said to be resonating?

    My first question is: At resonance frequency wo, can we immediately
    (without calculating time dependent expressions) say that when the
    stored energy in the capacitor reaches maximum, the stored energy in
    the inductor becomes 0 ?

    2)If we want voltage across R to reach its maximum, we calculate
    complex expression for Vr and if we let Vr's imaginary part to be 0,
    we come up with a new resonance frequency w1 = (-R/L)^0.5, which is
    meaningless. So can we immediately say or predict that voltage accross
    R will reach its maximum when we set w = wo = [1/LC - (R/L)^2]^0.5
    (which will reveal that finding resonance frequency has nothing to do
    with letting the imaginary parts equal to 0)? If wo is making both the
    voltage across Zo and R maximum, what is the reason for this?

    (calculating the frequency which will make abs(Vr) maximum seemed to
    be impossible, so i made a prediction)

    3) Is there a series RLC equivalence of this above circuit (for Vr or
    Vc)? If there is no such equivalence, how is parallel RLC equivalent
    of this circuit calculated (for Vc and Vr) ?
     
  2. CBarn24050

    CBarn24050 Guest

    Subject: Resonance
    NO, the enrgy in both components are equal but opposite.

    Maximum voltage accros R is at DC.

    Maximun impedence for a parallel tuned circuit is at resonance.

    Yes, but here the minimum impedence occurs at resonance.
     
  3. Spajky

    Spajky Guest

    could be driven also with voltage IMHO & rappresents a impedance step
    down transformation @ resonant frequency; R is just a load ...
     
  4. The Phantom

    The Phantom Guest

    What does it mean for the capacitor and inductor to have opposite
    energies?
     
  5. At any given time, they don't have 'opposite' energy, because energy is
    a scalar value.

    However, the energy that they have is passed back and forth between the
    two elements; thus, when the capacitor has maximal energy, the inductor
    has minimal energy.

    You could say that the energy for each is a sinusoidal wave above the x
    axis, 180' out of phase, and that the sum of the two sine waves is equal
    to the total energy in the system. That energy can be increasing if
    there is an impulse, decreasing if the oscillation is damped, or
    'constant' if the damping and impulse balance out.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  6. The Phantom

    The Phantom Guest

    CBarn24050 says they do. My question was directed at him; I was
    hoping he would explain about these 'opposite' energies.

    He also denies that the energy in the cap is max when the energy in
    the inductor is 0. I was hoping he could give some details showing
    just how this comes about.

    , because energy is
     
  7. Steve Evans

    Steve Evans Guest

    It is a fact that in a parallel resonant circuit, the impedance is at
    a maximum. Is this due to the fact that - at resonance - the energy
    flows between the cap and coil are so large as to be able to repell
    any current from the external energy source, thereby rendering its
    path effectively blocked?
     
  8. CBarn24050

    CBarn24050 Guest

    Subject: Re: Resonance
    Perhaps not the best choice of words, i'll try again. These components are
    reactive so there is no energy as such, ie it's not real energy. It does not go
    from 1 to the other and back again. When I said that they equal but opposite,
    what I meant was that they allways add up to zero at any time in the cycle.
     
  9. Steve Evans

    Steve Evans Guest

    Zero over an aeveage *whole* cycle, I think you mean.
     
  10. I dont' want to put you on the spot, but I think there is energy stored
    in resonant systems. Look at the film of the tacoma narrows bridge being
    torn apart due to resonant oscillations. Each little nudge from the wind
    stores more energy in the resonant system, until it collapses.

    And, in electrical systems, the energy does go from the inductor to the
    capacitor and back again. At any point, the energy stored in a capacitor is

    Uc = 1/2 C * V^2

    the energy stored in an inductor is

    Ul = 1/2 L * I^2

    Thus, the energy is maximum in the inductor when the current in the
    system is maximum, and the energy in the capacitor is maximum when the
    voltage across it is maximum. However,

    Uc + Ul = U

    which is the total energy at a given time, which is what I think you
    were trying to say.

    This is exactly analogous to a mass and spring system in basic physics,
    where the energy goes from kinetic energy to potential energy and back
    again.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
  11. The Phantom

    The Phantom Guest

    When the voltage across a capacitor is greater than zero, is there
    not an energy of .5*C*V^2 associated with that capacitor? And
    likewise given a current greater than zero in an inductor, is there
    not an energy of .5*L*I^2 stored in the inductor? Are these not
    *real* energies?

    I notice on some of your other posts that you seem to be familiar
    with circuit simulators such as Spice. If you simulate the circuit in
    the OP's post (I chose C=.01uF, L=25mH and R=1 ohm, for a resonance
    freq of 10.066 KHz, and a current source of 1 amp at that freq), and
    plot the current in the C and L for about 10 milliseconds, you will
    see that as the currents in C and L increase, they stay 180 degrees
    out of phase with each other. If you plot .5*C*v^2 and .5*L*i^2 (the
    instantaneous power in C and in L), you will see that these quantities
    are double frequency sinusoids with a DC component. Let the current
    source drive be applied for about 8 milliseconds, then reduced to
    zero; plot the *sum* of those instantaneous powers. You will see that
    after the 8 milliseconds, the result is very nearly a constant (if you
    make R=0 so the Q is infinite, then the sum of the two instantaneous
    powers will be dead constant after the current source drive is reduced
    to zero). If the current source drive stays on with R=1, it
    superimposes a little ripple on the sum of the powers as the stored
    energy in C and L ramps up.

    (In the low-Q case, where for example, I made R=10k, the sum of the
    instantaneous powers is not constant. But that is not the interesting
    case.)

    So, in fact, (in the infinite-Q case, where R=0) after the current
    source drive is turned off, the capacitor and inductor continue to
    exchange energy indefinitely. The voltage and current in each are
    sinusoids, and the sum of the instantaneous powers is constant. The
    law of Conservation of Energy requires this, because without loss
    components in the circuit, there is no mechanism for energy to be
    lost. The energy is sloshed back and forth between the capacitor and
    inductor forever (in the simulator, at least, where it is possible to
    have a circuit with R=0, including the wiring and parasitics).

    It does not go
     
  12. The Phantom

    The Phantom Guest

    Steve, I've been watching your postings as you strive to understand
    a somewhat difficult subject--AC circuit theory. I can see that you
    are struggling with it, but I admire your perseverance! Keep at it,
    and you'll eventually get it. I'll add my input to the help others
    have been giving you.

    I think it's helpful to realize that what are called two-terminal
    circuit elements (R, L, and C are the fundamental components)
    *enforce* a relationship between voltage and current. The voltage
    *across* a component and the current *through* it are not independent.
    It is what two-terminal components do; they establish a relationship
    between voltage and current, *for that component only*.

    So, if some two-terminal circuit elements are in series, the
    current in each of them *must* be identical. If they are in parallel,
    the voltage seen (applied across) by each of them *must* be identical.

    Thus, if they are in series, the currents must be the same and
    *only* the voltages across each can be different. If they are in
    parallel, only the *currents* in each can be different; the voltage
    seen by each is the same.

    For components in series, since the current in all of them is the
    same, it makes sense to use current as a reference, and speak of the
    phase of the voltages *across* (not to ground) each component with
    respect to the current through all of them.

    For components in parallel, it is appropriate to use the voltage
    *across* them as the reference, and speak of the phase of the current
    in each with respect to the voltage across all of them.

    Now, since for a C and L in parallel the voltage across the two
    components is the same, only the phase of the currents can differ.
    The current in one is 180 degrees out of phase with the other, and
    when those currents are added by the parallel connection, they tend to
    cancel. If the magnitude of the currents is identical, which is what
    happens at a frequency such that the reactance of each is the same
    (this is resonance), then we get complete cancellation of the currents
    (for ideal L and C). Thus the current into the parallel combination
    of the L and C is zero, even though we have applied some non-zero
    voltage to the two of them. When we have a circuit that has the
    property that no current (or very little) is produced with a finite
    applied voltage, we say that the impedance of that circuit is high.
    It's not that the parallel combination of L and C at resonance repel
    the applied voltage. In fact, a current does exist in both the L and
    C, but the two currents are 180 out of phase, and completely add to
    zero at the connection of the L and C.

    The same thing happens in a series resonant L and C circuit, but
    with the roles of current and voltage reversed. For a given current
    through the L and C, if the applied current is at a frequency where
    the reactance of the L and C is the same, then the magnitude of the
    voltage *across* the inductor and *across* the capacitor is the same
    and since these voltages are 180 degrees out of phase (for ideal
    components), they completely cancel (add to zero). Remember that the
    current in the L and in the C is the same, since they are in series.
    Thus we have a circuit with (almost) no voltage across it even though
    a current is passed through it. We say that such a circuit has a low
    impedance. This circuit doesn't repel the applied current; it's just
    that the voltage *across* one component cancels the voltage *across*
    the other, giving a resultant of zero *across* the series combination.

    I hope this helps.
     
  13. CBarn24050

    CBarn24050 Guest

    Subject: Resonance
    Just because the energy is not real doesn't mean that it does not exist, it's
    just the confusing terms (real & imaginary) that apply to complex (but not
    complicated) maths. You would think that they would have come up with something
    a bit better by now.

    What I should have said was that there is no CHANGE in the total energy level
    during the cycle.
     
  14. Steve Evans

    Steve Evans Guest

    tnx, phantom! Thats the clearest explanation i've come across so far.
    there are still a couple of outstandingpoints i need to clear up....
    I'll get back in a shrot while!

    Steve
     
  15. Steve Evans

    Steve Evans Guest

    Yeah, but5 the energy has to come at the *right* moment each time.
    it's a bit like pushing a child on a swing. you have to impart the
    force at the rpecise time in each arc to get the swing moving with
    minimal effort. i think the equivalent term in elctronics is the
    'flywheel effect'; get the timing spot-on and you can keep the
    oscillation moving with minimal energy input/maximal efficitency.
     
  16. The Phantom

    The Phantom Guest

    I think the problem here is that you are trying to analyze the
    problem with the *phasor* representation of the voltages on the L and
    C. But the question the OP asked about stored energy cannot be
    appropriately dealt with from that point of view. You must analyze
    the instantaneous time response of the circuit.
    This is better than saying they add up to zero, because it allows
    for the possibility that the sum is non-zero. But, as I explained in
    my long post, it is only when the Q of the circuit is infinite (and
    there is no drive from the current source) that there is no change in
    the total energy level during one cycle. If the current source drive
    is in operation, or if R is non-zero, then the total energy level
    varies somewhat during a cycle. This cannot be determined from a
    phasor point of view, however.
     
  17. Rich Grise

    Rich Grise Guest

    They're 90 degrees out of phase, because the capacitor stores energy as
    charge, and the inductor stores energy as current. The current hits its
    maximum as charge is going through zero, and vice-versa.

    But yes, it's the same energy, just being sloshed back and forth. Think of
    a spring and a weight. The spring's compression is the voltage, its
    spring constant is the capacitance, the weight is inductance, and its
    movement is current.

    And somebody said something about a series resonant circuit and a parallel
    resonant circuit - if you just look at the inner loop, a parallel resonant
    circuit is just a series resonant circuit with its ends tied together. So
    the difference really depends on your point of view. :)

    Hope This Helps!
    Rich
     
  18. Rich Grise

    Rich Grise Guest

    It's definitely real - it's just that it stays in the resonant circuit,
    rather than doing work or being thrown away as heat, which is the part
    that you pay for. :) This is why power companies like power factor
    correction.
    Yeah. :)

    Cheers!
    Rich
     
  19. Rich Grise

    Rich Grise Guest

    They've determined that the bridge wasn't resonating - it just got blown
    down:

    http://www.math.umbc.edu/~gobbert/teaching/math101.s2003/reports/Group1Tacoma.doc

    Cheers!
    Rich
     
  20. The Phantom

    The Phantom Guest

    If the L and C are all there is, and you don't make any connection
    to the pair, then the series/parallel distinction doesn't exist. See
    E. A. Guillemin's classic book on circuit theory. He talked about
    what he called a "soldering iron entry" or a "pliers entry" into a
    (sub) circuit. The L and C combination looks different to the
    observer depending on whether you open the loop (pliers entry) and
    make the connection, or leave the loop intact and tack on in parallel
    (soldering iron entry).
     
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