R-LC Parallel and series resonance

Discussion in 'General Electronics Chat' started by Rubberfrog, Dec 10, 2008.

  1. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
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    Hello guys,
    First off thanks for helping me with my last question-much appreciated!:D im curently having difficultly with a R-LC Parrallel and series resonance (L with winding R) question, having looked through previous threads and lessons im still not clear on the topic.Heres where i am so far;
    Series R-LC circuit resonates at 2251Hz
    Parallel R-LC circuit resonates at 2228Hz at 25Vrms I is found at 5mA
    Find vaules of components

    R=V/I
    R=5kΩ
    R=Ztotal
    I=V/R Branch IL =2.5mA Ic =2.5mA Itotal= 5mA
    Xc=V/Ic = 1kΩ
    At Resonance Xl=Xc
    ω=2∏f
    ω=14k rads
    Ic=ωCV
    C=7.14nF
    Xl=2∏fl
    L=14H

    Vaules of R L C have been found using the above(apolgises for the layout)

    When vaules subsited into F=1/2∏rootLC it isnt my resonance vaule which means my caculations must be out somwhere having worked through again im still non the wiser.
    Am im right in keeping Vs in rms vaule?
    Will Isupply be the same as the parrallel cct vaule?

    Any help would be greatly received!
    Many Thanks,
    :confused:Rubberfrog
     
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    Hello, Mr. Frog:

    Number one rule of electronics.....never make things more complicated than they are!

    Resistance doesn't figure into the resonant frequency at all....it just determines the Q (sharpness of resonance). A circuit containing only one inductor and one capacitor will have only one resonant frequency, whether series or parallel. When XL=Xc, the circuit is resonant. The only difference between series and parallel is that in series the impedance is lowest at resonance, and in parallel, the impedance is highest.

    There is ONE special case I only mention in passing...but it does not apply here. In a VERY LOW Q parallel circuit, there can be three different definitions of resonance...180 degree phase shift....maximum impedance....or equal reactance (xl=xc). But keep in mind this is a SPECIAL CASE. In 99% of all AC math problems you'll run across, resonance is resonance....Xl=Xc.

    Simple!

    eric
     
  3. davebee

    Well-Known Member

    Oct 22, 2008
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    The differential equations describing resonance show that resistance does slightly change resonant frequency.

    For small amounts of damping in a practical circuit it can usually be ignored, but you might need to take it into account if you're answering a question for a college course on vibration.
     
  4. KL7AJ

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    You bring up an interesting conundrum. Does damping actually change the resonant frequency (there is no difference between electrical an mechanical systems, by the way) or is the very DEFINITION of resonance changed with dampiing? With Fourier analysis, a pure sine wave assumes that its origin is infinitely far back in time (as well as forward in time). If a sine wave has ANY finite beginning, it is no longer a true sine wave....and hence has a finite bandwidth! And a RAPIDLY decaying damped sine wave has a rather undefined frequency...except at the zero crossing!

    Probably more info than Rubberfrog needs at this point, though. :)

    eric
     
  5. The Electrician

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    Oct 9, 2007
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    Rubberfrog says he was given this:

    "Series R-LC circuit resonates at 2251Hz
    Parallel R-LC circuit resonates at 2228Hz at 25Vrms I is found at 5mA
    Find values of components"

    This means that he is dealing with a low Q case, doesn't it?
     
  6. KL7AJ

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    Yes...except that he's already DEFINED his resonance as Xl=Xc...the one "type" of resonance in which R doesn't figure in...by definition.

    eric
     
  7. The Electrician

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    He defined resonance as Xl=Xc, but the problem statement didn't give any definition. I would say that part of what he must do to solve the problem is to determine what definition the problem uses.

    And if he can't get that information, what I would do is to solve the problem with all the definitions and see which ones will give the numbers from the problem for series and parallel circuits.
     
  8. KL7AJ

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    Excellent suggestion. Great minds run in the same gutter, eh? :D
     
  9. davebee

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    Oct 22, 2008
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    Rubberfrog, where did those frequencies come from?

    Is this a problem from a course in which you were taught formulas for how to calculate damped resonance? With different formulas for series versus parallel circuits?
     
  10. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
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    Thanks for your posts guys ,heres the question in full:

    "When placed in series a coil and capcitor resonate at a freq. of 2251Hz and when placed in parrallel they resonate at 2228Hz.At Parrallel resonance it is found that a current 5mA flows from the supply when the Vs is 25Vrms.Determine the vaule of the capcitor and the inductance and resistance of the coil."

    I think the question is trying to give an understanding that the resitance of the windings plays no part in series resonance but has an effect at parallel.
    To clarify -the resonace condtions of a resonance cct are: Xl=XC and Max Z?

    Are Ic and IL 180 out of phase at this condition? (Making the correct calcuation for It=Ic-Il rather than It=Ic+Il as i have used?!)
    Cheers
    Rubberfrog
     
  11. davebee

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    Oct 22, 2008
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    Wikipedia shows formulas for calculating the changed frequency due to damping; different formulas for series as for parallel circuits. A little complicated but not too bad. Look at damping factors and damped resonance at this link -

    http://en.wikipedia.org/wiki/RLC_circuit

    Given that your problem presents two different frequencies, and assuming that the wiki formulas are correct, then it seems that you could plug your numbers into the wikipedia formulas and calculate values of L and C that with 5K of resistance would produce those adjusted resonances.

    Wiki says:

    adjusted frequency = natural frequency * fudge factor

    series fudge factor : sqrt( 1 - ( R/2 * sqrt(C/L)^2))

    parallel fudge factor: sqrt( 1 - (1/(2R) * sqrt(L/C)^2))

    So you have:

    2251 = F * sqrt( 1 - ( R/2 * sqrt(C/L)^2))
    2228 = F * sqrt( 1 - (1/(2R) * sqrt(L/C)^2))

    where R is 5K,

    so if you find values of C, L and F that satisfy these, you're done.

    Ideally you'd solve these for C and L somehow and calculate your answer algebraically, but I'm not smart enough to tell you how to do that. But if all that was needed was a close enough answer, if it were me, I'd write a perl script that adjusts L and C values until the equations are satisfied to within some small error.

    Is this for a class? Have you been given formulas to solve a problem like this?

    I seem to remember, like eric also suggested, that when damping is involved the situation gets difficult to describe as a simple resonance because the frequency of 90 degree phase shift starts to deviate from the frequency of greatest amplitude.
     
  12. The Electrician

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    The wiki formula for parallel resonance is for 3 components, R, L and C, all in parallel, whereas rubberfrog says he has "...L with winding R...". In other words, I think his "parallel circuit" has L in series with R and that combination in parallel with C.
     
  13. KL7AJ

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    Yes....I caught that too. This shows the difference between "loaded Q" and "Unloaded Q". The resistance inside the closed loop (as in this case) sets the unloaded Q. A resistor in parallel with the tank will determine the loaded Q.

    In working with high powered R.F. transmitters, the efficiency is based on the ratio of unloaded Q to loaded Q. of the tank circuit. You want a high unloaded Q and a low unloaded Q for maximum efficiency (consistent with bandwidth). Just one of those real world applications of perhaps dubious interest here.

    Carry on! :)

    eric
     
  14. davebee

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    Oct 22, 2008
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    Good point! I missed that.

    But I would imagine that in either case, the resistance will damp the resonance.

    To get to rubberfrog's case, is there a formula to calculate its effect?
     
  15. The Electrician

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    The formulas aren't too hard to derive oneself. Notice however that the natural resonance frequency of both circuits is the same, because if nothing is connected externally to the "parallel" version of his circuit, the R, L and C form a loop. So, if the capacitor were initially charged, and then the circuit were left to oscillate on its own, with no driving stimulus, its natural frequency of oscillation would be the same as the series circuit, because if the ends of the series circuit were shorted to form a loop, it also consists of an R, L and C in series. I'm assuming both circuits will be driven by a voltage source (of zero internal impedance).

    Hence, the problem, with its two different resonance frequencies could not be solved if the definition of resonance were the natural resonance frequency. This leads me to believe that the definition of resonance the problem wants him to use is the case where the applied voltage and the resultant current are in phase. This occurs when the imaginary part of the impedance is zero. It's easy enough to derive the impedance of each circuit.

    The impedance of the "parallel" circuit is:

    j*w*L + R
    -----------------------------
    (j*w)^2 *L*C + j*w*R*C + 1

    Multiplying numerator and denominator by the conjugate of the denominator, we get for a numerator:

    -j * w^3 * L^2 * C + J*w*L - j*w*R^2*C

    Dividing out j*w and setting equal to zero, we get:

    w^2*L^2*C = L - R^2*C = 0, which can be solved for w:

    w = SQRT(1/(L*C) - R^2/L^2)

    This is the radian frequency (which the problem says is 2*Pi*2228) at which the impedance of the circuit has a zero imaginary part.

    Similarly, the frequency at which the series circuit has an imaginary part of zero can be found to be w = SQRT(1/(L*C)); this is at a radian frequency of 2*Pi*2251.

    This gives us two equations with 3 unknowns (R, L and C). We need another equation.

    The problem says that with 25 volts applied to the parallel circuit, a current of 5 mA is obtained. This means that the magnitude of the impedance of the parallel circuit is 5000 ohms at a radian frequency of 2*Pi*2228 (resonance).

    So, if we take the expression for the impedance of the parallel circuit:

    j*w*L + R
    -----------------------------
    (j*w)^2 *L*C + j*w*R*C + 1

    and get its magnitude by multiplying numerator and denominator each by its own conjugate and taking the square root, we can set this equal to 5000, with w equal to 2*Pi*2228, and this is our third equation.

    The three equations can be solved simultaneously. It worked out nicely for me, but since this looks like a homework problem to me, I'm not going to post the final answer until rubberfrog gives it a try.

    If he's still around and wants some more help, he should show his attempt at a solution.
     
    Last edited: Dec 12, 2008
  16. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
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    Cheers guys (im still around) thanks for you time and all your posts just been working through.so just to confirm (please tell me if im wrong!)

    At Cct is at resonance when Xl=XC
    Z=Infinate (Parallel)
    Z=0 (series)
    I is Phase with Vs

    The difference between the frequencys is because the windings Resistance damps resonance in the series cct whereas its not Affected in parallel?

    Thanks electrician for the formulas however i have not yet studied j numbers (but are aware of there exsistance).
    cheers all
    rubberfrog
     
  17. The Electrician

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    The Z of the parallel circuit will not be infinite, and the Z of the series circuit will not be zero. These values can only occur for ideal circuits with no resistance.

    The impedance of the series circuit can't be less than the series resistor as you can see by just looking at the circuit. The impedance (at resonance) of the parallel is somewhat more complicated to derive.

    What course are you taking? I'm surprised that they expect you to take advantage of the fact that the current into the parallel resonant circuit is 5 mA, without being able to calculate the impedance of that circuit. Or, at least, they should have given you a formula. And you really can't derive the formula for yourself without using complex arithmetic (the j operator).

    Can you ask your instructor what definition of resonance you should use?

    Using the natural resonance frequency of the two circuits won't work because they are the same, as I explained in another post.

    So, you're left with either the frequency of maximum impedance (for the parallel circuit; minimum for the series circuit), or the frequency where the current is in phase with the applied voltage. The latter would seem to be the appropriate one to me.
     
  18. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
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    Hello all,
    Having contacted my tutor ive been given the formula to solve for the component vaules however i need to know IC and IL in the parralel cct.None of the equations ive been given seem to give me this- i know the IC and IL cannot be equal due to the windings resistance in series with L.In the series cct I total hasnt be given and i have no vaules to calcuate Z without assuming Itotal is the same as the parallel cct. Once i find either Il or Ic (and knowing Itotal) im there.

    Thanks for your perseverance,
    Cheers
    M.wickett
     
  19. The Electrician

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    Do you still want help? If so, you should post the formulas and tell us what your tutor said was the resonance definition to use, and show your work so far.
     
  20. Rubberfrog

    Thread Starter New Member

    Nov 12, 2008
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    Heres my workings,sorry about the layout i hope viewing word .docs arent a problem.
     
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