Resonant Frequency

Discussion in 'Homework Help' started by PsySc0rpi0n, Jun 9, 2014.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hi once more...

    I'm starting to solve a few problems about parallel and series RLC circuits to find the Resonant frequency...

    For instance, the first problem I have is a series RLC circuit with R=5Ω, L=20mH, C=unknown and f=100Hz...

    I need to find C so that my circuit meets the Resonant Frequency condition which is when It's current is max!

    In this kind of circuit I need to meet the Resonant condition which I think it's Xc=Xl <=> 1/W.C = W.L<=>C=1/(W^2*L)<=> C=1/(4*∏^2*100^2*0.02)<=>C=3.97x10^-4C.

    How can I use LTSpice to check if my math is correct?



    Edited;
    If I "Add Trace" with I(L1)+I(C1) and get a straight line at zero Amps, means the math is correct?
     
    Last edited: Jun 9, 2014
  2. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Looks like not!:(

    How can I do it?
     
  3. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi.
    Have you considered using LTSpice AC analysis of the circuit.??
    Your answer for Cap is wrong.
    E
     
    Last edited: Jun 9, 2014
  4. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    How do I do that? What do I measure and at what points? Linear, Decade, etc?? What setup should I use?


    Edited;
    I tried to run a Decade AC analysis, starting from 1Hz to 100Hz with 20 points per Decade but I don't know what and where to measure or hoe to interpret the plots!
     
    Last edited: Jun 9, 2014
  5. ericgibbs

    AAC Fanatic!

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    hi,
    As LTSpice simulation is not part of the college question I will post how I would check the answer.

    You answer of 397uF is wrong,so rework the calc's, your premise of Xc = XL at resonance is correct, your maths are wrong:rolleyes:
    E
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I got the math right!

    So, as far as I can understand, to check if math is correct, that inverted peak at the given frequency means the the C cal is correct. If I change the Cap, that inverted peak goes off 100Hz, right?
     
  7. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Keep in mind that this is a series circuit so the current is everywhere the same. It's confusing that you show the sum of the current in two series elements leading to a zero value.
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Forget that... I realised immediately that that was obviously zero and was not what I was looking for!
     
  9. ericgibbs

    AAC Fanatic!

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    Give it a try using this asc file.
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    I tried it before and noticed that. If i change the Cap value, that inverted peak goes away from 100Hz!
     
  11. ericgibbs

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    hi,
    Thats what should happen, the resonant frequency of the R,L,C will change.

    What value of Caps do you think is required to half the Freq to to 50Hz and double the Freq to 200Hz.??

    E
     
  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    As the fraction denominator increases when we increase the frequency and vice-versa, the Cap value will be the other way around. It will decrease if frequency increase and vice-versa. The capacitor will be 506.61μF for 50Hz and 31.66μF for 200Hz if math is correct!
     
  13. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Another question.

    For another series RLC, the problem is asking us to calculate the resonant frequêncy and the Low limit and high limit of that frequency.

    Those 2 limits are (resonant frenquancy - 0.707*resonant frequency) and (resonant frenquancy + 0.707*resonant frequency). Is this correct?
     
  14. ericgibbs

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    hi,
    You are asked for the -3dB limits of the frequency bandwidth at 0.707 [ 70.7%] of the maximum amplitude.
    So measure the upper and lower freq when the amplitude is 0.707 of its max.

    E
     
  15. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    0.707 of max current, right? Or of max frequency?

    Edited;
    The Resonant Freq is 35.58Hz. I got that.

    Now, the lower limit is 35.58-35.58*0.707 and the upper limit 35.58+35.58*0.707??? Or should I calculate max current at Resonant Freq, then calculate 0.707 of max current and then calculate impedance at this current and then frequency at this impedance????
     
    Last edited: Jun 10, 2014
  16. ericgibbs

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    I thought you had been given the Freq resonance as 100Hz in the college question, so where has the 35.58Hz come from.?

    This is what I posted.
     
    Last edited: Jun 10, 2014
  17. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Sorry if I confused you...

    At my post #13 I said I was at another question.

    This time the RLC circuit has the folowing info from the problem:

    R=100Ω
    L = 0.5H
    C=40μF

    The problem asks us to calculate the Resonant Frequency, the lower and upper limit Frquency for -3dB and the Bandwith.

    So, for the Resonant Frequency I got the value of 35.58Hz.

    Now I need the lower and upper limit frequency at -3dB.

    I have the resolution done by the teacher but I can't understand it.


    The teacher resolution says:

    for W1 - lower limit angular frequency:

    I rms = 0.707*I max

    If the circuit is at the Resonant point we can say that Z = R = 100Ω and that Z' = R - jX' because at lower limit, the circuit is capacitive!

    Then we have:

    Z' = 100*(1/0.707) = 141|θ Ω (I don't understand this step)

    Then we calculate θ by:
    cos θ = 100/141 <=> θ = -45º

    So Z' = 141|45º Ω

    Finally for W1 we have:

    As at this point, the capacitor has more influence in the circuit, then:

    Xc-Xl = R

    which leads to:

    1/W1C - W1L = R

    W1 = 145rad/s.

    Doing the same to the upper limit we get

    Xl - Xc = R

    which leads to:

    W2 = 345rad/s
     
  18. ericgibbs

    AAC Fanatic!

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    Its where the circuit impedance has changed to 141R from the resonant 100R value.
    Note that 141R *0 .707 = 100R,, 0.707 the -3dB amplitude level.

    I have broken the circuit down in this Sim image.
    You can see that Xind increases with higher frequencies and Xcap increases with lower frequencies, where Xind = Xcap is the resonant Freq.
     
  19. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I think I'm clarified now!

    Another problem:

    Series RLC circuit again.
    R = 50Ω
    L = 0.7H
    Vin = 220V at 50Hz.

    a) Calculate Z and I for the folwoing Cap values:

    a1) C=2μF
    a2) C=12μF
    a3) C=20μF

    To do this one I've calculated Xc for each Cap value and Xl.
    Then I did Ztotal = R + Xc + Xl
    Then, assumed Vin angle is 0º because nothing about it is said and calulated the circuit current.

    I got the following results:

    a1)
    Xc = -1591.55jΩ
    Xl = 219.91jΩ
    Ztotal = 50 - 1591.55jΩ + 219.91jΩ = 50-1371.64jΩ
    I = Vin/Z = 0.1604|87.91A
     
    Last edited: Jun 10, 2014
  20. ericgibbs

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    Show me your maths for the Xind impedance,,, how do you make it 157R.?
    E
     
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