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Resonance in parallel RL-RC circuit

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data

    calculate R2, so the resonance is still posible ?
    C = 40 μF
    R1 = 30 ohm
    L = 9mH

    image : http://postimg.org/image/kampvczv5/



    2. Relevant equations
    ω = 1/sqroot(L*C)
    c = 1/(ω*C)
    XL = ω * L
    Z = (Z1*Z2)/(Z1+Z2)


    3. The attempt at a solution
    My approach was to first calculate the resonance frequency ω = 1/sqroot(L*C) = 1667 s-1 than
    XL = ω * L ; Xc = 1/(ω*C) == 15ohm, since they have to be the same to cancel out if I'm correct.
    And at last got the R2 = 30ohm using the equation Z = (Z1*Z2)/(Z1+Z2) but the right answer should be R2 > 15ohm, so I don't know how to tackle this problem and I'm asking you guys for some advice/solution.

    Thanks in advance !
     
  2. jcsd
  3. May 3, 2016 #2

    BvU

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    Hello Kamen, :welcome:

    Do us a favor and don't post pictures on sites where one gets smothered with pictures of hookers by russian criminals :smile:
    Drawing2.jpg
    Did you render the complete problem statement ?

    Did you already learn to work with complex impedances ? It seems to me that's required here....

    What is the criterion for resonance ? How do you find the resonance frequency when there are resistors present ?
     
  4. May 3, 2016 #3

    LvW

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    No - you cannot (blindly) assume that the resonant frequency is the same as for a lossless tank circuit.
    Yes - that is the approach pointing into the right direction:
    In words: At resonance, both imaginary parts of the resulting impedance (or admittance) must cancel each other.
    That means: You are looking for the frequency at which the resulting input impedance of the circuit is REAL.
    Therefore: Find the expression for the total admittance Y(jw) at the input (this is simpler than finding the impedance Z(s)=1/Y(s) ) and split it into real and imag. parts.
    Finally, set Im(Y)=0 and solve for w.
    This expression gives you the resonant frequency as a function of all 4 elements.
     
  5. May 3, 2016 #4
    Thank you BvU for the welcome and sorry for the hooker ads, didn't know how to post a picture :)
    Thanks for the advice "LvW" I used the admittance this time and set the im(Y) = 0 just like you said and I solved the problem, I will post the picture of the problem solved for anyone looking for the solution.

    Anyone looking for the basic equations used in this problem : https://en.wikipedia.org/wiki/Admittance
    HQ picture of solution : https://mega.nz/#!tJ0mXKII!qWFYuqfMc8Bsr1RLLWfeaWbTHXtF_LeVgjp77pMnUE4
     

    Attached Files:

  6. May 3, 2016 #5

    BvU

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    Bravo ! :smile:
     
  7. May 3, 2016 #6
    I suggest you sanity check your result before you hand in your answer.

    Given these values:
    C = 40 μF
    R1 = 30 ohm
    L = 9mH

    What do you calculate for a frequency of resonance? What is the value of the imaginary part of the admittance at that frequency?
     
  8. May 4, 2016 #7
    Sorry, but forgot to mention the results in my previous post, here are my "findings":

    After I've gotten the result that R2 > 15

    1) R2 = 10 ohm
    i took that R2 = 10ohm than inserted that into the equation for
    angular frequency (w), and found out I get negative number in the
    denominator and positive in numerator which gives me a negative number under a square root function.
    1) R2 = 10 000 ohm
    R2 = 10 000 ----> angular frequency (w) equation
    got angular frequency (w) = 790,57s-1
    and than inserting that into equation for imaginary part of the admittance I got a value of around
    0 (approximation error)
     
  9. May 4, 2016 #8

    LvW

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    Here are my results:
    After setting Im(Y)=0 we get an expression for the frequency w.
    This expression is a function consisting of a numerator N and a denominator D.
    Both must be positive for a positive frequency (N>0, D>0).
    From this we get:
    a) (R1)²>L/C=225 (V/A)² and
    b) (R2)²>L/C=225 (V/A)².

    R1=30 ohms can fullfill equation a) and for R2 we find R2>SQRT(225)=15 ohms.
     
  10. May 4, 2016 #9
    I derive an expression for the impedance of the circuit like this:

    ParResZ1.png
    Now, for a sanity check, I plot over a wide frequency range my expression for the imaginary part of Z (blue) and Kamen12's expression for the imaginary part of the admittance (red) for values of R1=5 and R2 = 5. I see plainly that the imaginary parts do become zero at a certain radian frequency:

    ParResZ2.png

    Now, let R1 become 30 ohms and let R2 remain at 5 ohms. A plot shows that the imaginary part is never zero, even though R2 is only 5 ohms. When R2 is 15 ohms, the situation is even worse; the imaginary part is never zero, so we have no resonance:

    ParResZ3.png

    But, if for example, we make C = 4 uF rather than 40 uF, with R1=30 and R2=15 we do get a resonance:

    ParResZ4.png

    It appears that with the given values for R1, L and C, there is no value of R2 which will give a resonance.
     
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