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RLC circuit

  1. Nov 15, 2015 #1
    1. The problem statement, all variables and given/known data
    See the picture. If at electrical resonance ##U_c=60 V## and ##U_L=40 V##, find the ratio of reactances ##\frac{X_L}{X_c}##.


    2. Relevant equations ##Z=\sqrt{R^2+(\omega L - \frac{1}{\omega C})^2}## , ## Y= \sqrt{\frac{1}{R^2} + (\omega C - \frac{1}{\omega L})^2}## , ##\phi = \arctan \frac{\omega L - \frac{1}{\omega C}}{R}## and so on


    3. The attempt at a solution
    I am not sure how exactly use given conditions, I somehow got that if there is electrical resonance , than
    ##R^2C=R^2C^2\omega^2L+L## but that's surely wrong, and not sure how to use the voltages given.
     

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  3. Nov 15, 2015 #2

    BvU

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    "I somehow got" is not very strong. What does your textbook or your notes say ?
     
  4. Nov 15, 2015 #3

    ehild

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    These equations are not relevant for the circuit in the problem.

    Your equation is correct.
    What is the current at resonance? What are the expressions for UL and Uc at resonance?
     
  5. Nov 16, 2015 #4
    Maybe not these in particular, but little modified,I used them to get ## R^2C=R^2C^2\omega^2 L + L ##. I got that in two different ways (then I thought that it might be correct).
    I first calculate ## Z_{RC}## (parallel) with ##Y_{RC}## and since I know ##\tan \phi##, i can calculate ## \sin \phi=\frac{\tan \phi}{\sqrt{\tan^2 \phi+1}}##, and since imaginary part of ##Z_{RLC}## has to be ##0##, then ##Z_{RC} \sin \phi = L \omega## . The other way was to calculate ##Z_{RC}= \frac{-jRC \omega}{R-jC\omega}## and so on. Is there maybe easier way to derive this formula?


    I am not sure, this isn't only serial or parallel circuit. Maybe
    ## U_L=I_{ef} \cdot X_{L} ##
    ## U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C}## ?
     
  6. Nov 16, 2015 #5

    ehild

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    I do not think there is an easier way. I calculated the imaginary part of ZRC and added to ωL and made the sum equal to zero. But your formula for ZRC is not correct. Remember that Xc=-j/(ωC).
    Knowing the resonance frequency, you can determine the impedance at resonance, and the current in terms of the generator voltage.



    It is all right. Use the result for the resonant frequency and current.
     
  7. Nov 16, 2015 #6
    Yes, you are right, I meant ##Z_{RC}=\frac{-jRX_{C}}{R-jX_{C}}##.
    I got that ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and from voltages that

    ## \frac{U_{L}}{U_{C}}=\frac{2}{3} = \frac{X_L(X_C+R)}{RX_C} ##. Don't know what to do next.
     
  8. Nov 16, 2015 #7

    ehild

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    You need to eliminate R somehow. Use the expression of the resonant frequency.
    At resonance, Z is real. Write up the complex impedance and take the real part. It becomes quite simple.
     
  9. Nov 16, 2015 #8
    Well,I did got ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}+j\frac{R^2X_L-R^2X_C+X_LX_C^2}{R^2+X_{C}^2} \Rightarrow Z=\frac{RX_{C}^2}{R^2+X_{C}^2} ##
    Now, is omega ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
    When you say resonant frequency do you mean ## f=\frac{\omega}{2\pi}## ?
    And if yes, how should I use that expression?
     
  10. Nov 16, 2015 #9

    ehild

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    Correct. Substitute the expression for Xc.
    I mean the angular frequency, ω, when I say frequency. Use the expression for ω to get the impedance at resonance.
     
  11. Nov 16, 2015 #10
    I am probably missing something obvious here,but how do I get impedance with ##\omega##?

    From those voltage equations I got that ##R=\frac{3X_LX_C}{2X_C-3X_L}## and I tried to use that in ##R^2X_L-R^2X_C+X_LX_C^2=0## but didn't lead me nowhere.
     
    Last edited: Nov 16, 2015
  12. Nov 16, 2015 #11

    ehild

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    How is XC and XL defined?
     
  13. Nov 16, 2015 #12
    Is ##\omega= \frac{1}{RC} \sqrt{\frac{R^2C}{L}-1} ##?
    I have impedance ##Z=\frac{RX_{C}^2}{R^2+X_{C}^2}## and (if thats useful at all) ##R=\frac{3X_LX_C}{2X_C-3X_L}##.
    ##X_L = L \omega## and ## X_C = \frac{1}{C \omega}##.
    I don't know how to find impedance with ##\omega##.
     
  14. Nov 16, 2015 #13

    ehild

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    yes.
    XC=-1/(ωC). Sustitute the expressions for XL and XC into the equation for Z. Then substitute the expression of ω.
     
  15. Nov 17, 2015 #14
    Okay I got ##Z=\frac{L}{RC}## . How do I use that ?
    Since ## \frac{X_L}{X_C}= LC\omega ^2 = 1-\frac{L}{R^2C} = 1-\frac{Z}{R}## , does it have to do something with this?
     
  16. Nov 17, 2015 #15

    ehild

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    You know Z . What is the generator current? With the current, you get an equation for UL and UC.
     
  17. Nov 18, 2015 #16
    You mean
    ##U_L=I_{ef} \cdot X_{L} ##
    ##U_C=I_{ef} \cdot \frac{R}{X_C+R} \cdot X_{C} ##

    where ## I_{ef}= \frac{U_{ef}}{Z} ##?
     
  18. Nov 18, 2015 #17

    ehild

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    You know that the effective voltages are UL=IeffωL=40 V and UC=60 V.
    But you get UC if you multiply Ieff with the magnitude of the impedance ZRC which is not RXc/(R+Xc).
    Write the equations for UC and UL in terms of C, L, R and Ueff.
    You have to get a numercal value for XC/XL.
     
  19. Nov 20, 2015 #18

    ehild

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    You can not get the complex impedance of the parallel resistor and capacitor correctly. The impedance of the capacitor is ##\frac{1}{j\omega C} ##, so ##Z_{RC}=\frac{1}{jωC+\frac{1}{R}}=\frac{R}{1+jωCR}##.
    Its magnitude is ##|Z_{RC}|=\frac{R}{\sqrt{1+(ωCR)^2}}##
    In #Post 8, you got that the impedance is
    ##
    Z=\frac{RX_{C}^2}{R^2+X_{C}^2}
    ## at resonance. Substitute ##X_C=-\frac{1}{ωC}##, you get ##
    Z=\frac{R}{(RωC)^2+1}
    ##
    Compare it with |ZRC|.
    You also derived that the impedance at resonance is ##Z=\frac{L}{RC}##, so you can write |ZRC| in terms of L and C. Use the equations for UL and UC and divide them. What do you get?
     
  20. Nov 22, 2015 #19
    Hi, sorry I wasn't active, I decided to practice these kind of problems more (easier problems) and then try to solve this one. I got by drawing vector diagram that:

    and ##U_C## vector has angle ##\sin \phi= \frac{-U_L}{U_C} \rightarrow \sin \phi = -\frac{2}{3}##.
    And ## \sin | \phi| = \frac{I_C}{I}## so ##\frac{I }{I_C} = \frac{3}{2} ##

    Then ## \frac{X_C}{X_L}= \frac{I}{I_C} \cdot \frac{U_C}{U_L} = \frac{9}{4}=2.25##

    Is this correct?
     
  21. Nov 22, 2015 #20

    ehild

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    It is correct, and a very nice short solution. Congratulation! I prefer using complex impedances instead of vector diagrams, but my solution was very long and complicated.
     
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