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Formula for calculating w0 (w at resonance)

  1. Sep 6, 2005 #1
    Hi dudes. sory if it is not the correct area for this question.

    A question. What is the formula for calculaing w0 (w at resonance) in a two-branch parallel circuit consisting of:
    A voltage source
    branch 1: R1 + L (in series)
    branch 2: R2 + C (in series)

    I thought it was the same as RLC series, but it actually is pretty different.

    BTW I got a parallel resonance exercise but cannot figure out how to solve it.
    Same circuit as above, but:
    r1 = 2 ohm
    r2 = 5 ohm
    C = -10j ohm
    W = 5000 radian

    Find out the value of L (both in ohms and Henry)
    Any help on this subject is really appreciated.
     
  2. jcsd
  3. Sep 6, 2005 #2

    SGT

    User Avatar

    1. Calculate the impedance [tex]Z_1(\omega)[/tex] of branch 1.
    2. Calculate the impedance [tex]Z_2(\omega)[/tex] of branch 2.
    3. Calculate the impedance [tex]Z(\omega) = X (\omega)+ iY(\omega)[/tex] of the two parallel branches.
    4. The resonant frequency [tex]\omega_0[/tex] is that which makes null the imaginary term [tex]Y(\omega)[/tex]
     
  4. Sep 6, 2005 #3
    Thanks SGT for your prompt reply.

    By Y(w) you mean the reactive component or admitance?

    somewhere i've read that for this kind of circuits:

    w0 = [ 1 / sqrt(LC) ] * [ sqrt(r^2 - L/C) ] / [ sqrt(r^2 + L/C) ] or something similar. This is what i do not understand.

    sorry for not posting in Latex!
     
  5. Sep 7, 2005 #4
    at resonance the circuit will act as pure resistive circuits, meaning the power factor is zero or the value of capacitive and inductive impedance is the same. thus:

    given:

    W= 5000 radians

    then:

    ZL=WL...............impdedance (Z) of inductive element
    ZC= 1/WC..........impedance (Z) of capacitive element

    since ZL=ZC (at resonance)

    then:

    WL= 1/WC or

    L= 1/(W^2*C)

    given, W=5000 rad, C=-j10

    C= -j10= -j/(WC)

    C= 20x10 exp (-6) or 20 microfarad

    therefore:


    L= 1/ {[5000^2]*[20 x 10 exp (-6)] }

    L= 2 x 10 exp (-3) or 2 mili Henry

    or, ZL= WL= [5000]*[2 x 10 exp (-3)]

    ZL= 10 ohms,,,,thats it
     
  6. Sep 7, 2005 #5
    at resonance the circuit will act as pure resistive circuits, meaning the power factor is zero or the value of capacitive and inductive impedance is the same. thus:

    given:

    W= 5000 radians

    then:

    ZL=WL...............impdedance (Z) of inductive element
    ZC= 1/WC..........impedance (Z) of capacitive element

    condition: ZL=ZC (at resonance)

    then:

    WL= 1/WC or L= 1/ (W^2*C)...this is the key formula

    given, W=5000 rad, C=-j10

    C= -j10= -j/(WC)

    C= 20x10 exp (-6) or 20 microfarad

    therefore:

    L= 1/ {[5000^2]*[20 x 10 exp (-6)] }

    L= 2 x 10 exp (-3) or 2 mili Henry

    or, ZL= jWL= [5000]*[2 x 10 exp (-3)]

    ZL= j10 ohms,,,,thats it

    answers:

    L= 2 MILIHENRY

    ZL= j10 ohms at an angular frequency of 5000

    .....at resonance
    :smile:
     
    Last edited: Sep 7, 2005
  7. Sep 7, 2005 #6

    SGT

    User Avatar

    [tex]Y(\omega)[/tex] is the reactive component.
    Using the set of rules I suggested you, we can conclude that [tex]\omega_0 = \frac{1}{\sqrt{LC}}[/tex] exactly like in a series or parallel circuit.
     
  8. Sep 7, 2005 #7
    Thanx a lot guys for your replys.

    About the solution by "young e.", a question:

    From the results you obtained, it means that Ztotal = 15.43 + 4.2857i.

    should not the reactive component on Ztotal be equal to 0, thus leaving only a resistive part <> 0 ? Even working it out as admittances, Y1L and Y2C are different.

    y1 = 0.019231 - 0.096154i
    y2 = 0.040000 + 0.080000i

    y1 + y2 = 0.059231 - 0.016154i

    Whats wrong here? I do not get it. AFIK, parallel resonance has a rather ****ty behaviour compared to series resonance.

    Best regarss.
     
  9. Sep 7, 2005 #8

    SGT

    User Avatar

    The imaginary part of the admittance is frequency dependent. What frequency have you used to calculate it?
     
  10. Sep 8, 2005 #9
    Hi SGT.

    I have used W = 5000 (which is 2*pi*796.178) as stated in the excercise. It supposed to be W0 = 5000

    Best Regardz.
     
  11. Sep 8, 2005 #10

    SGT

    User Avatar

    Forget my previous post. [tex]\omega_0[/tex] is not [tex]\frac{1}{\sqrt{LC}}[/tex]. Use the rules I suggested you to obtain the resonant frequency.
    If you want only to calculate the inductance, the admittance of branch 1 is:
    [tex]Y_1(\omega)=\frac{1}{R_1 + j\omega L}[/tex]
    The imaginary part of this admittance must be equal to the negative of the admittance of branch 1.
    [tex]imag(Y_1) = - 0.08[/tex]
    From this you get a second degree equation in L.
     
    Last edited by a moderator: Sep 9, 2005
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