(R+Xl)||(R+Xc) circuit

Discussion in 'Homework Help' started by Tommer Stein, Nov 27, 2015.

  1. Tommer Stein

    Thread Starter New Member

    Nov 27, 2015
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    0
    hello
    I need help with this circuit, i need to find the Q of this circuit
    Q=W0/Bw=?
    thanks for all who could help

    by the way how Q call in english?
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    1. Do you think the resonant frequency, ω0, depends on L and C alone, or do the resistors have any effect?
    2. Do you know how to find the bandwidth (BW)?
    3. In English, Q is just Q.
     
  3. hsazerty2

    New Member

    Sep 25, 2015
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  4. RBR1317

    Active Member

    Nov 13, 2010
    229
    48
    I would think the place to start is to excite the circuit with a voltage E, and then calculate the current flowing in the circuit. I=E/Z where Z is the parallel combination of the RL & RC branches. Z=(Rl+sL)║(Rc+1/sC)

    Plot the magnitude of the current vs frequency. At some point the current will be at a minimum - that point is ω0. At two nearby points the current will be 3dB above the minimum value. BW = ω2-ω1

    So Q = ω0/(ω2-ω1)
     
  5. Tommer Stein

    Thread Starter New Member

    Nov 27, 2015
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    i already have ω0. the problem is how i found the Bandwidth, its not the case of [im=re], i think its suppost to be with energies but i have some mistakes probably because i cant found an answer of wH and wL..
     
  6. Tommer Stein

    Thread Starter New Member

    Nov 27, 2015
    4
    0
    how i can i find the bw?
     
  7. Tommer Stein

    Thread Starter New Member

    Nov 27, 2015
    4
    0
  8. hsazerty2

    New Member

    Sep 25, 2015
    22
    1
    Why exactly would we need E and I ?
    Just plot the magnitude of Z vs frequency, and to get a plot, we need to have values for the components, no ?

    @Tommer Stein: can you give us ω0 that you found ?
     
  9. RBR1317

    Active Member

    Nov 13, 2010
    229
    48
    Because there seemed to be a conceptual issue. The concept of finding the bandwidth of the current response seemed like it would be easier to understand than finding the bandwidth of the impedance.
     
  10. e-learner

    New Member

    Apr 25, 2015
    28
    0
    did u try using the series-parallel equivalent circuit? I mean converting each of the series branch into a parallel ,which will give you two resistances , a capcitor and an induction all in parallel and u can easily combine two parallel resistances into single resistance. this will give a simple parallel RLC circuit.
     
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