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AC circuit problem

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/HTGycr8.png?1
    [tex]f = 248 Hz\Rightarrow \omega = 1558.23r/s\\ L = 34 mH\\R = 7\Omega\\\alpha = -6°\\C=? [/tex]
    Alpha is the angle between current and voltage (V - I).
    [tex]X_L=\omega\times L=1558.23 * 34 * 10^-3 = 52.9798 \Omega[/tex]
    2. Relevant equations
    [tex]\arctan{\frac{R}{X}} = \alpha[/tex]
    3. The attempt at a solution
    Basically I tried combining everything into one impedance, then I would find real and imaginary part:
    [tex]\frac{\Re{Z}}{\Im{Z}} = \tan{\alpha}[/tex]
    but I can't get past some maths and I know this is probably a problem; so far (jX_L = L, -jX_C = C):

    [tex]\frac{RL(C+R)}{(C+R)(R+L)+RL}[/tex] now if I were to plug everything in I would get something very complicated... Now I need to get this denominator to something simple, but I don't know how to do it the easy way... it's too tedious... is there a simpler way?

    Also, if the problem stated that current and voltage were in phase, I could combine R and L that are separate (in first and last branch) in one branch (a series) then use this formula: [tex]\omega_0 = \frac{1}{\sqrt{LC}} \times\sqrt{\frac{R_C^2 - L/C}{R_L^2 - L/C}}[/tex]? I tried to do that exactly and it didn't work, I checked several times with the given result...

    Thanks in advance.
     
    Last edited: Jan 10, 2015
  2. jcsd
  3. Jan 10, 2015 #2

    gneill

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    Staff: Mentor

    Hmm. Are you sure about that phase angle being negative 8 degrees? If you ignore the capacitor branch for a moment and calculate the phase angle associated with what's left (parallel R and L), then the result is -7.53°. That is, the current lags the voltage by 7.53°. That's as negative as it's going to get if all you can do is add capacitive reactance in parallel.
     
  4. Jan 10, 2015 #3
    Eh, right, it's -6°.
     
  5. Jan 10, 2015 #4

    gneill

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    Staff: Mentor

    Okay, that would make more sense.

    Try working with admittances rather than impedances. The admittance will have the same phase angle as the current since I = U*Y and U has only a real component. The math will still be nasty, but perhaps a bit less so. The admittance of the lone resistor branch is 1/R and purely real. The admittance of the inductor branch is -1/XL and purely imaginary (XL being the reactance). You should be able to write an expression for the admittance of the RC branch and split it into real and imaginary components without too much difficulty. After that the nastiness begins :nb)
     
  6. Jan 10, 2015 #5
    I did it now 3 times only to remember that the angle is of the opposite sign (+) and then the result was fine. Also i think i figured out why the formula for resonance didnt work -- i didnt change L later (i.e. i left it as it was, in parallel: which it wasn't)...

    Thank you very much. :)
     
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