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RLC circuit potential difference

  1. Mar 14, 2009 #1
    1. The problem statement, all variables and given/known data

    An L-R-C series circuit is constructed using a 175{ \Omega} resistor, a 12.5 muF capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V.


    At the angular frequency in part A, find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value.

    2. Relevant equations

    V= I Z
    Z^2 = R^2 + (XL-XC)^2
    XL= w.L
    XC = 1/ w.C


    3. The attempt at a solution
    for part A, i get the correct answer for angular frequency = 3160 rad/s
    here, i didn't understand what they meant by " the current is equal to one-half its greatest positive value." do they mean i have to find the total current and divide by two??

    i found the current through Ir=Ic=Il= 0.143
    by doing
    I = V/Z = 25/175
     
  2. jcsd
  3. Mar 14, 2009 #2

    rl.bhat

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    At resonance the current is maximum, and XL = XC.
    You have already found out that current.
    Find impedance Z at which the current is half the maximum current.
    Then ( Z^2 - R^2 )^1/2 = WL - 1/WC.
    Solve the quadratic equation to find the value of W. From that you can find the potential difference across them.
     
  4. Mar 14, 2009 #3
    then, after i found w??
    find the resistence on capacitor and inductor..
    then, times with I to get the potential different??
     
  5. Mar 14, 2009 #4

    rl.bhat

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    To find the potential difference you have to take I/2.
     
  6. Mar 14, 2009 #5
    hey..but here..
    Z=R
    so
    0= wL - wc..

    what did you mean by this equation??
     
  7. Mar 14, 2009 #6
    ahh..ic
    you meant
    Z = V/ (I half of maximum)

    rite??
     
  8. Mar 14, 2009 #7

    rl.bhat

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  9. Mar 14, 2009 #8
    i think i made a mistake :)
    thanks
     
  10. Mar 14, 2009 #9

    rl.bhat

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    Any one will do. When you take the other value, voltage across L and C will interchange.
     
  11. Mar 14, 2009 #10
    how about potential different on ac source and resistor?
     
  12. Mar 14, 2009 #11

    rl.bhat

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    VR = I/2*R, VC = I/2*XC and VL = I/2*XL
    Here V is not equal to VR + VC + VL
    So the potential difference on ac source is V itself.
     
  13. Mar 14, 2009 #12
    hi, why cant i get the correct answer for this??
    though i have done accordingly..
    after find the quadratic equation,
    i get w=38150 and w=(-261.875)
    i choose the positive one,
    and do the calculation to find XL and XC,
    after that multiply by the current(half of max)

    but, my answer is not correct..
    please advise!
     
  14. Mar 14, 2009 #13

    rl.bhat

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    How did you get 3160 rad/s?
     
  15. Mar 14, 2009 #14
    for part A the question is
    At what angular frequency will the impedance be smallest?

    i did
    f= 1/2*pi*sqr(LC)
    after i got the frequency, then calculated the angular speed w=2*pi*f
    ended up with 3160 rad/s
     
  16. Mar 16, 2009 #15
    help!
     
  17. Mar 16, 2009 #16

    rl.bhat

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    To get the current I/2, (WL - 1/WC) must be equal to (3)^1/2*R.
    Further simplification shows that W1 = WR - (3)^1/2*R./2L and W2 = WR + (3)^1/2*R./2L
     
  18. Mar 16, 2009 #17
    why it should be equal to (3)^1/2*R. ??
    and what is WR?? omega*R???
    i dont understand why it is WR??
     
  19. Mar 16, 2009 #18

    rl.bhat

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    Sorry. I should have explained the symbols. Here omega*R is the resonance frequency. At any other frequencies, I = E/Z, where Z^2 = R^2 + (XL - XC)^2. For I/2, Z' = 2Z. It is possible if (XL - XC)^2 = 3R^2.
    If you see the graph of current vs frequency, current will be I/2 for two frequencies. So W1 and W2 must be positive. In the above problem you are getting one positive and other negative. The reason may be either R value is wrong or instead of half the maximum current they might have asked half the maximum power. Just check it.
     
  20. Mar 17, 2009 #19
    hm...
    for this one, since we know that Z'=2Z, can i just find the current by I= V/ 2Z, then after find the current, multiply by each of the resistance to get the voltage., without finding omega first.?
     
  21. Mar 17, 2009 #20

    rl.bhat

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    If you don't know the omega, how can you find XL and XC?
     
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