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Finding R,L,C at resonant state

  1. Jul 21, 2013 #1
    Hello again. I hope someone could check if I got the answers correct.

    The magnitude of the current I in resonant state was 3[A]. When the angular frequency of the power supply was 1 [rad/s] and 2[rad/s], the magnitude of current I became 1/√2 of the value when resonance occurred. Find R, L, C.
     

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  3. Jul 21, 2013 #2

    gneill

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    When I solve your equations as you've given them I'm seeing -3/8 for the capacitance... which shouldn't be.

    You have to be a bit careful about 'losing' solutions when you eliminate squares.
     
  4. Jul 21, 2013 #3

    rude man

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    Your approach looks right, but I got a different result. Check your equations against mine & recheck your math:

    (L - 1/C)^2 = 16,
    (2L - 1/2C)^2 = 16
     
  5. Jul 21, 2013 #4
    Yes, at first I got -3/8 for the Capacitance but since it's an absolute value, I made it positive. Well, looks like I have to redo my calculation again :(
     
  6. Jul 21, 2013 #5
    I redid it and still got the same simultaneous equations. Can you tell me where I am wrong?
     

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  7. Jul 21, 2013 #6

    rude man

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    Do you agree with my equations or not? If so, what solutions do you get?

    Hint: there are multiple solutions but only one where L and C are positive, as they must be.
     
  8. Jul 22, 2013 #7
    Yes, yours seems to be the same as mine. But can we cancel out the squares by just adding a root to both sides?
     
  9. Jul 22, 2013 #8

    rude man

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    When you take a square root you have to realize there are two polarities involved. Here we are taking two roots so be prepared to get four sets of answers for L and C.

    But only one of them has L > 0 and C > 0.
     
  10. Jul 22, 2013 #9
    Oh, I totally forgot about that. So that's what "losing solutions" means. So C=1/8 and L=4.
    Thank you soooo much!!!! Could you take a look at my new thread? I need help.
     
  11. Jul 22, 2013 #10

    rude man

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    Yes, you have got it!
    Will check your new thread.
     
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