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Self induction

  1. Jul 12, 2010 #1
    A 50.0mh inductor is connected to an ac voltage source of 24.0V.The current flowing in this circuit is 0.60A.The frequency of the voltage source is :
    a)50Hz
    b)60Hz
    c)1.27*10^3 Hz

    My Work so far ! :
    f = w/2pi.

    But here we have L , E and i.

    Any help will be welcomed.
     
  2. jcsd
  3. Jul 12, 2010 #2

    kuruman

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    You can do better than this. How will you bring in L, E and i? Think impedance.
     
  4. Jul 12, 2010 #3
    So,impedance is : z=Lw = L*2pi*f ....

    A guess , z = V/I = E/i !! but V(max)/I(max)
     
  5. Jul 12, 2010 #4

    kuruman

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    If z = V/I, can you relate ω to V and I?
     
  6. Jul 12, 2010 #5
    So is it , V/I=Lw then V/LI=w=2pi*f .
     
  7. Jul 12, 2010 #6

    kuruman

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    Correct.
     
  8. Jul 12, 2010 #7
    Thank you, kuruman.
     
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