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Inductor Maximum Energy

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Assuming steady-state sinusoidal ac.A voltage of 240V at 50Hz applied to an inductor yields a current of 4A.Assume negligible resistance.Find inductance, max stored energy.



    2. Relevant equations

    E=0.5Li^2

    X=2*pi*f*L


    3. The attempt at a solution

    V/I=X Which I found to be 60

    L=60/2*pi*f
    L=0.191H

    E=0.5*0.191*4^2
    E=1.528J

    However the answer given is 3.06J , which has left me confused.

    Thanks
     
  2. jcsd
  3. Feb 13, 2014 #2

    gneill

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    Staff: Mentor

    The given voltage and current values are no doubt RMS values...
     
  4. Feb 13, 2014 #3

    PhysicoRaj

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    Gold Member


    Maximum energy is stored when maximum current flows through it.
     
  5. Feb 13, 2014 #4
    Of course, I need to use I pk,

    so E=0.5*0.191*(4*2^1/2)^2=3.06J

    Thanks :)
     
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