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Power in a circuit with an inductor and integrals

  1. Apr 28, 2009 #1
    1. The power into a circuit element in the product of the voltage across the element and the current through the element. Assuming a voltage v(t) = V_p cos(omega*t) across inductor L, integrate the power over one cycle and show that the net energy into the inductor is zero.

    V_p is the peak voltage

    2. Relevant equations:
    P=IV
    I=(V_p/(omega*L))sin(omega*t)



    Ok, so I'm really bad at calculus. I combined the three above equations to solve for Power, P and got: P=((V_p)^2/(omega*L))cos(omega*t)sin(omega*t). I assume I have to take the integral of this, but I'm not even sure what one cycle means. Can anybody point me in the right direction?

    My best guess of the integral would be something like (since integral(sinxcosx)=.5sinx^2):

    .5sin(omega*t)^2*V_p/(omega*L)

    Thanks
     
  2. jcsd
  3. Apr 28, 2009 #2

    rl.bhat

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    In one complete cycle omega*t changes from zero to 2*pi. Take the integration between these limits.
     
  4. Apr 29, 2009 #3
    Can you help me out a little bit more about how to take that integral?
     
  5. Apr 29, 2009 #4

    rl.bhat

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    P= Vp^2/wL*sinwt*coswt = (Vp^2/2wL)*sin2wt.
    Now find the integration with limits.
     
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