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Finding the average energy in an RLC circuit

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-6-14_1-6-44.png

    2. Relevant equations
    Q(t) = Aei(wt+Φ); dQ/dt = i*w*Q(t); E = (L/2)(dQ/dt)2 + Q2/2C

    i = √-1 E above is average energy

    3. The attempt at a solution
    When I plug in Q(t) & dQ/dt into equation above (E) I get:

    A2L/2(w02-w2)cos[2(wt+Φ)]

    w02 = 1/LC

    After I plugged both of them in I simplified & took the real part. Why aren't I getting the answer?
     
  2. jcsd
  3. Jun 15, 2017 #2

    scottdave

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    I don't think the Real part represents the amount of energy which is stored in the cap and inductor. Note that resistors do not store energy.
     
  4. Jun 20, 2017 #3
    Well in the answer the teacher gave, he took the real parts of Q(t) & dQ/dt. He then plugged them into the E formula above. Why does this work and my approach doea not (I take the real part AFTER equations are plugged in).
     
  5. Jun 20, 2017 #4

    scottdave

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    Do you have the notes from what the teacher did from an example problem? Could you post how that worked?
    Suppose that there is not a resistor at all. There will not be any real part, yet there is energy stored (and transferring back and forth) between inductor and capacitor.
     
  6. Jun 20, 2017 #5
    upload_2017-6-20_22-39-21.png

    Here's the entire answer. Why doesn't my approach work?
     
  7. Jun 20, 2017 #6

    scottdave

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    The screenshot is too blurry for me to make out. Is it possible to upload (or link to) the PDF? Perhaps Slideshare.net if this site does not support it.
     
  8. Jun 20, 2017 #7
  9. Jun 21, 2017 #8

    scottdave

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    I understand taking the real part, now. They found it easier to work with e^(iwt), rather than cos(wt) then just took the real part, because e^(ix) = cos(x) + i*sin(x).
     
  10. Jun 22, 2017 #9
    Yes, I know that they plugged in Q(t) = Aeiwt into the differential equation & found A. Then they took Re(Aeiwt) & plugged that into energy equation for part a. I still don't understand this: Why can't you plug in Aeiwt into the energy equation as opposed to its real part?
     
  11. Jun 23, 2017 #10

    collinsmark

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    There's a little more to it than that. Recognize that your instructor's solution first converted/transformed the differential equation from the time domain to the phasor domain by replacing [itex] V' \cos(\omega t)[/itex] with [itex] V' e^{i \omega t} [/itex]. The steady-state solution of the differential equation was then found in the phasor domain. The solution was then converted back to the time domain (from the phasor domain) by taking the projection of the solution onto the real axis, [itex] Q_0 A e^{i \omega t + \phi} \rightarrow \Re \left[ Q_0 A e^{i \omega t + \phi} \right] = \Re [Q_0 A] \cos(\omega t + \phi) [/itex].

    In summary, the process is:
    1) Convert (i.e., transform) to the phasor domain. (Convert from real sinusoidal to complex spiral)
    2) Solve the special case, steady-state solution to the differential equation (ignoring all transients).
    3) Convert back to the time domain by finding the real projection of the complex solution.

    Why do this? Why convert to the phasor domain in the first place? Answer: it just makes the math a little easier. It's easier to work with [itex] e^{i \omega t} [/itex] than it is with [itex] \cos(\omega t) [/itex]. That's all.

    You could have solved the differential equation in terms of [itex] cos(\omega t + \phi) [/itex] from the beginning, but it would have been a lot harder. It's easier to first transform [itex] V' \cos(\omega t) \rightarrow V' e^{i \omega t} [/itex], solve for the steady-state solution, then transform back via [itex] Q_0 A e^{i \omega t + \phi} \rightarrow \Re \left[ Q_0 A e^{i \omega t + \phi} \right] = \Re [Q_0 A] \cos(\omega t + \phi) [/itex].

    By the way, this process of solving the steady-state solution by way of the phasor domain transformation is done by electrical engineers all the time. They even have special notation where [itex] Ae^{i \omega t + \theta} = A \angle \theta [/itex]. Your instructor's solution does not use this notation, but mathematically it's the same thing.

    [Edit: The use of the "phasor domain" as described here can be seen as a stepping stone toward solving steady-state solutions of differential equations by means of the "frequency domain" and Fourier analysis. But I'm getting ahead of myself here. Full conversion to the frequency domain transformation was not done in your instructor's solution; all that was done was to transform from real sinusoidal to complex spiral and back.]
     
    Last edited: Jun 23, 2017
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