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Damped Electromagnetic Oscillations

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    This problem was given in my physics test, and my physics teacher was unable to provide me with an answer for it. Given a circuit made up of a generator of variable frequency, i=Im(sin2∏ft-∅) and voltage u=Um(sin2∏ft), capacitor of capacitance 1μF, resistor of resistance 120Ω, and a coil of inductance 0.06H with negligible resistance. We were asked to find the expression of Im^2 in terms of Um and f using the law of addition of voltages, and by taking the values of time as 0 and 1/(4f).


    2. Relevant equations
    Uc=q/C
    U of coil = Ldi/dt
    Ur=Ri


    3. The attempt at a solution
    I was unable to think past finding the law of addition.
     
  2. jcsd
  3. Jun 3, 2013 #2
    I think it would be easier to help you if you wrote the problem statement here exactly like it was worded on your test. Maybe just include a picture of it, if possible.
     
  4. Jun 3, 2013 #3
    Unfortunately, I don't get my exam back till Friday.
     
  5. Jun 3, 2013 #4

    rude man

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    Gold Member

    So probably the circuit consisted of the inductor, capacitor and resistor in series with the voltage generator U.

    You've written the voltage drops across these 3 components above, so how about equating their sum to the voltage generator for openers?
     
  6. Jun 5, 2013 #5
    I did that, let me show you how far I got. Ug= Uc+Ucoil+Ur

    Uc= q/C, i=-dq/dt, thus q=-∫i.dt=(Im/2∏ft)cos(2∏ft-∅)
    Ucoil= Ldi/dt = L2∏ftcos(2∏ft-∅)
    Ur= Ri= Rsin(2∏ft-∅)

    Umsin(2∏ft)=(Im/2∏ft)cos(2∏ft-∅) + L2∏ftcos(2∏ft-∅) + Rsin(2∏ft-∅)

    substituting t by 0 yields: (Im/2∏ft)cos(∅) + L2∏ftcos(∅) = Rsin(∅)

    substituting t by 1/4f yields: Um= -(2Im/∏)sin(∅) - (L∏/2)cos(∅) + Rcos(∅)

    however, I was unable to find an expression of Im^2 in terms of Um and f but without the ∅.
     
  7. Jun 5, 2013 #6

    rude man

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    There's a problem here. You're solving for just the steady-state solution. Unfortunately, judging by the wording of the question, they seem to want the complete solution which includes a transient as well as a steady-state solution.

    If at t = 0 you apply the source voltage Umsin(wt) to the circuit you will get a time reponse of the current I that includes a transient part which includes time-decaying exponentials, and a steady-state part, which goes as Im*sin(wt + ψ). If they wanted just the steady-state part then your assumption that I = Imsin(wt + ψ) would be justified. Unfortunately, the fact that they asked for the current at t = 0 and t = 1/4f indicates that they wanted you to find the complete solution, which as I say includes a transient part and a steady-state part.

    In which case you have to start with the differential equation as you state but change it so it's all in I. You can do this by differentiating your equation once with respect to time to get
    dU/dt = L dI2/dt2 + R dI/dt + (1/C) I. You then solve for I(t).

    This may be overwhelming to you now so I suggest you post the exact problem when you get it back Friday and we can discuss it some more then if you want to.

    Meanwhile you might look at the following link: http://en.wikipedia.org/wiki/RLC_circuit
     
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