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LC circuit (Differential Equation)

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    An inductor with value L and a capacitor with value C are connected in series to a power source. At time t, the voltage of the power source (i.e. the voltage across both the inductor and capacitor) is given by ## v(t)=Asin(\frac{2t}{\sqrt{LC}}) ##. If the voltage across the capacitor at time 0 is 0 and at time ##\frac{\pi \sqrt{LC}}{2}## is B, what is the voltage u(t) across the capacitor?

    2. Relevant equations
    ##I = \frac{dV}{dt}C##
    ##V = \frac{dI}{dt}L##

    3. The attempt at a solution
    Honestly, I'm pretty stuck. I've tried plugging the expression for current from the capacitor into the inductor equation, and then ## v(t)=Asin(\frac{2t}{\sqrt{LC}}) = \frac{dI}{dt}L+\frac{1}{C} \int_{t_0}^{t} I dt##, but neither approach got me very far. Any help would be greatly appreciated!
     
  2. jcsd
  3. Sep 22, 2015 #2

    NascentOxygen

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    Staff: Mentor

    Hi rakhil11. :welcome:

    From exercises similar to this that you have worked on in class, is it concerned with just steady state conditions, or is it including transient conditions, too? That is, are voltages and currents each a pure sinusoid, or is each a mixture of sinusoids?

    Does the textbook give the answer?
     
  4. Sep 25, 2015 #3

    rude man

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    Homework Helper
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    @nascent: there are no transients in this circuit as it comprises a pure inductor and a pure capacitor. No matter when the voltage source was applied, at t = 0 or t → -∞, the current will always comprise two sinusoids, one at ω = 1/√(LC), the other at ω = 2/√(LC).

    @rakhil11: you're almost OK with your last equation, but given the problem's initial condition, what is t0?
    Hint: change your integro-differential equation into a second-order ODE and solve conventionally for I(t) and then VC(t). You will need your initial conditions on I and dI/dt of course.

    The problem statement is somewhat misleading: VC = B cannot be assigned arbitrarily. When you solve the ODE the value of VC is determined uniquely by setting t = π√(LC)/2.
     
  5. Sep 26, 2015 #4

    NascentOxygen

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    Staff: Mentor

    This being an engineering question, it's possible that the instructor intends the circuit's resistance be considered negligible (i.e., as good as zero) rather than precisely zero. I await clarification from OP.
     
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