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Non-homogenous Diff EQ, LRC circuit

  1. Jun 18, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Consider an LRC circuit. L = 3/5, R = 10, C = 1/30, E(t) = 300, Q(0)=0, I(0)=0
    i) Find charge and current.
    ii) Find maximum charge on the capacitor.

    2. Relevant equations

    LQ'' + RQ' +(1/C)Q = E

    3. The attempt at a solution

    (3/5)Q'' + 10Q' +30Q = 300

    For the roots of characteristic equation I got:

    [itex]m = -\frac{25}{3} \pm \frac{5\sqrt{7}}{3}[/itex]

    This leads to the solution to the homogenous part, and the particular solution is simply Q=10.

    So, after applying initial conditions to find the constants, the full general solution came out to

    [itex] Q(t) = (5- \frac{25}{\sqrt{7}})e^{(-\frac{25}{3} + \frac{5\sqrt{7}}{3})t} + (\frac{25}{\sqrt{7}}-15)e^{(-\frac{25}{3}-\frac{5\sqrt{7}}{3})t} +10[/itex]

    Differentiating to find the current gives me:

    [itex] I(t) = (-\frac{25}{3} + \frac{5\sqrt{7}}{3})(5- \frac{25}{\sqrt{7}})e^{(-\frac{25}{3} + \frac{5\sqrt{7}}{3})t} + (-\frac{25}{3}-\frac{5\sqrt{7}}{3})(\frac{25}{\sqrt{7}}-15)e^{(-\frac{25}{3}-\frac{5\sqrt{7}}{3})t}[/itex]

    So that's part i) finished.

    But I'm confused on part ii)....

    It's asking for the maximum charge on the capacitor, and I assume charge is given by Q(t).
    So to find the maximum charge, do I set I=0 and solve for t? Since it says I(0)=0, then wouldn't t=0, which means Q=0?

    I'm pretty sure there is more to it then that, but I'm confused.
     
  2. jcsd
  3. Jun 18, 2012 #2

    vela

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    You should try plotting Q(t) to see what's going on.

    The extrema of Q(t) may occur at points where I(t)=0. You don't know if they're maxima or minima without looking closer, though. The extrema can also occur at other places as well.
     
  4. Jun 18, 2012 #3

    LCKurtz

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    I didn't check your numbers. But the capacitor is going to charge up and everything will come to a halt, won't it? You need to look what happens as ##t\rightarrow\infty##.
     
  5. Jun 18, 2012 #4

    ElijahRockers

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    According to my trusty graphing calculator, The graph increases sharply until it levels off at Q=10, at approximately t=3.

    OH. Ok I think I get it... so the lim as t-> inf of Q = 10. duh. For some reason my mind just didn't want to register that extra +10 on the end of the equation, that has been giving me trouble the whole time.

    Thank you.

    So since I(0) = 0, then Q(t) has a minimum at t=0.

    What about the maximum? It doesn't really have a formal max because it is an open interval in the positive direction, but it converges to Q=10... is that enough to say the max=10?
     
  6. Jun 18, 2012 #5
    From the distinct real roots, you know that the 2nd order system is overdamped, so there is no over shoot in the step response. Hence, the max Q occurs at steady state while t->infy.

    On the other hand, if you have a complex conjugate roots (underdamp), the max Q is the peak overshoot of the step response.
     
  7. Jun 18, 2012 #6

    vela

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    A mathematician would say it doesn't have a maximum. Everyone else would say Qmax=10.
     
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