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RLC circuit

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data

    j9ouuh.jpg

    Question and part 1 as above. The second part involves solving this equation where [tex]L = 8R^2 C[/tex]. The system is kept in steady state by maintaining V(t) = -Q/C (constant). V(t) is then set to 0 at t=0.

    It also says "Note that V(t)=0 for t>0 and that appropriate initial conditions
    at (or just after) t=0 are that q=Q and dq/dt= −Q/CR."

    3. The attempt at a solution

    The first part is just very tedious math, but I managed to get it.

    The second part is just a second order ODE, but I am unable to get the answer which is

    255rvd5.jpg

    Given the differential equation above, and substituting V = -Q/C, dV/dt = 0, the right hand side becomes a constant. This means that there is a particular integral (q = k = Q), but the answer does not have the form q = Q +... !

    I did it a few times but keep getting back to the same problem. However, if I do attempt a solution that omits the particular integral, I do get the answer needed. Why is this the case?
     
  2. jcsd
  3. May 8, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi bigevil! :smile:

    Yes, the given solution for q is a "general" solution for the original equation, and should have a Q added to it, but
    I don't completely understand what's going on here, but it seems to be changing the original RHS to zero (instead of Q/LC), so no particular solution is needed …

    (perhaps they're charging the capacitor using the battery until t = 0, and then turning the battery off?)

    does that make any sense? :redface:
     
  4. May 9, 2009 #3
    Yes, I think so, tim, thanks =)
     
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