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Over-damped Circuit Equation

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    For an overdamped series RLC circuit, determine the coefficients [itex]\beta_1[/itex] and [itex]\beta_2[/itex] in the equation [itex]V=Ae^{-\beta_1t}+Be^{-\beta_2t}[/itex] for the case where [itex]R=600 \Omega[/itex], [itex]L=100 \mu H[/itex], and [itex]C=.01 \mu F[/itex]. Also determine the ratio of [itex]B[/itex] to [itex]A[/itex].


    2. Relevant equations
    For a series RLC circuit, the general differential equation is [itex]\large \frac{d^2V}{dt^2} + \frac{R}{L} \frac{dV}{dt} + \frac{1}{LC} V = 0 [/itex].


    3. The attempt at a solution
    I think I can just assume that [itex]V=Ae^{-\beta t}[/itex], take the necessary derivatives, and then solve the quadratic equation that results for [itex] \beta_1 , \beta_2 [/itex].

    After canceling the exponential terms, I get [itex] \beta^2 - \beta \frac{R}{L} + \frac{1}{LC} = 0 [/itex].

    The quadratic equation is [itex]\large \frac{\frac{R}{L} \pm \sqrt{\frac{R^2}{L^2} - \frac{4}{LC}}}{2} [/itex].

    My numbers end up being [itex]\beta_1 = 5.828 * 10^6 sec, \beta_2 = 1.715*10^5 sec[/itex]. Is this the correct answer? I don't know why they would be an entire order of magnitude separate...

    Was it ok to assume the solution was not the sum of exponentials, but a single exponential which I take both roots of? I don't have much experience with differential equations.

    Then for the A to B ratio, I know that [itex]\frac{dV}{dt} = 0 [/itex], so I can use the [itex]\beta 's[/itex] and I end up getting [itex] \frac{A}{B} = \frac{-\beta_2}{\beta_1} [/itex]... is that right as well? I end up getting a huge ratio, like [itex] -33.98 [/itex].

    Thanks for the help!
     
  2. jcsd
  3. Nov 13, 2012 #2

    rude man

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    Having problems here:
    1. What is V?
    2. I suspect you mean a parallel, not a series, RLC circuit, for which the equation is
    d2V/dt2 + (1/RC)dV/dt + V/LC= 0.
    Got a diagram of your circuit handy?
     
  4. Nov 13, 2012 #3

    rude man

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    Assuming L/R = RC for some reason having to do with your definition of "overdamped", what you did to get the two betas looks right, except the units of beta is 1/sec, not sec.

    Anyway, without a diagram including indication of how the initial conditions (capacitor voltage and inductor current at t = 0+) are determined I don't know what else to say.
    I don't know where you got dV/dt = 0 at t = 0+ or any other time.
    ]
     
  5. Nov 13, 2012 #4
    V refers to the voltage across the capacitor. It really is a series circuit. The differential equation I wrote is exactly correct for a series RLC (undriven) circuit. The circuit is given in Purcell's textbook, Figure 8.4a on page 303. It is essentially a series RLC with charged capacitor and switch closed at time t=0. So there is no need to say L/R=RC, because the RC comes from the parallel differential equation with which are not concerned here.

    So the voltage is instantaneously equal to the initial voltage of the capacitor at the t=0, and at t=0 dV/dt (proportional to current) is 0 because the inductor prevents any flow of current at that instant.
     
  6. Nov 13, 2012 #5

    rude man

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    p;pp
    My gleanings from your post: C is pre-charged, then the switch throws C across the series L-R network. The current thru L zero at time t = 0. If this is the case the results you got for A/B agree with mine.

    I can't verify your values of β1 and β2. The reason, which probably means nothing to you but might to some of my Laplace-savvy friends, is that the Laplace of v(t) is V(s) = v(0)(s + R/L)/[s+a)(s+b)] where (s+a)(s+b) = s2 + (R/L)s + 1/LC. In other words, I can't vouch for Aexp(-β1t) + Bexp(-β2t) as being the correct solution without inverse-transforming V(s) which I am not in a position to do now. However, if you were given the correct form of v(t) then what you did looks all right, ignoring the possibility of numerical errors. Your equation for finding β1 and β2 (the quadrature in β) would be correct.
     
  7. Nov 14, 2012 #6
    The roots of a second order equation like this will lie on a semicircle of radius 1/√(LC) in the left half of the complex plane when the roots are complex. Keeping LC constant but manipulating other components (R) will cause the roots to move, following a fixed path. Just as the roots leave the semicircle and become real, they will both be located at -1/√(LC) = 1e6 /s on the -ve real axis in this case. This happens when the stuff under the square root in your β equation equals zero. After this, one root will move toward zero on the real axis (but never get there) and the other root will move toward -∞. You can see you found one root to the right of 1e6 and one root to the left. The root that moves toward -∞ is dominated by the root that moves toward zero. This is because the exponential term for the root toward -∞ settles more quickly.


    What you did was right. You assumed the solution was exponential and found two exponentials that will satisfy the equation. The differential equation is second order so there must be two linearly independent functions in the solution.

    That's right too. You will still need one more initial condition to find values for A and B.
     
    Last edited: Nov 14, 2012
  8. Nov 14, 2012 #7

    rude man

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    For what it's worth, I finished my computations. I agreed with your characteristic equation. But now for A/B I got
    A = E(R/L-β1))/(β21)
    B = E(β2-R/L)/(β21)
    so that A/B = (R/L-β1)/(β2-R/L).
    Note that this is the complete solution to the problem assuming initial L current = 0 ad initial C voltage = E.
     
    Last edited: Nov 14, 2012
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