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ResonantW
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Homework Statement
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].
Homework 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].
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!