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Re: resistance in an LRC circuit used to calculate time constant
Hi Alfred,
I forgot to do the derivation in my last reply sorry. I have
done it several times before so it must be in the list archives
somewhere:
> Original Poster: "Alfred C. Erpel" <alfred-at-erpel-dot-com>
<snip>
> > Not strictly true. In fact it is 1/[2*PI*SQRT{L*C + (R^2/4L^2)}] or
> > something pretty close. R has to be factored in because... consider
> > the case where it is very large for instance.
OK - my memory's failing. The correct equation is:
f = SQRT( 1/L*C - R^2/4*L^2 )/2*PI
<snip>
> > a single resistor. Note that you can derive the value of Q required
> > for critical damping from the formula given above (turns out to be
> > 0.5).
>
> I don't know enough to do this derivation yet. If it is easy enough, please
> explain.
Below is a paste of an old post, somewhat paraphrased) that I wrote
several years ago:
> Here is the mathematical derivation for the critical value for
>Q in a resonant system.
> We start by saying we want to find the value of Q at which the
>frequency of a system has dropped to 0 Hz. This is true for all
>systems as outlined in my Q post.
(I was referring to a mechanical system in that post).
>The natural (unforced) resonant frequency of an LCR circuit is:
>f = SQRT( 1/LC - R^2/4L^2 )/2PI
You can see that if R=0, the familiar equation pops out of the mix.
>We equate frequency to 0, square both sides of the equation, and
>algebraically re-arrange to give: 1/LC = R^2/4L^2
>Multiply both sides by L^2 gives: L/C = R^2/4
>We now introduce Q into the equation using the identity:
>Q = SQRT(L/C)/R (this is easily derived using Q = wL/R = 1/wCR)
>Squaring both sides and multiplying each side gives
Q^2 x R^2 = L/C
>Substituting for L/C in our original equation gives:
Q^2 x R^2 = R^2/4
>Dividing each side by R^2 gives Q^2 = 1/4
which => Q = 1/2 = 0.5
>(negative square root discarded).
Regards,
Malcolm