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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