A question about LCR circuits

To: tesla-at-poodle.pupman-dot-com
Subject: A question about LCR circuits
From: Tesla List <tesla-at-stic-dot-net>
Date: Fri, 09 May 1997 01:08:31 -0500
Approved: tesla-at-stic-dot-net

Subject:       A question about LCR circuits
       Date:   Fri, 9 May 1997 11:22:38 +1200
       From:   "Malcolm Watts" <MALCOLM-at-directorate.wnp.ac.nz>
Organization:  Wellington Polytechnic, NZ
         To:  tesla-at-pupman-dot-com


Hello all,
            Since this relates to resonant circuits, I thought this 
might be of interest to list members. I posted a version of this on 
SCI-TESLA about 5 months ago. I was explaining the significance of Q
to some list members and relating it to oscillations in mechanical 
and electrical systems.

FYI,
     Here is the mathematical derivation for the critical value for
Q in a "resonant" system - you will see why I have highlighted
resonant in a moment. We start by saying we want to find the value of 
Q at which the frequency of a system has dropped to 0 Hz. I start 
with an electrical system since I am familiar with the necessary 
equations. However, this is true for all systems which can store 
energy.

The natural (unforced) resonant frequency of an LCR circuit is:
f = SQRT( 1/LC - R^2/4L^2 )

Now 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

[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 by R^2 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).

This is the value of Q, at and below which the circuit can no longer 
ring. In other words, a circuit with this value of Q loses energy at 
the same rate as it can take it up.

Think that's all OK but comments welcomed as always,
Malcolm