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RE: A question about LCR circuits
Subject: RE: A question about LCR circuits
Date: Mon, 12 May 1997 08:10:34 -0400
From: Heinz Wahl <hwahl-at-jtc-campus.moric-dot-org>
To: "'Tesla List'" <tesla-at-pupman-dot-com>
Malcolm,
If you equate f=0 your talking D.C.. "at and below which" your circuit
can't ring.
Heinz
----------
From: Tesla List [SMTP:tesla-at-pupman-dot-com]
Sent: Friday, May 09, 1997 2:09 AM
To: tesla-at-poodle.pupman-dot-com
Subject: A question about LCR circuits
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