Re: resistance in an LRC circuit used to calculate time constant

To: tesla-at-pupman-dot-com
Subject: Re: resistance in an LRC circuit used to calculate time constant
From: "Jan Florian Wagner" <jwagner-at-cc.hut.fi> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
Date: Mon, 13 Mar 2000 06:40:29 -0700
Approved: twftesla-at-uswest-dot-net
Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme


> The resonant frequency of this LRC circuit is 1 / [2 * PI * SQRT(L * C)]
> regardless of the total resistance in the circuit.
 
> a)   The time constant of a capacitor is C * R.

It doesn't matter! And neither does L/R.

>     In the context of this resonant circuit, when you calculate the time
> constants of each device, how do you figure R?  Is R just the resistance
> internal to the device (inductor or capacitor) or do you add up the total
R for

As far as I understand this is the time to charge (or discharge) the cap
to 63% of voltage through a resistor R. In the tank circuit, R is due to
the wiring (there exists no "cap internal resistance"). 

So, you should take the negliblibly (sp?) small total resistance of the
wiring or copper braid/straps and add to that the resistance of the
primary coil.

If you have a cap of 10nF and a total resistance of 0.001 Ohm connected
to the cap, you get

t = RC = 0.001 ns = 1 pico-second

which is again very very short, compared to one fourth of a cycle
(comparing 1/4 cycle to charging cycle) in a, say 500kHz, tank circuit,
which would be

T/4 = 2/4 us = 0.5 micro-seconds

The charging up time doesn't matter! So neither does the resistance.


The same applies for the inductance time constant: the capacitor is an
open circuit having a resistance approaching infinity, so L/R approaches 
0, and t ~= 0


Only part where the resistance comes in is as a damping factor for the LC
oscillator. It causes a damped oscillation, because power is lost in the
resistance.


greetz,
Jan

--
*************************************************
 Jan Florian Wagner
 http://www.hut.fi/~jwagner