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

Hi Malcolm,

    thanks for your response and please see comments interspersed below.


> Original Poster: "Malcolm Watts" <malcolm.watts-at-wnp.ac.nz>
>
> Hi Alfred,


> > Original Poster: "Alfred C. Erpel" <alfred-at-erpel-dot-com>
> >
> > Hello,
> >
> >
> >     An LRC circuit has three components of resistance; the internal
resistance
> > of the inductor, the internal resistance of the capacitor, and
resistance in
> > the wiring connecting the inductor and capacitor.
> > The resonant frequency of this LRC circuit is 1 / [2 * PI * SQRT(L * C)]
> > regardless of the total resistance in the circuit.


> 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.


    None of my reference books include the "(R^2/4L^2)" section you cite in
the equation above. Is this a special case scenario, or does it apply to all
resonant circuits? Also, in using this formula, would you assign some value
to the R of the spark gap in the circuit?


> > a)   The time constant of a capacitor is C * R.
> > b)   The time constant of an inductor is L / R.


> Both of those assume that R is present in the circuit.

    This seems to be a good assumption in the real world arena of tesla
coiling.


> >     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
> > the circuit (all three components) to determine R for the equations
above? How
> > do you account for the R in the circuit external to both devices?


> R is simply the ESR (effective series resistance) of the resonant
> circuit and encompasses all circuit resistances suitably modelled as
> 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.


Regards,
Alfred