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


> Original Poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>


> > Original Poster: "Alfred C. Erpel" <alfred-at-erpel-dot-com>



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




> Assuming that the 3 resistors, the inductor, and the capacitor are in
> series, forming a closed loop, R is just the sum of the three resistances.
> The time constant of a series RLC circuit is 2*L/R, regardless of C.
> This means that if you put an initial charge in C, you will have an
> oscillation at the frequency f=1/(2*Pi*sqrt(L*C)), decaying with a
> time constant 2*L/R. This if Q=sqrt(L/C)/R > 0.5. Otherwise the
> capacitor discharges without oscillation, with two time constants, one
faster
> than 2*L/R, other slower.

> An additional R in series, as a gap resistance in a primary coil in
> a Tesla transformer (assumed as approximately linear), you just add
> to the others. Resistances in parallel with the elements add some
> complications, and it's better to see first exactly what is the problem.
>
> Antonio Carlos M. de Queiroz


So far, what I have read regarding time constants is (one time constant) =
L/R for inductance and (one time constant)= RC for capacitance. Could you
explain why in the context of the series resonant circuit you described
above, that 1t = 2*L/R and that it is regardless of C? Are you also saying
that RC is irrelevant in this circuit under any circumstances? If it is not
too involved, please explain.  Thank you.

Regards,
Alfred Erpel