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Re: resistance in an LRC circuit used to calculate time constant
Tesla List wrote:
>
> 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