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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>
> 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.
I will try, but this requires some calculus with Laplace transforms and
some circuit theory. (I could also use differential equations, and much
greater complication.)
Suppose the following model:
I
+----L-->-+
+ | |
Vc C R
- | |
+---------+
Vc(0) is the initial voltage at the capacitor. It can be modeled as a
step
voltage Vcs in series with an uncharged capacitor.
In Laplace transforms:
Vcs(s) = Vc(0)/s.
The current in the series circuit is this voltage divided by the sum
of the three impedances in the circuit. In Laplace transforms:
Z(s) = 1/(s*C)+s*L+R
I(s) = Vcs(s)/Z(s) = Vc(0)/s/(1/(s*C)+s*L+R) = Vc(0)/(1/C+s^2*L+s*R) =
=(Vc(0)/L)/(s^2+s*R/L+1/(L*C))
This is in the form:
I(s) = k/((s+b)^2+a^2)
that, inverting the Laplace transform, corresponds to the waveform in
the time domain:
I(t) = (k/a)*exp(-b*t)*sin(a*t)
where:
k=Vc(0)/L
a=sqrt(1/(L*C)-R^2/(4*L^2))
b=R/(2*L)
The current is a sinusoid with a frequency that is -approximately-
(I forgot this) 1/(2*pi*sqrt(L*C)), decaying with a time constant 2*L/R.
This is valid as long as "a" is positive.
C doesn't affect the term multiplying s in I(s), and so disappears
from the time constant of the exponential decay.
The complete theory about this is normally found in books about circuit
theory.
Antonio Carlos M. de Queiroz