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Re: time constant in resonant circuit
Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>
Hi Jim and Alfred,
The time constant L/R of a series L R network is not pertinent to the
original LCR network. If the C was replaced by a DC source, 1 Vdc for
example, then the computed 5 time constants or 1.2 sec would be the time for
~10000 amps to build up in the inductor. With the original C in place, the
voltage on C drops as the current in L increases. Max current in the
inductance (neglecting losses) is when all the capacitive energy is
transfered to the inductance. Otherwords:
L*I^2 = C*V^2 or
I = V / sqrt (L/C) in the form of I = V / Z
Gerry R
> Original poster: Jim Lux <jimlux-at-earthlink-dot-net>
>
>
> At 09:44 PM 1/4/2004 -0700, you wrote:
> >Original poster: "Alfred Erpel" <alfred-at-erpel-dot-com>
> >
> >Hello All,
> >
> > ASCII art warning -- be sure to view with fixed width font (e.g.
Courier
> >New)
> >
> >For purposes of discussion it is assumed that no resistance exists except
as
> >specified.
> >In the circuit above, when the switch is closed, it will take 5 time
> >constants of L/R before the inductor is conducting at essentially 100%.
> >
> >L/R = 1 time constant
> >1 time constant = .00024318 H / .0001 ohms = .243 seconds
> >5 time constants = 1.215 seconds
>
>
> I think you have the equation wrong.. Increasing either R or L will
> increase the time constant it takes for the current to reach 63% of the
> eventual value, right? Therefore, the TC must be the product of R and
L...
>
>
>
>
>
> >************************************************************
> >
> >In this circuit let's assume no resistance except as specified and a
fully
> >charged capacitor just before the switch closes. When the switch closes
the
> >circuit will ring at its resonant frequency of 250,000 hz.
> >
> >resonant frequency = 1 / 2 * pi * sqrt[L * C]
> >resonant frequency =~ 250,000 hz
> >
> >The quarter cycle time (the first 90 degrees) is ( 1 / 250,000 ) second /
4,
> >so the quarter cycle time is .000001 seconds
> >
> >If the inductor in figure B above was subject to charging at the same
rate
> >as the identical inductor in figure A there couldn't be resonance since
the
> >time constant of .243 seconds is orders of magnitude away from .000001
> >seconds. It appears that the time constant does not come into play in a
> >resonant circuit. Is this what is happening? Could someone please
explain
> >this? Thank you.
>
>
>
> You've got a classical 2nd order differential equation here, and the
> behavior will depend on whether it is over or under damped. If under
> damped, it will ring at some frequency (approximately(!) 1/(2*pi *
> sqrt(L*C) => Damping changes the resonant frequency (makes it lower))
>
> If over damped, it won't ring, and it will, in the limit, look just like a
> RC or RL.
>
>
>
>
>