Re: time constant in resonant circuit

["Tesla list" <tesla-at-pupman-dot-com>]

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.
 >
 >
 >
 >
 >