Re: time constant in resonant circuit

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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.