Re: A Question About A Simple LRC circuit

[Roger Haar <haar@PHYSICS.ARIZONA.EDU>]

Bob,
        This system is the damped ( UNDRIVEN )
oscillator, with solutions that are underdamped,
critically damped, and overdamped.  I usually
think in terms of current through the circuit but
voltage across the resistor is basically a measure
of the current.  For the underdamped case for
charge on the capacitor ( Take the derivative to
get the current, but this only inserts a constant
factor.), this gives:

        q = q_initial  e^{-bt}   sine{ omega t + phi}

        b = R/2L

        Omega = sqrt( (1/LC) - (R/2L)^2  )

        The above is straight from PHYSICS  3rd ed
Halliday and Resnick Chapt 38  Electromagnetic
Oscillations , page 848.

        The underdamped solution is a sine in an
exponentially decaying envelope.  The mechanical
version is a damped spring and mass system.

Thanks
Roger Haar


*********************************************
Robert B Zannelli wrote:
> If we have a simple RC series circuit and we apply an instantaneous voltage
> across this circuit the equation which describes the voltage across the
> resister as a function of time is:
>
>       V(t)=Va*exp[-t/(R*C)]   Where Va is the applied voltage etc.
>
> For a simple LR circuit we have (replacing the capacitor with an inductor):
>
>       V(t)=Va*(1-exp[-(t*R)/L])
>
>  My question is, if you have a series LCR circuit is there a simple equation
> which will give the voltage across the resister as a function of time?
>
> Bob Zannelli