Series RLC Circuitry

 

 

 

Suppose the circuit of Figure 1 on the home page now has a series resistance, R, in the loop.  Suppose, further, that all circuit conditions & descriptions are as before, except for this one change.  Additionally, R is perfect (no inductance, no capacitance, and is invariant with time, temperature, and voltage).

The Analysis

If we write Kirkhoff’s voltage equation around the loop, we get

 

 

   VBatt – iR – L(di/dt) – (1/C)idt = 0

 

If this equation is differentiated with respect to time, re-ordered, then multiplied by -1, and if the now-familiar shorthand for a derivative (‘) is applied, we see that

 

                   Li” + Ri’ + i/C = 0

 

This is another classic second-order differential equation.  In this case, however, the form of the general solution is dependent on the relative values of L, R, and C, as follows (see References):

If R2– 4L/C > 0, then

 

                                   = e-[R/(2L)]t(A1e[+SQRT(R*R-4L/C)/(2L)]t + B1e[-SQRT(R*R-4L/C)/(2L)]t)

 

If R2– 4L/C = 0, then 

 

……………………………i = e-[R/(2L)]t(A2 + B2t)

If R2– 4L/C < 0, then

 

 

……………………………i = e-[R/(2L)]t(A3cosθt + B3sinθt)

where θ is given by

 

 

…………………………….θ = [SQRT(|R2-4L/C|)]/2L

 

and where, in the limit, as R becomes vanishingly small, θ approaches ω=1/SQRT(LC), as expected.

 

 

 

Determining A and B

An A and a B appear in each of the three general solutions.  Determining A and B is essential.  This is accomplished by first recognizing two initial conditions:

 

  1. Immediately after the switch is thrown i=0 because current cannot change instantaneously in an inductor.  Therefore i=0 at t=0.

  2. Immediately after the switch is thrown vL=VBatt because vC=0 (a declared initial condition, and voltage cannot change instantaneously on a capacitor) and vR=0 (because iLoop=0 at t=0).  Thus, Li ‘=VBatt at t=0.

 

1 & 2 will yield two equations in two unknowns (always manageable).

The General Solutions

                                 vR = iR
                                                              vL = Li’
                                                                       vC = VBatt – iR – Li’

 

 

where the applicable i is used (as determined by the value of R2-4L/C).

 

The Next Possible Challenge

It may be that the input source is not a simple step voltage (which nicely differentiates to 0 for t>0).  If that is the case, an appeal to one of the References will lead the serious practitioner to the concept of the “particular” integral (solution).  Or, applying an online search engine to one of the following strings may lead an investigator to a circuit with the input source of interest.

LC Circuits
RLC Circuits