**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

** **V_{Batt} – i_{R} – 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 R^{2}– 4L/C > 0, then

##### i = e^{-[R/(2L)]t}(A_{1}e^{[+SQRT(R*R-4L/C)/(2L)]t} + B_{1}e^{[-SQRT(R*R-4L/C)/(2L)]t})

If R^{2}– 4L/C = 0, then

**……………………………**i = e^{-[R/(2L)]t}(A_{2} + B_{2}t)

If R^{2}– 4L/C < 0, then

**……………………………**i = e^{-[R/(2L)]t}(A_{3}cosθt + B_{3}sinθt)

where θ is given by

…………………………….θ = [SQRT(|R^{2}-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:

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

Immediately after the switch is thrown v

_{L}=V_{Batt}because v_{C}=0 (a declared initial condition, and voltage cannot change instantaneously on a capacitor) and v_{R}=0 (because i_{Loop}=0 at t=0). Thus, Li ‘=V_{Batt}at t=0.

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

**The General Solutions**

##### v_{R} = iR

##### v_{L} = Li’

##### v_{C} = V_{Batt} – iR – Li’

where the applicable i is used (as determined by the value of R^{2}-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