Parallel RLC Circuitry

Here we have four ideal circuit elements in parallel:

• current source (iSource = 0 for all t < 0, iSource = I for all t > 0)
• An inductor, L
• capacitor, C
• resistor, R

Note:  For purposes of this discussion, R is ideal (no inductance, no
capacitance, and is invariant with time, temp., and voltage).

Initial Conditions

The charge on the capacitor is 0. Therefore its voltage, VC, is 0V.
No current is flowing anywhere in the network, nor is any current changing.

The Analysis

Applying Kirkhoff’s current law to one of the two nodes shows that

………….I – v/R – i L – Cdv/dt = 0

And differentiating yields

………….0 – v/R – i L – Cv = 0

Recognizing that iL = v/L results in

………….0 – v/R – v/L – Cv = 0

Multiplying through by -1 and rearranging gives

……………..Cv + v/R + v/L = 0

which is another classic second order differential equation.  And we have three general solutions (see References), determined by the values of C, R, and L, as follows:

If 1/R2-4C/L > 0, then
……………………………v = e-t/(2RC)(A1e{+SQRT[1/(R*R)-4C/L]/(2C)}t + B1e{-SQRT[1/(R*R)-4C/L]/(2C)}t)

If 1/R2-4C/L = 0, then
……………………………v = e-t/(2RC)(A2 + B2t)

If 1/R2-4C/L < 0, then
……………………………v = e-t/(2RC)(A3cosθt + B3sinθt)

where θ is given by
……………………………θ = SQRT(|1/R2-4C/L|)/2C

and where, in the limit, as R approaches infinity, 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. v=0 at t=0, because voltage cannot change instantaneously across a capacitor.

2. Cdv/dt=I at t=0, because iR=0 (voltage is 0) and because iL=0 at t=0 (iL had been 0 for t<0, and cannot change instantaneously).

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

The General Solutions

…………………………..iR = v/R

…………………………..iL = (1/L)∫vdt

…………………………..iC = Cdv/dt

where the applicable v is used [as determined by the value of (1/R)2-4C/L].

In the case of iL, an alternative approach (if the integration appears challenging) would be to recognize that

…………………………..iL = I – iR – i
C

The Next Possible Challenge

It may be that the input source is not a simple step current (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