**Parallel RLC Circuitry**Here we have

*four*ideal circuit elements in parallel:

- A
**current source**(i_{Source }= 0 for all t < 0, i_{Source}= I for all t > 0) - An
**inductor**, L - A
**capacitor**, C - A
**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, V

_{C}, 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 i

_{L}

**‘**= 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/R

^{2}-4C/L > 0, then

**……………………………**v = e

^{-t/(2RC)}(A

_{1}e

^{{+SQRT[1/(R*R)-4C/L]/(2C)}t}+ B

_{1}e

^{{-SQRT[1/(R*R)-4C/L]/(2C)}t})

If 1/R

^{2}-4C/L = 0, then

**……………………………**v = e

^{-t/(2RC)}(A

_{2}+ B

_{2}t)

If 1/R

^{2}-4C/L < 0, then

**……………………………**v = e

^{-t/(2RC)}(A

_{3}cosθt + B

_{3}sinθt)

where θ is given by

**……………………………**θ = SQRT(|1/R

^{2}-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:

v=0 at t=0, because voltage cannot change instantaneously across a capacitor.

Cdv/dt=I at t=0, because i

_{R}=0 (voltage is 0) and because i_{L}=0 at t=0 (i_{L}had been 0 for t<0, and cannot change instantaneously).

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

**The General Solutions**

**…………………………..**i_{R} = v/R**…………………………..**i_{L} = (1/L)∫vdt**…………………………..**i_{C} = Cdv/dt

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

In the case of i_{L}, an alternative approach (if the integration appears challenging) would be to recognize that**…………………………..**i_{L} = I – i_{R} – 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