The Physics
Opus in profectus

LC Circuits

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lc circuit

Begin with Kirchhoff's circuit rule.

V = L  dI  +  q
dt C

Take the derivative of each term.

dV  = L  d2I  +  1   dq
dt dt2 C   dt

The voltage of the battery is constant, so that derivative vanishes. The derivative of charge is current, so that gives us a second order differential equation.

0 = L  d2I  +  1  I
dt2 C

Rearrange it a bit…

d2  I = −  1  I
dt2 LC

and then pause to consider a solution.

We need a function whose second derivative is itself with a minus sign. We have two options: sine and cosine. Either one is fine since they're basically identical functions with a 90° phase shift between them. Without loss of generality, I'll choose sine with and arbitrary phase angle (φ) that could equal 90° if we let it. Or it could be equal to some other angle. The other parameters in a generic sine function are amplitude (I0) and angular frequency (ω).

The basic method I've started is called "guess and check". My guess is that the function looks like a generic sine function…

I = I0sin(ωt + φ)

and the check is to pop it back into the differential equation and see what happens.

d2  I0sin(ωt + φ)  = −  1  I0 sin(ωt + φ)
dt2 LC
− ω2I0sin(ωt + φ)  = −  1  I0 sin(ωt + φ)

Basically everything cancels but one parameter — angular frequency.

ω =  1

An LC circuit is therefore an oscillating circuit. The frequency of such a circuit (as opposed to its angular frequency) is given by…

f =  ω  =  1

So what? How is this useful?

The author holding a crossover circuit

An audio crossover circuit consisting of three LC circuits, each tuned to a different natural frequency is shown to the right. The inductors (L) are on the top of the circuit and the capacitors (C) are on the bottom. On the left a "woofer" circuit tuned to a low audio frequency, on the right a "tweeter" circuit tuned to a high audio frequency, and in between a "midrange" circuit tuned to a frequency in the middle of the audio spectrum.

RC circuits are basically filters.

rcl circuit

I need to write this part.