LC Circuits
Discussion
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 an 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 = I0 sin(ωt + φ)
and the check is to pop it back into the differential equation and see what happens.
d2 | I0 sin(ωt + φ) | = − | 1 | I0 sin(ωt + φ) |
dt2 | LC | |||
− ω2I0 sin(ωt + φ) | = − | 1 | I0 sin(ωt + φ) | |
LC |
Basically everything cancels but one parameter — angular frequency.
ω = | 1 |
√LC |
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 |
2π | 2π√LC |
So what? How is this useful?
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.