Crossover

Apply a sinusoidally varying voltage

V[t_] = V0 Cos[ω t]

across a resistor R and the current will be

J[t_] = V[t]/R

which means the instantaneous power consumed in the resistor is I*V, or ...

P[t_] = J[t]*(J[t]*R)

Averaging this power a cycle we get (using one of our favorite integrals)

(ω/(2π))*Integrate[P[t],{t,0,2π/ω}]

i.e., just like the DC case with an extra (1/2).
BTW it is common in the AC case to talk about the “RMS” (root mean square)
voltage, V_RMS = V0/Sqrt[2].  When we say our household voltage is 120V, we are
talking V_RMS, and likewise for the oscillating current.  Then P=V_RMS*I_RMS.

Now put an inductor in the way and the current oscillates with amplitude

JL[ω_] = V0/Sqrt[(R^2 + (ω L)^2)]

The average power in the resistor is then

PL[ω_] = JL[ω]^2/(2R)

Choosing some numerical values (e.g. V_RMS=1), we plot:

p = {R->1,L->1,C->1,V0->Sqrt[2]}

Plot[PL[ω]/.p,{ω,0,4}]

It is common to plot the power on log-log plot

LogLogPlot[PL[ω]/.p,{ω,.001,1000}]

This is the classic low-pass filter, with the cutoff frequency tuned to 1 here.
The power is usually quoted in decibels, which you recall is 10*Log_10(power),
so that for example a factor of 2 in power is 3 dB (since Log10[2] = .301).
Thus the slope of the fall off is often quoted as “6 dB per octave,” which is
an audio engineer’s way of saying that if you double the frequency, you lose
a factor of 4 in the power.  Physics-speak is to say it falls off as 1/ω^2.

If we replace the inductor with a capacitor we get a high-pass filter:

JC[ω_] = V0/Sqrt[(R^2 + (1/(ω C))^2)]

PC[ω_] = JC[ω]^2/(2R)

(PC[ω]+PL[ω])/.p

LogLogPlot[PC[ω]/.p,{ω,.001,1000}]

A cross-over is a circuit which takes as its input your audio source (containing many different frequencies) and
produces two outputs, a low frequency signal for playing on your woofer speaker and high frequency signal
for the tweeter.  It is desirable that the response of our tweeter/woofer system be flat, i.e. that we get the
same total power independent of frequency.  If we pick the R’s to be the same (most speakers are 8Ω),
and pick parameters so that cut-off frequencies match, and two powers add nicely,
i.e. the sum is indeed independent of ω.

PC[ω]+PL[ω]/.p//FullSimplify

What happens if you don’t match the parameters?  Then the response is not flat, but maybe that is what you want.
That is, in some applications you want to remove (or enhance) certain frequencies.

q = {R->1,L->1.1,C->1,V0->Sqrt[2]}

LogLogPlot[(PC[ω]/.p)+(PL[ω]/.q),{ω,.001,1000}]