Electrical – Voltage Division when we have a capacitor and resistor in series

So, starting point and ending point are easy to compute for a constant input source. At start the capacitor shunts the resistor and you basically get vo = vi (vo is output voltage and vi is input voltage). At steady state there is no current through the resistor so you get a simple voltage divider vo = 10/110 * vi

You can find the transient behavior by solving a differential equation. Let's take the output node. The current entering the output node has to be the same as the current leaving it so we could write the equation 10e-6*d(vi-vo)/dt + (vi-vo)/100 = vo/10. Simplifying, we have 1e-3*dvo/dt + 11*vo = vi. From the characteristic equation, we know vo has to be of the form vo = A*e^(-11e3*t)+B for this differential equation to be satisfied.

Given the steady state condition, vo = 10/110*vi=A*0+B, then B=10/110*vi and vo = A*e^(-11e3*t)+10/110*vi. If we use the initial condition vo=vi=A+10/110*vi, then A=100/110*vi. Thus, vo = 100/110*vi*e^(-11e3*t)+10/110*vi.

If vi is not constant then dvi/dt is not zero and the output will also be dependent on the time-varying behavior of the input. You will need to solve a non-homogenous differential equation to get the answer depending on vi as a function of time.

You don't say what kind of buzzer it is, but if I had to guess then I suspect it is a three-tab piezo coupled with a simple BJT circuit like this:

^{simulate this circuit – Schematic created using CircuitLab}

This might account for your multimeter measurements and also account for the relatively low voltage measurement across it when active through a voltage source and added resistor. It's really hard to say for sure, though.

Have you considered the idea of opening things up and looking to see?