Why is the resistor voltage initially equal to supply voltage? Is it because there is no voltage going across the capacitor yet? Therefore, as there is no voltage drop across the capacitor, all the voltage from the battery is across the resistor?
Sum of voltages on the passive elements must add up to the supply voltage.
$$
V_{supply}(t) = V_{switch}(t) + V_{resistor}(t) + V_{capacitor}(t)
$$
Because of the fact that \$V_{switch}(t) = 0 \$ and \$V_{capacitor}(0) = 0 \$, \$V_{resistor}(0)\$ must be equal to \$V_{supply}(0)\$.
2.What exactly does "the voltage developed as the capacitor charges" refer to?
When you apply a voltage difference between capacitor plates, one plate has more positive potential with respect to the other one. This initiates an electric field field between the plates, which is a vector field, whose direction is from the positive plate the negative one.
There is an insulating material (dielectric material) between these capacitor plates. This dielectric material has no free electrons, so no charge flows through it. But another phenomenon occurs. The negatively charged electrons of the dielectric material tend to the positive plate, while the nucleus of the atoms/molecules shift to the negative plate. This causes a difference in the locations of "center of charge" of electrons and molecules in the dielectric field. This difference create tiny displacement dipols (electric field vectors) inside the dielectric material. This field makes the free electrons in the positive plate go away, while it collects more free electrons to the negative plate. This is how charge is collected in the capacitor plates.
3.Am i correct in assuming that the resistor voltage drops because the capacitor's voltage is increasing? (kirchoff's law where volt rise = volt drop).
As the capacitor voltage increases, the voltage across the resistor will decrease accordingly because of the Kirchoff's Law, which I formulated above. So, yes, you were correct.
1.If the capacitor's voltage is dropping(due to it being discharged), shouldn't the resistor's voltage be increasing due to kirchoff's law? Also,this should therefore INCREASE the current instead of decreasing it, which would then cause the capacitor to discharge even faster?
You are missing the fact that, the source voltage is zero (i.e.; the voltage source is missing) in the discharge circuit. Substitude \$V_{supply}(t)=0\$ in the formula above. The capacitor voltage will be equal to the resistor voltage in reverse polarities during the discharge. Together, they will tend to zero.
"Basically, this low pass filter passes low frequencies but blocks high frequencies."
At first, no lowpass filter can "block high frequencies" - and, in particular, not the most simple first order RC lowpass. Each lowpass only can - increasingly with rising frequencies - attenuate the amplitudes of frequencies above the corner frequency. Please note that the attenuation at this corner frequency is 3db (attenuation factor 1.414).
Secondly, regarding the squareroot expression: Please note that the capacitive impedance is an imaginary expression (Xc=1/jwc) - and, hence, application of the classical voltage divider rule (between R and Xc) leads to a COMPLEX expression for the transfer ratio T(jw) consisting of a real (Re) and an imaginary part (Im).
Then, the magnitude of the transfer function T(jw) (output-to-input ratio) is |T(jw)|=SQRT(Re²+Im²) and the phase is Phi=arctan(Re/Im).
Setting |T(jw)|=1/1.414=0.7071 and solving for w gives the corner frequency fc=1/2Pi*RC.
Regarding C-R highpass:
The above described calculation (based on SQRT(2)=1.414) reveals that at the corner frequency w=wc the magnitude of the capacitive resistance (1/wC) is equal to the real resistance (R). Therefore, the first-order C-R highpass has the same corner frequency as the R-C lowpass.
Best Answer
Both correctly describe how an RC filter will react at DC. The reason you see a contradiction is because you're actually describing two different situations: loaded, and unloaded. If there's a load, there's current flow from the source, through the resistor, to the load. The voltage drop across the resistor will be \$V=IR\$. If there's not a load, there's no current flow, so there's no voltage drop across the resistor.
Now, when you say "at low frequencies", as long as there's any frequency content, the capacitor will have some effect on the output. But for DC, you seem to understand the analysis correctly.