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.
capacitors are kind of like rechargable batteries. if you increase the voltage feeding them they charge up some, they absorb some of the difference between their voltage and the voltage source, if the voltage source drops they give some back to the circuit, esp if the voltage source goes away all together.
it goes like C dv/dt using calculus the capacitance times the change in voltage over time. doesnt matter if that change in voltage is from 10 to 100 or 3 to 7 or 27 to 13 volts.
When the source has a step change, the capacitor does not instantly step, that is what C dv/dt tells you, there is some period of time for that capacitor and that step change for the capacitor to store up (or release) enough charge to match the source, if it can. if your step changes are far enough apart then you can forget about step changes before, but if you change it fast enough it may not have caught up before it has to change again a fast enough square wave on the source and the capacitor makes it look like a sawtooth or like sharks fins...
the "resistance" is this capacitance times the change in voltage over time. be it an increase or decrease.
Best Answer
Talk about "current leading voltage" or "phase difference" only applies to AC analysis. In the more general case, one could say what a capacitor really does is differentiate voltage, according to:
$$ i = C\frac{dv}{dt} $$
From this, you can derive all sorts of well-known things about capacitors. Such as, if you want a linearly changing voltage across a capacitor, you must apply a constant-current source to it. As an example, consider a 1 ampere current source connected to a 1 farad capacitor:
$$ \require{cancel} \begin{align} 1A &= 1F \frac{dv}{dt} \\ 1A &= \frac{1 A \cdot s}{V} \frac{dv}{dt} \\ \frac{1\cancel{A}\cdot V}{1\cancel{A}\cdot s} &= \frac{dv}{dt} \\ \frac{1V}{s} &= \frac{dv}{dt} \end{align} $$
If you consider the case where the applied voltage is sinusoidal, then so too is the current:
$$ \begin{align} i &= C\frac{dv}{dt} \\ i &= C\frac{d\sin(t)}{dt} \\ i &= C\cos(t) \end{align} $$
because \$\cos\$ is the derivative of \$\sin\$.
You will also see if you graph these functions, that \$\cos\$ (current) leads \$\sin\$ (voltage) by 90 degrees, as an electrical engineer would put it: