Electronic – What happens to the circuit Voltage, Energy

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.

For a shared current into a resistor and capacitor you might be tempted to say: -

\$V_{SUPPLY} = V_R + V_C\$ (incorrect)

This would not be true because the voltage across a capacitor does not rise and fall sinusoidally as the current rises and falls sinusoidally. For a capacitor the current and voltage looks like this: -

_{(source: electronics-tutorials.ws)}

In other words it is 90 degrees out of phase with voltage. This is because the basic formula for a capacitor is

\$I = C\dfrac{dV}{dt}\$

And, if V is a sinewave voltage then I has to be a cosine current.

If instead of real waveforms we drew them as phasors we would represent the voltages and current like so: -

So now if we want to "relate" Vsupply to the individual voltages of the capacitor and resistor we have to add them using pythagorous i.e.

\$V_{SUPPLY} = \sqrt{V_R^2 + V_C^2}\$.

It follows from this that impedances also add this way.