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.

No current flows, and the number of electrons on each plate remains the same. So now you have 2V across the new capacitance.

One view of the question is, how does the new voltage 2V across both capacitors square with the same charge Q on the top plate of C1, and the equation Q = CV? There does seem to be an inconsistency.

However, this is resolved by the formula for series connection of capacitors 1/C = 1/C1 + 1/C2 so the the capacitance has changed to C/2.

So with the same charge Q available from the resulting capacitance C/2, the voltage across it must be 2V.

This also squares with the conservation of energy E = 0.5 * C * V^2.

With both C1 and C2 storing energy E, the result of connecting both capacitors in series is 2E.

## Best Answer

This is standard technique and quite a good way to subtract voltages, as long as they are static/slow moving.

You use analog switches to first connect your capacitor between V1 and V2, now VCAP = V1-V2. Then you switch the capacitor to ground. Now you have V1-V2 referenced to ground. C2 holds the voltage on the output, while the capacitor is sampling. It takes some cycles for the voltages to reach equilibrium.

It is really useful if you are using an ADC in a micro, your ADC can sample the voltage when you switch to single ended, and your micro can control the switches. (C2 is not needed)

This has a lower parts count and low supply current, and lower cost than differential amplifiers. Matched resistors are not needed. It can also have far better balance/common mode performance than a simple opamp differential arrangement.

CD4053 can be used for voltages up to 18V, and 74HC4053 for voltages up to 6V (and both will work with -ve voltages) Other analog switches allow higher voltages, and you can use discrete fets, or opto-fets to do this at hundreds of volts.

^{simulate this circuit – Schematic created using CircuitLab}This technique is also very good with tiny input voltages, e.g thermocouples, as you can use an AC amplifier to eliminate drift and offset. e.g here we use an LM358 (filthy cheap, big offset and drift) to measure thermocouple voltages, in the presence of a large offset V2. Even though the gain is 1000, the offset is only uV at the input. This can be subtracted by the micro.

^{simulate this circuit}