The answer is simple, the resistor doesn't know what voltage drop to provide, and to some extent it doesn't "care" either (unless it's so high that the current fries it but that's another issue).
The "voltage drop" is only governed by the potential difference created by the generator (battery or other). If the circuit is open, the electrons have no path so there is no flow. As soon as the circuit is closed, the electrons have a path and start to flow across the conductor (a resistor is also a conductor).
The only thing the resistor will govern, is how strong the flow of electron will be from A to B (from one potential to another, but remember this potential was created by the generator, the resistor doesn't need to know it).
I often do the analogy between electricity and hydraulic flow. Consider the difference of potential (the voltage) created by the generator as a height, and the resistor as a slope. The flow of water will be different in the same way as the electrons in a circuit.
Consider the 3 scenario in the following image:
- Scenario
A:
No resistor to close the circuit => no current flow possible.
- Scenario
B:
Low resistor. The electrons (or the water) fall from the high potential to the low one (rather rapidly).
- Scenario
C: The electrons (or the water) fall from high to low potential, still from the same height (your Voltage drop is identical). Except this time the higher resistance (flatter slope) makes it harder to reach the low potential (=> lower current).
Now imagine the high potential is 24V, or even 10,000, the mechanics are still the same, the electrons will flow from one potential to another as soon as they have a path, regardless of the resistor value. The only difference made by the resistor is how strong they will flow (how strong the current will be).
note: The analogies with the water flow are easily arguable and quickly reach their limits with complex circuits, this is not the exercise here. They still are a great way to explain electric current in the simple cases
I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that electromagnetism is a conservative force, and that means that if you go all the way around you have to get back to where you started.
Think of it like a roller coaster that goes up and down hills--if the roller coaster starts at the top of a hill, rolls down without friction, and comes back around, it should get back to the top of the hill without any "extra" speed. Otherwise the rollercoaster could just go around again and again and get more speed for free with no input! In this analogy, resistors don't work like a friction force, they work more like the hills. You can put a charge on top of a hill (a high voltage place) and it runs out of steam (gets to 0 V) right before it is pushed back "up the hill" (i.e. crosses the voltage source)
Best Answer
In this case, the voltage $V_2$ is a node voltage which means that is the voltage between the output node and the ground node. By definition, the ground node voltage is zero
$$V_0 = 0 \mathrm V $$
But, by inspection, the voltage across the resistor $R$ in the left-most circuit is simply
$$V_R = V_2 - V_0 = V_2 - 0 = V_2$$
Similarly, the voltage across the capacitor in the right-most circuit is simply
$$V_C = V_2 - V_0 = V_2 - 0 = V_2$$
Keep in mind that the output voltage must be taken across two nodes of a circuit. If there is just one circuit element connected between the two nodes, the output voltage is simply the voltage across that circuit element.
However, there may be a complex network connected between the output nodes so it isn't generally true that the output voltage is just the voltage across a single circuit element.