Voltage drop due to a resistor I'm afraid this question is going to sound rather dumb but here it is:
Suppose I have a very simple circuit: one battery and one resistor. The sum of the voltages in this circuit has to equal zero. My question is, how does the resistor "know" how much voltage drop to provide? How is it that it always manages accommodate a voltage drop that exactly equals to voltage provided by the battery? 
Because of V=IR, I can see that if the resistance is kept constant then the current flowing across the resistor will increase or decrease according to the voltage drop needed. But how and why does this "accommodation" occur?
 A: It's not a dumb question.
The electrons at the negative side of the battery have energy, and will look for "any path" to get rid of that energy - move in the direction of an electric field. When you create a circuit, you move the potential of the positive terminal closer to the negative terminal. If you assume that "wire" has no resistance (for simplicity) and the only resistive element is the resistor, then the electric field (potential) along the wire must be zero, so the entire potential of the battery ends up across the resistor.
At this point, electrons say "hey, we can start to move across this resistor", and they start moving. But as they move, they experience resistance - basically they are knocked off their path from A to B, and keep having to accelerate again. This "being knocked off course" is experienced as resistance (which results in heating of the resistor) and the time it takes the electron to pick up speed again (and be knocked off course again) is a function of the electric field - the potential difference.
Thus, there is a feedback mechanism; and it turns out that "resistance" is a convenient quantity with which to describe this "losing energy" as the electron traverses the resistor. So the "accommodation" you are asking about is really happening at the microscopic level.
Does that make it clearer for you?
A: 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
