[Coda]

Originally Posted by Potironette

I think I was having trouble coming to terms with what voltage dropping means...does "all elements" include the battery?

No. If you INCLUDE the battery, then the net voltage drop across the circuit must be zero (because the battery has a negative voltage drop because it's a source of potential).

So far, what I think I understand is that in an imaginary circuit with one resistor, say a lightbulb, and where the wires are perfect conductors, the change in voltage, or the potential difference, across the lightbulb will be exactly that of the battery, because in V = IR, the current doesn't have to be split up among more than one resistor's resistance. It's like the wires didn't exist...so:

This looks correct.

The thing that might be throwing your intuition is the idea that the same resistor is causing a different voltage drop depending on the voltage of the battery. The resolution to that is to realize that it means that the current (that is, I) is affected -- less electricity is flowing through the circuit altogether because of the resistance. What the resistor does is it inhibits the ability for that voltage to be translated into work elsewhere in the circuit. (This is a very desirable trait, because otherwise you're shoving excess power through things that AREN'T designed to deal with it -- connecting an LED directly to 5V is a good way to blow it out, but if you put a load resistor ahead of it in the circuit then the voltage feeding into it will be lower.)

On the other hand, when the wires do contribute to the total resistance of that circuit, then the current is affected by all "resistors," which includes the wires. This means voltage keeps dropping steadily all the way to the end.

Yep.

I'm still confused about the battery though. Setting the negative terminal to zero, and moving across to the positive terminal, the voltage increases.

This goes back to correlations not being a cause-and-effect relationship, which we were discussing earlier. Yes, you observe that the voltage is higher, but that doesn't explain anything about WHY that's true.

Although, it has to increase in order for the circuit to move electrons around. I guess chemical stuff inside the battery creates a potential difference from the negative end to the positive end mostly through the wires and mostly not within the battery, and that provides the potential difference required for resistors to turn potential energy of electrons into heat/light/etc. Except isn't it more like potential difference creates some sort of potential energy for electrons --> electrons move and so that's kinetic energy --> electron kinetic energy turns into heat/etc. in resistors?

Ideally, NONE of it within the battery -- that would cause batteries to discharge themselves when not connected to anything, which is definitely not what you want to have happen. :P

And yes, you are completely correct about the energy conversions.

And I guess that whole thing is related to series circuits (my class started on that immediately after the quiz) which says that the current across the entire circuit is the same, and the circuit's resistance (equivalent resistance) is made up of all the resistors' resistances, and the battery voltage is split up among all the resistors.

Yup. Goes along with what I said above.

Err, so from what I remember when voltage drops it just means electrons came up to the resistor, and slowed down.

Mm... Vaguely, but mostly no.

The entire circuit is already populated with electrons before you ever connect a power source. They're just not really moving. So when you add a voltage, that's applying a force to one end, but you can't really say that the electrons have to slow down when they get to the resistor, because they can't pile up on each other -- there's only so much room for them. This is why the current through the entire circuit is constant: the electrons can only move through the circuit as fast as they can move through the slowest part, because they're always being pushed into the flow.

Think about a circular tube filled completely up with marbles. If you push on one of the marbles, that's applying a voltage, and you can feel the resistance of friction, but all of the marbles move at the same speed. If you were to put a narrower bit of tubing in there somewhere, you'd have to apply extra force to shove a marble through it, but that wouldn't change the fact that they all have to stay at the same speed -- it just slows the whole thing down.

The friction in that narrower tubing has a normal force that acts against your attempts to move the marbles. The net force available to move marbles through the rest of the tube is, correspondingly, exactly that much less. That's the voltage drop of our marble-circuit resistor. And just like the normal force of friction is proportional to the force being applied, the voltage drop across the resistor is proportional to the voltage being applied to the whole circuit!

Now there's one side of the resistor that's more negative, and one side that's more positive, and so now there's a difference..? But how does the battery and overall current fit into this? The difference in voltage across the entire circuit just happens to be that of the battery because that's what's connecting the battery terminals? With the voltage drop being greater on resistors having more resistance because they just take more potential energy to power and slow electrons down more like that so the difference created across those resistors is greater? And all the while the current just accommodates for the resistors?

The battery is the reason that one side is more negative and one side is more positive. The difference in voltage across the entire circuit is equal to the voltage of the battery BECAUSE the voltage is defined by the work necessary to push a charge across the circuit! It's not a coincidence; it's a definition.

But yes, the resistors use up energy, leaving less energy available to do other things.

Also, for parallel circuits, whatever the "equivalent resistance," and the current and voltage for the circuit as a whole is, am I basically treating all the resistors in series as one resistor for the circuit? And it just so happens that because I_b = I_1 + I_2 +..., and I = V/R, V_b = V_1 + V_2+..., it works out that 1/R_eq = 1/R_1 + 1/R_2 +...?

I'm not quite sure what the question is. Can you try rephrasing it?

And, do parallel circuits affect everything just because if resistors aren't very resistant (whatever that mean) or incredibly not-resistant (whatever that means), and metals have free moving electrons, then electrons in all wires will be affected?

When you're mixing parallel and serial circuits, you can treat multiple elements on a branch of a parallel circuit as a single resistor, and you can treat the whole of a parallel group of elements as a single resistor in a bigger serial circuit.

So yes, that means that the effective resistance across the parallel part of the circuit DOES affect the current of the whole thing.