I think I was having trouble coming to terms with what voltage dropping means...does "all elements" include the battery?
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:
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.
I'm still confused about the battery though. Setting the negative terminal to zero, and moving across to the positive terminal, the voltage increases.
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?
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.
Err, so from what I remember when voltage drops it just means electrons came up to the resistor, and slowed down. 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?
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 +...?
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?