How Does Voltage Influence Current Flow in Circuits?

[defetey]

So I'm sort of confused by voltage and I hope these will clear it up for me;

1) Why can parallel circuits only have equal voltage drops on both sides? Say a voltage of 5 is entering the parallel circuit, and one side uses 5 and the other uses 4, why is that not possible?

2) If voltage is what moves current by giving it energy, how come the current stays the same? Why doesn't voltage dropping it cause it to lessen and even stop near the end of the circuit? Is it cause the current coming in from behind the old current is pushing it?

3) Is voltage just basically charge attracting electrons from one point to another? If not, how is it created?

4) What happens if Kirchhoff's 2nd law is not followed, and all the voltage is not used up by the time the current reaches the end of the circuit and enters the battery? Does it just overheat?

 

[sophiecentaur]

I think you have some misconceptions about how Electrical Quantities are defined. (Probably not your fault because there is a high level of midguided B/S talked about them in many Schools)
Your Point 1 is like saying "Why does one stone fall 2m whilst a different stone falls 3m, when they are both dropped from the same height onto the same floor?"
The Voltage is the energy transferred per Coulomb as a charge is moved between two points. It's the same, whatever path it follows.

Point 2. No problem if you regard the Current (moving charges) as being like a bicycle chain; the same number of links just go round and round, not being 'used up'.

Point 3. The 'Volts per metre', or Field is what causes a force the electrons. The Volts aren't actually any kind of Force, as is popularly taught by non-Physicists - Voltage is, essentially, energy.

Point 4. K2 is NEVER violated, if you actually measure Every emf and Every IR, they will add up to Zero. We're talking Conservation Law. How can you suggest that the Voltage is not "used up" by the time you get back to the battery negative terminal? It will always measure as Zero (the two measuring probes will be side by side on a perfectly conducting piece of metal).

Does that help?