bobkolker
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Mr Davis 97 said:I've been studying circuits, but I can't find any resources that really answer my specific questions. So far I understand that a battery creates a potential difference between the two terminals, one of which is positive while the other is negative. There is a buildup of positive charge on the positive side and buildup on the negative side, and this creates an electric field which can accelerate electrons from a location of high potential to one of low potential. When you connect a conducting wire between positive and negative terminals, this creates a path of least resistance which electrons can flow through, where the electrons are accelerated by the electric field that the battery creates. Hopefully all of this correct so far. My main question arises when it comes to voltage drops. What really is a voltage drop? If we connect one resistor to the circuit, a circuit with 12 V, then there must be a voltage drop across the resistor of 12 V. However, why is this the case? And also, how are the electrons still able to move if on the other side of the resistor there is 0 V, since there is a voltage drop?
I know that most of this is probably wrong, so I need someone to really help me understand how circuits and specifically voltage drops work.
Here is a "must read" Please read http://science.uniserve.edu.au/school/curric/stage6/phys/stw2002/sefton.pdf
Sefton demolishes the "garden hose" analogy and shows power is transmitted as a field effect. The electric and magnetic fields produce the Poynting Vector field which is really what carries the power from the energy source to the resistance.
See also http://cq-cq.eu/Galili_Goihbarg.pdf and http://amasci.com/elect/poynt/poynt.html
http://amasci.com/elect/poynt/poynt.html
http://www.furryelephant.com/conten...tric-current/surface-charges-poynting-vector/
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