|Feb26-13, 12:24 PM||#1|
I'm just trying to get a conceptual grounding for the circuit problems in our book.
So I understand that a circuit maintains a potential difference via redox reactions. The electrons have a desire to flow, so they do so.
The potential difference and the conducting wire allow the electrons to follow a path and reach the higher potential - conventionally opposite, but it doesn't matter.
The electric field inside the circuit is related to the current density by j = σE. From this, we can derive ohm's law.
This really is my question. The resistors are going to cause voltage drops due to the loss of kinetic energy by the electrons, correct? Afterward, what happens to electric field? There's no difference in potential between the bottom of the resistor and the lower potential of the battery. If so, the electric field must be 0 then? And if that, what pushes the charges to the other electrode? The repelling force of the other charges? And then, the electric field isn't constant in the circuit. I'm pretty confused, and would be very appreciative if you could enlighten me. Thanks for taking time to read.
|Feb26-13, 01:03 PM||#2|
Ohm's law reads as j=σE.
This is true locally, that is, for any infinitesimally small cross sectional area element da and thickness dl,
j.da = σ.E.da;
dI = (σ.da)/dl * E.dl.
When you sum this over the whole of the length of the container, you get I = V/ R,
where R equals the total resistance of the circuit
. Now this R includes the resistance of the resistor, the battery and pretty much everything present in the circuit
The answer to your question is that no, the Electric field between the bottom of the resistor and the anode is NOT 0 and there is a certain non-zero potential drop( or rise in conventional sense). However this is negligible to the potential change across the resistor and therefore as the description suggests, neglected.
But now that you are talking about ideal circuits, since they are an imaginary concept, why not just connect the resistor across the battery in the circuit diagram? Ugly right? hence the wires..
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