Considering this simple circuit :
The potential is supposed to be constant along each wire (because they're conductors), such that the left wire in its entirety is at the same potential as the positive terminal of the battery and the same goes for the right wire and the negative terminal of the battery.
So what exactly happens to the potential inside the resistor ?
Answer
Let me first take a little detour away from this circuit to particle accelerators.
If you have some electrons in vacuum and a potential set up between two points (exactly the same as saying you have an electric field set up) you can accelerate your electrons. If you move a single electron through $1V$ of potential the electron gains $1eV$ of energy where $e$ is the electron charge. Moving through $2V$ gives the electron $2eV$ of energy and so on. So you can see that moving charge through a potential changes the energy of the charges. So in some sense you can say that the energy you get from your accelerator is $E=qV$ where $V$ is the voltage you have set up and $q$ is the total charge of all your electrons.
So going back to your circuit you can see that you have $1V$ on the left of the resistor and $0V$ on the right hand side.
As the electrons leave the battery they each have $1eV$ of energy. In the idealised circuit nothing happens until the electrons reach the resistor. At this point the electron collides with ions in the resistor (or phonons if you know what these are but it doesn't really matter) and gives up some of its energy to the lattice of ions. This happens enough that the electron basically has no energy by the time it leaves the resistor. (This isn't strictly true but all that really matters is that the difference in potential between the two sides of the resistor is $V$. You really have $V+\delta$ going to just $\delta$).
So from what I said earlier about the energy gained when you move charges through potential we got the equation $E=qV$. So if the electrons lose energy there must be a drop in voltage as the charge can't change (you can't lose or gain electrons from the wire).
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