Resistors in Series: Why Don't Electrons Use All Energy?

[||spoon||]

Say there is a circuit that is extremely basic. Say a battery with two resistors in series (kind of pointless but anyway). If there was only one resistor the electrons would lose all their energy over that single resistor. Why (or how) when there are more components in the circuit, do the electrons "know" not to use all of their energy over the first resostor/component?

Hopefully that makes sense, had to write quickly so sorry for any confusion.

||spoon||

 

[nicksauce]

You should never, ever think about physics in terms of what particles 'know'. It is important to realize that physics MODELS various physics systems mathematically, and may not actually reflect how various atoms or electrons behave. If the models make accurate predictions, then they are good models. The V = IR model in this case, where R = R1 + R2, if a very good model. It is irrelevant to think about what the individual electrons are doing.

 

[russ_watters]

Agreed - if you drop a rock, how does it "know" how to fall?

Resistors resist - so when you put two together, you get more resistance. Physically, it really is that simple. Consider water in a pipe trying to pass a restriction. If you add another restriction, does it "know" to flow slower?

 

[||spoon||]

I know you shouldn't think about what particles "know" hence why I used " ". I only used know for lack of thought of a better word. I know that v=ir is a model. What i wish to understand is why it is the case for circuits that electrons will not lose all of their energy over a resistor only if their are more components within that circuit, whereas if there were no other components they would. Why is this the case? Is there some change in E-field or some other phenomenna which can account for this behaviour?

||spoon||

 

[russ_watters]

It is best to think of the problem in terms of flow rather than energy. It is easier to see that if the resistance increases, the amperage must decrease and that via conservation of eneregy, if a certain number of amps are flowing through one, they must flow through the other. From that, you can see that the energy dissipated must be the same.

Again, it is like the energy loss in a pipe due to resistance. Very similar concept.

 

[uart]

Say there is a circuit that is extremely basic. Say a battery with two resistors in series (kind of pointless but anyway). If there was only one resistor the electrons would lose all their energy over that single resistor. Why (or how) when there are more components in the circuit, do the electrons "know" not to use all of their energy over the first resostor/component?

Hopefully that makes sense, had to write quickly so sorry for any confusion.

||spoon||

The electrons lose their energy more slowly when there are two series resistor because the electric field is lower (as the voltage is shared across the two resistors), thus they reach a lower terminal velocity and lose less energy with each collision.

 

[rbj]

I know you shouldn't think about what particles "know" hence why I used " ". I only used know for lack of thought of a better word. I know that v=ir is a model. What i wish to understand is why it is the case for circuits that electrons will not lose all of their energy over a resistor only if their are more components within that circuit, whereas if there were no other components they would. Why is this the case? Is there some change in E-field or some other phenomenna which can account for this behaviour?|

independent of the resistance, the electrons "lose their energy" as a function of sort of how far they are getting from the "-" terminal and how close they are getting to the "+" terminal of the battery. if they get halfway there (where the voltage is halfway between the "-" and "+" terminals) they have lost half of their energy. it doesn't matter what the resistance was in between.

what the resistance does mean is best considered in terms of the reciprocal, the conductance. for a given voltage drop across a particular conductance, the E field inside that conducting material is proportional to the voltage (pretty much the voltage difference divided by the length between the terminals where the "perfectly" conducting wires are attached to the resistive material, E field is "volts per meter"). now the current through the conductor is proportional to the current density, j (pretty much the current density is equal to the current divided by the cross-sectional area of the resistive material that is at a right angle to that length between terminals previous alluded to). that current density is

$$j = \rho \ e \ v_d$$

where $\rho$ is the density of free electrons in the material (typically one or two free electrons per atom), e is the electron charge, and v_d is the average "drift velocity" (electrons in a conductive material are floating around in all directions, but if there is a general trend to move more in one direction than the others, you have a current flow in the conductor). now, these electrons flying around all over the place have a sort of constant average time (let's call that t_e) between when they collide with each other or the atoms (think of all these electrons in a sort of cloud or "electron gas"). each time they collide, they will on average have their drift velocity go to zero (after a collision, they are just as likely to be flying around in any direction equally likely, so their average velocity in the direction of the "+" terminal is zero after a collision). but between collisions, these electrons will accelerate (toward the "+" terminal) at an acceleration of

$$a = \frac{F}{m_e} = \frac{e E}{m_e}$$

where m_e is the mass of an electron and E is the E field strength (voltage divided by distance). so in classical mechanics, if a particle experiences a constant acceleration, their velocity increases linearly, and in time t_e, the velocity they will reach is

$$v = a t_e$$ .

but the average velocity, v_d, will be the average of that maximum average velocity v, and the minimum average velocity, 0, so that maximum velocity will be twice their average velocity which is their average drift velocity v_d. so

$$v_d = \frac{1}{2} v = \frac{1}{2} a t_e = \frac{1}{2} \frac{e E}{m_e} t_e$$.

that results in a current density of

$$j = \rho e \frac{1}{2} \frac{e E}{m_e} t_e = \frac{\rho e^2 t_e}{2 m_e} E$$

this is Ohm's Law on a microscopic scale. first of all, the stronger the E field (the more applied voltage), the greater the current density (and the greater the current). the term $\frac{\rho e^2 t_e}{2 m_e}$ is the "conductivity" of the material. the denser the free charge carriers (electrons), the more current density. the more charged those carriers are, then doubly the more current density. the more average time they are allowed between collisions, the fast they get and the more current. the heavier electrons are, the slower they accelerate, then less speed, and less current.

but, to anthropomorphize, these electrons are sort of like a bunch of drunk and mindless people in a big room wandering around in all directions bumping into each other and the walls and column supports. but there is this E field that is gently nudging them in a certain direction, and, if you were to draw a line on the floor perpendicular to the E field, the average number of people crossing that line per second is proportional to the strength that E field (how "pushy" it is).

 

[||spoon||]

ok thanks for that I think i understand what is happening a little better now.

Cheers,

||spoon||