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What affects the speed of electrons in a resistor?

If two resistors are connected in series, they both have the same current; same number of electrons passing at a point per second.

Suppose one resistor is a thin wire and one resistor is a fat wire. Do electrons travel faster in the thin wire or a fat wire of same material and external conditions.

Is the speed affected by the electric field strength? So Voltage / Length of resistor.

If so, then

$${\text{current} \cdot \text{resistance} \over \text{length}} = \text{current} \cdot \text{resistivity} \cdot {\text{length} \over \text{area} \cdot \text{length}} = {\text{current} \cdot \text{resistivity} \over \text{area}}$$

so thinner wires have a higher electric field strength hence electrons travel faster, is this true?

Note: it feels like the oppsoite of what you would think as it feels like electrons will have a easier time travelling through wires of large cross-sectional area.

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Edit after rereading the quesiton.

Agree that it is (all else being equal) easier for electrons to travel down a wire with a greater cross sectional area. (Not sure why this is not intuitive - for me it is easier to walk down a wide pavement than a narrow passage between two buildings - particularly if other people are about to provide some 'resistance' - also it is perhaps a bit like putting two resistors in parallel reduces the resistance - e.g. for two resistors of $R$ resistance in parallel the combined resistance is ${1 \over 2} R$)

Agree that thinner wires have higher electric fields (if compared to thicker wires with the same resistivity). Your analysis is correct if resistivity is constant.

One factor that you have not considered that might be important with some resistors is that the average speed of electrons will also depend on the number of free electrons per unit volume. This density of free electrons will contribute to the resistivity, but it is not the only factor that determines resistivity. In fact, another way to do the analysis in your question would be to think about the number of free electrons passing a point in unit time - for a narrower wire with less electrons per unit length of wire the speed has to be higher, but really this is just restating what you have in your question.

(from original answer)

the exact electron speed is harder to pin down. It is more accurate to think of 'average drift velocity' because the electrons are moving about so fast in the metal in all directions all the time that applying a voltage just slightly shifts the average speed over a bit in one direction so that there is a relatively small net drift velocity, which is the current we can measure. (Here small is used because teh net drift velocity will be very much smaller than the mean speed of free electrons in the metal).

Finally in one special case you want a high cross-section - at very high frequencies the 'skin effect' means that current only really goes through close to the surface so you need wires with high surface area (but generally they are hollow).

You might also want to have a look at this question about how electrons behave in a metal when a potential difference (or voltage) is applied.

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Suppose a conductor with cross sectional area $A$ has $n$ mobile charge carriers per unit volume, each carrying a charge $q$, which moves through the conductor at an average drift velocity of $v_d$.

Now the total charge in a segment $\Delta x$ is:

$\Delta Q = (nA\Delta x)q$

Now, the charge carriers moving at an average drift velocity of $v_d$ will move a distance $\Delta x$ in the time interval $\Delta t$, so the current through the conductor can be expressed:

$I_{avg}=\frac{\Delta Q}{\Delta t}=nqv_dA$

That is, the drift velocity is:

$v_d=\frac{I_{avg}}{nqA}$

So for two resistors in series, carrying the same current and made from the same material, but differing only in cross-sectional area, would have charge carriers with different average drift velocities. That is, the smaller cross-sectional area (higher resistance) will have charge carriers moving at higher drift velocities.

The thinner wire with higher resistance will have a higher voltage across it, which (asuming both resistors are the same length) means higher electric field strength in the higher resistor.

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