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Let us consider a fixed voltage. $ I = nAve $ is the relation between current, number of free electrons in unit volume $(n)$, cross-sectional area $A$ and electron drift velocity $v$. ($e$ is the charge of electron)

We know how to halve the current by modifying the wire, just by halving the cross sectional area $A$ of the original wire. Neither the $n$ quantity, nor $v$ is changed since we did not change the material of the wire.

Now how the Ohm's law fits here? It is said that halving the cross sectional area will double the resistance. But where $n$ is unmodified, where the material is same dense; how the extra collision comes that give rise to the resistance?

(Essentially, I am asking that, why increasing that area gave rise to the current? Because of there are twice many electrons to have a rush, or because of half less collision will take place? Resistance goes down indicates the second choice, and I think that should not be because our material is same dense. After all, resistances just lists rate of collision, isn't it?)

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  • $\begingroup$ What does v depend on? $\endgroup$ – M. Enns May 8 '16 at 21:22
  • $\begingroup$ v does not change here either. Some law states that already I is proportional to A if the material of wire is fixed, this also appears to be the reason behind R's increment. Fundamentally, I fall in a vicious circle. R should be changed if ohm had the right law, if the drift velocity equation was true than R should not be changed. $\endgroup$ – Tony Marshle May 8 '16 at 21:48
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The resistance of a given wire is more than the rate of collisions alone. It also depends on the wire cross section. If you have twice as many electrons drifting you have twice the current for a given potential over the wire. In other words the resistance is half.

You might be confused with conductivity or resistivity which is an intrinsic material property.

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$n$ and $e$ are properties of the material and they do not change.
Suppose you do halve the area to $\frac A 2$.
Think of the situation before when it was area $A$ equivalent two conductors in parallel each of area $\frac A 2$ and each part carries a current $\frac I 2$

Now what is it about the conductor with area $\frac A 2$ which makes you think it has twice the resistance?
It is the fact that for the same voltage the current is halved.
For what reason is the current halved?
It is because you have half the number of charge carries $n \frac A 2 L$, where $L$ is the length of the conductor, to carry the charge where before for the conductor of area $A$ it was $nAL$.

So it is not the number of collisions which has doubled rather it the number of charge carries which has halved which leads to a reduced current through the conductor.

Think of an analogy with two identical teams of people with buckets emptying a swimming pool.
There is a certain rate of emptying - volume of water per second - which is equivalent to the current.
One team's rate of emptying is half of that for two teams.

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