"Resistance of an electrical conductor is proportional to it length"

The intuitive explanation I found in many articles was that the greater the length of the conductor, such as a wire, the greater the number of collisions of the electrons with ions and therefore greater resistance. But how would this greater number of collision do anything to the drift velocity and therefore to the current $I$? Between any two collisions, the average speed of the electrons is the same and this affects the current? How would a greater length change this average speed between two collisions?

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    $\begingroup$ Imagine walking through a crowd of people standing still in a corridor of length $\ell$. You will be twice as tired from collisions if you need to walk through a crowd in a corridor of length $2\ell$, irrespective of your walking speed. Each individual collision remains equally probable, but the total number of collisions in the longer corridor can be expected to be twice what it is in shorter one. $\endgroup$ – ZeroTheHero Jul 9 '17 at 17:32
  • $\begingroup$ Incidentally, the resistance is proportional to the length of the wire. $\endgroup$ – ZeroTheHero Jul 9 '17 at 17:38
  • $\begingroup$ But if you change the resistance and keep the voltage same, you change the current? $\endgroup$ – Samama Fahim Jul 9 '17 at 17:40
  • $\begingroup$ $I = \frac{V}{R}$ and so a wire of greater length should decrease the current? Which means decreasing the drift velocity or 'walking speed' in the analogy? $\endgroup$ – Samama Fahim Jul 9 '17 at 17:46
  • $\begingroup$ Yes but the analogy is more like keeping the current constant and observing a larger drop in potential, i.e. you arrive at the end with less energy. $\endgroup$ – ZeroTheHero Jul 9 '17 at 17:47

Suppose we apply the same pd across a wire of twice the length. In that case the potential gradient halves, so the electric field strength halves. [This follows from $work=force \times distance$; the pd gives the work per unit charge; the field strength is the force per unit charge.]

If the electric field strength halves, the acceleration of the electrons between collisions halves, so the electrons' mean drift velocity halves. [The mean time between collisions isn't affected much, as the electrons' speed is almost entirely thermal, and far greater than the mean drift velocity due to the field. We make the crude assumption that on average each time it collides with the lattice, the electron loses all the velocity it has acquired from the field, and starts accelerating all over again.]

If the drift velocity halves, the current halves and the resistance doubles.

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  • $\begingroup$ nicely done using the gradient! $\endgroup$ – ZeroTheHero Jul 9 '17 at 18:46
  • $\begingroup$ Length affecting the field strength is the reason why the current would fall not the greater number of collisions as compared to that in a shorter wire? $\endgroup$ – Samama Fahim Jul 9 '17 at 20:21
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    $\begingroup$ I suspect they're equivalent ways of looking at it, but in my answer I tried to get away from a hand-wavy idea of resistance and discussed resistance as defined by $R=\frac{V}{I}$. $\endgroup$ – Philip Wood Jul 9 '17 at 21:01
  • $\begingroup$ What you mean is that no matter how we see it, we get the same results? Still the greater number of collisions has nothing to do with slowing down the current. $\endgroup$ – Samama Fahim Jul 10 '17 at 8:45

The fact that resistance is proportional to length and inversely proportional to area can be intuitively shown with a bit of thought experiment.

Say you have a conductor of length $L$ and when you apply $5 V$, a current of $3 A$ flows through it. So the resistance is $\frac{5}{3} \Omega$. Now take another exactly similar conductor. Now if you join them end-to-end, you have a conductor of length $2 L$. Now if you have to pass $3 A$ of current through this new conductor, how much voltage do you need?

Well, since each of the conductors required $5 V$ to pass $3 A$ through them, it will take a total of $(5 + 5)V = 10 V$ across the combined conductor to pass the same amount of current. Hence the resistance becomes $\frac{10}{3} \Omega$, which is exactly $2$ times the original conductor. If you took $n$ conductors and joined them end-to-end, you would have a new resistance of $n$ times the original resistance i.e. resistance is proportional to it's length.

You could think of doubling (or $n$ times) the area in same way as before. Applying same $5 V$ across two conductors joined side-by-side will create $3 A$ current in each conductor, total of $6 A$ (or $3n\ A$) through combined conductor. Making the new resistance being $\frac{5}{6} \Omega$, which is half (an n-th) of the original resistance. So resistance is inversely proportional to area.

Note: Now, you may say that this method just proves it for integer multiples of original length/area. To solve that, we could take help of differential lengths/areas. Let's consider the original conductor of length $L$ and area $A$. You could just say this conductor is actually stacked up vary tiny conductors of length $\frac{L}{N}$ and area $\frac{A}{M}$ where $M$ and $N$ are huge numbers. Rigorously speaking, we're talking about differential length and differential area. So now, you could just talk about any integer multiple of these differential conductors. That'd prove the proportionality for any length/area of the original conductor, not just the integer multiple ones.

N.B. What I did here is just simply thinking of increasing length as connecting more conductors in series increases equivalent resistance. And increasing area as connecting in parallel configuration which reduces equivalent resistance.

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