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This is for the case of wires only (pipes are easy). I see that if I place a 9-Volt (constant) battery across any wire, the current is proportional to resistance. I don't understand why, for a constant potential difference, a current would be slower for certain lengths of wire, aren't all the electrons experiencing the same potential difference as before and therefore moving at the same rate?

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    $\begingroup$ Why would you think resistance has something to do with how fast the electrons are? $\endgroup$
    – ACuriousMind
    Jul 16 '14 at 17:36
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/126485/2451 $\endgroup$
    – Qmechanic
    Jul 16 '14 at 17:39
  • $\begingroup$ @ACuriousMind temperature affects resistance because temperature is really just average kinetic energy. Think superconductors(cold right?) But I really mean to imply that since the potential energy difference is the same, shouldn't the current also be the same in the case of two different-lengthed (but equal cross-sectional) resistors? $\endgroup$ Jul 16 '14 at 17:43
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    $\begingroup$ Just to throw a monkey wrench into the discussion: imagine a wire whose linear resistivity is zero but whose end-face junctions have a specified resistance. Then the resistance of the circuit is independent of the wire length but linear in the number of connected segments of this unobtanium wire. $\endgroup$ Jul 16 '14 at 18:29
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    $\begingroup$ @AndresSalas: "unobtainium" is a play on the word "unobtainable," see this site for more information. $\endgroup$
    – Kyle Kanos
    Jul 16 '14 at 19:54
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When you place a battery across any wire, the electrons on it starts to move. When electrons start to move, they get scattered from the nuclei present in the material which is the wire made from. This process creates the resistance. Thus, when the length of the wire increases, the amount of particles scattered from the nuclei increases which also increases the resistance. I hope this helps.

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    $\begingroup$ Thanks, but hmmm, I thought the electric field affected all particles equally. Isn't that why increasing the surface area actually decreases the resistance, since more particles are available to react to the call? If more particles always hurt, then the surface area increase wouldn't decrease resistance, see what I mean? $\endgroup$ Jul 16 '14 at 18:09
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Here's one way to look at it:

The electrons don't experience a potential difference: the experience a field (potential difference per unit length). Double the length = half the field.

Another way of thinking about this: if the total potential difference was 9 V for a 2 m length of wire, then we know that the potential at the mid point was 4.5 V. Thus, the situation of 2 meters of wire with 9 V is equivalent to two sections of 1 m wire with 4.5 V across them. For each of these shorter sections of wire it's easy to see that the current is half of what it would be if the voltage was double.

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  • $\begingroup$ You and qwartz I believe know what is right and where I'm wrong. I thought that the potential energy of the electrons was so quickly converted to kinetic energy that an electron that starts in an end of the loop will have attained its terminal velocity in picoseconds and therefore have no potential energy. $\endgroup$ Jul 16 '14 at 18:25
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    $\begingroup$ @AndresSalas - the electron drift velocity in a conductor is very small. In a vacuum tube an electron might convert all potential energy to kinetic energy - but not in a conductor. See for example recent discussion at physics.stackexchange.com/a/126094/26969 - seems relevant to help your understanding of the subject. $\endgroup$
    – Floris
    Jul 16 '14 at 18:41
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The electron moving from negative to positive terminal has to suffer lot of collisions with other electrons and has to also interact with the kernels present.Hence greater the length,, greater is resistance.

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For a constant potential difference applied across two ends, $\Delta V=\int\vec{E}.d\vec{l}$ or for simple cases, $V=E.l$

When you increases the length of the wire, electric field $\vec {E}$, decreases, thus driving the charges less and decreasing the current.

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  • $\begingroup$ Here is perhaps the flaw in my understanding. I thought that voltage and the electric field were one and the same thing. $\endgroup$ Jul 16 '14 at 18:22
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    $\begingroup$ no.they are not. $\endgroup$ Jul 16 '14 at 18:25
  • $\begingroup$ Well done. Well answered, I see where I went wrong $\endgroup$ Jul 16 '14 at 18:31
  • $\begingroup$ How does the electric field "know" to reduce exactly by the amount necessary to maintain a constant potential? Why is the potential constant? $\endgroup$ Jul 17 '14 at 22:08
  • $\begingroup$ because you are applying constant potential difference. when the distance between two points with a constant potential difference increases, the electric field or the driving force decreases. you can see the derivation of the relation V=E.l $\endgroup$ Jul 17 '14 at 22:16
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@AndresSalas -

The following is a bit of a simplification, but may help you understand.

When current flows through a wire, the electron that enters the wire doesn't immediately come out the other end. Rather, it's own electric field pushes on another electron, and that electron's electric field pushes on another electron, and so on until you get to the end of the wire and an electron "pops" out. The longer the wire, the more electrons have to be pushed against the resistance of the wire to push an electron out the other end, thus the higher the resistance of the wire.

In answer to other questions you've posed in comments:

How does the electric field "know" to reduce exactly by the amount necessary to maintain a constant potential? Why is the potential constant?

The field doesn't have to "know". Using the same simplified example above, a little bit of the potential (force) is used to overcome the resistance against the first electron moving, then a little bit for the next electron, and so on. By the time you get half-way through the wire, half of the force has been used. If more had been used there would be less force along the second half, and the entire train of electrons wouldn't be able to move as fast. Because less force is needed to move the train through the first half at the slower pace (lower current), the amount of force used in the first half reduces, thus the potential at half-way self-adjusts.

How do we know that the battery will exude the constant potential difference? Why doesn't it vary?

It actually does vary, however I'll cover that later.

The chemical reaction pulls electric charges, in the form of ions (charged atoms or molecules), across the electrolyte with a fixed strength, depending on the chemical nature of the anode and cathode materials. These charges keep moving across until the voltage (potential) difference on the electrodes exactly matches the pull of the chemical reaction, stopping the reaction from pulling any more charge across. This voltage is the potential which pushes the electrons through the circuit. As the electrons leave the cathode, the voltage difference between the electrodes drops, allowing the reaction to pull more ions across, which increases the voltage back to where it was. This all happens so quickly that the voltage doesn't seem to change.

As I said, the potential difference does vary. The battery has a certain resistance itself, which forms part of the circuit "using" the potential difference (voltage). The more current is drawn from the battery, the more of the voltage is used up in the battery's resistance, so the less potential (voltage) there is on the battery's terminals.

In some batteries, such as lead-acid batteries, the resistance increases substantially as the battery gets "flat", so that a large current drain will cause a substantial drop in battery voltage. This is why a flat car battery may be able to turn over the car engine, but not start the car - the starter motor draws a lot current turning over the engine, and this current combined with the high internal resistance of the battery causes the battery voltage to drop so low that the ignition system can't work to ignite the fuel and start the car.

Note: Saying that the chemical reaction pulls the ions across is a simplification.

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