I am struggling to understand the nature of the relation between resistance, heat and the length of the cable. In the scenario where a battery is short-circuited, a short cable will heat up significantly faster than a long cable.

I understand the mathematics behind it: combining Ohm and Joule's law, we are getting that the heat is inverse proportional to resistance given constant voltage. Resistance is then proportional to the length of the cable hence longer cable generates less heat.

I am struggling to understand the physical nature of the process. Heat is generated due to collisions between free electrons and atoms in a conductor. Intuitively, twice as long cable contains twice as much free electrons, hence there should be twice as many collisions and twice as much heat generated (with constant voltage).

Why is then a longer cable results in less heat?

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    \$\begingroup\$ Maybe what is missing is a conception of the electric field (voltage gradient) in the wire. In the shorter wire the voltage gradient is much steeper so the collisions are more energetic. Not sure if that is physically accurate. But it is intuitive. And the power dissipation in a segment of wire is proportional to both voltage and current. So it works out OK mathematically, too. \$\endgroup\$
    – user57037
    Sep 17, 2021 at 5:30
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    \$\begingroup\$ Longer wire has more resistance, so less current flows given that there is same voltage being applied to the wire, thus power dissipated in the wire is less, and so it does not heat up so fast? \$\endgroup\$
    – Justme
    Sep 17, 2021 at 5:37
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    \$\begingroup\$ Just to keep your mind active, see today's article in Phys.org. \$\endgroup\$
    – jonk
    Sep 17, 2021 at 6:11
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    \$\begingroup\$ twice as long cable contains twice as much free electrons, It is not about the amount of free electrons but about the electric field they they're in, because the field is what makes the electrons move (current). In a longer cable, the field becomes weaker as the same batter voltage is divided across a longer wire. \$\endgroup\$ Sep 17, 2021 at 6:56
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    \$\begingroup\$ You should also not forget that a longer cable has more mass and thus a greater capacity to absorb heat. \$\endgroup\$
    – Bart
    Jun 21, 2022 at 13:28

3 Answers 3


By increasing the wire length you are not only increasing resistance R, and therefor lowering I and P (for constant voltage), but the surface over which the power is dissipated increase as well.

Heat, or to be more specific, wire temperature, is a result of how difficult or easy it is to dissipate the resistive power loss. For a wire in open air the main contributes are:

  • conduction, power increase linearly with temperature difference ΔT, neglectable unless your wire is insulated
  • convection, power increase quadratically with ΔT
  • radiation, power increase ^4 with ΔT, neglectable unless your wire starts glowing

You can find more details here: https://en.wikipedia.org/wiki/Heat_transfer_coefficient


Perhaps you need to look at this from the perspective of a voltage which is not constant, but where we keep current constant instead.

I'll represent a short length of cable as a single resistor, with value R. We can model longer and longer cables as multiple sections of the original short cable chained together, which is like placing lots of resistances in series:


simulate this circuit – Schematic created using CircuitLab

I have endeavoured to keep the current in each chain the same, 1A, by increasing the voltage across the whole chain. Notice that, by the potential divider rule, the resulting voltage across each section of cable must be 1V for this situation to occur.

Potential difference between two points is the amount of potential energy (in electronvolts) lost by each charge as it travels between those points. As an electron moves from one end of each section to the next, it travels across 1V of potential difference. Each electronvolt of potential energy the electron started with at the beginning of each section, became kinetic energy for a short while, then became heat energy after its interactions ("collisions") with stationary charges in the atomic nuclei of that section.

Each section will heat by the same amount for each individual charge that makes the journey from one end to the other end of that section. Thus, assuming uniform charge distribution throughout the metal of the cable, each section will heat by the same amount per second. This rate of heating is "power', given by \$ P = I \times V \$

Notice that section current I is the same in all instances (1A) and section potential difference V is the same in all instances (1V), so power dissipation in each section (heating in this case of pure resistance) is the same.

Consequently, then, from the perspective of constant current in a uniform conductor, it's easy to see why the total power-per-unit-length is also constant. You can increase the length as much as you like, and as long as the current doesn't change, the power dissipated as heat in any given millimeter of cable doesn't change. In other words, one end of the cable is heating exactly as much as the other, or any point in between, and you don't get any hot-spots.

However, if you keep the voltage across the cable constant, and you vary the length of the cable, the result is very different:


simulate this circuit

In this scenario we see that the power dissipated within the single length of cable is five times greater than the power dissipated over the entire length of the longer cable. Each individual section on the long cable is dissipating 5 times less than the single section in the short cable! Each electron making the entire journey in the top circuit is imparting 5eV of heating energy per 1m of cable, whereas each electron in the bottom circuit is imparting its 5eV over the entire 5m length.

That's why a shorter cable might get very hot, while a longer one remains relatively cool, but it's because the conditions under which it is operating are very different. This time we have allowed current to vary, as a function of total resistance of the cable itself.

Different real-life scenarios will be some mix of these two situations, constant current or constant voltage. For example, in domestic power distribution we rely on the power delivery cables having a very low resistance, so that they "drop" a very small percentage of the full 240V AC (or 120V). We want as big a chunk of that 240V as possible across the appliance.

When you are powering a kettle from the mains, the length of cable to that kettle won't matter much. Even by doubling its length, and doubling the total cable resistance, that resistance is tiny compared to the kettle's heating element. The voltage across the kettle will have changed very little, and the current through it will remain almost unchanged, and for the reasons I described in the "constant current" explanation above, the heating per unit length of cable will also remain the same.

However, when the appliance's resistance is comparable to the resistance of its supply cables, the story becomes more like the second scenario, in which current would be very much dependent on those cables, and shorter cables could potentially get very hot indeed.


Heat is generated due to collisions between free electrons and atoms in a conductor.

Not quite. All free electrons have thermal energy, and hence significant speed, and are "colliding" constantly. They are similar in this way to the molecules in air, which are moving rapidly, even when the air is "still".

Heat is created when a net motion of electrons is randomized by "collisions". That is, the collisions of randomly moving (I.e. no net current) electrons does not produce heat. To continue the analogy with air, the net movement of electrons is like a wind or breeze in the air. The electrons themselves, like the molecules in the air are moving much faster than the net drift velocity.

Now you are correct that a longer wire has more free electrons. But it isn't the total number of electrons that is important for heating, but their current. The longer wire will have more resistance, hence less current (other things being equal.)

  • \$\begingroup\$ You can also include the concept of drift velocity and how it changes as a function of the produced Electric Field which in turn can vary with the voltage. \$\endgroup\$ Sep 17, 2021 at 12:26

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