OK, lets consider an artificial example with numbers made up to keep the maths simple.
Lets say you have a cable that has a resistance of 1 ohm/metre (lets just assume that's the loop resistance, i.e. the sum of 0.5R down one wire and 0.5R up it's partner), and you pass a steady 1 amp through it. It will therefore have a volt drop of 1 volt/metre, and generate 1 watt/metre of heat.
So if the cable is just 1m long, you'll have 1W of heat to dissipate. Make that 10m long but still pass the same 1A, you now generate 10W of heat, but have 10m of cable to dissipate that heat - i.e. still the same 1W/m. Make that 100m and the same holds - now 100W of dissipated heat, but over 100m of cable, and so still 1W/m.
Same cable, same 1W/m of heat to dissipate = same temperature rise in the conductors.
As I said, a very artificial example with "quite lossy" cable. You can see that if your supply is 240V, then at the end of a 10m length of cable, you are down to 230V - a loss of 4.1%. But if it were 100m of cable, you'd be down to just 140V left.
It also illustrates what's also been said about the effect on cable protection. Lets say you had a hard fault at the end of a 10m cable, you now have 240V dropped across 10R, and will pass 24A. If you had a 6A MCB protecting the cable, then it's possibly going to take quite a while to trip - and until it does, then the cable will be dissipating 5760W (I^2R=24*24*10), or 576W/m. Asking the same cable to dissipate 576W/m instead of 1W/m absolutely will make it heat up more
Were the cable only (say) 1m long, then under fault conditions the current would be more like 240A (assuming zero source impedance) - so the cable would generate over 57kW of heat (I^2R = 240 * 240 * 1). But the 6A MCB would trip "almost instantaneously" with such an overload.
Now, as already hinted at, there is the complication that the load drawn may well vary with voltage. A resistive load (such as a heater or incandescent light bulb that's lit) will decrease power and current as the voltage decreases. A switch mode PSU will typically increase current to maintain constant output power as input voltage reduces. In extreme (such as the 100m long cable above), an incandescent light bulb will initially take less power and current, but at some point the resistance of the filament will decrease and current draw may go up - typically you can expect an inrush of around 10x the "lit" current for incandescent light bulbs when switching on from cold.
Some railway modellers use that characteristic of light bulbs to protect sections of track - with DCC*, you could power (in theory) the whole layout from one supply/controller, but without any protection/sectioning, a fault anywhere on the layout would kill the whole thing. By using a low voltage (i.e. 12V) bulb in series with a track section, the normal train current won't light the bulb and it's cold resistance will be low enough not to cause problems. But should there be a short, that will light the bulb, it's resistance will increase and limit the current (while allowing other sections of track to continue working), and as a bonus the light will indicate both the fault and it's location.
* Digital Command and Control. Power is applied to the rails continuously, with a control signal superimposed upon it. Multiple items can run non one section, all fed from one power supply but controlled independently by the control signal.
