Sorry to contradict you, my friend, but that's not quite right :-)
The cable and the load form a series circuit (see diagram); so the total resistance of the circuit is
R = rL + 2 x rc
Now to get the total circuit current we just use Ohm's Law:
I = V / R
So if
rc increases (longer cable, thinner conductors
etc), then
R gets larger; and of course V remains the same (240V).
Therefore for a longer or thinner cable, the current drawn from the mains
, I, gets
smaller.
This also follows intuitively: imagine using a
very high resistance cable (say
rc = 1MΩ). You would rightly expect the current drawn from the mains to be
very low.
View attachment 65598
The reason you might need to use a heavier duty cable if
rc increases, is because the power dissipated in each conductor of the cable (due to Ohmic heating) is
isquared x
rc.
If you work this out using the above formula for
i , and assume that
rc <<
rL, you will get this power as (
V /
rL)
squared x
rc (see below).
So if you double the length of the cable (or halve the conductor cross-sectional area) then, regardless of the fact the current is lower, you lose twice as much power in the cable.
Note in fact that as you increase the length of the cable, the power dissipated
per metre remains roughly the same, so the surface of the cable doesn't get any hotter... UNLESS it's all grouped together on a cable reel ... which is why extension reels tend to melt if you don't fully unreel them.
View attachment 65599