IIRC With Clapton it doesn’t matter so much because the outer wire doesn’t carry any current.
*much current. I wouldn't say it doesn't carry
any. To understand it better, think in terms of conductance, the inverse of resistance.
In a typical Clapton wire, you have a core wire with comparatively low resistance, while the outer wrap, which is super-thin and
reaally long, provides a much higher resistance. Let's view an example, with simple numbers, for clarity. Let's say the core wire is .1 ohm, in isolation, and the outer wrap is 10. For conductance, we take the inverse of those numbers. So the core has 1 / .1 = 10 Siemens (or "mhos") of conductance while the outer wrap is 1/10 or .1 Siemens. When you add another route to the current path, it becomes more conductive rather than less. So putting these two together, the completed Clapton wire's conductance is simply the sum of its parts. You add the 10 and .1 together and you get 10.1 Siemens for the finished wire. Taking the inverse of this number, we see it has a resistance value of .099, which is very close to -- but not exactly the same as -- the .1 of the core wire by itself. Adding the outer wrap didn't affect the current path
very much, but it's definitely not zero.
