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Yes a tone put does make a difference to the very high end (as I stated earlier) a no-load pot effectively gives you infinite resistance so it is out of the circuit. It goes from 0 to 250K and then jumps to 1000000...K.

I have a no-load pot on my Tele but I don't like it in that position as its just too sharp sounding. I've often thought of swapping to out for a standard pot.

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Found a quote from this page in regards to removing the Tone pot, it seems to assume that all the pots are in Parallel though, is a Telecaster tone pot wired in series on a standard Tele therefore making no difference to overall resistance?
The more I research the more differing opinions I seem to uncover, I think just experimenting with different resistors until my ear tells me it sounds right is still going to be my best option.

http://www.guitarelectronics.com/catego ... g_faqs/#q6

What is a Fender No Load tone control and how does it work?
The Fender No Load Pot is used on some USA Strats, Teles and Fender basses and is wired like a standard tone control. From settings 1-9 it works like a standard tone then clicks in at 10 (full clockwise/ bright setting) and removes the pot and capacitor from the circuit. This eliminates the path to ground that exists with standard pots even in the full treble position. By eliminating the path to ground thru the pot, the only load on the pickup is the volume pot. So if 250K pots are used, the load is reduced from 125K to 250K and if 500K pots are used, the load is reduced from 250K to 500K (high resistance = low load) The reduced load allows more power output from he pickup and reduces the amount of high frequencies that bleed off to ground. This gives a noticeable increase in brightness and output in the full treble setting. The no load pot can be used in place of any standard tone control on any guitar or bass.

That article is wrong too. Just experiment with the resistors.

For combining parallel resistors, just turn each of the resistance values into reciprocals and add them up, and then turn the sum back into a reciprocal again.

For example, for three 5 ohm resistors in parallel, the reciprocal of 5 is 1/5, so:

1/5 + 1/5 + 1/5 = 3/5

The reciprocal of 3/5 is 5/3.

5/3 = 1.666 Ohms


For an 8 ohm resistor and a 4 ohm resistor:

1/8 + 1/4 = 3/8.

8/3 = 2.333 Ohms


For 4 x 8 ohm speakers

1/8 + 1/8 +1/8 + 1/8 = 4/8

8/4 = 2 Ohms


for a 250k resistor and a 500k resistor

1/250 + 1/500 = 3/500*

500/3 = 166.6k

*(when adding different fractions, find the lowest common denominator, in this case 500. 1/250 = 2/500)

...or use one of those new fangled calculator thingys

yeah like one of these...

Conceptually I used to find it weird that that in parallel even though you're adding a 'resistor' you end up with less resistance. until someone explained it with a flowing water analogy that if you split the flow more ways into channels, even though each path has resistance more flow gets through the system.

That reciprocal thing is good though - easy to remember.