If one takes a -- 100K ohm 10 turn Clarostat, and places a 1K resistor of equal wattage, in parallel to the untapped element, that reduces the 10 turn to a less-than-1K 10 turn ??? yes / no ???

I am aware that if you parallel two 100 ohm resistors, you get an equivalent of around 50 ohms, but this represents a dividend between equals...

How far can one push the uneven balance ... a 100 K and a 1K and still have a useable variable element ?? how does this affect the current of the load... Which takes more of a load, or does the Ohms Law dictate that the current splits evenly across both...

Maybe I should burn up a few packrat components, and answer the question that way.... If I go that route, I'll publish results...

 

100K in parallel with 1K is 990 ohms.

 

I have something about how that distorts the curve...

(scratching sounds from the back room...)

Here it is:

#$%^ Those only apply if you connect the wiper to the helper resistor. With no wiper connection, it's probably linear. Let us know how it "turns" out.

 

Pictures added to the first post... the first shows what I meant to say... paralleling the internal fixed resistor, which has the fringe benefit of increasing the wattage of the unit. Power / voltage will be drawn as normal off the wiper tap...

The one showing two leads-to-leads, shows what I do not intend to do, someone misconstrued the text, thought I was trying to parallel as shown... not the case... the pots one could parallel in that manner, are stacked on a common shaft...-- as stated, the setup as shown would be an impossible fire hazard...

 

Just do the math, and you can plot yourself a curve of total resistance versus knob position.

Without going to that level of detail, just realize that once the pot resistance is >5K or so, additional turning to higher resistance will have very little effect on the total. So a 100K pot in parallel with a 1K resistor will have a very small range of effect.

[update] Just re-read your first post and realized I misunderstood the question. My first sentence still applies - that's the way I do these problems. Math is your friend. I may be the source of that pdf on this site. I might be able to convert it to another format if needed.

 

Without going to that level of detail, just realize that once the pot resistance is >5K or so, additional turning to higher resistance will have very little effect on the total. So a 100K pot in parallel with a 1K resistor will have a very small range of effect.

I may be the source of that pdf on this site. I might be able to convert it to another format if needed.

The range of effect, is the fact I was fishing for... Thank you very much, I shall dust off my calculator, and quit being so lazy

 

Here is what happens if you parallel a 100K pot with a 1K resistor: Nothing. If the pot is driven from a voltage source, paralleling another resistor just wastes some power.

If you connect two resistors as shown, you can modify the Vout vs pot shaft position, and reduce the effective source resistance on the output side of the pot. The sim shows what happens as a function of the resistance values that shunt the pot wiper.

 

Good! That will help Packrat until he can open the pdf in post #3.