MIMF

Stupid question, but I just need confirmation (or rejection) of my thinking. I've been working on a friend's guitar amp and have it back to working condition. However, one of the gain pots has a broken shaft, and can't be adjusted. Unfortunately its physical size is small - about 1/2" diameter - and I can't fit a regular sized pot in the space. Its value is 100K, and I don't want to spend $10 (price and shipping) on a replacement, since he's financially destitute, and got the amp free at the dump.

I do have a 250K pot that will fit. I was thinking to parallel a 180K resistor across it, to keep the same resistance in the circuit. I know it will change the effective taper, but otherwise shouldn't have any effect on the circuit. Correct?

Thanks for any thoughts.

If that trick works it would be easy to test out with a VOM.
A 250K is pretty close to a 100K anyway so I doubt you'd hear much difference between them.

I finally got around to checking it out with a meter. It works, but with a reverse log taper. The resistance initially climbs rapidly, then more slowly, especially after the midpoint of rotation. I drew out the circuit and calculated and plotted resistances for several points. The pot is linear.

Pretty good for a simpleton like me.

I'm impressed Peter.
You can order the little 1/2" sealed pots from China on ebay, they run about $3 ea including postage.