Just looking at the paths current takes from the pot to ground, I redrew the schematic and untangled it a bit.

Now to simplify:

1- If the power supply is large enough, you can ignore the small voltage drop of the diodes, so lets short those in the drawing.

2- Since we know the voltage across the coil when the pot is turned all the way in one direction, lets draw the schematic, with a 10k ohm resister and the source placed at one side. (Imagine the slider moving all the way to the left.)

3- Since 10k ohm is much large than 60 ohms, you can assume that any current flow in the 10k won't affect the current flow through the 60 much, so let's just cut that path out.

4- Since one of the mystery R's is just a path to ground, it won't affect current through the 60 ohm

*as long as it doesn't load the power supply much.*
so let's ingore that current path, but remember later to check that the supply isn't loaded much.

Redawing with all that gives me:

So, now it's pretty simple. We just need to know the supply voltage.

If V+ is 12V and the 60 ohms drops 3.8V, then the mystery resister drops the other 8.2V

3.8V / 60 ohm is about 63mA and 8.2V / 63mA is about 130 ohm.

So that would be 130 ohm for each of the mystery R's.

But still consider the draw on the source.

One path through 130 ohm draws about 90mA

And one path throgh 210 ohm (60+130) 60mA

(the third path through the 10K can still be ignored)

So the source needs to handle about 150mA. Can it?

That's just an example with a 12V supply. Plug in a different supply voltage and you get different values for the R's and also a different total current draw.

If the supply can't easily hadle the current then some of my earlier assumptions are no longer valid.

Also if the supply is much lower, (like arround 6V) then my assumption that the diodes can be ignored is invalid also.

--tim