Thanks for the reply and explanation. I believe the input impedance of the CEM3340 is negligible.
Just so I'm clear; for ease of calculating, would it be satisfactory to say that pins 1 and 3 on the pot can be treated as 0V and 24V respectively, instead of -12V and +12V?
Therefore, assuming negligible impedance in the wire and into the IC, the total current across the pot i.e. pins 1 and 3 is:
I = V/R = 24V/100K = 0.24mA.
Therefore, when the tap is at the 100% position, as there is no resistance between pins 2 and 3, the current output through pin 2 would be 24A, however, this would have a mixing resistor of 100K in series with the output, so the current into the IC would also be 0.24mA, as per the above.
When the tap is at the 0% position with the 100K mixing resistor again included in series with the pot, the current would be,
I = 24V/(100K + 100K) = 24V/(200K) = 0.12mA.
Does this sound correct or am I being stupid?
Many thanks,
Chris