I have a 10 kilohm potentiometer being powered by a twelve volt battery. After I run the circuit, the potentiometer begins to overheat. I thought that a 10 kilohm potentiometer could take 12 volts. Was I wrong or was it an issue with the potentiometer?
simulate this circuit – Schematic created using CircuitLab
Power is equal to the square of the voltage divided by the resistance, or $$P=\frac{V^2}{R} $$
For 12 volts and 10k, P = .0144 watts, which certainly will not overheat.
The thing is, you are not applying the voltage across the fixed leads of the pot. Instead, you are applying voltage to the wiper, and as you change the wiper position you change the resistance. As the resisistance goes down (the shaft approaches one limit) the power will go up. In the worst case, if you turn the pot all the way to its limit, the resistance will nominally become zero, and the power (nominally) infinite.
So don't do it. Or, if you must, put a fixed resistor in series with the pot so that the minimum resistance produces an acceptably low power.
Your circuit shows that as you move the wiper towards the bottom position that it will progressively decrease the resistance until it is a short circuit at the bottom. In that case the current through the wiper will be limited only by the battery's internal resistance and the wiper contact resistance. Most likely this current will be high enough to destroy the potentiometer.
If you edit your question to explain what you are trying to make we can help further.
Figure 1. Uxcell Ceramic Tube Adjustable Rheostat Variable Resistor.
If you really want to use a variable resistor for your circuit you may need a rheostat. You need to work out what the maximum current your "custom wire" will draw on a 9 V power supply and get a rheostat that can handle that current and has a resistance high enough for your application.
