Okay, I guess I miss my instructor days. Consider this diagram of four different circuits.
All circuits have 12 volts at the top and ground at the bottom, so they will all drop a total of twelve volts on each leg or path to ground. That cannot change.
Circuits A and B are series circuits. The total resistance of both circuits is 2.4 ohms, so they will both carry 5 amps of current, even though each resistor in circuit B drops an equal voltage that totals 12 volts.
Circuits C and D are parallel circuits. Each of the parallel legs has the same resistance as the single leg in A or B, but the circuit will carry 10 amps of current because current in each leg adds up instead of staying the same, as voltage does. You'd have to start changing resistance values to keep current under control.
Now let's add rheostats (or two leads of a potentiometer) into the mix.
Circuit E will carry all the amps you put into it when turned down to zero resistance, and pop your breaker, but will still carry 5 amps when turned up to 2.4 ohms. But that circuit doesn't control the voltage to a load like we need it to do.
Circuit F will now carry 10 amps when the rheostat is turned down to zero resistance because the overall resistance is now only 1.2 oms (0 on the rheostat and 1.2 on the load). It will carry 5 amps when the rheostat is turned up to full resistance, because overall resistance then goes up to 2.4 ohms. You are varying the voltage drop on the load as desired, but you're also radically changing the current through the load.
Now it gets really interesting. Circuit G uses a potentiometer instead. For the sake of demonstration, we'll leave the straight through resistance of the pot at 2.4 ohms on the right side. The resistance of the left leg is the load resistance added to the resistance you're getting off the pot in the right leg, giving you anywhere from 2.4 ohms (arrow all the way up) to 4.8 ohms (arrow all the way down) of total resistance through the left leg depending on where the pot is set. That still gives you a maximum of 10 amps in the overall circuit, but not through the load. It gives you anywhere from 2.5 to 5 amps through that leg, assuming my math is correct. Actually I think it's even less than that (I hate math, which is why I never became an engineer). In any case, the resultant current through the load is much better than the 10 amps we saw in circuit F, and that's far less damaging to sensitive electronic components. And you still control the voltage to the load exactly as you intended.
This gets even more interesting if you factor in that your load resistance isn't going to equal your pot value, and there are likely more components in the actual circuit than there are in these simplified diagrams. A 10k pot like you used radically alters the dynamics too. That's mainly why I always stick with potentiometers when engineers put them into a circuit.
Any of you engineer types out there can feel free to chime in here and tell me what I royally screwed up.