If a potentiometer is basically a voltage divider, why are they made in different values?
It is true to say that the output voltage from similar potentiometers at the same position with different overall resistances will be the same into an infinite impedance. That is the same as saying "if we draw absolutely no current out of it".

Now often that is the case (or so nearly the case that it makes no difference). Let's say that we're using it to set a parameter on a large piece of control gear. It is almost certain that the first thing that control gear will do is to buffer the signal - that is to pass it to an amplifier that doesn't amplify it, but draws nearly no current and makes a higher current version of the signal available. In this case, it really doesn't matter much what the overall value of the potentiometer is, and you'll likely find it is whatever is easiest and cheapest to manufacture.

However, let's say we chose not to do that. Take a look at the circuit below:







simulate this circuit &ndash; Schematic created using CircuitLab

Let's say that we're trying to vary the brightness of a 100 Ohm light bulb powered by 12V. We could try to do this as in the schematic. Let's say that the potentiometer is at half-way. We're expecting 6V across the light bulb, which gives a light bulb current of 6 / 100 = 60mA. However, what we actually get is different. The current for the light bulb has to flow through the "top half" of the potentiometer - 5kOhms and then splits itself between the bottom 5kOhms and the 100 Ohm light bulb. The next drawing shows this redrawn as separate resistors to show you what's going on.







simulate this circuit

So let's try and calculate the voltage at the across the lamp, and hence the current through it. We're going to use the potential divider equation but with the bottom resistance calculated using the resistors in parallel formula, calculated taking R2 and the light bulb in parallel. This resistance = \$ \frac{1}{\frac{1}{5000}+\frac{1}{100}}\$ = 98 Ohms  (or thereabouts). 

So now, we do our potential divider calculation:

\$V_{out} = V_{in} \left(\frac{98}{98+5000}\right) = 12 \left(\frac{98}{98+5000}\right) = 0.23 Volts !!!\$

That's way off what we wanted, and our current (which we wanted to be 60mA) is now 0.23/100 = 2.3mA. So what happened? Well, the equations I used show it, but effectively the current that we pulled into our load partially "shorted out" the bottom half of the potentiometer. 

If you repeat this equation with a 1 Meg Ohm load, you will see that it makes almost no difference and the output voltage is pretty much bang on 6V. Larger resistors pull less current and so load the output less, leaving the output voltage exactly where we wanted it. However, you can't just say "let's make all our lightbulbs 1 Meg Ohm resistance", or the voltages required to get any power out of them would be enormous. Remember that \$ Power = \frac{V^2}{R}\$, so for the same voltage, the higher the resistance of the load, the less power it draws.

Another thing to notice from this is that when you're calculating resistors in parallel, the answer is always smaller than the smallest resistor in the set - this is handy when checking results yourself.