Changing a Pot’s Adjustment Range

Short answer: what you want can not be done this way.

As you assumed, a potentiometer itself forms a resistor divider. The two parts (let's call them upper and lower) divide the voltage proportionally to the two resistor values.

Now hen you connect something (in your case the heater) to the output of the potentiometer you in effect connect it in parallel to the lower resistor. Two resistors in parallel have a total resistance of 1/( 1/R1 + 1/R2). When one of the parallel resistors has a very low value this means that you can ignore the other one.

This is what happens in your setup: in effect you have a resistor divider consisting of

• your heater, and

• the upper part of the potentiometer.

Your heater has a rather low resistance, so when you turn your potentiometer 'up' almost nothing happens, until the very end, when the resistance of the upper part of the potentiometer becomes comparable to that of the heater.

I don't know the resistance of the heater, but to create any heat it must be low. Combined with 24V this will result in a large current, for which your potentiometer is probably not designed. Don't be surprised when you see smoke coming out of your potentiometer.

Without switching, there is no way to do what you want effectively (that is, without wasting around 9 times the heat you get in your heater in the rest of the circuit). A simple solution using Pulse Width Modulation (PWM) could consist 555 chip circuit and a switching series transistor or FET.

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Based on new information:

You have a 24V DC source, a 10k potmeter, and you want it to produce 0 .. 2.5 V DC output? That's a very different ballgame, and an easy one.

The trick is not to reduce the output of the potmeter (because that suffers from the same load problem as I described above, although somewhat less because the resistors involved have a higher value), but the input. You can use the potentiometer itself as the lower resistor, you only need to add a top resistor (between the potentiometer and the 24V source).

The value is easy to calculate. You want 10k (your potentiometer) to get 2.5V. That leaves 24 - 2.5 = 21.5 for the top resistor. Each 1k drops 0.25V, hence 21.5V requires 21.5 / 0.25 = 86kOhm. 82k is a standard value.

A potentiometer works in the same way as a simple two resistor potential divider, so here on the left is the equivalent of the schematic on the right showing a potentiometer. The middle output being the wiper of the potentiometer. On the physical potentiometer itself the middle pin is usually the wiper but I would recommend that you check that with a data sheet for the potentiometer.

I hope that this clears up the connections and just for extra knowledge here is a formula which will work to calculate the output voltage from the resistance of the two resistors either side of the output:

$$\mathrm{Voltage\space Out} = \mathrm{Voltage\space In} \times \left(\frac{R_2}{R_1+R_2}\right)$$

^{simulate this circuit – Schematic created using CircuitLab}