EDIT: Oops, you wanted to measure resistance and not voltage. But pretty much the same thing still works- just use the resistance being measured as one of the resistors in theh resistive divider and the voltage measured will be proportional. YOu have to be able to apply a known voltage across the resisitive divider though.
Sorry guys. I think I am in over my head. I can get 5v or 12v to the 5K pot but not sure how to make resistance show on a LCD screen yet. I will do some more reading!!!
Ok so far I have figured out that I will need two resistors. One a constant resistance and the other the variable. power supplied to the resistors in parallel, the output of variable resistor to ground, measure the output voltage from constant resistor use formula to find the resistance of variable.
You can buy panel meters that display voltages between 0 and 10V, see Google for places where you can buy them from.
You can create a constant current source using an op-amp.
RL is your 0 to 5k variable resistor.
Rsc is a sense resistor.
Vref is a stable reference voltage.
[Latex]I_{RL} = \frac{Vref}{Rsc}[/latex]
In the example below.
Vref = 1V
Rsc = 1k
Therefore the current flowing through RL::
[Latex]I_{RL} = \frac{1}{1000} = 0.001A = 1mA[/latex]
Ohm's law states that V = IR which means that when the resistance is 5k 0.001 * 5000 = 5V so all you need to do is connect your panel meter's input across RL and (assuming it's input impedance is very high, >10M) you'll read 0 to 5V which corresponds to 0 to 5k for Rl.
To get V_ref I'd power the whole circuit of a 12V regulated supply and use a 2K 22k potential divider to get 1V.
Not that the op-amp needs to be able to accept input voltages down to 1V and output voltages down to 1V, an LM358 or MC33171 (I recommend the latter) will do fine. Also Rsc can be made smaller by adding another non-inverting gain stage or the output of the whole circuit can be amplified using a differential amplifier if needs must.