4 Wires

First off, there are $4! = 24$ different ways the wires could be connected up, and every time you make a measurement, you get one bit of information (connected or not connected).

So, the lower bound on the number of times you'd need to hook up the potentiometer is:

$\lceil \log_2(24) \rceil = 5$

And, you can do it with five measurements as follows:

Call the wires on one end A1, B1, C1, and D1, and the wires on the other end A2, B2, C2, and D2.

(1) Short A2 and B2 together, and hook the potentiometer between them and A1.

If it shows a connection, you know A1 connects to A2 or B2. Otherwise, you know that A1 connects to C2 or D2.

(2) Connect the potentiometer between A1 and one of the two that you know it has to be connect to.

Now you have determined which one A1 connects to with 2 hook ups.

So, now you have 3 wires you need to figure out the connectivity of, lets rename them E1, F1, and G1 on one end, and E2, F2, and G2 on the other end.

This can be done, as follows:

---

3) Measure E1 vs E2:

   If they connect:
          4) Measure F1 vs F2
                   If they connect:  E1-E2, F1-F2, G1-G2
                   If they don't:  E1-E2, F1-G2, G1-F2
   If they don't:
          4) Measure F1 vs E2:
                   If they connect:  
                          5) Measure E1 vs F2:
                                 If they connect:  E1-F2, F1-E2, G1-G2
                                 If they don't:  F1-E2, E1-G2, G1-F2
                   If they don't:   
                          5) Measure F1 vs F2
                                 If they connect:  E1-G2, F1-F2, G1-E2
                                 If they don't:  E1-F2, F1-G2, G1-E2

So, the most it should take is $\boxed5$ measurements.