3 Wires

For showing that 2 is the lower bound, see the other solution. For seeing if it is acheivable, Here is 1 easy possible answer Connect $A_1$ and $A_2$ with each each other and connect $B_1$ and $B_2$ to the potentiometer.

Case 1: Current flows. $A_3$ and $B_3$ Are corresponding ends. Check $A_1$ and $B_1$ , If they connect, or do not connect , either way We have anwer in 2 moves.

Case 2: Current does not flow. Check $A_3$ with $B_1$ . If it does connect, we know that they are corresponding ends . If they do not, $A_3$ and $B_2$ are corresponding ends . Check $A_2$ and $B_3$ If the match or do not, either way we have Answer in 3 moves

First off, there are six 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:

$\lceil \log_2(6) \rceil = 3$

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

Call the wires on one end A1, A2, and A3. And, call the wires on the other end B1, B2, B3.

1) Measure A1 vs B1:

   If they connect:
          2) Measure A2 vs B2
                   If they connect:  A1-B1, A2-B2, A3-B3
                   If they don't:  A1-B1, A2-B3, A3-B2
   If they don't:
          2) Measure A2 vs B1:
                   If they connect:  
                          3) Measure A1 vs B2:
                                 If they connect:  A1-B2, A2-B1, A3-B3
                                 If they don't:  A2-B1, A1-B3, A3-B2
                   If they don't:   
                          3) Measure A2 vs B2
                                 If they connect:  A1-B3, A2-B2, A3-B1
                                 If they don't:  A1-B2, A2-B3, A3-B1

If you liked this problem, try its harder counterpart, this one.