Hello there,
If the resistance of r in the diagram is set to 1 ohm and the pot is a higher value pot like 100ohms then when the pot is adjusted to the far right side you'll get about 600ma current limit (approximately).
There are a couple problems with this circuit as it is however. One is that they pot selection has a huge bearing on the adjustment of the circuit current limit. Even with a 10 ohm pot the resistance varies very strangely and so the adjustment isnt that good. For a few adjustments here's what we get:
r=1
Rp=10, I=0.66
Rp=5, I=0.72
Rp=1, I=1.2
Rp=0.5, I=1.8
Rp=0.25, I=3
Here's what we get using a 1 ohm pot:
r=1
Rp=1, I=1.2
Rp=0.5, I=1.8
Rp=0.25, I=3
Rp=0.125, I=5.4
The other problem is that if there is a high current surge on the output the transistor be junction can blow out, so a 100 ohm resistor is connected in series with the base of the transistor (the transistor that connects to the pot right now).
Another idea that works is to use a 1 ohm resistor for 'r', and wire the pot in parallel but connect the arm to the base of the transistor. That gives you a decent adjustment range even with a 1k pot, but you have to be watchful of the power dissipation in the 1 ohm resistor.