yeah, but as drawn, that resistor has no current, and thus no voltage drop. once it stops being a partial circuit and is attached to other things, then that will change.
I guess I should have made it more clear. This is a problem in which we are supposed to find the Thevenin equivalent. I had omitted redrawing the circuit on the farthest left of the diagram ( which is shown by dotted lines). The farthest right of the circuit (where Voc is shown) is supposed to be V Th. Do I still need to count the resistor to the right of the current source in my calculation? I thought there is no current floeing through that resistor since it is an open circuit situation?
You're right about Voc, sorry for the misinformation. Voc is the same as the voltage across the current source. The resistor does need to be counted in the Rth calculation, though. I deleted my previous post so I wouldn't confuse anyone else. Thanks for straightening me out Proton Soup.