The Thevenin voltage isn't 10 V. Even with no load connected, there will be current flowing through the resistors so there will be voltage drops across them.
To find the resistance, just short the left end of the circuit, where the 10-V supply would have been, and find the equivalent resistance seen from A-B. It may help to redraw the circuit so it's easier to see what's in series and what's in parallel.
Hmmm I see. The nature of the drawing is a bit confusing to me. Would the 1 K and 2 K resistors be in series with each other, as the 4 K and 4 K would be as well? Or are they parallel. Normally, this is easy to identify, but I am struggling with this diagram for some reason. If I trace the current from +10 V, it would imply that 1 K and 2 K are in parallel, but I am unsure.
Once the left side is shorted, those pairs are neither in series nor in parallel. To be in series, they need to be connected end to end with nothing branching off in the middle, which isn't the case because of the shorted source. To be in parallel, they have to be connected to the same two nodes in the circuit, which they aren't. The 1K and 2K resistors are both connected to the top node, but one's connected to node A and the other to node B.
Look at the 1K and 4K resistor below it. You should be able to see they're in parallel even though they don't look that way as drawn.
Oh ok so after it's shorted, then the 1 K and 4 K are in parallel because they both connect to the unlabeled node, as well as the other side due to the short. wouldn't that logic still apply to the 2 K and the 4 K below it as well?
Ok so that means I can replace the 1 K and 4 K by a 4/5 K equivalent resistor, and the 2 K and 4 K by a 4/3 K equivalent resistor. Now these two are neither in parallel or series, so how can a Thevenin circuit be constructed? Don't I need a Thevenin circuit to be able to figure out the V and R?
Oh ok I think I see it now. So the ultimate equivalent resistance is 32/15 ohms. So I need to figure out how to use that to find I, right? I don't think V=IR applies here though.
No, to find the Thevenin voltage, you have to go back and analyze the original circuit with the sources. You could short A and B and calculate the resulting current Isc. Then since Rth=Vth/Isc, you could solve for Vth. Alternately, you could just find the open-circuit voltage drop from A to B, and that will be equal to Vth.
Take the original circuit. Put your hand over everything to the right of terminal A and pretend it's gone. What's the voltage at terminal A with respect to the 0V line?
Hint: It's a simple voltage divider.
Now disappear terminal A and the 1K and 4K resistors. What's the voltage at terminal B?
