When the emitter is connected directly to 0V it is at that potential... 0V ... it cant be at anything else cuz of the direct connection to the 0V line.
Now when you turn the transistor on you are now directly connecting the +6V line directly to 0V via the collector - emitter junction now you have a short circuit actoss the 6V power supply and as others said earlier the transistor is going to get very hot and probably die quickly.
The 4k7 resistor in the emitter line (leg) to 0V "lifts" the emitter / Vout point above the 0V potential.
Na... no hijack guys :) ya all chimed in whilst I was sleeping through sunday nite and working through monday morn :) thanks for the added input.
I'm not sure how many other ways I can explain this so Saladsamurai finally grasps what is happening :)
I know, I know guys ... I suck. Haha! And FYI, as infuriating as this is (my lack of understanding), I really am enjoying this too! This is how I learn something new. I struggle through it. I am not gifted by any means (hence the high post count consisting of all questions like this ). But in the course of the struggle, I pick up so much more information through these interactions than I ever could through a text. I could just as easily accept all of what you are saying at face value and go ahead and build some circuits in the lab and be done with it. But I think there is something more to be had from the learning process.
Ok ... back on topic. Let's discuss this part, since this exact image has crossed my mind during our discourse (added some labels to the image):
In Image A:ok... forget about the transistor being there for a minute, consider these 3 ccts (A, B, and C) below ....
what do you think the voltage in cct A will be at point 1
what do you think the voltage in cct B will be at point 2
what do you think the voltage in cct C will be at point 3 and point 4 (Vout)
imagine in cct C the first resistor @ 0.5 Ohms is the transistor junction resistance
explain why you will get the voltage readings you do
At point 1 I would have to say that the voltage is 6V relative to the negative terminal. If we start at point d and travel upward through the loop, there is a voltage rise of 6V and then no sink. So at every point in the wire, the voltage is 6V.
In Image B:
At point 2, again the voltage is 6V. If we start at point h and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 2. So any point in the wire from the + terminal to the resistor has a voltage of 6V.
In Image C:
At point 3, same deal of 6V for the same reason, start at point m and travel upward through the loop, there is a voltage rise of 6V and then no sink before point 3. At point 4 (Vout) the voltage would be 6V - I1R1 or from Ohm's Law Vout = 6V - 6V*R1/(R1+R2).