Ok so not directly an OPAMP question but rather a very simple one and I just can't seem to understand it. For V1=-40E3(i2) where does the -40E3 come from? It's been two semesters since I took my first Networks class and I'm embarrassed to say I can't recall how to arrive there, Voltage Division maybe? That doesn't seem to work though. Thanks guys.

 

Got it, it's because i2 = -v1/R because of it being an inverting amp, makes sense now

 

Hello,

I have turned the picture over and changed the perspective.
This way the others can have a better look at the picture.

Bertus

 

Thank you, and I was wrong with my original assumption, it looks like it equals that because let's call the note next to the 2k resistor Vs. Then i2 comes from (Vs-Vi)/40k.
But why does Vs = 0? That's the only thing that seems to make sense but I am having trouble seeing why.

 

But why does Vs = 0? That's the only thing that seems to make sense but I am having trouble seeing why.

Because of Virtual Ground, note the use of capital letters.

The none inverting input of op amp is connected to ground. There is 0 volts at none inverting input.

The inverting input of the op amp MUST be equal to the none inverting input. So, V inverting = V none inverting, therefore V inverting = 0 volts because V none inverting is connected to ground which means it is 0 volts.

If you want more background, google: op amp virtual ground.

That is also how they found i3.
V none inverting = 0 volts.
V inverting = V none inverting = 0 volts.
i3=(V source - V inverting)/2k=(150 mV - 0)/2k = 75 microA
Your basic nodal analysis.