I answered this question in my post you reply to. It is just a convention. This is an inverting amplifier. If you show both currents in the same direction then you need to invert Vout.

is it not current flowing in = current flowing out? so I1 = I2 + 0

It depends on what “don’t have the resistor” means. If the resistor is missing, you replace it with a large value, say 100 Mohm. If the resistor is replaced by a simple wire, you replace it with a 0 ohm value. Keep in mind that a zero ohm value for R1 does not make sense for this circuit.

Correct. I2 = -I1. I will update the equation.

The currents are shown in opposite direction conventionally, each one leaving the source Vin and Vout. Of course, physically, they both flow in the same direction. I1 and I2 flow towards output, if Vin is positive, and flow towards input if Vin is negative. if you want, you can write I2 = I1 (and invert I1 in the figure) and then apply Kirchhoff equations: (Vout-V)/R2 = (V-Vin)/R1. The result is the same, Vout = -Vin * R2/R1.

You can also write I2 = -I1 as in the article. One can choose the current direction as he wants. In that case the equation becomes (Vout-V)/R2 = – (Vin-V)/R1. As you can see, the negative sign is now in front of the parenthesis.

Should it be I2 = -I1 because they are flowing in opposite directions? Where does the minus suddenly come from?

Also since the op amp has high input impedance wouldn’t the current take the path of least resistance and bypass the thing by going through R2 to the output?

If I2 flows backwards and I1 forwards, summing at the node and none goes into the op amp, wouldn’t charge accumulate at the node too?

Thanks! 🙂

Read this article, part two: MasteringElectronicsDesign: The Virtual Ground. As for resistor biasing, let me know which resistor you are talking about in reference to an op amp in an inverting configuration.

Well, I disagree. If you read this article again, you will notice that I used an ideal op amp. In an ideal op amp, the inverting input is at virtual ground. The virtual ground concept is well established in electronics design. The op amp inputs are considered equal in literature and everywhere else due to the op amp high gain and the feedback provided by R2. It is a physical consequence of feedback when the op amp output is not saturated. It is a shortcut used in finding the transfer function of circuits with op amps in DC domain. If you would start with finite gains for circuits with 3 or 4 op amps the equations would become cumbersome, so you have to use this shortcut.

To be clear, I included a link to my article The Virtual Ground, published some time ago, which shows, with equations, why the inverting pin is at a potential zero. In the same article I also show, how to derive the inverting amplifier transfer function when the operational amplifier gain is finite, and the result when the gain is large (click here). So, for a complete answer, read that article as well.