To gain insight into what is happening, replace the op-amp with an ideal voltage amplifier model (we assume the gain \$A \rightarrow \infty\$):

^{simulate this circuit – Schematic created using CircuitLab}

Now it's easy to see two important points

- \$R\$ can only change the current through the 5V source - it has no
other effect
- there is no path for output current thus the output current is zero.

Thus, in this odd circuit, the output voltage adjusts to be 5V less than the voltage applied to the non-inverting terminal which, in this case, implies

$$V_O = -1.7\mathrm V$$

and the resistor is irrelevent to this result.

(Added to address edited and expanded question)

As I understand it voltage is simply current pressure measured with
respect to some reference point (usually ground). In this case, we
have Iin producing Vin "pressure"

I'm not sure what you mean by the "current pressure" but, in this circuit, it is commonly understood that the voltage \$V_{in}\$ is an independent variable - *a given* - which means that \$V_{in}\$ isn't 'produced' by \$I_{in}\$ but, rather, produced *externally* to the circuit.

To make this clear, one can explicitly add the external source to the circuit, e.g.,

^{simulate this circuit}

Now it's clear that \$I_{in}\$ depends on \$V_{in}\$ but \$V_{in}\$ is *fixed* by the voltage source, i.e., changing the value of \$R_{in}\$ will change the value of \$I_{in}\$ but not the value of \$V_{in}\$.

Intuitively, I'm thinking that the output pin "sinks" some current to
reduce the voltage at the summing point. But that sinking of current
would reduce Iin (since no current flows through the inverting pin).
The result would seem to be that Vin drops. But is this the case?

The voltage at the output of the ideal op-amp, if negative feedback is present, *will be whatever it needs to be so that the inverting input voltage equals the non-inverting input voltage*.

Now, this might mean that the output must *sink* current or it may mean that the output must *source* current.

In my opinion, the most intuitive, straightforward way to think about this is to apply voltage division.

By voltage division, the voltage at the inverting input is given by

$$V_- = V_{in}\frac{R_F}{R_{in} + R_F} + V_{out}\frac{R_{in}}{R_{in} + R_F}$$

This result is *elementary* and holds even if the op-amp is removed from the circuit and \$V_{out}\$ is produced by an *independent* voltage source.

So, at this point, we can ask the question

*What must \$V_{out}\$ be such that the inverting input voltage, \$V_-\$, equals the non-inverting input voltage, \$ V_+\$?*

A little bit of quick algebra yields the answer

$$V_{out} = V_+\left(1 + \frac{R_F}{R_{in}} \right) - V_{in}\frac{R_F}{R_{in}}$$

Thus, *if* \$V_{out}\$ equals the above, the inverting input voltage will equal the non-inverting input voltage.

just one more thing: in the case where Vout is positive what effect
does this have on Iin?

We can straightforwardly write the equation for \$I_{in}\$ as follows:

$$I_{in} = \frac{V_{in} - V_{out}}{R_{in} + R_F}$$

**But**, under the assumption that *\$V_{out}\$ is whatever it needs to be so that the inverting input voltage equals the non-inverting input voltage*, we have

$$I_{in} = \frac{V_{in} - V_+}{R_{in}}$$

Carefully note that, under the above assumption (which is the same as assuming an ideal op-amp), \$I_{in}\$ does not depend on \$V_{out}\$ *period*. This is a consequence of the constraint \$V_- = V_+\$.

In summary, assuming an ideal op-amp, there is no instant in which \$V_- \ne V_+\$.

For physical op-amps, we must add additional circuit elements to model the departure from non-ideal behaviour and that is beyond the scope of this answer.

## Best Answer

The left op-amp is working in inverting mode.

^{simulate this circuit – Schematic created using CircuitLab}Figure 1. Positive going input.^{simulate this circuit}Figure 2. Negative going input.It took me a while to see what was happening. I hope that helps.