Electronic – the use of an op-amp whose output and inverting input are connected to ground

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.

How do you know the system is first order, was this given in the assignment? I ask because you said the box contained impedances, not any energy storage elements (capacitors or inductors). If you can confirm that it does not have any energy storage elements you can have a looked at the theory for two-port networks. The two-port theory can still apply if the circuit contains capacitors and inductors but the source would then have to be AC.

Basic two-port network is defined as:

The two port method to determine the impedance parameters are based on the following two equations:

We analyse one port at a time, if we open circuit the output port then I2 will be zero and we have solutions for Z11 and Z22.

Similarly if we open circuit the input port I1 will be zero and we have solutions for Z22 and Z12:

Whether the impedances are due to capacitors, inductors or just plain resistors can be determined from the phase difference between the resultant currents and the applied voltage. Capacitors will make the current lag behind the voltage, inductors will make the current lead the voltage and resistors will create no phase shift.