The traditional approach for explaining the inverting configuration, particularly "R1 phenomenon", is by using the virtual ground concept. My observation is that this is a rather formal and abstract concept and it is not well understood what it really is. For example, when we say that "the inverting and non-inverting inputs of an ideal op-amp are virtually shorted" (tonal_123), an obvious question arises: And how are they shorted? Maybe they are connected by a "piece of wire"? It should be explained in details what and where this "wire" is... since beginners often think that op-amp inputs are internally connected through it:)
That is why, I decided to explain the phenomenon of this connection without mentioning any ground - neither real nor virtual (both they still exist but not used in the circuit diagrams below). Thus my explanation will be valid even in the case of a "floating" circuit; it is universal. In short, my idea is:
Since the question is about the value of resistance R1, we can use more direct way of explaining it - in terms of resistances.
1. Ideal input voltage source (Rs = 0). Let's replace the specific op-amp circuit with a simpler equivalent electric circuit - Fig. 1a, where the op-amp output is represented by a variable voltage source with voltage VOA = VR2.
Fig. 1. An equivalent electric circuit of the op-amp inverting amplifier driven by an "ideal" voltage source
The result is surprising - it turns out the four elements of this configuration (VIN, R1, R2 and VOA) are connected in a loop; the same current flows through them and the sum of voltages across them is zero (KVL).
Trying to grasp the idea of this connection, we can distinguish three pairs of elements here:
> Composed voltage source (VIN + VOA). First, we can see the two sources VIN and VOA at the lower part of the circuit diagram. They are connected in series, in the same direction, so that their voltages are summed up. So we can think of them as of one composed voltage source (circled in yellow) with effective voltage Veff = VIN + VOA.
> Composed resistor (R1 + R2). Also, at the upper part of the circuit diagram, we see two resistors in series that can be thought as of one composed resistor with resistance Reff = R1 + R2. Thus the current is I = Veff/Reff.
> Composed zero voltage source (VR2 - VOA = 0). Then, we look to the right of Fig. 1 where there is another intriguing pair of elements - R2 and VOA (circled in blue). The current flowing through R2 "creates" a voltage drop VR2 = I.R2 across it. The op-amp creates, by the mechanism of the negative feedback, equivalent voltage VOA = VR2. These voltages are summed in series so that their sum is zero. Thus the voltage drop VR across R2 is compensated by the additional voltage VOA and the current is I = VIN/R1 as though there is no resistor R2 connected. Figuratively speaking, the combination of R2 and the op-amp output is equivalent to a "piece of wire" (outlined in pink).
> Composed zero resistor (R2 - R2 = 0). To make our (and OP's) life more interesting... and to show that circuitry can be not only a boring craft but also a romantic art... we can explain the same phenomenon through the mystic concept of negative resistance. Thus, besides understanding something very important in circuits, we will have fun too. We can reason in the following way:
The current I flows through the resistor R2 and, as a result, a voltage drop VR2 = I.R2 appears across the resistor. The op-amp "copies" this voltage so the voltage VOA = VR2 appears at its output. Both voltages are proportional to the current (Ohm's law) but while the first is subtracted from the input voltage VIN (since it is a drop), the second is added to the input voltage (since it is a voltage). So, we can make a conclusion that the op-amp output behaves as a kind of "inverted resistor". It is accepted to say that it is a resistor with negative resistance... in short, a negative resistor.
Figuratively speaking, the combination of the positive resistance R2 and the negative resistance -R2 of the op-amp output is equivalent to a "piece of wire" with zero equivalent resistance - Fig. 1b... and we can think of this network as a sort of "active superconductor" (outlined in pink). Another interesting thing about this artificial superconductor is that we can make its resistance not only zero, but even more than that...
This magic will continue until the op-amp is saturated. From this moment, the positive resistance R2 will come to picture. For example, if R1 = 0, should a short circuit to become since the total loop resistance is 0 + R2 - R2 = 0. But the op-amp will saturate and the current will be limited by R2 (0 + R2 - 0 = R2).
Of course, some input resistance (R1, Rs or both) is still needed to decouple the input voltage source from the op-amp inverting input and this way, to provide a negative feedback. If you connect an "ideal" voltage source directly to the op-amp input, the op-amp output will not be able to confront it through R2 and the negative feedback will not function. The gain will be maximum op-amp open loop gain... and the op-amp will be saturated. So both Vin and Vout have to impact the negative input through resistors, in opposite directions.
2. Real input voltage source (with Rs). When the input voltage source has some internal resistance Rs - Fig. 2, it is added to the other resistances. So there are a total of four resistors in the loop - three positive (Rs, R1 and R2) and one negative (-R2). But the last two are mutually destroyed and the current in the circuit is determined only by the first two - I = VIN'/(Rs + R1).
Fig. 2. An equivalent electric circuit of the op-amp inverting amplifier driven by a real voltage source
3. Without op-amp (VOA = 0). To appreciate something in life, you have to remove it for a moment. So let's imagine what the circuit would look like if the op-amp was gone and the right end of R2 was directly connected to ground - Fig. 3.
Fig. 3. Equivalent electric circuits of the op-amp inverting "amplifier" without op-amp (VOA = 0)
The negative resistance -R2 disappears and the positive resistance R2 comes to picture. The total resistance is a sum only of positive resistances - R1 + R2 or Rs + R1 + R2. As a result, the current decreases. This situation will happen when the op-amp reaches the supply rails (saturates).
But still... what is the idea behind these four elements in a circle (Fig. 1a)?
The idea is to put the input and output voltages in a steady proportion VOUT/VIN (= 1, > 1 or < 1) set by the ratio R2/R1 between two constant resistors R1 and R2. How is this acomplished? Here is another fresh explanation:
Think of the combination VIN and R1 as of a current source (sourcing voltage-to-current converter)... and of the combination VOA and R2 as of a current sink (sinking voltage-to-current converter). The current source and sink are connected in a loop so the same current should flow through both. The source is a "master" and the sink is a "slave" in this "tug of war" named inverting amplifier. So, when the input source increases its voltage to "push" more current, the op-amp output sink decreases its voltage to "suck" the same current...
The OP question was an illustration of how a delusion can be useful. It was obvious that an ideal voltage source (Rs = 0) cannot be connected directly to the inverting op-amp input (R1 = 0)... but the persistence with which OP defended this thesis made us think of the circuit idea... which was the most important thing. I hope my thoughts have helped you to realize the problem in depth.