Electronic – Flow of current in a simple inverting amplifier circuit

Assume V= 0 for now. Since the op-amp is ideal (other than noise), the inputs are always at the same potential (due to the the output voltage changing and through presumed negative feedback). If you have a current I_n then, in order for the inputs to be at the same potential, a current must be flowing through R3 equal to I_n. The resistors R1 + R2 across the inputs do not have any effect since the voltage across them is always zero (if R1 and R2 were both shorts you'd have a 0/0 situation).

However, the way you have drawn this circuit (floating V) no current can possibly flow through R3 because you don't have a ground anywhere. So no matter how high the output voltage gets, no current flows. With an ideal op-amp and ideal parts this creates another paradox.

The output current from the op-amp (as depicted in the picture in the question) is that current needed to keep the inverting input at ground potential.

So, with 1V at R1 (left hand side), there has to be -1V at the output to make the inverting input zero volts.

This means the current is -1V/100R = -10 mA.

If R2 were (say) double the value (200 ohms), the voltage at the output is -2V but the current remains the same.

In fact the current flowing in R2 is that current flowing thru R1 due to the input voltage. Remember the op-amp forces the inverting node always equal the non-inverting node and, if the non-inverting node is 0V then R1's current is simply input volts/R1. This is what is meant by a virtual earth amplifier.