Consider the circuit shown below, which consists of an ideal op-amp and two resistors R1 and R2. Resistor R2 is connected from the output terminal of the op amp back to the inverting or negative input terminal. We speak of R1 as applying negative feedback. Iif R1 were connected between the output terminal and the non-inverting input terminal, we would have called this positive feedback. Note that R2 closes the loop around the op-amp. In addition to adding R2, we have grounded the non-inverting terminal and connected a resistor R1 between the inverting terminal and an input signal with an input voltage VIN. The output (VOUT) of the overall circuit is taken at the output terminal, i.e. between the output terminal and ground). Since the output impedance is ideally zero, the voltage will not depend on the value of the current that might be supplied to a load impedance connected between terminal 3 and ground.
Because the open-loop gain A approaches infinity, the V1 approaches V2. The input resistance of the closed-loop inverting amplifier is simply equal to R2. The output of the inverting configuration is taken at the terminals of the ideal voltage source A (VIN+ - VIN-), it follows that the output resistance of the closed-loop amplifier is zero. Whatever voltage is at node 2 will automatically appear at node 1 because of the infinite open-loop gain A, which means V1=V2. Because terminal 2 is at ground, terminal 1 is also ground and this ground is referred to as virtual ground.
Putting all of the above together, we obtain the equivalent circuit model of the inverting amplifier configuration under the assumption that the op amp is ideal.