I need help understanding why this op-amp's gain calculation behaves in a certain way.
Please assume the non inverting input is connect to the ground ( V+ = 0).
My issue is, that when assuming this is an ideal op-amp, I have been told that the 10k resistor going to ground does not affect Vout/Vin or Rin. So Rin is 10k and vout/vin is -10. I don't understand why this is the case.
But now, if we say this op-amp is no longer ideal, for some reason, I've been told that actually that extra 10k resistor going to ground does matter when determining vout/vin, and treating it as potential divider network, using superposition, we get V- as a sum of two potential divider circuits given that A = 1000.
I really can't understand why it didn't matter when the case was ideal but in non-ideal case it made a difference... could anyone please explain what actually is taking place in both cases..?