simulate this circuit – Schematic created using CircuitLab
I am trying to reduce the T feedback network(R2 R3 R4) into a Thevenin equivalent circuit looking into the network from the v- (inverting) terminal. The Thevenin equivalent circuit I get is this:
simulate this circuit It is easy to see that Voc is just a voltage divider. To calculate the equivalent resistance a test source would be inserted then Isc (short circuit current) would be calculated and then the equivalent resistance is given by $$R_{th} = \frac{V_{oc}}{I_{sc}}$$ This bring me to my question: Is it possible to short the dependent source since $$ A(V_{+}-V_{-})$$ will equal some finite value that is independent of the thevenin equivalent; thus the VCVS is actually independent of this equivalent circuit and can be shorted like an independent source?
The book I'm reading gives the equivalent resistance as $$R_{th} = R_2+R4 || R3$$ Which would suggest that the dependent source has been shorted. To my knowledge we have to use a test source to determine equivalent parameters when there are dependent sources. Is this just a coincidence or can someone explain why we are able to short this VCVS.