I want to analyze this circuit and find \$\frac{V_{out}}{V_{in}}\$. But how should I handle the four resistors in the square?
simulate this circuit – Schematic created using CircuitLab
Can I redraw the circuit as following?
Are now \$R1\$, \$R3\$ (and \$R2\$, \$R4\$) in series? I assume it is not the case because I end up with:
So my question is: How should I handle the four resistors?
Just look at this circuit. what I have done is that since both R3-R1 and R4-R2 are going to ground , I split it into two separate path flowing to ground. Now you can just re draw it and observe the working of this circuit. You can further reduce it by finding R1/R5.
EDIT: you can also calculate the Thevenin equivalent and find the voltage at +ve terminal of Opamp.
Your first transformation is correct (it's exactly the same circuit, nothing changed), your second is not. In general, electronic components are only "in series" when there's nothing connected between them.
R5 and R1 both go from V2 to ground, so they're in parallel and you can simplify them to a single resistor (let's call it R51). Next, R3 and R51 form a voltage divider. You can replace it with its Thevenin equivalent.
R4 and R2 also form a voltage divider that you can replace with its Thevenin equivalent. This means that the Thevenin equivalent voltage source is Vin*R2/(R2+R4) and its impedance is R2||R4. It connects to the inverting input of the OpAmp. R6 stays where it is because you can't include it in the Thevenin calculation due to it being involved in the feedback path.
After these transformations, you're left with a voltage source connected directly to the non-inverting input of the OpAmp. The OpAmp itself is connected as an inverting amplifier.
Keeping it simple: R4 is irrelephant. Ignore/remove it. It just makes extra heat. All it does is change the input impedance which does not matter since the circuit is driven by a voltage source.
