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This problem is asking for I/o function, i.e V2/V1. I tried to solve this problem by first figuring out the Thevenin equivalents of input voltages and resistance before op-amp's input terminals but this approach is being discouraged by the question. Any insights on solving this problem will be appreciated.

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    \$\begingroup\$ This is a very special or "trick" circuit example that is designed to be able to be solved by inspection. Most circuits like this will not be able to be solved so easily "in your head". Note the classic R:2R ladder network and the fact that the last horizontal resistor in the ladder is 2r, and not R like the others. Note also that as Elliot says, the two opamp inputs will be at the same potential so the rh 2R goes to ground. Number the resistors from the left R1 R2 R3 ... . || R7//R8 = R = Ra. || Ra in series with r5 = r = rB. || rB IN PARALLEL WITH r4 = r = rC. | ... \$\endgroup\$ – Russell McMahon Jul 18 '19 at 11:39
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    \$\begingroup\$ You may notice a pattern here :-). | When you get to R1 R2 junction R1 sees R to ground so R1:R2 junction voltage is V1/2. Then ... . | You can work along the chain by eye and it soon becomes obvious what the voltage to the left of R7 is so what V2 is going to be. |||| Mentoring suggestion: Start by answering questions and/or asking good ones. Try to avoid petty editing of other people's questions that is main stylistic and/or just plain erroneous. Once you get a good feel for the technical aspects then technical error edits may be sensible. Nit picking may well annoy potential mentors :-). \$\endgroup\$ – Russell McMahon Jul 18 '19 at 11:41
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Use the Golden Rules of op amp circuit analysis. Since you have negative feedback, the voltages at the inverting and non-inverting inputs are (effectively) the same. With that information you can figure out how much current is flowing through the right-most 2R resistor. You could use any common circuit analysis method, but if you redraw that part of the circuit an easy approach may become obvious.

The other Golden Rule is that no current flows into the inverting or non-inverting input. So, where does the current flowing through the right-most 2R have to go? You now know the current through the feedback resistor as well as the voltage at one end of it, so finding an expression for the output voltage should be easy.

Rearrange this expression to get V2/V1 on one side. Done!

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  • \$\begingroup\$ Yes. But in this special case the classic R:2R ladder input is fun. A shame that not all circuits like this can be so easily solved by inspection. \$\endgroup\$ – Russell McMahon Jul 18 '19 at 11:43

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