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Usually I don't care about algebra and just focus on the functionality. However, below circuit form appeared so many times in last few days in op amps that I thought of memorizing it by heart.

From my limited op amps study so far, it seems the entire op amp analysis boils down to figuring out the voltage at point \$\text{x}\$ shown in the diagram.

Working the equations using superposition gave a nice symmetry:
$$V(x) = \frac{V_1R_2 + V_2R_1}{R_1+R_2} = \frac{\left<V_1,V_2\right>\cdot \left<R_2,R_1\right>}{R_1+R_2}$$

If we think \$R_2, R_1\$ as weights, then \$V(x)\$ is the weighted average of \$V_1,V_2\$. It seems \$R_2\$ controls \$V_1\$ and \$R_1\$ controls \$V_2\$. This is beautiful and I'm dying to know if this symmetry has a nice physical explanation. Thanks!

schematic

simulate this circuit – Schematic created using CircuitLab


Few observations:

  1. When \$V_1=V_2=V\$ above formula gives \$V(x) = V\$. This means the entire branch including \$R_1, R_2\$ and every other point float to voltage \$V\$. Doesn't matter what \$R_1, R_2\$ are. This is used in instrumentation amplifiers to amplify AC while perfectly grounding common mode.
  2. \$x\$ will be virtual ground when the average value is \$0\$
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    \$\begingroup\$ I don't see how this question is related to op-amps. For the reason for the symmetry, see How do I use superposition to solve a circuit?. \$\endgroup\$
    – The Photon
    Mar 31, 2020 at 18:59
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    \$\begingroup\$ @beccaboo you can analize most resistor networks using weighed averages, so that's why this doesn't look related to op amps. Take a Y circuit for example \$\endgroup\$
    – FrancoVS
    Mar 31, 2020 at 19:27
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    \$\begingroup\$ I may be misinterpreting, but it looks like V1 is the input signal, V2 is the op amp output signal, and X is connected to the op amp inverting input (\$\small V_-\$). If that is the case, then \$\small V_X= V_- =V_+\$ That's what op amps do. In other words you can't use the voltage divider because X is not floating. It's unwise to 'not care about algebra' btw. \$\endgroup\$
    – Chu
    Apr 1, 2020 at 0:16
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    \$\begingroup\$ How so? I've been using for a while and getting correct answers.. \$\endgroup\$
    – across
    Apr 1, 2020 at 6:47
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    \$\begingroup\$ For example, to find \$V_{out}\$ of inverting opamp with noninverting input grounded, I solve \$\frac{V_{in}R_2+V_{out}R_1}{R_2+R_1} = V_x=0\$. \$\endgroup\$
    – across
    Apr 1, 2020 at 6:51

2 Answers 2

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I'm not sure where you're going with all the maths but the following may help. Often to get an intuitive feel for the various voltages in an op-amp circuit I tend to think of the circuit in proportional terms or, when looking at a virtual earth point, as a see-saw.

enter image description here

Figure 1. With the values shown in the OP's circuit the voltage at x can be graphically or mentally calculated as 1/3 of the way between 1 V and 10 V. The graph shows this to be 4 V.

enter image description here

Figure 2. The voltage see-saw about the virtual earth on an inverting op-amp.

When given the input voltage, V1, of the inverting op-amp circuit the output can be graphically or mentally calculated by using a see-saw or lever about the 0 V point with the lever arm lengths being in the same ratio as the resistor values.*

I hope that help.

Images: original.

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No, your assumptions are not valid, if the node you call "x" is the inverting input of an amplifier with negative feedback. If that is the case, and the (ideal) amplifier has not saturated, then the voltage at the inverting input (node x) will be the same as the voltage at the non-inverting input. But you haven't told us anything about how the non-inverting input is connected.

The analysis gets tricky if you want to make a KVL loop from the input to the output, because the voltage at x is determined by forces outside of that loop. You could include an ideal voltage source connected to x, with its value set to the value of the voltage at the non-inverting input, but you must also explicitly specify that no current flows into this ideal source.

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