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I am trying solve the question shown in the picture below

enter image description here

(I am assuming the switch is down in the picture, I can't tell)

The equation that I am using is $$\frac{V_{\text{out}}}{V_{\text{in}}} = - \frac{R_{\text{f}}}{R_{\text{in}}}$$ because there is negative feedback.

For 2a), I ignored \$R_4\$ because there is no current input to the op amp, so I am assuming \$R_4\$ can be neglected. I combined \$R_2\$ and \$R_3\$ so that the total resistance for \$R_{\text{f}}\$ is \$1000\Omega\$. I then substituted \$V_{\text{in}} = 20\text{ mV}\$, \$R_{\text{in}} =1\text{ }\Omega\$ and rearranged to get \$V_{\text{out}} = -20\text{ V}\$. I am unsure if I applied the right logic to this question and I am don't know how to solve 2b).

This circuit is different from other inverting op amp circuits I have seen which is why I am struggling to answer it. How do I approach a question like this?

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  • \$\begingroup\$ AFAIK your logic seems right in lower position, for switch in Upper position, that formula doesn't apply and you have to derive new one. Use ideal op-amp's equivalent circuit for that.. \$\endgroup\$ – Deep Oct 1 '17 at 15:41
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    \$\begingroup\$ Simply treat the opamp circuit and the resistive divider separately. \$\endgroup\$ – user_1818839 Oct 1 '17 at 15:47
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Redraw the circuit like this

schematic

simulate this circuit – Schematic created using CircuitLab

The questions you have are essentially the same as finding the voltage at point a and b.

You worked question a correctly.

Now you can easily find out the current through R3 and use Ohm's law to find the drop across it. From the known voltage at a and the drop across R3, you can find the voltage at b.

Alternately, you know the voltage at the op-amp inverting input, and you can use the voltage drop across R2 to find V(b).

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    \$\begingroup\$ Thanks very much for your informative answer, I really appreciate the diagram too. I have done what you said and calculated the voltage to be 8V at b, is this correct? \$\endgroup\$ – Coconut Oct 1 '17 at 16:15
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    \$\begingroup\$ @s.twenty you are wrong. It is not 8V. It is -8V. \$\endgroup\$ – Chupacabras Oct 1 '17 at 16:21
  • \$\begingroup\$ @Chupacabras Thank you for the correction, I forgot the sign must be inverted too \$\endgroup\$ – Coconut Oct 1 '17 at 16:24
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Or much simpler

schematic

simulate this circuit – Schematic created using CircuitLab

Any output resistance inside the feedback loop will be hidden away by "infinite open loop gain" a.k.a. ideal operational

You can revert to studying the well known inverting amplifier on the right side of the diagram

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