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I am given this circuit

enter image description here

The voltage at each inverting ends should be 0, since the non-inverting ends are grounded. However, given that VIN is positive, there's a current flowing between each of the non-inverting ends (through the capacitor and resistor: hence to me it's equivalent to saying that the voltage through a series capacitor-resistor circuit is 0 even though there's current through both components). How can this be the case?

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  • \$\begingroup\$ What inductor???? \$\endgroup\$ – Andy aka Dec 28 '16 at 11:22
  • \$\begingroup\$ @Andyaka sorry capacitor and resistor, not capacitor and inductor. \$\endgroup\$ – The First StyleBender Dec 28 '16 at 11:23
  • \$\begingroup\$ You're right, a certain Vin will cause a current Vin/R to flow into the Vin node. It cannot flow into the opamp's input. So the opamp has to do something to make that current flow. You seem to have a problem with the current flowing through a capacitor. But is the current through a cap always zero ? Think about what happens if you charge a capacitor. What can the opamp do to make that capacitor charge ??? \$\endgroup\$ – Bimpelrekkie Dec 28 '16 at 11:28
  • \$\begingroup\$ @FakeMoustache When the capacitor charges up, will the voltage between the capacitor and the inductor (i.e. the output of the op-amp on the left hand side) be negative? \$\endgroup\$ – The First StyleBender Dec 28 '16 at 11:34
  • \$\begingroup\$ You keep mentioning inductor, now what are you talking about? Also what does this mean "there's a current flowing between each of the non-inverting ends"? \$\endgroup\$ – Andy aka Dec 28 '16 at 11:41
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For reference and to protect against future edits, here is the circuit we are discussing:

This is simple to analyze if you break it into its two distinct parts. You left out component designators in the schematic, so it will be difficult to talk about. You'll just have to guess which ones I mean. Next time, draw the schematic properly.

The IC1 is just a classic integrator. Due to the feedback, the opamp does what it needs to so that the - input is kept at 0 V. This means the current flowing thru R1 is directly proportional to Vin. The only place that current can go is thru C1. By the nature of what a capacitor does, the voltage across C1 is the integral of the current thru it.

The IC2 is just a classic inverting amplifier. Ignore the V1 input for now. The gain from the output of IC1 is simply -R3/R2. Note that IC1 actually inverts while integrating. IC2 inverts the signal again, so Vout is the positive integral of Vin.

V1 is just another input added to that from IC1. The gain from V1 to Vout is -R3/R4. This is simply added to the previous result. In that sense, you can think of IC2 as a summing amplifier.

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  • \$\begingroup\$ Perhaps it is helpful to add that the circuit will NOT work for real opamps (resp. a real non-ideal opamp model) because of missing DC feedback for IC1. Due to the finite input offset voltage, the output of IC1 will go into saturation. This is not a problem in case the circuit is part of a larger system with overall negative dc feedback. \$\endgroup\$ – LvW Dec 28 '16 at 14:12

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