I'm trying to understand the purpose of the capacitor C which is in parallel with the resistor R3. Current sources IB are indicative of the current offset which was the part of the lecture we had where we tried to compensate the offsets and the op amp is ideal.

When analyzing the DC regime the capacitor is an open circuit and all of current IB goes through the resistor R3 and when analyzing the AC regime it's a short circuit and no current passes through it.

Was it it's sole purpose to short-circuit the R3 resistor? Why would that be so important if that's the case other than reducing the total resistance of the circuit?

The capacitors are thought of as having infinite capacitance and IBs are DC current sources.

Ideal Op Amp Amplifier with Current Offset

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    \$\begingroup\$ I suggest that you re-draw this circuit without the capacitors, that circuit defines the DC behavior of the circuit. Now draw it again but now the "inf F" capacitors are replaced by a short (optional: remove the IB sources as these are DC only) now that gives us the AC behavior of this circuit. What conclusions can be drawn about each of these circuits? \$\endgroup\$ – Bimpelrekkie Mar 5 '19 at 12:50
  • \$\begingroup\$ When you use the CircuitLab button on the editor toolbar an editable schematic gets saved with your post. We can then copy and edit it in our answers. No need for a CircuitLab account, screengrabs, uploads, etc. and no background grid. Note: your ground symbols should point to, um, the ground, i.e., downwards. \$\endgroup\$ – Transistor Mar 5 '19 at 13:38
  • \$\begingroup\$ "your ground symbols should point to, um, the ground" especially because on some schematics, an upward-pointing triangle means "VCC". Not that it makes a difference when you're using an idealized current source, but still... \$\endgroup\$ – TimWescott Mar 5 '19 at 15:54

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