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I took a guess at this multiple choice question and got it right, however I'm not sure how current through the 1 kOhm resistor is equal to \$i_o\$? enter image description here

I understand that for the ideal op amp, the voltages at both input nodes are equal to each other. However, because each of the voltages are divided across varying resistances, I do not understand how the 2 current values are equal. Can someone please help me understand this?

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    \$\begingroup\$ Well, does current flow into ideal op-amp inputs or not? \$\endgroup\$ – Justme Mar 24 at 16:58
  • \$\begingroup\$ Op amps are a new concept to me. So far, we have learned that 0 A flow into both inputs of an ideal op amp. So based on the wording of the question (current is flowing downwards through the 1 kOhm resistor) I imagined current flowing left to right through the 2 kOhm resistor and splitting at the node, with 0 A going into the op amp and the remaining current going into the 1 kOhm resistor. Since 0 A goes into the op amp, the 1 kOhm resistor takes all of the current coming from the 2 kOhm resistor. All of that being said, I still don't see how the 2 values are equal. \$\endgroup\$ – Vanidad Velour Mar 24 at 17:31
  • \$\begingroup\$ Exactly, no current flows in to or out from the op-amp inputs. So there is only one path where the current flows. What does Kirchhoff's law of currents (KCL) state about currents in a circuit? \$\endgroup\$ – Justme Mar 24 at 17:36
  • \$\begingroup\$ It is nicely illustrated in the answer below\ \$\endgroup\$ – Eugene Sh. Mar 24 at 17:38
  • \$\begingroup\$ Ah ok...It clicked! I see it now. Thanks! \$\endgroup\$ – Vanidad Velour Mar 24 at 17:48
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It's just a case of asking yourself what current flows into the input of an ideal op-amp then seeing what remains to flow in the resistors: -

enter image description here

The trouble is that it looks like the set answer is wrong because the current into the 1 kΩ resistor is upwards and not downwards. So, the right answer should be \$-i_0\$ as I see it.

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  • \$\begingroup\$ @EugeneSh. darn it I was too late to read the insults LOL \$\endgroup\$ – Andy aka Mar 24 at 17:23
  • \$\begingroup\$ I see. So, considering 0 A going into the op amp at both inputs, all the other components are in series, thus they all share the same current value, i_o. Is that how to interpret this? In class, we are told that for the ideal op amp, i_o goes into the op amp. I'm wondering if i_o splits at the far right node, with some of that current going into the op amp and some into the R resistor. That would throw off my calculation of i_o if I'm thinking the components are in series. \$\endgroup\$ – Vanidad Velour Mar 24 at 17:45
  • \$\begingroup\$ I upvoted but I'm still too new for my votes to count. Thanks! \$\endgroup\$ – Vanidad Velour Mar 24 at 17:49
  • \$\begingroup\$ \$i_0\$ exclusively feeds into the op-amp output. It is the value of \$i_0\$ that sets up the condition of the two op-amp inputs being ideally equal in voltage. \$v_0\$ dictates the current that flows in \$R\$. \$\endgroup\$ – Andy aka Mar 24 at 17:50
  • \$\begingroup\$ @VanidadVelour I suspect you may have misunderstood what was said in class. In an idea opamp circuit you consider the input impedance to be infinite. Since no current can flow under those circumstances that makes the analysis of the surrounding circuit elements easier. In the real world, most applications cause the input impedance to approach infinity so it can be ignored there also. \$\endgroup\$ – jwh20 Mar 24 at 17:51

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