Solving Op amp with nodal equations

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Alex Santos
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Homework Statement


Find V0 and i0 of the circuit.

Homework Equations


KCL

The Attempt at a Solution


I tried multiple times and never got the right answer.
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Hello, I am having trouble solving a problem. I will link a picture of the problem with answers, but it just seems I can't get it right. Also if someone has a good rule of thumb when applying direction of current it would be highly appreciated to get to know it.
 
Attachments
  • Circuit schematic of an op amp with 18 V source and five 9 kΩ resistors, output labelled v0 with i0 shown, titled Figure P 6.4-2; page is rotated sideways.
    20141019_164039.jpg
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Can you show us at least one of your attempts? It's a Forum requirement that before you can receive help you must show some attempt in order that helpers can evaluate the level of help you need.
 
Pleas show as your attempt. And do you know that if op amp works in linear region the voltage at inverting input will be 0V.
 
Yes of course I am trying to do this once again, I will post my results and show you.
 
Here is my attempted result, as you can see the current I0 is incorrect and therefor the rest of the problem is incorrect. Any tips on strategy how to solve those problems is welcome.
Thank you for taking the time to help me.
 
Attachments
  • Handwritten working on lined paper, rotated sideways, for problem P.8.4-2 of a circuit analysis exercise: a resistor circuit with a 1V source, nodal equations at V1 and V2, and steps giving V2 = 6V, I1 = 1.667e-4 A and I0 = 1.3167e-4 A, marked Incorrect.
    20141019_171619.jpg
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Here is another attempt, This time I got the current right but in the wrong direction, It confuses me how to determine the direction of current, It always seems like its flowing out of the node all the time at every node.
Handwritten nodal analysis on lined paper for an op amp circuit: KCL at Va gives (Va-18)/9k + Va/9k + (Va-Vo)/9k = 0, leading to Va = 6V; then I = (Va-Vo)/9k = 2/3 mA and KCL at Vb gives 2Vb/9k + Io = 0, with Io = -4/3 mA, marked wrong direction.
 
I don't see where you determined V3 to plug into the last equation. Otherwise you seem to be doing okay.
 
It's an inverting amplifier, so V3 is going to be -6V here.