MHB Op-Amps Circuits: Explaining Prof's Method

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The discussion revolves around understanding a professor's method for solving op-amp circuits without extensive nodal analysis. Participants highlight the use of Thevenin's theorem, which simplifies the circuit to a voltage source and resistor. One contributor notes that when the output node is open, the current through two resistors must be equal, leading to a relationship between the voltages across them. This approach allows for a more elegant solution compared to traditional nodal analysis. The conversation emphasizes the effectiveness of Thevenin's equivalent in simplifying circuit analysis.
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Can anyone explain to me what my professor did? I am able to get the same answer, but not without a page of nodal analysis. It seems quite elegant and none of my friends know what method she used. Thanks!
 

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Rido12 said:
Can anyone explain to me what my professor did? I am able to get the same answer, but not without a page of nodal analysis. It seems quite elegant and none of my friends know what method she used. Thanks!

The Thevenin equivalent of a linear electrical network is composed by a voltage generator $V_{T}$ in series with a resistor $R_{T}$...$V_{T}$ is the output voltage when the output node is open [output current is 0...], $I_{T}$ il the otput current when the output node il shorted to ground [output voltage is 0...] and finally $R_{T} = \frac{V_{T}}{I_{T}}$...

Kind regards

$\chi$ $\sigma$
 
Hi $\chi \sigma$,

I have done that and applied nodal analysis at node $V_T$ and at ground, but I am unable to follow the logic presented by my professor:

$$V_A=\frac{10}{40}V_s=\frac{V_s}{4}$$
How did she get this?
 
Rido12 said:
Hi $\chi \sigma$,

I have done that and applied nodal analysis at node $V_T$ and at ground, but I am unable to follow the logic presented by my professor:

$$V_A=\frac{10}{40}V_s=\frac{V_s}{4}$$
How did she get this?

If the output node is left open, then the current in the 10 k and 30 k resistors must be the same, and that means that the voltage on the 30 k resistor is three times the voltage on the 10 k resistor...

Kind regards

$\chi$ $\sigma$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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