In the image above, the subcircuit in the box in the top circuit is reduced to the subcircuit in the light box below it using Thevenin”s Theorem. I don”t really see how to get a Thevenin equivalent out of the first circuit. I spent a while on it to no avail – could someone please give me a hand? How can you make a Thevenin equivalent of the subcircuit highlighted in the top circuit?

There is a pretty much general method for finding the Thevenin equivalent circuit:

Find the open circuit output voltage. This means take the circuit you are trying to find the equivalent of, replace the load with an open circuit, and find the output voltage.

You are watching: For the attached circuit, find the open circuit voltage (voc) if the op amp is ideal.

In your circuit this means replacing everything to the right of node “B” with an open circuit. So the output voltage is given by the voltage divider rule,$$V_{oc} = v_sfrac{R_2}{R_1+R_2}$$

Find the short circuit output current. In your circuit this is given by$$I_{ss} = frac{v_s}{R_1}$$

Now the Thevenin source voltage is $V_{oc}$ and the Thevenin resistance is $frac{V_{oc}}{I_{ss}}$.

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edited Dec 7 “14 at 3:31

answered Dec 7 “14 at 3:13

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