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I had an homework to do for school. One of the questions of this homework was to find the equivalent of a circuit using Thevenin's theorem. Unfortunately, my teacher told me that I didn't get the right answer for this question. I asked him what I did wrong but he didn't really looked at what I did and said that this isn't the right answer. I would like to know what I did wrong. Here is the circuit and how I tried to find its equivalent circuit using Thevenin's theorem :

enter image description here enter image description here

Thank you!

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  • \$\begingroup\$ please label the resistors so that they can be talked about \$\endgroup\$ – jsotola Mar 16 at 16:00
  • \$\begingroup\$ so what's the right answer? \$\endgroup\$ – Mitu Raj Mar 16 at 18:55
  • \$\begingroup\$ @Mitu Raj The right answer is VTh = -0.5 V and RTh = 1 Ohm. As you can see in the answer I made, I was able to find these values using Ohm's law. \$\endgroup\$ – A.Lacasse Mar 18 at 1:06
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Your \$V_{th}\$ calculation is correct. But it's that funny that you don't see it cz of the notation \$V_{AB}\$ that you are using through out. You are trying to find the thevenin eq. circuit looking from the terminals A and B. And You have finally derived that: $$V_{th} = -V_{AB}-1$$ But \$V_{AB}\$ is nothing but the open circuit voltage which is = \$V_{th}\$

So that makes the equation: $$V_{th} = -V_{th}-1\implies V_{th} = -0.5V$$ However your \$R_{th}\$ calculation is wrong. You just simply wrote down whatever resistor is left in the circuit. That's not the way to calculate it. Here is one method to find it:

In the first circuit, short all independent voltage sources, open all independent current sources and put an imaginary current source of 1A, across the terminals AB (across which you wanna calculate \$R_{th}\$).

schematic

simulate this circuit – Schematic created using CircuitLab

You can solve for \$i_x= 1A\$ using KVL and Nodal analysis on V. Solve to get \$i_x= 1A\$

Finally, \$R_{th}\$ can be found out by calculating the voltage drop across our imaginary current source and applying Ohm's law:

$$R_{th} = V/i = V_{R2}/i = i_{x}R_2/1=1\Omega$$

You can now draw your thevenin ckt:

schematic

simulate this circuit

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Vth

Following along with what you already wrote, here's your schematic (on top) with part identifications added:

schematic

simulate this circuit – Schematic created using CircuitLab

By using the Norton to Thevenin conversions for both current sources, it quickly becomes the what you see on the bottom case above.

This is easy to solve using nodal analysis for \$V_A\$ now:

$$\begin{align*}\frac{V_A}{R_{\text{th}_1}+R_2}+\frac{V_A}{R_{\text{th}_2}}&=\frac{V_A-3\:\text{V}}{R_{\text{th}_1}+R_2}+\frac{1\:\text{V}}{R_{\text{th}_2}}\\\\\therefore\\\\V_A&=\frac{R_{\text{th}_1}+R_2-3\cdot R_{\text{th}_2}}{R_{\text{th}_1}+R_2}=-500\:\text{mV}\end{align*}$$

Therefore, \$V_\text{TH}=-500\:\text{mV}\$.

Rth

We can re-write this schematic in the following way, applying \$I_x=V_A-1\:\text{V}\$ so that we now know: \$I_x-2\:\text{V}=V_A-3\:\text{V}\$:

schematic

simulate this circuit

Since the voltage difference across \$R_{\text{th}_1}+R_2\$ is fixed (it cannot change and will always have the same magnitude), we can go further:

schematic

simulate this circuit

Node \$X\$ doesn't matter, as it is behind a current source which has \$\infty\:\Omega\$. Because a current source has infinite impedance, it doesn't affect the Thevenin impedance of this result. So this leaves just \$R_{\text{th}_2}\$ as representing the final impedance.

Therefore, \$R_\text{TH}=1\:\Omega\$.

Summary

So here is the result:

schematic

simulate this circuit

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I was able to find VTh and RTh using Ohm's law.

enter image description here enter image description here

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