Thevenin's theorem equivalent circuit

Question

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 :

Thank you!

Answer 1

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}\$).

^{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:

^{simulate this circuit}

Answer 2

Vth

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

^{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}\$:

^{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:

^{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:

^{simulate this circuit}

Answer 3

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