Find the Thévenin equivalent \$R_{eq}\$ and \$V_{eq}\$ between terminals A and B.

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The solution is this: $$R_{eq}=1~\Omega$$ and $$V_{eq}=0~\mathrm{V}$$

I tried to simplify the circuit in this way:

enter image description here

I tried to find the current in this circuit:

Mesh Method

$$\left[\begin{matrix} \frac{6}{5}+1+3 \\\\ \end{matrix}\right] \left[\begin{matrix} I_1 \\\\ \end{matrix}\right] = \left[\begin{matrix} \frac{-18}{5}+I_1 \\\\ \end{matrix}\right]$$

This solves to:

$$ \left[\begin{matrix} I_1=\frac{-6}{7}\:\text{A} \\\\ \end{matrix}\right]$$

I don't know how to calculate the voltage from B to A or the \$V_{eq}\$ and of course the \$R_{eq}\$.


1 Answer 1


I am not an expert but I will try to help you.

  1. I suppose that the mesh current that you drew is \$I_1\$.
  2. It is not the most important thing when solving a circuit, but it's good to start by trying to find the best way to draw it so that it makes the most sense for itself. This helps to guess by inspection what kind of a problem you have, which method is the best, etc. I've done that and I find it much better now:


simulate this circuit – Schematic created using CircuitLab

Note that I have unified the \$3 \, \Omega\$ resistor with the \$6/5 \, \Omega\$ one.

  1. Something important that should've been extracted to make the solution more simple is that $$i_3 = - I_1$$. That's because the only closed circuit through which \$i_3\$ and \$I_1\$ can circulate is the same.

That means that through \$R_2\$ circulates \$i_3\$. Whit that information, you can pay attention to another mesh you can form in this circuit. Remember that you can form a mesh just by giving a label to any voltage, even open circuits. So, we are going to do that:


simulate this circuit

You can completely ignore the rest of the circuit because (since the \$R_2\$ branch is in parallel with \$A\$ and \$B\$) \$V_{Th}\$ and the voltage of the branch are equal, no matter what. Remember, the voltage between \$A\$ and \$B\$ is \$ V_{Th} \$. So, now maybe you can see that

$$V_{Th} = R_2 i_3 - i_3 = 1 \cdot i_3 - i_3 = 0 $$

and we did not even need to know the "real" value of \$i_3\$.

  1. Normally, you could now find the Norton (or short circuit) current, \$I_{SC}\$, and then the quotient \$V_{Th}/I_{SC}\$ is going to be your \$R_{eq}\$. The problem here is that \$V_{Th}\$ is zero, which does not necessarily mean that \$R_{eq}\$ is zero nor infinite, so the "quick" method does not work. Generally, this is solved using test source. So, you connect an independent source between \$A\$ and \$B\$ with a known (or maybe not, as you will see) value of voltage or current. Then you try to find the other value (of current if you chose a voltage source, or of voltage if you chose a current source) and then calculate the quotient, \$R_{eq}\$.

I chose a test voltage source \$ V_p \$ of \$1 \, \mathrm{V}\$ and I need to find the current through it, which I called \$I_p\$:


simulate this circuit

Okey, so we will use nodal analysis, and for that we have one critic node which I called \$C\$ as you can see on the schematic. Its voltage, relative to the nodal reference marked with an arrow pointing downwards, is \$ V_C \$. We will write the nodal analysis equation followed by one equation describing \$i_3\$:

$$ \begin{align} [1]& \quad -I_p + \frac{V_C + i_3}{1 \, \Omega} + \frac{V_C - 3.6}{4.2} &= 0 \\ [2]& \quad i_3 = \frac{3.6 - V_C}{4.2} \\ \end{align} $$

Note that if you substitute \$i_3\$ (from eqn. 2) into eqn. 1 the two terms will cancel out and you will be left with

$$ [3] \quad -I_p + \frac{V_C}{1 \, \Omega} = 0 $$

Check the schematic and make sure you see why \$ V_p = V_C \$, which means, by substituting into eqn. 3, that

$$ -I_p + \frac{V_p}{1 \, \Omega} = 0 $$

Do the math and you will finally see that

$$ R_{eq} = \frac{V_p}{I_p} = 1 \, \Omega $$

I am so sorry for the longitude of the answer, I think it would be useful if we go through the whole process together.

Final note: I chose \$V_p = 1 \, \mathrm{V}\$ just to make it simpler and to make it easier for you to ignore for the moment, but if you take a quick look over the process you will see that we could have just left \$V_p\$ without a set value and everything works just fine anyways.

  1. Now you are ready to give the Thévenin model of the circuit. Since \$V_{Th}\$ is zero, and a voltage source of zero volts is the same as just a wire, your equivalent will be a simple \$1 \, \Omega\$ resistor:


simulate this circuit

Hope it helps.


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