Determine the Thevenin equivalent circuit

AI Thread Summary
The discussion focuses on determining the Thevenin equivalent circuit for a given circuit, involving calculations for open circuit voltage and Thevenin resistance. The user expresses uncertainty about the correctness of their solution and seeks clarification on current flow in a specific subcircuit with a dependent voltage source. They derive equations for the open circuit voltage and equivalent resistance, ultimately questioning how current behaves in relation to the dependent source. The conversation highlights the importance of applying Kirchhoff’s current law to understand current flow in the circuit. Understanding these concepts is essential for accurately determining the Thevenin equivalent.
zenterix
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Homework Statement
Determine the Thevenin equivalent of the circuit below.
Relevant Equations
##V=iR##
Here is the circuit that we need to find the Thevenin equivalent of
1707188885643.png

I am really not confident that the attempted solution below is correct.

Let's put a test current source between the terminals we're interested in
1707188942710.png


First let's compute the open circuit voltage. We set the newly introduced test current source to zero and compute the voltage between the terminals.

1707189010442.png

We have three unknowns ##e, i,## and ##i_2##.

We have three equations

$$I_0=i+i_2\tag{1}$$

$$i_2=\frac{e+\alpha i}{R_2}\tag{2}$$

$$i=\frac{e}{R_1}\tag{3}$$

Sub (3) into (2) to obtain

$$i_2=\frac{e(R_1+\alpha R_2)}{R_1R_2}\tag{4}$$

Now sub (2) and (4) into (1) and solve for ##e## to obtain

$$e=\frac{R_1R_2I_0}{R_1+R_2+\alpha}\tag{5}$$

Then

$$i=\frac{e}{R_1}=\frac{R_2I_0}{R_1+R_2+\alpha}\tag{6}$$

The open-circuit voltage is ##e+\alpha i## and this is

$$V_{oc}=e+\alpha i=\frac{R_2(\alpha+R_1)}{R_1+R_2+\alpha}I_0\tag{7}$$

Next, we set current source ##I_0## to zero and we solve for the voltage on the terminals we're interested in.

1707189332394.png


We have the same three unknowns as before, and the equations turn out to be the same as before except that instead of ##I_0## we have ##I## in the equations.

Thus, for this subcircuit we have

$$V_b=e+\alpha i=\frac{R_2(\alpha+R_1)}{R_1+R_2+\alpha}I\tag{8}$$

The Thevenin equivalent resistance is

$$R_{eq}=\frac{R_2(\alpha+R_1)}{R_1+R_2+\alpha}\tag{9}$$

By superposition, the voltage at the terminals of the original circuit is

$$V=V_{oc}+V_b=\frac{R_2(\alpha+R_1)}{R_1+R_2+\alpha}(I+I_0)\tag{10}$$

The Thevenin equivalent circuit is

1707189712443.png


Indeed, the voltage at the terminals of this circuit is

$$V=IR_{eq}+V_{oc}=V_b+V_{oc}=\frac{R_2(\alpha+R_1)}{R_1+R_2+\alpha}(I+I_0)\tag{11}$$

One of my questions is about the second subcircuit.

How does current flow in this subcircuit?

It seems that both ##i## and ##i_2## are positive.

$$i=\frac{R_2}{R_1+R_2+\alpha}I\tag{12}$$

$$i_2=\frac{R_1+\alpha}{R_1+R_2+\alpha}I\tag{13}$$

It seems that current is flowing from positive to negative terminals of the voltage source.

In fact, we seem to have

$$e+\alpha i=\frac{R_1R_2+\alpha R_2}{R_1+R_2+\alpha}\tag{14}$$

$$e=\frac{R_1R_2}{R_1+R_2+\alpha}\tag{15}$$

Thus, ##e+\alpha i>e##.

But then current would need to flow from the ##e## node to the ##e+\alpha i## node, right?

In summary, how does current flow through the dependent voltage source in the second sub-circuit above?
 
Last edited:
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After posting, I noticed an algebra mistake. Turns out it only affected the denominators in the expressions. It is now corrected above, and my questions are the same.
 
Last edited:
zenterix said:
In summary, how does current flow through the dependent voltage source in the second sub-circuit above?
There is a node between only two elements - the dependent voltage source and the resistor R1.
What does Kirchhoff’s current law state?
 
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