Exercise about Thevenin equivalent circuit

In summary, the conversation discusses the process of finding the Thevenin equivalent of a circuit. The speaker starts by calculating the equivalent resistance and then proceeds with finding the voltage at a specific node. They encounter some difficulties with setting up the equations and consider using node analysis or source transformation. Eventually, the correct solution is found with the current flowing through the 5 ohm resistor and the voltage at node 2 determined.
  • #1
Frank-95
52
1
Hi all :)

I'm trying to figure out the Thevenin equivalent of this circuit:

Circuito.jpg


I start by switching off every source and calculating the equivalent resistance:

RTh = RN = (6 + 14) || 5 = 4 Ω


Then I proceed with finding the voltage at 2.

I note that the nodes between 1 and 2 form a supernode:

Supern.jpg


So I set up the (super)nodal equation:

1 = v1/6 + 3 + v2/5

Then I proceed with mesh analysis. I note than the voltage source and the current source I1 form a supermesh.

Superm.jpg


And now I have some problems at setting up the equation :/

I know that:

-14 + 14*i1 + 5*i2 - 6(1 - i1) = 0

To bring this equation as a function of v1 and v2

I tried to think up about these equivalences, but I probably mistakened since the equation would become v1 = 0

14i1 = 14 - v2
5i2 = v2
6(1 - i1) = v1


Thanks in advance :)
 
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  • #2
If you have a problem with a mesh analysis why you do not proceed with the node analysis ? Also you can use a source transformation and convert a current source into a voltage source.
And your supernode equation is also wrong because V2 - V1 is not equal to 14V.

EDIT

The correct super mash look like this:
Circuitoa.JPG


So we have
I1 = 3A
I2 = 1A

and for loop I3 we have:

-14V + 14Ω*I3 + 5Ω*(I3 - I1) + 6Ω*(I3 - I2) = 0

And please notice that I3 is always in first place in parentheses.

And the solution is
http://www.wolframalpha.com/input/?i=-14+++14*x+++5*(x+-+3)+++6*(x+-+1)+=+0

I3 = 7/5A = 1.4A and the current flowing through 5 ohm resistor is equal to:
IR1 = I3 - I1 = 1.4A - 3A = -1.6A and this means that V2 voltage is equal to :
V2 = -1.6A*5Ω = -8V
 
Last edited:
  • #3
Thanks for the answer, I had mistakened with R1 resistance :)
 

Related to Exercise about Thevenin equivalent circuit

1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplification of a complex circuit that consists of a single voltage source in series with a single resistance. It is used to analyze and predict the behavior of a more complicated circuit.

2. Why is the Thevenin equivalent circuit useful?

The Thevenin equivalent circuit is useful because it allows us to analyze complex circuits using simpler calculations and assumptions. It also helps in designing and troubleshooting circuits by providing an equivalent circuit that has the same behavior as the original circuit.

3. How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, you need to first remove the load from the original circuit and then calculate the open-circuit voltage (Voc) and the equivalent resistance (Req). The Thevenin voltage is the open-circuit voltage, and the Thevenin resistance is the equivalent resistance.

4. Can the Thevenin equivalent circuit be used for non-linear circuits?

No, the Thevenin equivalent circuit is only applicable to linear circuits. Non-linear circuits have varying resistance depending on the voltage and cannot be represented by a single equivalent resistance value.

5. What are some limitations of the Thevenin equivalent circuit?

The Thevenin equivalent circuit assumes that the circuit is linear and that the behavior of the circuit is constant over time. It also does not take into account non-linear components such as diodes and transistors. Additionally, it may not accurately represent the behavior of the circuit at high frequencies.

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