Find Thevenin Equivalent for Circuit - Basic Principles

In summary, the student is trying to find the Thevenin equivalent voltage and resistance for a circuit with just one loop. Thevenin equivalents are difficult to calculate, so the student is asking for help. The student is also trying to find the short circuit current for the Norton equivalent.
  • #1
cepheid
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Homework Statement



Find the Thevenin equivalent of the circuit shown below

Homework Equations



Nothing but basic principles

The Attempt at a Solution



Alright, I realize that it will take some time for the attachement to be approved, but it's a really simple circuit. It has just one loop. Starting from ground, you go up through a 10 V voltage source, across a 1k resistor in series with it, and then back down to ground through a "controlled" current source, which means going in the opposite direction of the current indicated by the arrow ('up'). Note that it is a controlled source, even though it doesn't look like it in the diagram. This is because Circuit Maker is mildly retarded and doesn't represent controlled sources using diamonds, but using some other much more complicated circuit symbol. So I just threw an independent current source in there. Anyway, the output voltage is taken to be across this branch (the current source). Anyway, the current across the 2k is ix, and the current supplied by the dependent source is 3ix. I realize I haven't done anything towards a solution yet, but that's because I'm having conceputal difficulty with the controlled source here. So to be fair, for the time being I won't ask for help with how to solve it except to ask, "what do I do with the controlled source when I am trying to calculate the Thevenin equivalent voltage and resistance?"
 

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  • #2
cepheid said:

Homework Statement



Find the Thevenin equivalent of the circuit shown below

Homework Equations



Nothing but basic principles


The Attempt at a Solution



Alright, I realize that it will take some time for the attachement to be approved, but it's a really simple circuit. It has just one loop. Starting from ground, you go up through a 10 V voltage source, across a 1k resistor in series with it, and then back down to ground through a "controlled" current source, which means going in the opposite direction of the current indicated by the arrow ('up'). Note that it is a controlled source, even though it doesn't look like it in the diagram. This is because Circuit Maker is mildly retarded and doesn't represent controlled sources using diamonds, but using some other much more complicated circuit symbol. So I just threw an independent current source in there. Anyway, the output voltage is taken to be across this branch (the current source). Anyway, the current across the 2k is ix, and the current supplied by the dependent source is 3ix. I realize I haven't done anything towards a solution yet, but that's because I'm having conceputal difficulty with the controlled source here. So to be fair, for the time being I won't ask for help with how to solve it except to ask, "what do I do with the controlled source when I am trying to calculate the Thevenin equivalent voltage and resistance?"

Applying KCL to the circuit you get
[tex]i_x = -3i_x[/tex]. So, [tex]i_x = 0[/tex] and [tex]e_{oc} = V_s = 10V[/tex]
 
  • #3
Hmm, that's kind of weird. So this configuration is not really possible with non-zero current unless a load resistor is attached?

Also, I know that you're supposed to find the "Thevenin equivalent" resistance of the network by hooking up a "test source" across the terminals and calculating the current drawn from it. When you do so, you are supposed to "turn off" the network's internal sources. Again, how does this apply to the controlled current source? Do I leave it as is, and add its current to that drawn by the test source?
 
  • #4
cepheid said:
Hmm, that's kind of weird. So this configuration is not really possible with non-zero current unless a load resistor is attached?
Yes.
Also, I know that you're supposed to find the "Thevenin equivalent" resistance of the network by hooking up a "test source" across the terminals and calculating the current drawn from it. When you do so, you are supposed to "turn off" the network's internal sources. Again, how does this apply to the controlled current source? Do I leave it as is, and add its current to that drawn by the test source?
Only the independent sources are cancelled. The controlled ones remain in the circuit.
There is an alternative way to calculate the equivalent resistance. You can calculate the short circuit current for the Norton equivalent. Since the Thevenin and Norton equivalents use the same resistance, you can calculate it as the quocient of the open circuit voltage and the short circuit current.
 
  • #5
I'll try all that, thanks for the tips.
 

Related to Find Thevenin Equivalent for Circuit - Basic Principles

1. What is Thevenin's theorem?

Thevenin's theorem is a basic principle in circuit analysis that allows us to simplify complex circuits into simpler equivalent circuits. It states that any linear circuit can be represented by a voltage source in series with a resistance, known as the Thevenin equivalent.

2. How do you find the Thevenin equivalent for a circuit?

To find the Thevenin equivalent, we need to follow these steps:
1. Remove the load resistor from the circuit.
2. Calculate the open-circuit voltage (Voc) across the load resistor terminals.
3. Calculate the equivalent resistance (Req) seen from the load resistor terminals.
4. Draw the Thevenin equivalent circuit with the voltage source (Voc) in series with the equivalent resistance (Req).

3. Can Thevenin's theorem be applied to non-linear circuits?

No, Thevenin's theorem only applies to linear circuits. Non-linear circuits do not follow Ohm's law and cannot be represented by a simple equivalent circuit with a voltage source and a resistance.

4. Why is Thevenin's theorem useful in circuit analysis?

Thevenin's theorem allows us to simplify complex circuits into simpler equivalent circuits, making it easier to analyze and understand the behavior of the circuit. It also allows us to quickly find the voltage and current at any point in the circuit without having to go through lengthy calculations.

5. What are the limitations of Thevenin's theorem?

Thevenin's theorem is only applicable to linear circuits, so it cannot be used for non-linear circuits. It also assumes that the circuit is in a steady-state condition, meaning that all transients have died out. In addition, it only applies to circuits with two terminals, so it cannot be used for circuits with more than two terminals.

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