Thevenin Equivalent for circuits with dependent sources

In summary, the conversation is about solving a circuit with a dependent current source asking for a Thevenin equivalent. The person asking for help has tried using superposition theorem but their answers do not match with the given answers. Another person suggests using KCL and node-voltage analysis to solve for Vth. They also provide steps to follow in order to find the correct answer.
  • #1
shanrei
1
0
Hi!

Good day! I was studying Circuit Theorems when I stumbled upon this circuit that has a dependent current source asking for a Thevenin equivalent. Up until now, I was only familiar with solving circuits with independent sources so this has me having a hard time. I tried using superposition theorem to solve for Vth but my answers don't match with the given answers. Can anyone please help me with this? Here is the pic for the circuit.

http://img230.imageshack.us/img230/3359/picture1um7.th.jpg


Thanks in advance!
 

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  • #2
Did you do KCL at node X (where the dependent current source is going into)? Can you see the relationship of voltage at this node to the current going out of the node? (Ix going into 3 and 4 ohms)
 
  • #3
First I would remove the sources and solve for the Rth.

Next I would use Node-Voltage analysis on the original circuit:
-use the node on the negative side of the voltage source as your reference node
-now the only nodes with unknown voltages are the upper-middle node and upper-right node, label as V1 and V2, respectively.
-write the constraint equation for Ix
-write the KCL equations for nodes V1 and V2
-solving these 3 equations for the 3 unknowns should give you the information needed to find this circuit’s Vth


I hope that helps you get the correct answer, and that I described it correctly :bugeye:
 

FAQ: Thevenin Equivalent for circuits with dependent sources

1. What is thevenin equivalent for circuits with dependent sources?

Thevenin equivalent is a simplified representation of a circuit that contains dependent sources. It is a single voltage source and a series resistance that can replace the original circuit without changing the behavior of the dependent sources.

2. Why is thevenin equivalent useful for circuits with dependent sources?

Thevenin equivalent makes it easier to analyze circuits with dependent sources by reducing the complexity of the circuit. It allows us to use standard circuit analysis techniques, such as Ohm's law and Kirchhoff's laws, to solve the circuit.

3. How do you find thevenin equivalent for circuits with dependent sources?

To find thevenin equivalent, we need to follow these steps:
- Remove all the independent sources from the circuit.
- Find the open-circuit voltage at the output terminals by setting the output to open and solving for the voltage.
- Find the equivalent resistance by shorting all the voltage sources and open all the current sources in the original circuit and solving for the resistance.
- Thevenin equivalent is the open-circuit voltage in series with the equivalent resistance.

4. Can thevenin equivalent be used for circuits with dependent voltage sources?

Yes, thevenin equivalent can be used for circuits with dependent voltage sources. The open-circuit voltage and equivalent resistance are calculated using the same steps as for circuits with dependent current sources.

5. What are the limitations of using thevenin equivalent for circuits with dependent sources?

The limitations of using thevenin equivalent for circuits with dependent sources include:
- It can only be used for linear circuits.
- The equivalent circuit may not accurately represent the behavior of the dependent sources at all operating points.
- It cannot be used for circuits with time-varying dependent sources.

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