Norton's Theorem with Dependent Sources?

In summary, the conversation is about finding the Norton equivalent circuit for a given circuit. The person has correctly found the short circuit current and solved for the open circuit voltage, but is getting a different answer from the solution scheme. They wonder if they have made an error in setting the sources to 0, as they have been told that method only works for independent sources. They are seeking clarification on the issue and someone suggests using superposition to solve the problem.
  • #1
Master1022
611
117
Homework Statement
Find the short circuit current and hence the Norton equivalent circuit.
Relevant Equations
V = IR
My question is: What is wrong with my working/ method (in the attached pictures) to find [itex] i_{sc} [/itex]? I can get the Norton equivalent from there, but seem to get the same answer as the solution scheme.

Context: we are given the circuit depicted in the picture (initially with no connection between the [itex] V_{out} [/itex] nodes) and the question is building up to us finding the Norton equivalent circuit. From the previous parts, we have shown:
- [itex] V_1 = \frac{20}{9} V_{out} [/itex] for the open circuit condition
- [itex] V_{out} = 4.5 [/itex] Volts

The answer uses the method of setting the sources to 0, but we have been told (in lectures) that method is not valid when there are dependent current sources.

I have asked a peer and they suggested that the error may lie in the fact that I let [itex] V_{out} = 0 [/itex] volts, but I cannot see why that is the error.

I have attached two versions of the working, but hopefully, it is legible.

I would appreciate any help.

Scannable Document on 24 Apr 2019 at 20_54_13.png
Image-1.jpg
 
Last edited:
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  • #2
When doing Norton/Thevenin Equivalent circuits you need## i_{sc}## (short circuit current) and ##v_{oc} ##(open circuit voltage). You correctly found ## i_{sc}## for the short/closed circuit as or 3.84 A.

.

If you solve the open circuit condition you will get ##V_{out}## =4.5 V. Then to get ##R_{N}##, you simply use the equation:$$R_{N} =\frac{V_{oc}}{I_{sc}}=1.17\Omega$$

You seem to have everything right, so could you clarify why you believe you have an error.

Also, you can use superposition applying one independent source at a time. Treat the dependent sources as resistors and always leave them on.

 
  • #3
LeafNinja said:
When doing Norton/Thevenin Equivalent circuits you need## i_{sc}## (short circuit current) and ##v_{oc} ##(open circuit voltage). You correctly found ## i_{sc}## for the short/closed circuit as or 3.84 A.

.

If you solve the open circuit condition you will get ##V_{out}## =4.5 V. Then to get ##R_{N}##, you simply use the equation:$$R_{N} =\frac{V_{oc}}{I_{sc}}=1.17\Omega$$

You seem to have everything right, so could you clarify why you believe you have an error.

Also, you can use superposition applying one independent source at a time. Treat the dependent sources as resistors and always leave them on.

Thank you for your response. The solution to the problem gets a different answer to me and it sets the sources to 0 (the shortcut method). However, I was previously led to believe that method only worked for independent sources.

EDIT: sorry, I just re-read my first post. It should say "cannot get the same answer as the solution scheme..."
 

Related to Norton's Theorem with Dependent Sources?

1. What is Norton's Theorem with Dependent Sources?

Norton's Theorem with Dependent Sources is a circuit analysis technique that allows for simplification of a complex circuit with dependent sources. It states that any two-terminal, linear circuit with dependent sources can be replaced by an equivalent circuit with a single current source and a single resistor in parallel.

2. How is Norton's Theorem with Dependent Sources different from Norton's Theorem?

Norton's Theorem only applies to circuits with independent sources, while Norton's Theorem with Dependent Sources also takes into account circuits with dependent sources. This allows for a more accurate analysis of complex circuits.

3. What are some common applications of Norton's Theorem with Dependent Sources?

Norton's Theorem with Dependent Sources is commonly used in the design and analysis of electronic circuits, particularly in the fields of electronics and telecommunications. It is also used in the development of control systems and in the study of electrical networks.

4. How do you determine the equivalent Norton current and resistance in a circuit using this theorem?

To determine the equivalent Norton current, the circuit is simplified by short-circuiting the output terminals and calculating the current flowing through the short. The equivalent Norton resistance is then calculated by finding the resistance between the shorted output terminals.

5. Are there any limitations to Norton's Theorem with Dependent Sources?

Yes, Norton's Theorem with Dependent Sources can only be applied to circuits that are linear and have two terminals. It also assumes that the circuit is in a steady state, meaning that all capacitors are fully charged and all inductors are fully energized.

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