Solving Circuit Problems with Thevenin's Theorem

In summary, the conversation discusses the use of Thevenin's Theorem in circuit problems, particularly when multiple power sources are involved. The standard procedure for calculating Thevenin equivalent circuits is outlined, including the calculation of open circuit voltage and short circuit current. The concept of superposition is also mentioned, where a 2-source circuit can be broken into two 1-source circuits to simplify calculations. The conversation also touches on the purpose of having a separate homework section for specific problems.
  • #1
skiboka33
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I'm working on some circuit Problems involving Thevenin's Theorem. When there is only one power source it seems pretty straight forward, but I'm a little confused when another power sources is added. Does it matter if a second power source is facing the opposite direction of the first?

I kind of understood the idea that you consider only one powersource at a time, but I just need some clarification. Thanks.

There is also a more challenging circuit problem (at least to me) in which a third powersource is added, this one is on a branch apart from the circuit which leads to the point A (a dead end). This is part of the system of points A,B for which the thevenin circuit is being constructed. Does this mean the Vab considering only this powersource is zero? thanks again, sorry if my description is confusing.

EDIT: should have put this in help with homework section, feel free to move it. I'd do it myself but I don't know how
 
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  • #2
anything would help, thanks...
 
  • #3
The standard procedure for calculating Thevenin equivalent circuits are as follows;

1. Calculate the Open Circuit Voltage [itex] V_{OC} [/itex] - This is the voltage the appears between the terminals when no load is connected between them.

2. Calculate the Short Circuit Current [itex] I_{SC} [/itex] - This is the current that flows through the terminals when they are shorted (i.e. connected with no resistance in between).

3. The Thevenin equivalent resistance can then be calculated using

[tex] R_{TH} = \frac{V_{OC}}{I_{SC}} [/tex]

Claude.
 
  • #4
skiboka33 said:
I kind of understood the idea that you consider only one powersource at a time, but I just need some clarification. Thanks.

That's the idea. It's called superposition. If all your circuit elements are linear then you can break your 2-source circuit into two 1-source circuits and add the results for open circuit voltage (or short circuit current, if you are doing a Norton conversion). Just remember that deactivated independent voltage sources become short circuits and deactivated independent current sources become open circuits.

This of course can be extended to n-source circuits.

EDIT: should have put this in help with homework section, feel free to move it. I'd do it myself but I don't know how

Only a Staff member can do it. ZapperZ or Doc Al may decide to do it, but I wouldn't worry about it. The idea of having a Homework section is to reserve the Physics section for interesting discussions. You're asking about general principles (Thevenin and superposition) rather than specific problems.
 

1. What is Thevenin's Theorem and how is it used to solve circuit problems?

Thevenin's Theorem is a method used to simplify complex electrical circuits into simpler equivalent circuits, making it easier to analyze and solve problems. It states that any linear circuit can be represented by a single voltage source in series with a single resistor, known as the Thevenin equivalent circuit. This equivalent circuit has the same voltage and current characteristics as the original circuit, but is much simpler to analyze.

2. What are the steps for using Thevenin's Theorem to solve a circuit problem?

The steps for using Thevenin's Theorem are as follows:

  • 1. Identify the load resistor in the circuit and remove it.
  • 2. Calculate the Thevenin voltage by finding the open-circuit voltage across the terminals where the load resistor was removed.
  • 3. Calculate the Thevenin resistance by finding the equivalent resistance of the circuit when all voltage sources are replaced with short circuits and all current sources are replaced with open circuits.
  • 4. Draw the Thevenin equivalent circuit with the Thevenin voltage and resistance.
  • 5. Place the load resistor back in the circuit and solve using basic circuit analysis techniques.

3. Can Thevenin's Theorem be used for both DC and AC circuits?

Yes, Thevenin's Theorem can be used for both DC and AC circuits as long as they are linear circuits. This means that all components in the circuit have constant values and obey Ohm's Law.

4. What are the advantages of using Thevenin's Theorem to solve circuit problems?

Thevenin's Theorem offers several advantages for solving circuit problems:

  • 1. It simplifies complex circuits into equivalent circuits, making them easier to analyze and solve.
  • 2. It reduces the amount of calculations needed, saving time and effort.
  • 3. It allows for easy analysis of the circuit's behavior when the load resistance changes.
  • 4. It can be applied to both DC and AC circuits, making it a versatile tool for circuit analysis.

5. Are there any limitations or assumptions when using Thevenin's Theorem?

Yes, there are some limitations and assumptions when using Thevenin's Theorem:

  • 1. The circuit must be linear, meaning that all components have constant values and obey Ohm's Law.
  • 2. Thevenin's Theorem can only be applied to circuits with one independent source.
  • 3. It assumes that the load resistance will not significantly affect the behavior of the circuit.
  • 4. The Thevenin equivalent circuit is only valid for the specific load resistance used in the calculation.

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