Solving Circuit Problems with Thevenin's Theorem

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Discussion Overview

The discussion revolves around solving circuit problems using Thevenin's Theorem, particularly in scenarios involving multiple power sources. Participants explore the implications of adding additional power sources, the concept of superposition, and the calculation of Thevenin equivalent circuits.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the effect of adding a second power source, especially if it is oriented in the opposite direction to the first.
  • There is a question regarding the impact of a third power source on the voltage between points A and B, particularly if it leads to a dead end.
  • Another participant outlines the standard procedure for calculating Thevenin equivalent circuits, including steps for determining open circuit voltage and short circuit current.
  • A participant explains the principle of superposition, stating that linear circuit elements allow for the analysis of multiple sources by considering them one at a time.
  • Clarification is provided on how to deactivate independent sources during analysis, noting that voltage sources become short circuits and current sources become open circuits.

Areas of Agreement / Disagreement

Participants generally agree on the principles of Thevenin's Theorem and superposition, but there remains uncertainty regarding specific applications and implications when multiple power sources are involved.

Contextual Notes

Some assumptions about circuit linearity and the behavior of independent sources may not be explicitly stated, and the discussion does not resolve the specific questions raised about the effects of multiple power sources.

Who May Find This Useful

Students and practitioners working on circuit analysis, particularly those seeking clarification on Thevenin's Theorem and the application of superposition in circuits with multiple power sources.

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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anything would help, thanks...
 
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.
 
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.
 

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