Thevenin's Theorem

  • Thread starter skiboka33
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  • #1
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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 dont know how
 
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  • #2
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anything would help, thanks...
 
  • #3
Claude Bile
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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
Tom Mattson
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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 dont 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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