What is Thevenin's Theorem and How Can It Help Simplify Complex Circuits?

AI Thread Summary
The discussion revolves around understanding Thevenin's Theorem and its application in circuit analysis. The original poster sought help with a specific circuit problem but found the worked solutions confusing, particularly regarding voltage division and the role of resistors. Clarification was provided that the upper 2k ohm resistor can be treated as a short circuit since no current flows through it, allowing for accurate calculations of voltage across the 12k ohm resistor. This understanding simplifies the analysis and aligns with the principles of superposition. The conversation highlights the importance of grasping these foundational concepts in circuit simplification.
daniel.
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Homework Statement



[PLAIN]http://img21.imageshack.us/img21/2408/screenshot20101028at112.png

Hi,

I'm new to these forums, so hopefully this is the right place to post this question. I need some help in the question above. I'm not obtaining the right answer for this question and I'm not too sure whether my working out is correct as there there are no worked solutions for this question. If anyone can provide me some steps to solve this question, it will be greatly appreciated.

Thanks in advance.

Daniel
 
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Okay, fortunately I've found the worked solutions for this question in my portal. But I still do not understand this part of the solution:

[PLAIN]http://img507.imageshack.us/img507/6680/screenshot20101028at115.png

I know that we are using superposition, but I do not understand how they got the values for the voltage division. Why is the 12k Ohm resistor used in the numerator? Like why does the voltage drop across the 12k ohm resistor equate to Voc'? And in the denominator, which 2k ohm resistor are they referring to? And why is the other 2k ohm ignored?

Thanks in advance
 
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There is no current through the upper 2k resistor, so no voltage drop. You can substitute it by a short circuit. In this way Voc is the voltage across the 12k resistor.
 
Thanks. Got it now.
 

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