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

[daniel.]

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

 

[daniel.]

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

 

[CEL]

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.

 

[daniel.]

Thanks. Got it now.

 

[DeeNos]

I would like to offer some guidance on how to approach this problem. Thevenin's Theorem is a useful tool in circuit analysis that allows us to simplify complex circuits into simpler circuits that are easier to analyze. It is based on the concept of equivalent circuits, where we can represent a complex circuit with a single voltage source and a single resistor.

In this particular problem, we are asked to find the Thevenin equivalent of the given circuit. The first step in applying Thevenin's Theorem is to find the open circuit voltage (Voc) at the terminals A and B. To do this, we can simply remove the load resistor RL and use circuit analysis techniques to find the voltage at A and B.

Next, we need to find the Thevenin resistance (Rth) at the terminals A and B. To do this, we need to short-circuit the voltage source and find the equivalent resistance of the remaining circuit. This can be done using the series and parallel resistance formulas.

Once we have both the open circuit voltage and the Thevenin resistance, we can construct the Thevenin equivalent circuit by connecting a voltage source with magnitude Voc and a resistor with value Rth in series.

Finally, we can use this simplified circuit to analyze the original circuit. For example, if we wanted to find the voltage across the load resistor RL, we can simply use Ohm's Law (V=IR) with the Thevenin voltage and resistance.

I hope this helps you in solving the problem. Remember to always double check your calculations and make sure you understand the concepts behind Thevenin's Theorem before applying it to a problem. Good luck!