Basic Circuits - Thevenin's Thm explanation

[Beeorz]

I've read through a ton of material and still do not quite understand it. The way that we were taught was:
1) Redraw, with load removed
2) Redraw, find Voc
3) Redraw, find Rth
4) Redraw, using TEC with removed load

Most of my questions involve finding Voc and Rth. I understand the steps, but I still do quite understand why certain things were done. Here is some worked about problems:

(Prob 6 Questions) http://i37.tinypic.com/24nl5le.jpg
1) For Voc, they used KVL. Why did they use KVL for the perimeter and not the inner portion of the circuit?
Voc, perimeter: Voc + 5(0) - 20(4) - 35 = 0
2) When finding Rth, how do you specify which resistors contribute? What question do I need to be asking myself in order to determine Rth. For this problem I assume the the 3Ω resistor is negated due to it not being connected, i.e the resistor current cannot jump across the short.

(Prob 7 Questions) http://i37.tinypic.com/2ufr1x2.jpg
1) I understand how the 12Ω and 5Ω resistor have no current through them but how Vth determined? I don't understand what procedure or assumptions they made.
2) Why are only the 12Ω and 5Ω resistors contributing to Rth and not the 20Ω and 4Ω resistors as well?

(Prob 8 Questions) http://i35.tinypic.com/og038o.jpg
1) Again, what are the rules for determining Rth. Do you consider all directly connected resistors and indirectly connected resistors separated by a node?

Any help is appreciated, thanks!

 

[Mr.Green]

Q6/

1)

Why did they use KVL for the perimeter and not the inner portion of the circuit?

The KVL is written for a closed loop. So, choose a closed loop and go on. For the perimeter the KVL is written as it is seen in the solution. Let's Write KVL for the inner portion to see what problem we will encounter:

-35 - 20 * 4 - 3 * 4 + V_cs = 0
where: V_cs is the voltage across the current source.
The problem is here. Do you know the voltage across the current source? No.

2)

When finding Rth, how do you specify which resistors contribute?

Assume that you are walking on the circuit starting from one end of the element that you removed. You should be able to start from one end and then back to the other end without any jump. In other word you can't choose an open circuit for your travel. Any resistor that you visit in your path will contribute in the R_TH.

OK. It seems that things are going well. But let me give you a problem that you will face during your trip on the cicuit.
If there were more then one path what should we do? Just like question 7 were there are more than one path. In this case combine the paths until you get just one path. HOW??
Look at question 7. Before you can begin your trip combine the resistors (20 Ohm), (4 Ohm), and the short circuit in parallel. Then go on with just one path.

If you have problems in determining R_TH see point 4 in the 7th post at:

https://www.physicsforums.com/showthread.php?t=338092"

Q7/

1) I understand how the 12Ω and 5Ω resistor have no current through them but how Vth determined? I don't understand what procedure or assumptions they made.

The voltage across the open circuit (which is V_TH) is the same across the (20 Ohm) resistor (they are parallel) and the voltage across the (20 ohm) resistor is the same of the voltage source (also are parallel). In one step: the open is in parallel with the voltage source.

2) Why are only the 12Ω and 5Ω resistors contributing to Rth and not the 20Ω and 4Ω resistors as well?

By combining the 3 paths that I mentioned above you will have just one path, a short circuit, or understand that in this way:
The electrical current is lazy. It will choose the simplest way to flow and this way is the short circuit. Is clear?Q8/

1) Again, what are the rules for determining Rth. Do you consider all directly connected resistors and indirectly connected resistors separated by a node?

I think that now you know the rules. Don't you?

Just one important thing is remained...when you are to determine R_TH you should look at the circuit from the ends of the removed element.

 

[Beeorz]

ty very much, now I just to put it all into practice