# Need explanation/help with Thevenin's Thm

#### beeorz

Joined Sep 20, 2009
7
I've read through a ton of material and still do not quite understand it. The way that we were taught was:
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 examples:

(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!

#### Jony130

Joined Feb 17, 2009
5,475
Hmm,
Find Vth by any method you prefer to circuit analysis
to find Rth
1. Replace voltage sources with short circuits and current sources with open circuits.
2. Find equivalent resistance seen at AB point.

#### hgmjr

Joined Jan 28, 2005
9,027
Review the material in the AAC ebook on Thevenin. See if it helps you get a deeper understanding of the concept.

hgmjr

#### beeorz

Joined Sep 20, 2009
7
I have already read tons of material on Thevenin. I understand the steps but not the procedure to get particular answers. I need to try and understand what "find equivalent resistance seen at AB point" means in my own terms. That just doesn't mean enough to me to solve for Rth. For example,

Solving for Vth is relatively simple. You remove the load and then solve for Vth using any method you wish.
Solving for Rth confuses me the most. You are suppose to find the resistance that the load see but I don't quite understand what in world that means.

If anyone can please take the time to look at the links I posted and give a more reasonable explanation to my questions I would greatly appreciate it. I feel like I am missing out on some important concepts in order to correctly use this thm.

Especially Problem #7. They use assumption/procedures that I do not see at all.

Thanks, again.

#### hgmjr

Joined Jan 28, 2005
9,027
One way to think of Rth is what resistance would you measure with an ohmmeter connected across the terminals to which your load resistor is to be connected if you replaced all of the voltage sources with a short and all of the current sources with an open.

hgmjr

#### beeorz

Joined Sep 20, 2009
7
Here is what I am talking about:

This is the first step. Finding Vth. Here the 12 and 5 ohm resistors are 0 due to no current going through them. That is easy to distinguish. So, then how is Vth determined? Any procedure I use doesn't equate to 10V.

This is the second step. Finding Rth. Using your example hgmjr, placing the black end on the (-) and red end on the (+) the ohmmeter would read a voltage across that load. I can see how the 12 and 5 ohm plays in but why not the 20 and 4 ohm resistors as well?

#### hgmjr

Joined Jan 28, 2005
9,027
That looks good to me.

At least the Rth looks good. I need to look at the Vth and will get back to you.

The Vth looks good also.

hgmjr

#### Jony130

Joined Feb 17, 2009
5,475
(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.
Hmm, II Kirchhoff law "thinks connect in parallel have the same voltage across them".
So there is non current through 12Ω, 5Ω and the rest of elements are conect in parallel and that's why Vth=10V
2) Why are only the 12Ω and 5Ω resistors contributing to Rth and not the 20Ω and 4Ω resistors as well?
Because 4Ω, 20Ω are short (0Ω resistance) by " Replace voltage sources with short"

#### beeorz

Joined Sep 20, 2009
7
That looks good to me.

At least the Rth looks good. I need to look at the Vth and will get back to you.

The Vth looks good also.

hgmjr
This is a worked out problem that I am having difficulty understanding. I do not know how or why they get the answers they do.

@Jony130
My book has a different definition for Kirchhoff's 2nd law. So as a general rule, any open circuit load has the same voltage across it as a voltage source in parallel with it??

For the second step, I do not see the short that you are referring to. The 10V voltage source has a short and the 6A current source is opened up. How does this affect the 4 and 20 ohm resistors??

#### Jony130

Joined Feb 17, 2009
5,475
My book has a different definition for Kirchhoff's 2nd law. So as a general rule, any open circuit load has the same voltage across it as a voltage source in parallel with it??
Well, it's a general rule, all elements connect in parallel have the same voltage across them.

VR4=V3+V2 or VR4=VB-V1, VR5=VB1, V1+V2=VR2

For the second step, I do not see the short that you are referring to. The 10V voltage source has a short and the 6A current source is opened up. How does this affect the 4 and 20 ohm resistors??
The current will be flow through 12Ω--->"short" 0Ω resistance--->5Ω

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#### beeorz

Joined Sep 20, 2009
7
alright, thx for the help fellas. I will try working a couple of examples and see if I can gain more understanding of different situations. I am still not quite sure why the current would flow through the short and through any other path.

thanks again

#### Jony130

Joined Feb 17, 2009
5,475
If we have a "short" in the circuit then for sure, all current will flow through the short.

PS. the amp-meter act like a short.

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