Thevenin's Help

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1. Sep 14, 2016

magnetpedro

Hello guys,

Can someone guide me through this problems solution?
I need to find the Thevenin's Equivallent for this circuit, between A and B.

I'm terrible at Circuits and I'm eager for some motivation and help.
Thanks a lot in advance.

2. Sep 14, 2016

phinds

You HAVE to show some effort. That's the rule here. We don't spoon feed answers. If you don't even know how to get started, best to go back to your text and work on simpler problems before attacking this more complex one.

For example, start with just the upper left loop from A back around through the 4, 2, 1ohm resistors to the point above E4. Ignoring the rest of the circuit, find the Thevinin equivalent for just that circuit. Then maybe add the upper right loop. Just adding that will make for a fairly complex analysis.

Also, I'm not seeing any "B" node.

3. Sep 14, 2016

magnetpedro

Thanks a lot for your guidance.
For the upper left loop wouldn't it be:

Rth = 7 ohm
Vth = 10* (2/(2+4+1) ) V

?

(Unfortunately the "B" node got cutted from the print. It's the "R" (a B without the lower leg) under E1)

4. Sep 14, 2016

magnetpedro

Adding the other right loop:

Rth = 7,4 ohm

Vth = [ 10 * ( 2 / ( 2+ 4 + 1) ) ] + [ 2 * ( 2 / ( 2,5 + 0,5 + 3 + 2 ) ] V

?

Thanks once again for your awesome support.

5. Sep 14, 2016

phinds

yes
I think you need to go back to your text book, as this shows a fundamental misunderstanding about how these things work.

6. Sep 14, 2016

magnetpedro

Hum, let's see.
In that isolated upper left loop, Vth is equal to VAB = VB - VA , right? ( B being the node above E4)
I thought VAB would be the V of the 2 ohm resistor, but I'm clearly confused.

I believe it's a matter of not being able to visualize it correctly.

Sorry for the trouble, do you have any texts or guides I can study?
Can you point me in the right direction, without giving me the final answer?

Thanks again.

Last edited: Sep 14, 2016
7. Sep 14, 2016

phinds

It's not the value that you are coming up with, it's the fact that you do not instantly recognize that Vth = E5 that leads me to say you have a fundamental misunderstanding. You need to study the DEFINITION of Thevinin equivalent. I'm sure there must be simple examples and tutorials online if your text doesn't help you.

8. Sep 14, 2016

Averagesupernova

How can Vth be equal to E5? Am I missing something?

9. Sep 14, 2016

jim hardy

I was taught to find Thevenin in two steps:
1) Find open circuit voltage between the two points. That is VThevenin
2) Replace every internal voltage and current source with its internal impedance, then solve for impedance between the two points. That is ZThevenin .

First thing i'd do is redraw the circuit combining whatever resistances i could. That'll make it look less intimidating.
Example : in top right i see resistances of 3Ω , 0.5Ω, and two fiveΩs that can be replaced by a single 6Ω.

Then i'd use either loop current or mesh current method to solve for VBA with Kirchoff's laws.

It is not apparent to me that Vth is E5, though the algebra might turn out that way - i haven't solved it.
oops - see posts 10 & 11 jh

Last edited: Sep 14, 2016
10. Sep 14, 2016

phinds

Exactly. In this case, Vth = E5 (the reduced case I had him working on. The entire circuit in that case is a voltage source in series with some resistors. What kind of math do you need to figure out the Vth = the voltage source?

11. Sep 14, 2016

jim hardy

Ahhh, you were working on just one leg of the circuit,,,, from A to the middle node . Now i see what you meant. .......

old jim

12. Sep 15, 2016

magnetpedro

I believe see it now, thanks a lot!
Like this?

The VBA equals the VTh and VB = 10 V and VA = 0 V, right?

Now, I'll try to add the upper right loop.

13. Sep 15, 2016

phinds

Yep, you have the right idea, but you were to find the Vth from A to B so you have the sign wrong.

14. Sep 15, 2016

magnetpedro

What do you mean?
Thanks!

15. Sep 15, 2016

phinds

Exactly what I said. I once again advise you to go back and study some basics. You were asked to find Vth as Vab

16. Sep 15, 2016

magnetpedro

Sorry to bother but let me clarify something. In the picture I've posted (#12), isn't VA suppose to be zero since it's a point before the source?
And in point B, if I follow the current's path from A to B, we first have a voltage rise of 10 V (the source) and then a voltage drop of 10 V (in the resistor) so shouldn't VB also equal to zero?

I'm mixing everything up.

Thanks again.

17. Sep 15, 2016

phinds

I see NO reference point ("zero") in your diagram, but traditionally, if there is one, it is at the bottom of the circuit. I once again advise you to go back and study some basics.

Look, I know my harping on this is perhaps not "motivating" for you but it really is what you need to do. This asking semi-random questions on an internet forum and trying to learn that way is not a good idea. It may give you some short-term satisfaction in solving one particular problem but it the important thing is that it won't tell you what you don't know. You need to get systematic about this.

18. Sep 15, 2016

magnetpedro

Nevermind, you're probably right.
Thank you very much for your time!

19. Sep 15, 2016

Averagesupernova

Thevenin voltage comes down to removing the load and looking back into the where the load was connected and determining the voltage. That really isn't part of Thevenin. It is part of understanding basic circuit analysis. You know, voltage division, current division, all that dandy good stuff.

20. Sep 15, 2016

phinds

Exactly, which is why I keep harping on his need to get back to the basics in a systematic way.