Help Me Find Thevenin's Equivallent for the Circuit Impaired

[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.

 

[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.

 

[magnetpedro]

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.

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)

 

[magnetpedro]

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.

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.

 

[phinds]

Rth = 7 ohm

yes

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

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

 

[magnetpedro]

yesI think you need to go back to your textbook, as this shows a fundamental misunderstanding about how these things work.

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.

 

[phinds]

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.

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.

 

[Averagesupernova]

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

 

[jim hardy]

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

.

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

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 V_BA 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

 

[phinds]

s:
1) Find open circuit voltage between the two points. That is V_Thevenin

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?

 

[jim hardy]

Vth = E5 (the reduced case I had him working on. What kind of math do you need to figure out the Vth = the voltage source?

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

 

[magnetpedro]

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?

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

The V_BA equals the V_Th and V_B = 10 V and V_A = 0 V, right?

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

 

[phinds]

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

 

[magnetpedro]

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

What do you mean?
Thanks!

 

[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

 

[magnetpedro]

Exactly what I said. I once again advise you to go back and study some basics.

Sorry to bother but let me clarify something. In the picture I've posted (#12), isn't V_A 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 V_B also equal to zero?

I'm mixing everything up.

Thanks again.

 

[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.

 

[magnetpedro]

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.

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

 

[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.

 

[phinds]

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.

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

 

[Averagesupernova]

When I was in school and we were shown thevenin I had mixed feelings. I could see it was a powerful tool but not so sure how often a person would need to reach for that tool. Nonetheless, it adds for more practice in circuit analysis. I am glad I had hard-nosed teachers that drilled and drilled and drilled.

 

[The Electrician]

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.

View attachment 105915

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

Don't give up yet. What circuit analysis techniques have you learned? Do you know superposition, nodal analysis, mesh and loop analysis?

 

[phinds]

When I was in school and we were shown thevenin I had mixed feelings. I could see it was a powerful tool but not so sure how often a person would need to reach for that tool. Nonetheless, it adds for more practice in circuit analysis. I am glad I had hard-nosed teachers that drilled and drilled and drilled.

In my early days as a working EE when it was most likely to be called for I had little occasion to "reach for that tool" (but not quite never). I, like you, came to see it as just an exercise that helped me learn circuit analysis which was of course very useful.