Dependent only sources - Thevenin eq.

Thread Starter

KevinEamon

Joined Apr 9, 2017
261
Hi guys
I'm new to the forum. I have what looks to be a relatively simple circuit and asked to find the Thevenin equivalent.



I'm really struggling with both of them. But I just want to talk about the first - number 10.
As far as I can gather, with dependent only sources - you should add a current or voltage source. I added a 1 amp source - at Voc, between a and b, pointing toward a.

After that you have to get the voltage drop across the amp source (not sure if that's correct)

Anyways, I've tried -
nodal analysis on this sucker.
KVLs,
tried writing the currents in terms of ix and substituting values in and all sorts of fun stuff currently 5 pages long.

The math just keeps coming out with zero values and all sorts of gooey nonsense / and / or / too many unknowns.

I'm sure this is something really profound, that I'm supposed to realize with this, but I sure don't have a clue what it is. Can someone just please, push me in the general direction, I should be taking with this.
 

WBahn

Joined Mar 31, 2012
24,848
It's hard to tell where you are going wrong since you haven't shown us what you did.

Pick whichever method you think you did the best with and show your work. Then we can see where you are going right and where you are going wrong.

You will be tempted to kick yourself when you see the answer to #10, but don't. It's not as intuitive as the answer will make it seem. We can talk about that more, later.
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
261
Thanks Wbahn. Ok now I think i've got this right. How to upload / convert this file.

So "attached" is some of my attempts.
 

Attachments

xxxyyyba

Joined Aug 7, 2012
289
Hi guys
I'm new to the forum. I have what looks to be a relatively simple circuit and asked to find the Thevenin equivalent.



I'm really struggling with both of them. But I just want to talk about the first - number 10.
As far as I can gather, with dependent only sources - you should add a current or voltage source. I added a 1 amp source - at Voc, between a and b, pointing toward a.

After that you have to get the voltage drop across the amp source (not sure if that's correct)

Anyways, I've tried -
nodal analysis on this sucker.
KVLs,
tried writing the currents in terms of ix and substituting values in and all sorts of fun stuff currently 5 pages long.

The math just keeps coming out with zero values and all sorts of gooey nonsense / and / or / too many unknowns.

I'm sure this is something really profound, that I'm supposed to realize with this, but I sure don't have a clue what it is. Can someone just please, push me in the general direction, I should be taking with this.
I don't know why the give "pathological" circuits to test knowledge of students. They can give for exercise some circuit which makes sense.
I didn't try to solve this problem nor I checked your work. This is how it looks to me at first glance. If short-circuit ix looks strange, you can add resistor Rx in branch wit current ix. You don't need to assign to it any value, just variable Rx. To find equivalent Thevenin's resistance, connect voltage source E between points a and b. You also don't need to assign any value to E, just use it as variable E in calculations. Calculate Rth as Rth=E/I, where I is current through E. That expression Rth will be function of Rx. Put Rx=0 in that expression to get eqivalent Thevenins resistance for original circuit.
You can say directly what is value for Eth.
ps. for every circuit with n nodes and m branches which connects nodes, you can write n-1 equations using KCL and m-(n-1) equations using KVL. Here you have one current source (current controlled current source) so number of equations using KVL is reduced to m-(n-1)-1.
 
Last edited:

WBahn

Joined Mar 31, 2012
24,848
Thanks Wbahn. Ok now I think i've got this right. How to upload / convert this file.

So "attached" is some of my attempts.
Edit_2017-04-10_1.png

First, you are being extremely sloppy with your units. You have currents equal to voltages, for instance. You need to learn to track your units properly.

In the circle equation, you are making one of the classic mistakes -- you are grabbing the voltage on one side of a resistor and dividing it by the resistance and then claiming that that is the current in the resistor. You need to use the voltage ACROSS the resistor. Since you did this on the next line, I'm assuming this is just a slip up.
 

WBahn

Joined Mar 31, 2012
24,848
I don't know why the give "pathological" circuits to test knowledge of students. They can give for exercise some circuit which makes sense.
What's pathological about it? It's not at all uncommon for the small signal equivalent for transistor circuits to yield circuits not unlike this one.
 

WBahn

Joined Mar 31, 2012
24,848
How is that a violation? Just solve for ix.

What does ix need to be in order to satisfy KCL/KVL in the above circuit?

You have a circuit that has no independent sources, so what do you expect ix to be?

Do these agree?

If so, where is the problem?
 

WBahn

Joined Mar 31, 2012
24,848
Why can't the circuit exist? It would be a trivial matter to take a physical current controlled current source and build the circuit. Yes, most such physical sources have some burden resistance in the sense portion of the circuit, but that can be made arbitrarily small compared to the other resistances in the circuit. You would find that, to the outside world (i.e., if the world is looking at the box containing this circuit with only terminals a and b accessible), the V/I characteristic would be indistinguishable from a fixed resistor of a certain size. The question merely asks you to find the size of that resistor (it's between 5 Ω and 10 Ω).

Do you have a problem with the following circuit:

Edit_2017-04-10_2.png

It doesn't have this (supposed) current division problem that is bothering people.

If not, then what is the open circuit voltage seen at a-b?
 

Thread Starter

KevinEamon

Joined Apr 9, 2017
261
Hi guys so I've tried to fix the error and reinterpret the circuit it terms of Ix.

Managed to derive 3 equations but they are just coming out as garbage again. I've attached the attempts.
 

Attachments

Thread Starter

KevinEamon

Joined Apr 9, 2017
261
Lol everything's politics! Even electronic theory apparently. Well this seems to have kicked off the fun. I'm glad to know i'm not the only one struggling with this. So what are we saying here? That the thing is indeterminate? At this point I'm ready to drop this dang question like a hot potato.
 

WBahn

Joined Mar 31, 2012
24,848
Hi guys so I've tried to fix the error and reinterpret the circuit it terms of Ix.

Managed to derive 3 equations but they are just coming out as garbage again. I've attached the attempts.
You are still being extremely sloppy with your units.

What does it mean for the length of a table to be equal to the weight of a chair?
6b
What does it mean for a current to be equal to a voltage?

What does it mean for one person to be one taller than another person?

What does it mean for one current to be one greater than another current?

Aside from that (and this sloppiness WILL get you into trouble, repeatedly, as you made mistakes that you could have trivially caught had you tracked units) your basic problem is that you have seven unknowns (Ix, Iv, Iy, Iz, V1, V2, V3) but only six equations.

What do you know about V1 and V3?

Also, keep in mind what you are looking for. You want Rth, which is the ratio of V3/(1 A), or given the sloppy units, just V3 (meaning that you are claiming that the resistance is a voltage).

Where you have "Three equations", you have three equations in four unknowns. Think again about what V1 is in relation to V3.
 

WBahn

Joined Mar 31, 2012
24,848
Lol everything's politics! Even electronic theory apparently. Well this seems to have kicked off the fun. I'm glad to know i'm not the only one struggling with this. So what are we saying here? That the thing is indeterminate? At this point I'm ready to drop this dang question like a hot potato.
This question is perfectly fine and this circuit is completely realistic, particularly as the small signal equivalent of a circuit fragment. YOUR circuit is NOT indeterminate (the other one that I gave is and was intentionally made so -- sorry if it caused any confusion for you).
 
Last edited:

WBahn

Joined Mar 31, 2012
24,848
Which is the more unrealistic of two unrealistic options? Is that a political question?
Again, why is the first one unrealistic? Its behavior is completely defined and it completely satisfies KVL and KCL and every law of physics. Consider a lone resistor between terminals a and b with a designator for the current in the resistor of Ix. Is THAT an unrealistic circuit fragment because Ix has no where to go as it exits the resistor? Of course not, but in that circuit fragment the only solution for Ix happens to be Ix=0. It is NO DIFFERENT for THIS circuit fragment. As an isolated fragment, the only solution for Ix happens to be Ix=0.

Connect EITHER of them to other components, and Ix can be non-zero. In the TS's circuit with the 1 A test source, Ix happens to be 2A.
 
Top