Impedance across supernode

[ramuna]

Hello,
This is my first post to this forum, so please make allowances for any protocol slip-ups
I would like some help in solving a MATLAB mesh analysis problem. Referring to the attached partial circuit diagram, I need to get an expression for the net (as in total) impedance across the voltage source (ie impedance between nodes A and B). I thought that this is V/(I3 - I4), but on plugging actual impedances into the MATLAB solution of the mesh equations, the net impedance differs from the simulation results in LTSpice. Incidentally, M1 and M2 in the diagram, are the mutual impedances, between the windings of the associated transformers.

Many thanks in advance for your help & advice,
Best regards
Ramuna

 

[MrAl]

Hi,

I dont see any attachment (maybe attachments dont work yet though).

 

[ramuna]

Hi MrAl,
Something funny is happening. My uploaded image, which was there all of last night, mysteriously disappeared today.
I tried replying to your post twice before, each time re-uploading the image. On refreshing the page after posting my reply, neither my reply nor the uploaded image, registered, on both occasions.

If this, my third attempt, fails, then I will PM you.

 

[MrAl]

Hi again,

Ok it is showing up now.

So by supernode do you mean you want to use something like VA=VB+V to specify the top node voltage?
The impedance across V is zero because it is a voltage source.

If you like you can show what work you did so far and we can take a look.

 

[ramuna]

Hi MrAl,
Thank you. Please allow me a day or so to write up my work on this issue, in a form suitable for upload.
Be back soon.

 

[MrAl]

Hi again,

Ok sure, no problem. Please make it legible so it isnt too hard to read. I see lots of scribbles sometimes that are very hard to deal with.

 

[ramuna]

Hi MrAl,
I have written up and uploaded my work on this problem for your inspection. I hope that its legible. If anything is unclear then please let me know and I'll sort it out. Regards.

 

[MrAl]

Hi again,

Well, first we might want to clear something up here.

If we take a resistor of R1=1k and connect it in series with two other resistors R2=1k and R3=1k, we have to total of 3k.
Now if we replace R3 with a 2k resistor we have a total of 4k. We have 4k partly because R1=1k and it stayed that way even when we changed the value of R3.

Now if we put that 1k back so R3=1k again but this time replace R1 with a voltage source of 2 volts. We have v=2 and R total = 2k so the current is 1ma, and if we look at the voltage source we see 2v with 1ma so 2/0.001=2k, so the voltage source looks sort of like 2k.

Now we replace R3 with 2k again, keeping the voltage source of 2v in place of R1. Now we have a voltage source 2v and two resistors R2 is 1k and R3 is 2k so the total resistance is 3k and 2/3000=6.666e-4 amps. Now 2v/6.666e-4 amps equals 3k, so NOW the voltage source looks like a 3k resistance.

So which is it...is the voltage source like a 2k resistor or like a 3k resistor?

Now in a another experiment, we use two separate voltage sources V1=1v and V2=2v, and some resistors.
Connecting the circuit up, we measure the voltage at some node and find that the voltage is 2.4 volts.
Now we short out V1, and find the voltage at that node is 1.9 volts, and unshort that source and instead short the other source V2 and find the node measures 0.5 volts. Now we add the two voltages, 1.9+0.5 and we get 2.4 volts, the same voltage that we get when we had both sources unshorted. We do this with 100 other combinations of resistors and voltages sources and we always find that the voltage with two sources equals the sum of the voltages with each source taken alone, with the other voltage source shorted out.

So the conclusion is that each source contributes exactly the same amount as if the other source was zero impedance.
Thus we say that the impedance of a voltage source is zero.

Maybe what you are after is the impedance of the circuit with the source removed?

Also, i see no source impedance or load impedance, which would affect the outcome.

Does this all make sense to you?

 

[ramuna]

Hi again,

Well, first we might want to clear something up here.

If we take a resistor of R1=1k and connect it in series with two other resistors R2=1k and R3=1k, we have to total of 3k.
Now if we replace R3 with a 2k resistor we have a total of 4k. We have 4k partly because R1=1k and it stayed that way even when we changed the value of R3.

Now if we put that 1k back so R3=1k again but this time replace R1 with a voltage source of 2 volts. We have v=2 and R total = 2k so the current is 1ma, and if we look at the voltage source we see 2v with 1ma so 2/0.001=2k, so the voltage source looks sort of like 2k.

Now we replace R3 with 2k again, keeping the voltage source of 2v in place of R1. Now we have a voltage source 2v and two resistors R2 is 1k and R3 is 2k so the total resistance is 3k and 2/3000=6.666e-4 amps. Now 2v/6.666e-4 amps equals 3k, so NOW the voltage source looks like a 3k resistance.

So which is it...is the voltage source like a 2k resistor or like a 3k resistor?

Now in a another experiment, we use two separate voltage sources V1=1v and V2=2v, and some resistors.
Connecting the circuit up, we measure the voltage at some node and find that the voltage is 2.4 volts.
Now we short out V1, and find the voltage at that node is 1.9 volts, and unshort that source and instead short the other source V2 and find the node measures 0.5 volts. Now we add the two voltages, 1.9+0.5 and we get 2.4 volts, the same voltage that we get when we had both sources unshorted. We do this with 100 other combinations of resistors and voltages sources and we always find that the voltage with two sources equals the sum of the voltages with each source taken alone, with the other voltage source shorted out.

So the conclusion is that each source contributes exactly the same amount as if the other source was zero impedance.
Thus we say that the impedance of a voltage source is zero.

Maybe what you are after is the impedance of the circuit with the source removed?

Also, i see no source impedance or load impedance, which would affect the outcome.

Does this all make sense to you?

Down the rabbit hole with MrAl! Upon my word, who would have thought that mere resistors and batteries could produce such mind-boggling complexities !

Yes, what you say does make sense MrAl. I humbly suggest that our differences are simply semantics rather than anything deeper. My definition of a voltage source has infinite internal impedance between the terminals (ie an open circuit, no current flow through the voltage source). The internal resistances of real-life batteries are between the internal cathode (or anode) of the battery and the cathode (or anode) stud on the battery case. Yet another view of this situation is that the battery produces electrons, rather than acting as a conductor for them (ie its a source rather than an impedance). Because of its own infinite impedance, the impedance measured across a battery's (or other voltage source's) terminals must be the impedance of the load connected to the voltage source.

The current mesh diagrams, such as the ones I uploaded, have mesh cells which include branches which are voltage sources. But even though such cells are drawn with circulating currents, this is simply a mathematical abstraction...for in reality, the current flow starts from one voltage source terminal and ends at the other, without flowing through the source.

This being the case, the impedance across the voltage source MUST always be that of the external load supplied by that source.

And now, its my turn to ask you, does all of this make sense to you ?

 

[MrAl]

Hi again,

Well first we need to get something else straight...
When you say:
"Down the rabbit hole with MrAl!",
does that mean i am down here all by myself or does that mean that you are down here with me? <ha ha>

Ok so it appears that you want to redefine circuit analysis. So you say that you want to state that a voltage source impedance is equal to the external network impedance rather than zero. This makes some sense because if we have a voltage source 2v and 2k resistor we have 1ma and 2v/1ma is 2k, so the voltage source appears to have a resistance (impedance) of 2k. But what happens when we change the 2k resistor to 4k is we get 0.5ma, so now the voltage source looks like a 4k resistance. So your theory works out

So what you have to do now is show how this viewpoint can help us analyze a circuit because after all if it does not help us then it is of no use.

For example, we have a 2v voltage source and a 1v voltage source. The negative terminals are connected together, and a resistor of 1k is connected between the positive terminals. Show how you solve this circuit for the most necessary quantities using your new theory. We'll take a look at your solution and go from there.

BTW you've made this very interesting

 

[The Electrician]

Ramuna,

You haven't shown the relative polarity between the L2-L3 pair, nor between the L4-L5 pair. You could use the dot convention:

https://en.wikipedia.org/wiki/Dot_convention

or some other means.

I assume you want the impedance across the A-B terminals when the source V is not present. In other words, you want the ratio V/I, where I is the current supplied by the source V; is this the case?

 

[ramuna]

Th

Ramuna,

You haven't shown the relative polarity between the L2-L3 pair, nor between the L4-L5 pair. You could use the dot convention:

https://en.wikipedia.org/wiki/Dot_convention

or some other means.

I assume you want the impedance across the A-B terminals when the source V is not present. In other words, you want the ratio V/I, where I is the current supplied by the source V; is this the case?

Thank you Mr Electrician Sir! You have correctly identified what I did wrong. I have uploaded the circuit diagram with the transformer poles drawn using the dot convention. Based on this, I think you will agree that my (eq4) on my worksheet page 2of4 should be -L4*i4 + M2*i1 - V = 0. And yes I do want the ratio V/I where I is the current flowing between B and A. Please confirm that with the correction to my eq4, I should correctly derive expressions for i1 - i4 and that the required impedance is V /(i3 + i4).

 

[ramuna]

Hi again,

Well first we need to get something else straight...
When you say:
"Down the rabbit hole with MrAl!",
does that mean i am down here all by myself or does that mean that you are down here with me? <ha ha>

Ok so it appears that you want to redefine circuit analysis. So you say that you want to state that a voltage source impedance is equal to the external network impedance rather than zero. This makes some sense because if we have a voltage source 2v and 2k resistor we have 1ma and 2v/1ma is 2k, so the voltage source appears to have a resistance (impedance) of 2k. But what happens when we change the 2k resistor to 4k is we get 0.5ma, so now the voltage source looks like a 4k resistance. So your theory works out

So what you have to do now is show how this viewpoint can help us analyze a circuit because after all if it does not help us then it is of no use.

For example, we have a 2v voltage source and a 1v voltage source. The negative terminals are connected together, and a resistor of 1k is connected between the positive terminals. Show how you solve this circuit for the most necessary quantities using your new theory. We'll take a look at your solution and go from there.

BTW you've made this very interesting

Thank you MrAl for your kind remarks
The aggregation of voltage sources, under the infinite impedance voltage source convention, follows the rules of vector addition. In the example you give, both voltage sources have a common ground, and the potential difference between their free terminals is 1 volt. This then is the voltage across the 1k resistor. Consequently the current running through it is 1 mA by the usual application of Ohm's law.

 

[The Electrician]

Th


Thank you Mr Electrician Sir! You have correctly identified what I did wrong. I have uploaded the circuit diagram with the transformer poles drawn using the dot convention. Based on this, I think you will agree that my (eq4) on my worksheet page 2of4 should be -L4*i4 + M2*i1 - V = 0. And yes I do want the ratio V/I where I is the current flowing between B and A. Please confirm that with the correction to my eq4, I should correctly derive expressions for i1 - i4 and that the required impedance is V /(i3 + i4).View attachment 80673

I believe that the worksheet equations are correct. Don't forget that you reversed the direction of I4 (as mentioned on page 1).

Perhaps you have the polarity wrong in the simulation, and that is why you don't get agreement between the matlab result and the simulation result.

 

[MrAl]

Thank you MrAl for your kind remarks
The aggregation of voltage sources, under the infinite impedance voltage source convention, follows the rules of vector addition. In the example you give, both voltage sources have a common ground, and the potential difference between their free terminals is 1 volt. This then is the voltage across the 1k resistor. Consequently the current running through it is 1 mA by the usual application of Ohm's law.

Hello again,

Well the way you are wording it makes it sound like we 'need' to consider some infinite impedance for the voltage source in order to be able to add them. I could say that anything has infinite impedance, like a short length of wire, simply because it is not connected to anything. So i dont see where you are trying to go with this.

What would you say if we had two voltage sources, one 2v and one 1v, connected by two 1k resistors in series instead of one 1k resistor as before, and we want to calculate the impedance at the center of the two resistors. Or simpler yet, one voltage source of 10v and two 1k resistors, what is the impedance at the center junction of the two 1k resistors. With the voltage source impedance being infinite, that would make the junction impedance infinite also.
So in other words we want to know the effect at the node of another external voltage with it's own true series resistance, by knowing the impedance at that node (as when we have to drive that node with an amplifier with some output impedance of it's own).

 

[ramuna]

Hello MrAl,

I'm afraid I am having difficulty following what you exactly mean. Please upload a circuit diagram, on which the points of impedance measurement are marked.

 

[MrAl]

Hi again ramuna,

Ok sure no problem. I also realized that there is a simpler question too, which i have included here as CIRCUIT 2.

What is the impedance at the nodes marked 'v' in each circuit, looking from that node to ground GND ?

What i am trying to get at is how you are using the 'infinite' impedance of a voltage source to help you analyze a circuit. I'd like to see how this is helping you understand the circuit, and these two i think are illustrative enough.

This also gives me a chance to try out the 'code' sections with the new forum software.

[FONT=Courier New]


CIRCUIT 1:

             v
  +----R1----+----R2----+
  |                     |
  E1                    E2
  |                     |
  +----------+----------+
             |
             o
             GND

CIRCUIT2:

      v
      o----R1----+
                 |
                E1
                 |
  GND o----------+

 

[ramuna]

Hello MrAl,

These diagrams make things much clearer !

I'll start with circuit 2. Please refer to the second paragraph of post #9 in this thread, the relevant part of which, I quote below for your convenience:
The internal resistances of real-life batteries are between the internal cathode (or anode) of the battery and the cathode (or anode) stud on the battery case.
Here, R1 acts as just such an internal resistance, and hence the voltage V is equal to E1.

The problem in circuit 1 is very similar to the question you posed in post #10, to which I replied in post #13. Just as we did in that problem, using the addition of vectors rules, we can replace the two voltage sources with its equivalent, a single source with EMF (E1 - E2). But to solve this current problem, all we need to do is to visualise the two vectors E1 & E2 drawn in one-dimensional Cartesian space (ie a single straight line). The needed value v, is simply the distance from the origin to the tip of the E1 vector, and is equal to E1.

 

[MrAl]

Hi,

Yeah but this time i asked for the "impedances" at the nodes labeled 'v' in the two circuits. We want to know the impedances (resistances in this case) from node 'v' to ground GND.

Also take a another look at the two circuit diagrams in my previous post because i had to edit one because the "code" tags dont work right yet.

 

[ramuna]

Hi,
My error. The impedances of the points v, w.r.t. ground, in both circuits, would be infinity, under the infinite impedance voltage source convention.