Electrical circuit Source transformation Question

  • #26
The Electrician
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Here's the setup. I use the convention that currents leaving a node are positive. I also follow the convention of ordering the rows of the matrix according to the node numbers; that is, node 2 is row 1, node 3 is row 2, node 4 is row 3, etc.

When I solve this with Mathematica, I am warned that the output is extremely large, and asks if I really want to display it.

?temp_hash=82ef94d1d1698933283bdd86f4347b66.png
 

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  • #27
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I don't know how it's structured in Matlab. It could set up by a 12x11 matrix, so that the first equation is written:

[0-N4] / R15 + [N5-N4] / R1 + [N7-N4] / R3 + [N8-N4] / R8 = -B1

But if the setup is split in two matrices, it must be a 11x11 and a 1x11 matxrix, so that the first row in the 11x11 matrix must be:

[ 0 0 -(1/R1 + 1/R3 + 1/R8) 1/R1 0 1/R3 1/R8 0 0 0 0] ( first zero removed. It would be the coefficient to N1, but you have already decided that N1 = 0, so the first column will be the coefficients to N2 ).
Sorry, as usual right you are, yes, 12x11 matrix, woops.

This is where I'm still a bit hazzy, specifically the phrase "so the first column will be the coefficients to N2". So the first column is set to zero, are you just saying that it goes from N2 to N12?

If I put some values in for the resistors and sources, would this give me values for the nodal voltages? And if so how do I check if they're correct? Also after I get the values am I then able to then work out the potential say between N4 to N6 or N2 to N5? So I can in affect replace the current sources with those potentials?

Here's the setup. I use the convention that currents leaving a node are positive. I also follow the convention of ordering the rows of the matrix according to the node numbers; that is, node 2 is row 1, node 3 is row 2, node 4 is row 3, etc.

When I solve this with Mathematica, I am warned that the output is extremely large, and asks if I really want to display it.
I appreciate your less ad hock approach than mine, I do like those as conventions to take. I really want to dip a toe in Mathematica, I've never gotten around to it.
So you're saying the warning indicates that the symbolic result would be so large it's not practical to re-model the circuit (from post #20) as voltage equivilant sources? Because I'm afraid of that.
 
  • #28
Hesch
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This is where I'm still a bit hazzy, specifically the phrase "so the first column will be the coefficients to N2". So the first column is set to zero, are you just saying that it goes from N2 to N12?
Having decided that N1 = 0, there is no need to solve the value with 12 equations instead of 11. Well, you could do it by adding an extra equation:

1 0 0 0 0 0 0 0 0 0 0 0 = 0

but then you will have to enter a number of extra coefficients = 12*13 - 11*12 = 24 coefficients. No need to do that.
If I put some values in for the resistors and sources, would this give me values for the nodal voltages?
Yes, N2 . . N12 with 11 equations.
And if so how do I check if they're correct?
Calculate the currents by ohms law. Having the equation

[0-N4] / R15 + [N5-N4] / R1 + [N7-N4] / R3 + [N8-N4] / R8 = -B1

. . you know that the current through R1 = (N5-N4)/R1, with positive direction toward N4. The potential between N5 and N4 is (N5 - N4).
Use KCL to check that the sum of current toward a node = 0.
So I can in affect replace the current sources with those potentials?
No, because a current source has the inner impedance = infinite. Regarding a node as a source, it will have a finite inner impedance. An ideal voltage source will have an inner impedance = 0.
 
  • #29
The Electrician
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I appreciate your less ad hock approach than mine, I do like those as conventions to take. I really want to dip a toe in Mathematica, I've never gotten around to it.
So you're saying the warning indicates that the symbolic result would be so large it's not practical to re-model the circuit (from post #20) as voltage equivilant sources? Because I'm afraid of that.
You saw the complicated expression in post #22? That was from the solution where only the first 5 rows were symbolic. For the full symbolic solution, the corresponding expression would probably be 100 times larger. I don't see how anything could be done with such a thing.

You can solve the network numerically:

?temp_hash=d961fab1ddbbc20fd74fc298dbf5db52.png


But even then, what do you do with the result? Just knowing the node voltages doesn't allow you to replace the current sources directly with voltage sources.
 

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  • #30
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I wrote a reply on the train a few days ago but apparently it didn't go, I can't really remember what I said, so I'll start a'fresh.
Having decided that N1 = 0, there is no need to solve the value with 12 equations instead of 11. Well, you could do it by adding an extra equation:

1 0 0 0 0 0 0 0 0 0 0 0 = 0

but then you will have to enter a number of extra coefficients = 12*13 - 11*12 = 24 coefficients. No need to do that.

Yes, N2 . . N12 with 11 equations.

Calculate the currents by ohms law. Having the equation

[0-N4] / R15 + [N5-N4] / R1 + [N7-N4] / R3 + [N8-N4] / R8 = -B1

. . you know that the current through R1 = (N5-N4)/R1, with positive direction toward N4. The potential between N5 and N4 is (N5 - N4).
Use KCL to check that the sum of current toward a node = 0.

No, because a current source has the inner impedance = infinite. Regarding a node as a source, it will have a finite inner impedance. An ideal voltage source will have an inner impedance = 0.
You saw the complicated expression in post #22? That was from the solution where only the first 5 rows were symbolic. For the full symbolic solution, the corresponding expression would probably be 100 times larger. I don't see how anything could be done with such a thing.
But even then, what do you do with the result? Just knowing the node voltages doesn't allow you to replace the current sources directly with voltage sources.
Well gentlemen, I'm sad to hear that the rearrangement is untenable. I greatly appreciate your help.
So without any 'clever' use of a trick like some sort of symmetry there is not conventional way to replace current sources with voltage sources, I find that surprising but not unbelievable. One thing I feel still needs clarification is the why? Is it because having multiple resistors in parallel with the current sources, each with a different potential (so they can't be combined) is the issue, and this is a known limitation/difficulty of the process?
Say if one was to investigate this circuit and (try) devise a method of approximating an equivilant circuit using voltage sources, if sucessful would this be a good topic for writing a scientific paper?

Cheers
 
  • #31
Hesch
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So without any 'clever' use of a trick like some sort of symmetry there is not conventional way to replace current sources with voltage sources,
In #6 I wrote: You cannot convert a current source into a voltage source, but you can convert a Norton equivalent into a Thevenin equivalent.

I've read the whole thread:

In #1 the figure shows types of blocks in series: Type 1 - 2 - 3 - 4 then 1 - 3 ? That's not infinite repeated blocks. How to continue?

You admit that, but then suddenly in #5 the circuit becomes a loop.

The discussion continues with Thevenin/Norton equivalents, but since the circuit now has become a (finite) loop, KVL or KCL will be a more simple solution. Now the dicussion concerns whether KCL or KVL is to be used.

The thread is a mess. You must clearify what is meant.

If you want the circuit in #1 to be of infinite lenght, you must use Thevenin/Norton equivalents in series, then make a mathematical series of voltages, currents, whatever.
 
  • #32
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In #6 I wrote: You cannot convert a current source into a voltage source, but you can convert a Norton equivalent into a Thevenin equivalent.

I've read the whole thread:

In #1 the figure shows types of blocks in series: Type 1 - 2 - 3 - 4 then 1 - 3 ? That's not infinite repeated blocks. How to continue?

You admit that, but then suddenly in #5 the circuit becomes a loop.

The discussion continues with Thevenin/Norton equivalents, but since the circuit now has become a (finite) loop, KVL or KCL will be a more simple solution. Now the dicussion concerns whether KCL or KVL is to be used.

The thread is a mess. You must clearify what is meant.

If you want the circuit in #1 to be of infinite lenght, you must use Thevenin/Norton equivalents in series, then make a mathematical series of voltages, currents, whatever.
In my last post (#30) I suppose I meant like 'having the source like that with those parallel resistors isn't a norton/thevenin equivalent?'

I sort of was concidering infinite length, but then when that was difficult I realised that a finite length network (in post #5) would be better. Sorry for the lack of clarity.
But I thought the electrician showed that even if we use KCL or KVL for the finite series, the solution (unless numerical) is huge...unusable. So I can't figure out how to redraw it with the sources switched.


Thanks
 
  • #33
Hesch
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if we use KCL or KVL for the finite series, the solution (unless numerical) is huge...unusable.
Of course you must solve the equations numerically. Manually you will solve the equations within a day, included 11 calculation errors. In #29 the Electrician has shown you a setup as for the coefficients ( I have not checked it ). You will have to enter about 40 coefficients different from 0. Write them in a file and let a program read them from the file. In this way a forgotten minus sign is easy to correct/edit.

The program/computer will solve the equations within 5 seconds with no errors. This is not unusable
 
  • #34
864
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Of course you must solve the equations numerically. Manually you will solve the equations within a day, included 11 calculation errors. In #29 the Electrician has shown you a setup as for the coefficients ( I have not checked it ). You will have to enter about 40 coefficients different from 0. Write them in a file and let a program read them from the file. In this way a forgotten minus sign is easy to correct/edit.

The program/computer will solve the equations within 5 seconds with no errors. This is not unusable
Yes I see, oh well.
What's the theoretical version of a solution, the word for it, like the opposite of a 'numerical' solution, like if I was able to find a solution that was the equivilant without it being set numbers?

Also one other question, I have is, I've seen an ideal inductor modeled as a resistor in series with a leakage inductance, in series with the ideal inductor. Couldn't I model this as the resistor in series with the ideal and leakage inductors in parallel with eachother?

Thanks
 
  • #35
Hesch
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What's the theoretical version of a solution, the word for it, like the opposite of a 'numerical' solution, like if I was able to find a solution that was the equivilant without it being set numbers?
Well, you are the one to know, it's your language :smile:.
Even in my language (danish), I hate these "words". The important matter is to know, what I'm doing rather than to know the word.
The opposite of a numerical solution could be an algebraic solution, but "building" equations and solve them numerically are somehow mixed disciplines.
I've seen an ideal inductor modeled as a resistor in series with a leakage inductance, in series with the ideal inductor. Couldn't I model this as the resistor in series with the ideal and leakage inductors in parallel with eachother?
You should start a new thread here.

As for a known voltage/current supplied ( 60Hz, sinusoidal ): Yes. Maybe you could find the impedance = R + sL = 10Ω + j100Ω
But say you lower the frequency to 0Hz, then the impedance in the series connection will be R + sL = 10Ω + j0Ω = 10Ω.
Now say you by 60Hz have found an equivalent in parallel. Then by lowering the frequency to 0Hz, the impedance will be R || j0Ω = 0Ω.
So generally you cannot convert a model in series to a model in parallel when inductors/capacitors are included in the models.
 
  • #36
The Electrician
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Yes I see, oh well.
What's the theoretical version of a solution, the word for it, like the opposite of a 'numerical' solution, like if I was able to find a solution that was the equivilant without it being set numbers?
Thanks
It would be called a "symbolic" solution, such as the one in post #22.
 
  • #38
864
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It would be called a "symbolic" solution, such as the one in post #22.
Analytic solution
Ah yes, Analytic was the word I was looking for, was on the tip of my tongue.
 
  • #39
864
17
Well, you are the one to know, it's your language :smile:.
Even in my language (danish), I hate these "words". The important matter is to know, what I'm doing rather than to know the word.
The opposite of a numerical solution could be an algebraic solution, but "building" equations and solve them numerically are somehow mixed disciplines.

You should start a new thread here.

As for a known voltage/current supplied ( 60Hz, sinusoidal ): Yes. Maybe you could find the impedance = R + sL = 10Ω + j100Ω
But say you lower the frequency to 0Hz, then the impedance in the series connection will be R + sL = 10Ω + j0Ω = 10Ω.
Now say you by 60Hz have found an equivalent in parallel. Then by lowering the frequency to 0Hz, the impedance will be R || j0Ω = 0Ω.
So generally you cannot convert a model in series to a model in parallel when inductors/capacitors are included in the models.
Ah-ha, yeah 'analystic' was the word.

Yeah that's not what I meant by what was in parallel, I took your advice and started it as a new thread:
https://www.physicsforums.com/threads/modeling-an-inductor.825613/
 

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