# Electrical circuit Source transformation Question

• tim9000
In summary: R1 ), all of the currents entering that part of the circuit will be the same.In summary, the circuit has some symmetry due to the fact that all the vertical resistors are the same value, and because all the current sources are connected in parallel. Kirchhoffs current law can be used to calculate voltages in this circuit.
tim9000
I'm extreeeeamly rusty at basic circuit theory.
I want to transform a sort of infinite arrangement of this circuit:

The values themselves aren't important, just the representation of what's happening, all the verticle resistors are the same value as each other, and all the horizontal resistors are the same value as each other.
I just made up 1 and 3 ohms as a random example. You could say that all the current soruces are 1 A if you like.
I'd like to figure out how to transform all the current sources into voltage sources using source transformation:
https://en.wikipedia.org/wiki/Source_transformation

But my issue is that since each current source has multiple resistors in parallel with it, how can you do this?

Thanks a lot!

tim9000 said:
I want to transform a sort of infinite arrangement of this circuit:
I suggest that you at first make a Thevenin/Norton equivalent of a "repeated group" in the circuit. Then connect these equivalents in series/parallel.

Hesch said:
I suggest that you at first make a Thevenin/Norton equivalent of a "repeated group" in the circuit. Then connect these equivalents in series/parallel.
I looked up Thevenin/Norton, repeated group equivalent circuits, but I can't find anything for 'repeated groups'. Did you have a link to a site or a textbook or anything? Sounds like that's what I need to do, so you don't think there's a way to do it for a single component with two parallel resistors because you can't isolate them? (you need to do it as a group?) The reason I ask is, say that it wasn't like an infite pattern, but for instance if it ended with shorts or a resistive branch on either side of the diagram. OR if both the ends looped back and were connected together with a resistive branch goind down between them (then I imagine the group thing wouldn't apply?).

tim9000 said:
I looked up Thevenin/Norton, repeated group equivalent circuits, but I can't find anything for 'repeated groups'.
Having a quick look at your circuit, I assumed it were build by repeated groups. For example from left to right the current sources point east, north, west, south and then repeated this "pattern", so that a "group" included 4 current sources. But by closer look I see it's not repeated, because it continues: east, west, . . . . .

Anyway I hope you understand what I mean as for the direction of current sources: If there are no repeated groups, it will be very difficult to predict how the network will continue to the right towards infinity.

If the decimals of the constant π were repeating themselves in "groups", say
π = 3.1415926514159265... it would be easy to "calculate" π with a million decimals.

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Hesch said:
Anyway I hope you understand what I mean as for the direction of current sources: If there are no repeated groups, it will be very difficult to predict how the network will continue to the right towards infinity.
I'm not sure what you mean by that statement.

I actually realized I made a mistake myself, I just fixed it in (the following picture) and an example of one of the hypertheticals of what I meant (is also in that picture):

So is there any way to convert the current sources to voltage sources?

Thanks alot.

tim9000 said:
I actually realized I made a mistake myself, I just fixed it in (the following picture) and an example of one of the hypertheticals of what I meant (is also in that picture):
Well, now you have fixed the mistake (B7), but you have also made a closed loop. You have to decide: Closed loop or infinite straight network?

You will never divide the last circuit into "groups", now there is only one group.
tim9000 said:
any way to convert the current sources to voltage sources?
No, you cannot convert a current source into a voltage source, but you can convert a Norton equivalent into a Thevenin equivalent.

Example: ( IN = 4A, RN = 5Ω ) → ( VT = 20V, RT = 5Ω )

What do you want to find? Voltages, currents?

Hesch said:
Well, now you have fixed the mistake (B7), but you have also made a closed loop. You have to decide: Closed loop or infinite straight network?
Well closed loop than. But How would a straight network be different?
Hesch said:
You will never divide the last circuit into "groups", now there is only one group.

No, you cannot convert a current source into a voltage source, but you can convert a Norton equivalent into a Thevenin equivalent.

Example: ( IN = 4A, RN = 5Ω ) → ( VT = 20V, RT = 5Ω )

What do you want to find? Voltages, currents?
Yeah exactly, I was still referring to the transformation of before, like here:

I'm after voltages.

tim9000 said:
I'm after voltages.
There is some symmetry in the circuit, due to all vertical resistors are the same, etc.

But generally you could use Kirchhoffs current law (KCL).
If you look at a part of the circuit: ( R3 . . R8, B2 ), there are 6 nodes. Naming these as V1, V2 . . V6, starting clockwise at the lower left corner, and saying that V6 = 0V ( reference ), you could make 5 equations, thereby finding 5 remaining voltages. Here is an example as for V2:

( V1-V2 ) / R4 + ( V3 - V2 )/R3 + IB2 = 0

In this way you can find all the voltages in this isolated part of the circuit.

To find all the voltages in the circuit as a whole, you will have to make 11 equations. Note that any values for the resistors and source-currents may be used.
So name all the 12 nodes, set one of the nodes = 0V, make 11 equations and solve them.

I suggest you make the first 3 equations, then post them and I ( or others ) will have a look at them.

Hesch said:
There is some symmetry in the circuit, due to all vertical resistors are the same, etc.

But generally you could use Kirchhoffs current law (KCL).
If you look at a part of the circuit: ( R3 . . R8, B2 ), there are 6 nodes. Naming these as V1, V2 . . V6, starting clockwise at the lower left corner, and saying that V6 = 0V ( reference ), you could make 5 equations, thereby finding 5 remaining voltages. Here is an example as for V2:

( V1-V2 ) / R4 + ( V3 - V2 )/R3 + IB2 = 0

In this way you can find all the voltages in this isolated part of the circuit.

To find all the voltages in the circuit as a whole you will have to make 11 equations. Note that any values for the resistors and source-currents may be used.
So name all the 12 nodes, set one of the nodes = 0V, make 11 equations and solve them.

I suggest you make the first 3 equations, then post them and I ( or others ) will have a look at them.
Thanks for the advice, so if I give that a crack will I have just the voltages or will they be a function of current? Because that won't really suit my purpose, which is to figure out what the circuit will transform to when the current sources become equivalent voltage sources.

Thanks again

By solving the 11 equations, you will find 11 voltages, but when knowing all of these, you can easy calculate the currents by means of ohms law: I = ΔV / R

EDIT: To find some equivalent for the circuit, you must define some "output". You can load this output by some current source say 1A, including this in your equations and solve them again. The voltage of the output will then change an amount of ΔV. Output impedance = ΔV / 1A.

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Hesch said:
By solving the 11 equations, you will find 11 voltages, but when knowing all of these, you can easy calculate the currents by means of ohms law:

I = ΔV / R
Hi and thanks,
but that's not quite what I mean, I kind of don't want to know the current. So say I followed your advice and generated some equations using KCL to got 6 vertical voltages and 6 horizontal voltages, could I then just substitute those voltages in as voltage sources in place of the current sources?

Read the "EDIT" in #10. That's how you can calculate an equivalent.

Hesch said:
By solving the 11 equations, you will find 11 voltages, but when knowing all of these, you can easy calculate the currents by means of ohms law: I = ΔV / R

EDIT: To find some equivalent for the circuit, you must define some "output". You can load this output by some current source say 1A, including this in your equations and solve them again. The voltage of the output will then change an amount of ΔV. Output impedance = ΔV / 1A.
Why was I making one node equal to 0V again? (why 11 equs not 12?)

Gee, that's going to be hard, I didn't quite realize the whole 'output' thing, yeah, ok, I vaguely remember the 'load with an arbitrary 1A' thing, I had no idea what you were talking about at first, lol. But all I can really think to do is take a source across two nodes and call that the whole circuit connected to that the output. I can't really get my head around this change. All I really remember doing is when you have some circuit with a source of current with one resistor parallel to that and say that the rest of the circuit is the output (or a voltage source with one resistor in series) but here there isn't a single entity in parallel I can isolate with a current source to take the rest of the circuit as output.
Do you have any idea what would actually be an output?

tim9000 said:
Why was I making one node equal to 0V again? (why 11 equs not 12?)
If you have some battery with an emk = 4.5V, and I asked you: What is the voltage of the anode? You could answer: It's 4.5V.

I could tell you: No, it's not, because I've connected the cathode to 115Vac, thus the voltage of the anode will be:

√2*115V*sin(ωt) + 4.5V

You have to know about some reference voltage, and you are completely free to choose a such. So if you have 12 unknown voltages, and define one of these as a reference, there is only 11 unknown voltages left. Thus only 11 equations are needed.
tim9000 said:
But all I can really think to do is take a source across two nodes and call that the whole circuit connected to that the output.
No, you must regard the two nodes to be the output, and regard the current source to be the load.

A thevenin/norton equivalent has two terminals, which I call the output ( not the input ).

Hesch said:
If you have some battery with an emk = 4.5V, and I asked you: What is the voltage of the anode? You could answer: It's 4.5V.

I could tell you: No, it's not, because I've connected the cathode to 115Vac, thus the voltage of the anode will be:

√2*115V*sin(ωt) + 4.5V

You have to know about some reference voltage, and you are completely free to choose a such. So if you have 12 unknown voltages, and define one of these as a reference, there is only 11 unknown voltages left. Thus only 11 equations are needed.
AAh, yeah ok, it's a ground I get ya. I'm starting to remember NR and Gauss-seidal, like having the slack bus. We need the ground reference, ofcourse.

Hesch said:
No, you must regard the two nodes to be the output, and regard the current source to be the load.
Hmm, if the two nodes across the current source is the load, don't you usually rearrange accroding to ohms law the remaining sourse (which in this case is the whole rest of the circuit)? Is this your way of telling me that I can't rearrange it, I can only do it via equations?

tim9000 said:
Is this your way of telling me that I can't rearrange it, I can only do it via equations?
Of course you can rearrange it, give it a try and do a lot of calculations. When you end up with a result, how will you check that there are no errors in your calculations?

Having not rearranged the circuit, you can check everything by means of ohms law ( resistors, currents, voltages ).

Having made the 11 equations, you will just have to add a "+" or "-" Iload to ( at most ) two of the nodes. Then recalculate. ( Your computer will do it within a second. )

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Hesch said:
To find all the voltages in the circuit as a whole, you will have to make 11 equations. Note that any values for the resistors and source-currents may be used.
So name all the 12 nodes, set one of the nodes = 0V, make 11 equations and solve them.
Alright, I just got a free moment after moving house.
These are very slapped-up so tell me if they're completely wrong. (I just did all the current loops anti-clockwise so I might have got the conventions wrong) But. How's this for a start:
B5/2*(R1 + R15 + R19) - B1/2*R1 - B7/2*R19 = 0
B5/2*(R2 + R16 + R20) - B1/2*R2 - B7/2*R20 = 0
B1/2*(R1 + R2) - B5/2*R1 - B5/2*R2 = 0
B1/2*(R3 + R4) - B2/2*R3 - B2/2*R4 = 0
B2/2*(R3 + R8 + R5) - B1/2*R3 - B4/2*R5 = 0
B2/2*(R4 + R7 + R6) - B1/2*R4 - B4/2*R6 = 0
B4/2*(R5 + R6) - B2/2*R5 - B2/2*R6 = 0
B4/2*(R9 + R10) - B3/2*R9 - B3/2*R10 = 0
B3/2*(R9 + R12 + R13) - B4/2*R9 - B7/2*R13 = 0
B3/2*(R10 + R11 + R14) - B4/2*R10 - B7/2*R14 = 0
B7/2*(R13 + R14) - B3/2*R13 - B3/2*R14 = 0
B7/2*(R19 + R20) - B5/2*R19 - B5/2*R20 = 0

CALLING THE THIRD SOURCE ON THE LEFT "B2'',
[WORKING FROM TOP LEFT DOWN AND TO THE RIGHT]

P.S Which computer programme would you recommend to do such calculations?

Cheers!

tim9000 said:
I just did all the current loops anti-clockwise so I might have got the conventions wrong
Well, now you are using Kirchhoffs Voltage Law( KVL with current loops ). I was speaking of his Current Law ( KVL with nodes ).

Anyway, your your equations are wrong: KVL uses loop currents but for example in your first equation:

B5/2*(R1 + R15 + R19) - B1/2*R1 - B7/2*R19 = 0

you are using physical currents ( B5, B1, B7 ) which are not loop-currents. Using KVL you must define al loop-current within every loop in your circuit, and yes, there is 12 loops with the loop-currents L1 . . L12 in your drawing in #5, thus you must have 12 equation to solve them.
tim9000 said:
Which computer programme would you recommend to do such calculations?
I think there are a lot of different programs that can solve 12 equations ( MatLab, MathCAD, etc. ), but I prefer to write such a program myself in good old Borland Pascal . Then I know know what I'm doing and I don't have to pay for a license .

If you are in doubt how to use KVL/KCL then post a simple circuit with an attempt to make the equations for that. I ( or others ) will have a look at it.

tim9000 said:
I'm extreeeeamly rusty at basic circuit theory.
I want to transform a sort of infinite arrangement of this circuit:View attachment 85252
The values themselves aren't important, just the representation of what's happening, all the verticle resistors are the same value as each other, and all the horizontal resistors are the same value as each other.
I just made up 1 and 3 ohms as a random example. You could say that all the current soruces are 1 A if you like.
I'd like to figure out how to transform all the current sources into voltage sources using source transformation:
https://en.wikipedia.org/wiki/Source_transformation

But my issue is that since each current source has multiple resistors in parallel with it, how can you do this?

Thanks a lot!
It probably wouldn't be too hard to convert a repeating segment of your circuit into a two-port (https://en.wikipedia.org/wiki/Two-port_network).

Would you care about what's actually in the two-port, or is it sufficient that the two-port contains different circuitry, maybe containing only voltage sources, maybe not, and the two port behavior at the terminals would be the same as your circuit? If that's ok, then you could cascade an infinite number of the two-ports and get an infinite circuit with the same behavior as your original circuit carried to infinity.

Hesch said:
If you are in doubt how to use KVL/KCL then post a simple circuit with an attempt to make the equations for that. I ( or others ) will have a look at it.
Oh yeah, good point(s). And thanks for the quick reply. Its starting to come back to me. Well here is a labeled picture For KCL:

let N1 = 0V
[0-N4] / R15 + [N4-N5] / R1 + B1 + [N7-N4] / R3 + [N8-N4] / R8 = 0
B5 + [N5-N4]/R1 + [N5-N6]/R2 = 0
[N6 - N5] / R2 + [N6 - N3] / R16 - B1 + [N6 - N7] / R4 + [N6 - N10] / R7 = 0
[N7 - N4] / R3 + B2 + [N7 - N6] / R4 = 0
[N8 - N4] / R8 + [N8 - N9] / R5 + [N8 - N1] / R12 - B4 + [N8 - N11] / R9 = 0
[N9 - N8] / R5 - R2 + [N9 - N10] / R6 = 0
[N10 - N6] / R7 + [N10 - N9] / R6 + B4 + [N10 - N11] / R10 + [N10 - N3] / R11 = 0
[N12 - N1] / R13 + B3 + [N12 - N3] / R14 = 0
[N1 - N2] / R19 + B7 + [N1 - N8] / R12 = 0
[N2 - N1] / R19 - B5 + [N2 - N3] / R20 = 0
[N3 - N2] / R20 - B7 + [N3 - N6] / R16 = 0

How's that? So what do I rearrange for now?

Hesch said:
I think there are a lot of different programs that can solve 12 equations ( MatLab,...
Right, I'll use MATLAB than.

THANKS

The Electrician said:
It probably wouldn't be too hard to convert a repeating segment of your circuit into a two-port (https://en.wikipedia.org/wiki/Two-port_network).

Would you care about what's actually in the two-port, or is it sufficient that the two-port contains different circuitry, maybe containing only voltage sources, maybe not, and the two port behavior at the terminals would be the same as your circuit? If that's ok, then you could cascade an infinite number of the two-ports and get an infinite circuit with the same behavior as your original circuit carried to infinity.
I'm so rusty that what you proposed more or less went over my head, could you humour me? Maybe with a picture? Cheers! (I'd like it to be finite, although an infinite circuit would still be of interest to me)

tim9000 said:
I'm so rusty that what you proposed more or less went over my head, could you humour me? Maybe with a picture? Cheers! (I'd like it to be finite, although an infinite circuit would still be of interest to me)

I want to tell you that if you actually solve the 11 equations which you have formulated with symbolic variables, you will get results for the node voltages which will be totally unusable. Each node voltage expression will take many lines, maybe pages. Here's what I mean, only many times bigger:

On the other hand, if you attempt a numerical solution you will get a usable solution.

What is your ultimate purpose in all this? What will you do with the solution?

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tim9000 said:
How's that? So what do I rearrange for now?
I've only checked the first equation:

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

Starting with the first parenthesis [0-N4] you are calculating the sum of currents that flows towards the node N4, thus all the currents must have this direction:

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

You don't have to "rearrange". Just setup a matrix: As for this first equation, the coefficient in the first row will be:

N4: -1/R15 - 1/R1 - 1/R3 - 1/R8
N5: 1/R1
N7: 1/R3
N8: 1/R8
= -B1

The solution will be all the node-voltages: N2 . . N12

The Electrician said:
What is your ultimate purpose in all this? What will you do with the solution?
I want to model a permanant magnetic array as an electromagnetic array made from amp-turns.

Hesch said:
Starting with the first parenthesis [0-N4] you are calculating the sum of currents that flows towards the node N4, thus all the currents must have this direction:
Is this something like the (12x12) matrix you were talking about (in Matlab):
[0-N4] / R15 + [N5-N4] / R1 + B1 + [N7-N4] / R3 + [N8-N4] / R8 = 0
B5 + [N4-N5]/R1 + [N6-N5]/R2 = 0
[N5 - N6] / R2 + [N3 - N6] / R16 - B1 + [N7 - N6] / R4 + [N10 - N6] / R7 = 0
[N4 - N7] / R3 + B2 + [N6 - N7] / R4 = 0
[N4 - N8] / R8 + [N9 - N8] / R5 + [N1 - N8] / R12 - B4 + [N11 - N8] / R9 = 0
[N8 - N9] / R5 - B2 + [N10 - N9] / R6 = 0
[N6 - N10] / R7 + [N9 - N10] / R6 + B4 + [N11 - N10] / R10 + [N3 - N10] / R11 = 0
[N1 - N12] / R13 + B3 + [N3 - N12] / R14 = 0
[N2 - N1] / R19 + B7 + [N8 - N1] / R12 = 0
[N1 - N2] / R19 - B5 + [N3 - N2] / R20 = 0
[N2 - N3] / R20 - B7 + [N6 - N3] / R16 = 0
Becomes:

resist =
[0 0 0 -(1/R1 + 1/R3 + 1/R8) 1/R1 0 1/R3 1/R8 0 0 0 0
0 0 0 1/R1 -(1/R1 + 1/R2) 1/R6 0 0 0 0 0 0
0 0 1/R16 0 1/R2 -(1/R2 + 1/R16 + 1/R4 + 1/R7) 1/R4 0 0 1/R7 0 0
0 0 0 1/R3 0 1/R4 0 -(1/R3 + 1/R4) 0 0 0 0 0
1/R12 0 0 1/R8 0 0 0 -(1/R8 + 1/R5 + 1/R12 + 1/R9) 1/R5 0 1/R9 0
0 0 0 0 0 0 0 1/R5 -(1/R5 + 1/R6) 1/R6 0 0
0 0 1/R11 0 0 1/R7 0 0 1/R6 -(1/R7 + 1/R6 + 1/R10 + 1/R11) 1/R10 0
1/R13 0 1/R14 0 0 0 0 0 0 0 0 -(1/R13 + 1/R14)
-(1/R19 + 1/R12) 1/R19 0 0 0 0 0 1/R12 0 0 0 0
1/R19 -(1/R19 + 1/R20) 1/R20 0 0 0 0 0 0 0 0 0
0 1/R20 -(1/R20 + 1/R16) 0 0 1/R16 0 0 0 0 0 0];current = [-B1 -B5 B1 -B2 B4 B2 -B4 -B3 -B7 B5 B7]';Voltage = inv(resist)*current

thanks

tim9000 said:
Is this something like the (12x12) matrix you were talking about (in Matlab):
[0-N4] / R15 + [N5-N4] / R1 + B1 + [N7-N4] / R3 + [N8-N4] / R8 = 0
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 ).

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.

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Hesch said:
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?

The Electrician said:
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.

tim9000 said:
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.
tim9000 said:
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.
tim9000 said:
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.
tim9000 said:
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.

tim9000 said:
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:

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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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.
Hesch said:
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.

The Electrician said:
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.

The Electrician said:
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

tim9000 said:
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.

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.

Hesch said:
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.

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 realized 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

tim9000 said:
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

Hesch said:
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 each other?

Thanks

tim9000 said:
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 .
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.
tim9000 said:
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 each other?
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.

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