Currents/voltages graph for an electrical circuit The 2019 Stack Overflow Developer Survey Results Are InSolving “Resistance between two nodes on a grid” problem in MathematicaCircuit drawing in MathematicaMerging (combining) tables of graph relationships (2-mode to 1-mode network)Reduce distances between vertices of graph to minimum possible?How can I sequentially apply different graph embeddings?Plotting a network or a graph with given coordinates for verticesEquivalent of RadialOutside for Graph VertexLabelsInvisible graph edgesHow to make the vertices move with the VertexRenderingFunction graphics when clicking and dragging vertices of a layered graphHow to delete unnecessary resistances in a resistance networkEvaluate the electrical resistance between any two points of a circuit
Geography at the pixel level
What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?
Does HR tell a hiring manager about salary negotiations?
Deal with toxic manager when you can't quit
How do I free up internal storage if I don't have any apps downloaded?
Why didn't the Event Horizon Telescope team mention Sagittarius A*?
Can a rogue use sneak attack with weapons that have the thrown property even if they are not thrown?
Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?
Is bread bad for ducks?
What to do when moving next to a bird sanctuary with a loosely-domesticated cat?
How to charge AirPods to keep battery healthy?
How much of the clove should I use when using big garlic heads?
The phrase "to the numbers born"?
Relationship between Gromov-Witten and Taubes' Gromov invariant
Mathematics of imaging the black hole
Why isn't the black hole white?
Correct punctuation for showing a character's confusion
If I can cast sorceries at instant speed, can I use sorcery-speed activated abilities at instant speed?
How to type a long/em dash `—`
How to support a colleague who finds meetings extremely tiring?
Button changing its text & action. Good or terrible?
Can we generate random numbers using irrational numbers like π and e?
A word that means fill it to the required quantity
Flight paths in orbit around Ceres?
Currents/voltages graph for an electrical circuit
The 2019 Stack Overflow Developer Survey Results Are InSolving “Resistance between two nodes on a grid” problem in MathematicaCircuit drawing in MathematicaMerging (combining) tables of graph relationships (2-mode to 1-mode network)Reduce distances between vertices of graph to minimum possible?How can I sequentially apply different graph embeddings?Plotting a network or a graph with given coordinates for verticesEquivalent of RadialOutside for Graph VertexLabelsInvisible graph edgesHow to make the vertices move with the VertexRenderingFunction graphics when clicking and dragging vertices of a layered graphHow to delete unnecessary resistances in a resistance networkEvaluate the electrical resistance between any two points of a circuit
$begingroup$
I am trying to design the network graph for an electrical circuit. I have done this by hand so far. Here is the code for the voltages graph
g = Graph[0, 1, 2, 3, 4,
0 -> 1, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 3, 0 -> 4, 4 -> 0, 1 -> 3];
PropertyValue[g, VertexLabels] = Table[i -> StringForm["(`1`)", i], i, 0, EdgeCount[basic]];
PropertyValue[g, 0 -> 1, EdgeLabels] = Placed[2, 1/2, 1/2, 0];
PropertyValue[g, 1 -> 3, EdgeLabels] = Placed[10, 1/2, 3/2, 1/2];
PropertyValue[g, 1 -> 2, EdgeLabels] = Placed[2, 1/2, -1/2, -1/2];
PropertyValue[g, 0 -> 4, EdgeLabels] = Placed[3, 1/2, -1/2, 1/2];
Here is the result:
I have a similar one for currents and now I need to apply Kirchhoff's laws and the only way out of this right now is doing it by hand. So my question is: is there any other more efficient way around what I am doing here?
graphs-and-networks physics
New contributor
$endgroup$
add a comment |
$begingroup$
I am trying to design the network graph for an electrical circuit. I have done this by hand so far. Here is the code for the voltages graph
g = Graph[0, 1, 2, 3, 4,
0 -> 1, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 3, 0 -> 4, 4 -> 0, 1 -> 3];
PropertyValue[g, VertexLabels] = Table[i -> StringForm["(`1`)", i], i, 0, EdgeCount[basic]];
PropertyValue[g, 0 -> 1, EdgeLabels] = Placed[2, 1/2, 1/2, 0];
PropertyValue[g, 1 -> 3, EdgeLabels] = Placed[10, 1/2, 3/2, 1/2];
PropertyValue[g, 1 -> 2, EdgeLabels] = Placed[2, 1/2, -1/2, -1/2];
PropertyValue[g, 0 -> 4, EdgeLabels] = Placed[3, 1/2, -1/2, 1/2];
Here is the result:
I have a similar one for currents and now I need to apply Kirchhoff's laws and the only way out of this right now is doing it by hand. So my question is: is there any other more efficient way around what I am doing here?
graphs-and-networks physics
New contributor
$endgroup$
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago
add a comment |
$begingroup$
I am trying to design the network graph for an electrical circuit. I have done this by hand so far. Here is the code for the voltages graph
g = Graph[0, 1, 2, 3, 4,
0 -> 1, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 3, 0 -> 4, 4 -> 0, 1 -> 3];
PropertyValue[g, VertexLabels] = Table[i -> StringForm["(`1`)", i], i, 0, EdgeCount[basic]];
PropertyValue[g, 0 -> 1, EdgeLabels] = Placed[2, 1/2, 1/2, 0];
PropertyValue[g, 1 -> 3, EdgeLabels] = Placed[10, 1/2, 3/2, 1/2];
PropertyValue[g, 1 -> 2, EdgeLabels] = Placed[2, 1/2, -1/2, -1/2];
PropertyValue[g, 0 -> 4, EdgeLabels] = Placed[3, 1/2, -1/2, 1/2];
Here is the result:
I have a similar one for currents and now I need to apply Kirchhoff's laws and the only way out of this right now is doing it by hand. So my question is: is there any other more efficient way around what I am doing here?
graphs-and-networks physics
New contributor
$endgroup$
I am trying to design the network graph for an electrical circuit. I have done this by hand so far. Here is the code for the voltages graph
g = Graph[0, 1, 2, 3, 4,
0 -> 1, 1 -> 2, 2 -> 1, 3 -> 2, 4 -> 3, 0 -> 4, 4 -> 0, 1 -> 3];
PropertyValue[g, VertexLabels] = Table[i -> StringForm["(`1`)", i], i, 0, EdgeCount[basic]];
PropertyValue[g, 0 -> 1, EdgeLabels] = Placed[2, 1/2, 1/2, 0];
PropertyValue[g, 1 -> 3, EdgeLabels] = Placed[10, 1/2, 3/2, 1/2];
PropertyValue[g, 1 -> 2, EdgeLabels] = Placed[2, 1/2, -1/2, -1/2];
PropertyValue[g, 0 -> 4, EdgeLabels] = Placed[3, 1/2, -1/2, 1/2];
Here is the result:
I have a similar one for currents and now I need to apply Kirchhoff's laws and the only way out of this right now is doing it by hand. So my question is: is there any other more efficient way around what I am doing here?
graphs-and-networks physics
graphs-and-networks physics
New contributor
New contributor
edited 4 hours ago
Szabolcs
164k14448946
164k14448946
New contributor
asked 5 hours ago
VictorVictor
1134
1134
New contributor
New contributor
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago
add a comment |
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.
Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix
.
Let $v=(v_1, dots, v_n)$ be the vector of voltages at each node, $j=(j_1, dots, j_m)$ the vector of currents through each edge, and $c = (c_1, dots, c_m)$ the conductance of each resistor.
Let us put a voltage between nodes $s$ and $t$.
Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.
Ohm's law tells us that $j = c (B^T v)$.
Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.
Now in Mathematica,
edges =
1 -> 2,
1 -> 2,
1 -> 3,
2 -> 4,
4 -> 3,
5 -> 6,
6 -> 4,
5 -> 1
;
SeedRandom[42];
conductances = RandomReal[0.1, 1, Length[edges]]
g = Graph[edges]
b = IncidenceMatrix[g]
c = DiagonalMatrix@SparseArray[conductances]
s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)
Now we can get the voltages at each node.
voltages =
LinearSolve[
b.c.Transpose[b],
ReplacePart[
ConstantArray[0, VertexCount[g]],
s -> -totalCurrent, t -> totalCurrent
]
]
This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.
Get the current through each edge:
currents = conductances (voltages.b)
Get the effective resistance between s and t:
effectiveResistance = (voltages[[t]] - voltages[[s]])/current
Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.
Let this be our graph:
ug = Graph[GraphData["GreatRhombicuboctahedralGraph"],
GraphStyle -> "BasicBlack", VertexSize -> 0.5];
We orient edges in an arbitrary way:
g = DirectedGraph[ug, "Acyclic"]
Then use the above code, but set the same conductance for all edges,
conductances = N@ConstantArray[1, EdgeCount[g]];
and choose
s = 1; t = 12;
Visualize voltages:
Graph[
ug,
VertexStyle ->
Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
VertexSize -> s -> 1, t -> 1
]
Visualize current magnitudes though each edge:
Graph[ug,
EdgeStyle -> Prepend[
Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
Thickness[0.02]],
VertexSize -> s -> 1, t -> 1
]
$endgroup$
$begingroup$
This is really clever! Thank you so much! One question: you added two edges1 -> 2
in theedges
list. Is that correct?
$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Victor is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195035%2fcurrents-voltages-graph-for-an-electrical-circuit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.
Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix
.
Let $v=(v_1, dots, v_n)$ be the vector of voltages at each node, $j=(j_1, dots, j_m)$ the vector of currents through each edge, and $c = (c_1, dots, c_m)$ the conductance of each resistor.
Let us put a voltage between nodes $s$ and $t$.
Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.
Ohm's law tells us that $j = c (B^T v)$.
Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.
Now in Mathematica,
edges =
1 -> 2,
1 -> 2,
1 -> 3,
2 -> 4,
4 -> 3,
5 -> 6,
6 -> 4,
5 -> 1
;
SeedRandom[42];
conductances = RandomReal[0.1, 1, Length[edges]]
g = Graph[edges]
b = IncidenceMatrix[g]
c = DiagonalMatrix@SparseArray[conductances]
s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)
Now we can get the voltages at each node.
voltages =
LinearSolve[
b.c.Transpose[b],
ReplacePart[
ConstantArray[0, VertexCount[g]],
s -> -totalCurrent, t -> totalCurrent
]
]
This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.
Get the current through each edge:
currents = conductances (voltages.b)
Get the effective resistance between s and t:
effectiveResistance = (voltages[[t]] - voltages[[s]])/current
Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.
Let this be our graph:
ug = Graph[GraphData["GreatRhombicuboctahedralGraph"],
GraphStyle -> "BasicBlack", VertexSize -> 0.5];
We orient edges in an arbitrary way:
g = DirectedGraph[ug, "Acyclic"]
Then use the above code, but set the same conductance for all edges,
conductances = N@ConstantArray[1, EdgeCount[g]];
and choose
s = 1; t = 12;
Visualize voltages:
Graph[
ug,
VertexStyle ->
Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
VertexSize -> s -> 1, t -> 1
]
Visualize current magnitudes though each edge:
Graph[ug,
EdgeStyle -> Prepend[
Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
Thickness[0.02]],
VertexSize -> s -> 1, t -> 1
]
$endgroup$
$begingroup$
This is really clever! Thank you so much! One question: you added two edges1 -> 2
in theedges
list. Is that correct?
$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
add a comment |
$begingroup$
A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.
Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix
.
Let $v=(v_1, dots, v_n)$ be the vector of voltages at each node, $j=(j_1, dots, j_m)$ the vector of currents through each edge, and $c = (c_1, dots, c_m)$ the conductance of each resistor.
Let us put a voltage between nodes $s$ and $t$.
Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.
Ohm's law tells us that $j = c (B^T v)$.
Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.
Now in Mathematica,
edges =
1 -> 2,
1 -> 2,
1 -> 3,
2 -> 4,
4 -> 3,
5 -> 6,
6 -> 4,
5 -> 1
;
SeedRandom[42];
conductances = RandomReal[0.1, 1, Length[edges]]
g = Graph[edges]
b = IncidenceMatrix[g]
c = DiagonalMatrix@SparseArray[conductances]
s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)
Now we can get the voltages at each node.
voltages =
LinearSolve[
b.c.Transpose[b],
ReplacePart[
ConstantArray[0, VertexCount[g]],
s -> -totalCurrent, t -> totalCurrent
]
]
This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.
Get the current through each edge:
currents = conductances (voltages.b)
Get the effective resistance between s and t:
effectiveResistance = (voltages[[t]] - voltages[[s]])/current
Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.
Let this be our graph:
ug = Graph[GraphData["GreatRhombicuboctahedralGraph"],
GraphStyle -> "BasicBlack", VertexSize -> 0.5];
We orient edges in an arbitrary way:
g = DirectedGraph[ug, "Acyclic"]
Then use the above code, but set the same conductance for all edges,
conductances = N@ConstantArray[1, EdgeCount[g]];
and choose
s = 1; t = 12;
Visualize voltages:
Graph[
ug,
VertexStyle ->
Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
VertexSize -> s -> 1, t -> 1
]
Visualize current magnitudes though each edge:
Graph[ug,
EdgeStyle -> Prepend[
Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
Thickness[0.02]],
VertexSize -> s -> 1, t -> 1
]
$endgroup$
$begingroup$
This is really clever! Thank you so much! One question: you added two edges1 -> 2
in theedges
list. Is that correct?
$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
add a comment |
$begingroup$
A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.
Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix
.
Let $v=(v_1, dots, v_n)$ be the vector of voltages at each node, $j=(j_1, dots, j_m)$ the vector of currents through each edge, and $c = (c_1, dots, c_m)$ the conductance of each resistor.
Let us put a voltage between nodes $s$ and $t$.
Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.
Ohm's law tells us that $j = c (B^T v)$.
Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.
Now in Mathematica,
edges =
1 -> 2,
1 -> 2,
1 -> 3,
2 -> 4,
4 -> 3,
5 -> 6,
6 -> 4,
5 -> 1
;
SeedRandom[42];
conductances = RandomReal[0.1, 1, Length[edges]]
g = Graph[edges]
b = IncidenceMatrix[g]
c = DiagonalMatrix@SparseArray[conductances]
s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)
Now we can get the voltages at each node.
voltages =
LinearSolve[
b.c.Transpose[b],
ReplacePart[
ConstantArray[0, VertexCount[g]],
s -> -totalCurrent, t -> totalCurrent
]
]
This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.
Get the current through each edge:
currents = conductances (voltages.b)
Get the effective resistance between s and t:
effectiveResistance = (voltages[[t]] - voltages[[s]])/current
Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.
Let this be our graph:
ug = Graph[GraphData["GreatRhombicuboctahedralGraph"],
GraphStyle -> "BasicBlack", VertexSize -> 0.5];
We orient edges in an arbitrary way:
g = DirectedGraph[ug, "Acyclic"]
Then use the above code, but set the same conductance for all edges,
conductances = N@ConstantArray[1, EdgeCount[g]];
and choose
s = 1; t = 12;
Visualize voltages:
Graph[
ug,
VertexStyle ->
Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
VertexSize -> s -> 1, t -> 1
]
Visualize current magnitudes though each edge:
Graph[ug,
EdgeStyle -> Prepend[
Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
Thickness[0.02]],
VertexSize -> s -> 1, t -> 1
]
$endgroup$
A resistor network can be represented with an undirected (multi-)graph. We are going to orient each edge to obtain a directed graph, so we have a reference for which direction the current is flowing.
Let $B$ be the vertex-edge incidence matrix of the oriented graph. This can be obtained with IncidenceMatrix
.
Let $v=(v_1, dots, v_n)$ be the vector of voltages at each node, $j=(j_1, dots, j_m)$ the vector of currents through each edge, and $c = (c_1, dots, c_m)$ the conductance of each resistor.
Let us put a voltage between nodes $s$ and $t$.
Kirchoff's current law tells us that the sum of currents is zero at each node except $s$ and $t$ where it is some $i$ and $-i$ respectively. In matrix notation, the sum of currents at each node is $Bj$.
Ohm's law tells us that $j = c (B^T v)$.
Putting the two together we get the sum of currents at each node as $B C B^T v$ where $C$ is a diagonal matrix obtained from $c$.
Now in Mathematica,
edges =
1 -> 2,
1 -> 2,
1 -> 3,
2 -> 4,
4 -> 3,
5 -> 6,
6 -> 4,
5 -> 1
;
SeedRandom[42];
conductances = RandomReal[0.1, 1, Length[edges]]
g = Graph[edges]
b = IncidenceMatrix[g]
c = DiagonalMatrix@SparseArray[conductances]
s = 1; t = 6; (* index of sink and source node *)
totalCurrent = 1 (* total current from s to t *)
Now we can get the voltages at each node.
voltages =
LinearSolve[
b.c.Transpose[b],
ReplacePart[
ConstantArray[0, VertexCount[g]],
s -> -totalCurrent, t -> totalCurrent
]
]
This system is underdetermined (corresponding to the fact that there's no reference for the voltages and only voltage differences make sense), but luckily Mathematica is smart enough to deal with that.
Get the current through each edge:
currents = conductances (voltages.b)
Get the effective resistance between s and t:
effectiveResistance = (voltages[[t]] - voltages[[s]])/current
Unfortunately, Mathematica is not capable of styling parallel edges differently. Below I'll use a simple graph (no multi-edges) to illustrate how to visualize the result.
Let this be our graph:
ug = Graph[GraphData["GreatRhombicuboctahedralGraph"],
GraphStyle -> "BasicBlack", VertexSize -> 0.5];
We orient edges in an arbitrary way:
g = DirectedGraph[ug, "Acyclic"]
Then use the above code, but set the same conductance for all edges,
conductances = N@ConstantArray[1, EdgeCount[g]];
and choose
s = 1; t = 12;
Visualize voltages:
Graph[
ug,
VertexStyle ->
Thread[VertexList[ug] -> ColorData["Rainbow"] /@ Rescale[voltages]],
VertexSize -> s -> 1, t -> 1
]
Visualize current magnitudes though each edge:
Graph[ug,
EdgeStyle -> Prepend[
Thread[EdgeList[ug] -> (ColorData["Rainbow"] /@ Rescale@Abs[currents])],
Thickness[0.02]],
VertexSize -> s -> 1, t -> 1
]
edited 3 hours ago
answered 3 hours ago
SzabolcsSzabolcs
164k14448946
164k14448946
$begingroup$
This is really clever! Thank you so much! One question: you added two edges1 -> 2
in theedges
list. Is that correct?
$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
add a comment |
$begingroup$
This is really clever! Thank you so much! One question: you added two edges1 -> 2
in theedges
list. Is that correct?
$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
$begingroup$
This is really clever! Thank you so much! One question: you added two edges
1 -> 2
in the edges
list. Is that correct?$endgroup$
– Victor
3 hours ago
$begingroup$
This is really clever! Thank you so much! One question: you added two edges
1 -> 2
in the edges
list. Is that correct?$endgroup$
– Victor
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
@Victor Yes, I did that on purpose to illustrate that the method works even with multigraphs. These two edges represent two resistors in parallel. In general, Mathematica does not deal well with multigraphs, so it's good to know that there's no problem here (IncidenceMatrix can handle them). Visualization will be troublesome though, so I added a non-multigraph example for that.
$endgroup$
– Szabolcs
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
$begingroup$
Thank you so much! It became clearer!
$endgroup$
– Victor
3 hours ago
add a comment |
Victor is a new contributor. Be nice, and check out our Code of Conduct.
Victor is a new contributor. Be nice, and check out our Code of Conduct.
Victor is a new contributor. Be nice, and check out our Code of Conduct.
Victor is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195035%2fcurrents-voltages-graph-for-an-electrical-circuit%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Please note that I have just started using Mathematica a few weeks ago!
$endgroup$
– Victor
5 hours ago
$begingroup$
A circuit of resistors is not a directed graph. It is undirected. It does make sense to orient edges so that we can distinguish between currents flowing in opposite direction (there's a frame of reference), but you have edges going in both directions between the same vertices. It's not clear what you are trying to represent with that.
$endgroup$
– Szabolcs
4 hours ago
$begingroup$
@Szabolcs edge 4->0 may represent a resistor (with current flowing from 4 to 0) and edge 0->4 may represent a voltage source.
$endgroup$
– Victor
4 hours ago
$begingroup$
@Szabolcs although not in English, please take a look at the two diagrams in the middle of page 8 of this document. That's an example of what I have to do here, or the figures at the top of page 23 in this document
$endgroup$
– Victor
4 hours ago
$begingroup$
"Sagetile de pe laturi indica sensurile referinta ale curentilor si tensiunilor", thus the edge directions do not have physical meaning, they only serve as a reference for the current values and voltage difference values. (I can read Romanian.)
$endgroup$
– Szabolcs
3 hours ago