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The generality of working of my designed circuit is based on this principle.

Thanks in advance!

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- Thread starter SmritiB
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- #1

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The generality of working of my designed circuit is based on this principle.

Thanks in advance!

- #2

cnh1995

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That can be found out by observation. However, if the resistances are not in series, that 'sum' is meaningless.The path along which when traversed, the sum of resistances that we come across in that very path is the least.

If you combine all the series resistors in this circuit, it will become compact with only parallel or star-delta configurations. Your original "paths" are not disturbed since the paths for possible division of current are kept as they are(parallel and star-delta). Since there are no series resistors left, current in every resistor will be different. So, comparing the resistances in parallel, you can get "shorter" paths. For star-delta, I think it'll be slightly complicated. In some circuits, currents can be predicted using symmetry arguments. It may not be possible to predict the currents in every circuit. You may have to solve the circuit first and then see where current is maximum and where it is minimum.

- #3

anorlunda

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- #4

jim hardy

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I tell beginners at circuit analysis to immerse themself in the circuitThe path along which when traversed, the sum of resistances that we come across in that very path is the least. How to find out this path?

Think of a hamster in a maze, but a hamster who can add and carries with him a notepad and pencil .

- #5

The Electrician

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If you want a visual demonstration of which resistors are carrying the most current, you could take advantage of the fact that all your resistors are a single value, 1000 ohms. Build the circuit, apply the 30 volts and observe the whole network with an infrared camera. The resistors carrying the most current will be hotter. At each node the brighter resistor (in infrared) will be the resistor carrying the most current. Just follow the path of most brightness at each node.

The dissipation in the resistors isn't all that great, so you might need to apply more than 30 volts to get a good result, or use lower value resistors.

- #6

anorlunda

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I push that because the only reason I can think of wanting to know which path had the least resistance would be if I planned to jump to the conclusion that most of the current flows through the path of least resistance. As I said in #3, that's not necessarily true.

Suppose the path of least resistance carries 1% more current than the path of second least resistance. How would that knowledge be useful?

- #7

Merlin3189

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And the current will take them ALL at the same time. Current does not take the "path of least resistance."I... Now there are many paths the current can take from A to B.

There is an equivalent resistance between A and B, which is calculated using common circuit analysis techniques. This is not the same as the sum of all the resistances along the path of least resistance.And there is an equivalent resistance computed as well. But how to calculate the least resistive path? I.e. The path along which when traversed, the sum of resistances that we come across in that very path is the least.

If you want to find this minimum sum path, look up shortest path algorithms in a discrete maths book. (I can't be bothered to do it for such a fool's errand. But just looking at the circuit, you can see that there are two shortest paths of 6k.)How to find out this path?

- #8

Baluncore

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Maybe you should consider this as a transport network. You want to find the one route with the least total cost. Each resistor is a road that has a cost. The nodes are junctions. Your start and end nodes will be points A and B.But how to calculate the least resistive path? I.e. The path along which when traversed, the sum of resistances that we come across in that very path is the least. How to find out this path?

Now you must find an algorithm that will identify the single least cost path across the network.

Is that the sort of solution you are looking for ?

- #9

berkeman

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I think it's a repost of his previous thread...

Thank you for the suggestion,

But i need to "show" the path. some sort of a detectable illumination (or even numerical values would do, provided the path is shown instantaneously, not manually moving with meters) like, any person seeing the circuit should know where all the larger portion of the current from the required node flows.

Actually, the deeper requirement is to follow from the first node, the branch with more current, then the same follow-scheme at the node this branch comes across and so on : This is what i mean by "path".

Sorry couldnt phrase it properly the first time.

- #10

Baluncore

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Do you want a physical circuit with LEDs to show the path, or is a computer simulation OK?

Can you please draw us an example network showing nodes and resistors with values. You can drag and drop it onto your post.

Does “complex” have a mathematical meaning where nodes are located in an orderly way at x + iy, where x and y are integers?

Or does “complex” mean many random nodes in space, randomly connected by resistors of different values?

Do resistors only connect to immediate neighbouring nodes? or do some resistors leap across the mesh?

If the mesh was an infinite grid of nodes with each node connected to it's 4 neighbours by the same value resistor, then the solution would be the resistors that are under a straight line passing between A and B. If it was not infinite, then the sides would bias the solution to one side of that straight line.

If you had fully specified the challenge in the OP then you could have answered it yourself. Where our answers are questions, we are actually helping you to better understand and specify your challenge. Once it is specified it will be solved.

- #11

The Electrician

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He gave a schematic in his other thread, and it is reproduced in post #2 of this thread.

- #12

Merlin3189

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I relented, because of the interest from others. Try Dijkstra's algorithm.

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