Find the current through a complicated circuit

Much appreciated.In summary, the circuit has an overlapping path and the current doesn't flow in one of the two paths in the middle.
  • #1
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Homework Statement
Six 100 ohm resistors are soldered together to form a regular tetrahedron. If the terminals of a 9V battery are wired to two of the vertices, how much current in A (amps) will flow from the battery?
Relevant Equations
V = IR
Screenshot (98).png

Yeah so I'm confused about this circuit because it has an overlapping path. My guess is that the current doesn't flow in one of the two paths in the middle but I don't have a good reason for it (I don't know why). I suppose there is symmetry but I haven't been able to find good videos on symmetry on YouTube. Can you help me solve the middle portion of this circuit? Much appreciated.
 
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  • #2
Do you know about Kirchhoff's laws?
 
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  • #3
Jaccobtw said:
Homework Statement:: Six 100 ohm resistors are soldered together to form a regular tetrahedron. If the terminals of a 9V battery are wired to two of the vertices, how much current in A (amps) will flow from the battery?
Relevant Equations:: V = IR

View attachment 299054
Yeah so I'm confused about this circuit because it has an overlapping path. My guess is that the current doesn't flow in one of the two paths in the middle but I don't have a good reason for it (I don't know why). I suppose there is symmetry but I haven't been able to find good videos on symmetry on YouTube. Can you help me solve the middle portion of this circuit? Much appreciated.
Where is the battery connected? What in the world does this mean for any connections?

1648509810833.png
 
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  • #4
1) As berkeman pointed out, the problem statement is meaningless until you specify WHAT vertices get the 9V
2) I understand exactly what the 3D circuit looks like but you need to redraw it in 2D so it is more amenable to normal circuit analysis.
 
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  • #5
phinds said:
1) As berkeman pointed out, the problem statement is meaningless until you specify WHAT vertices get the 9V
Not so hasty! Being a tetrahedron, it doesn't matter which two vertices you pick. (Any pair of vertices is related to any other pair by a rotation.)

To OP: once you've picked two vertices arbitrarily, can you use the symmetry to suggest i) which resistors have equal currents flowing through them and ii) which resistors have no current flowing through them? Follow-up: what's a simpler equivalent circuit?
 
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  • #6
ergospherical said:
Not so hasty! Being a tetrahedron, it doesn't matter which two vertices you pick. (Any pair of vertices is related to any other pair by a rotation.)
egg_large.jpg
 
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  • #7
ergospherical said:
Not so hasty! Being a tetrahedron, it doesn't matter which two vertices you pick. (Any pair of vertices is related to any other pair by a rotation.)

To OP: once you've picked two vertices arbitrarily, can you use the symmetry to suggest i) which resistors have equal currents flowing through them and ii) which resistors have no current flowing through them? Follow-up: what's a simpler equivalent circuit?
Nuh-uh! it depends on where the dot goes in that middle bit! :wink:
 
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  • #8
berkeman said:
Nuh-uh! it depends on where the dot goes in that middle bit! :wink:
Mike, you are misreading the 3D circuit. I think it's quite clear. Or are you joshing us? Again. Still. You need to stop that dammit ; you confuse me ! Besides you name's Mike, not Josh.
 
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  • #9
Well, I've never, ever seen a schematic with a fade to try to explain the connectivity. So you just stand down there young feller, util the OP can post a proper schematic diagram for us to help him on... :smile:
 
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  • #10
A regular tetrahedron is a 3D figure. The fade is just because one resistor goes behind a front one.
 
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  • #11
haruspex said:
But the circuit can be laid out as a 2D figure with no crossings, which would be a good start.
See item #2 in post #4
 
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  • #12
phinds said:
See item #2 in post #4
Yes, saw that after I posted, so deleted my post, but too late.
Anyway, the clarification "with no crossings" might help.
 
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  • #13
berkeman said:
Nuh-uh! it depends on where the dot goes in that middle bit! :wink:
Do you mean the dot that doesn't exist?

That style of wires crossing each other without connecting at least used to be somewhat common, and well understood.
And it sure cut down on the time in the drafting department!
 
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  • #14
Either way, the statement that the resistors form the edges of a tetrahedron should be sufficient wthout the image.

As for the problem at hand, it should be quite clear that the circuit is a parallel coupling of a single resistor (the one on the edge between the connection points) and the rest. The ”rest” form a circuit that can be easily flattened and analysed.
 
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  • #15
Jaccobtw said:
Homework Statement:: Six 100 ohm resistors are soldered together to form a regular tetrahedron. If the terminals of a 9V battery are wired to two of the vertices, how much current in A (amps) will flow from the battery?
Relevant Equations:: V = IR

View attachment 299054
Yeah so I'm confused about this circuit because it has an overlapping path. My guess is that the current doesn't flow in one of the two paths in the middle but I don't have a good reason for it (I don't know why). I suppose there is symmetry but I haven't been able to find good videos on symmetry on YouTube. Can you help me solve the middle portion of this circuit? Much appreciated.
Pick any vertex and look straight into it and what would you see?
 
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  • #16
Guys does it matter if it is a tetrahedron or just a square with all its sides and its diagonals resistors? I don't think it matters, it is the lumped model in circuit theory that makes it not matter.
 
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  • #17
Delta2 said:
Guys does it matter if it is a tetrahedron or just a square with all its sides and its diagonals resistors? I don't think it matters, it is the lumped model in circuit theory that makes it not matter.
I think a square with all its sides and its diagonals resistors would be a different configuration.
 
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  • #18
I think I figured it out. I connected the terminals to the left most and right most vertices, got rid of the middle vertical resistor due to symmetry, and the rest is just adding up the resistors in series and parallel.
 
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  • #19
bob012345 said:
I think a square with all its sides and its diagonals resistors would be a different configuration.
Yes it would be a different configuration IF you going to do complete field analysis, but if you are interested only for voltages and currents I think it doesn't matter.
 
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  • #20
Delta2 said:
Yes it would be a different configuration IF you going to do complete field analysis, but if you are interested only for voltages and currents I think it doesn't matter.
I can see that both have four vertices each with three connections and in that sense they should be the same electrically but I can't see how one morphs into the other. But then I'm not a topologist. Also, the tetrahedron projects into a 2D plane with no overlap while the square does not.
 
  • #21
phinds said:
A regular tetrahedron is a 3D figure. The fade is just because one resistor goes behind a front one.
And because the person who wrote this problem is a physicist, not an EE. Granted the words are clear, but the schematic doesn't follow normal standards. Components don't fade, wires don't cross through/over components. For bonus points you follow the old Mil-Stds, where wires only cross like "X" when they don't connect. If they connect, they are only drawn as "T" with a dot.

In any case, whenever you see this sort of problem, the first thing you want to look for are simplifications due to symmetry arguments as @ergospherical said. There are 4 vertices, two are connected to the source, what do you know about the voltage at the other two vertices?
 
  • #22
Jaccobtw said:
I think I figured it out. I connected the terminals to the left most and right most vertices, got rid of the middle vertical resistor due to symmetry, and the rest is just adding up the resistors in series and parallel.
If you get rid of a resistor even if it has no net current flowing, I think that changes the circuit.
 
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  • #23
bob012345 said:
I can see that both have four vertices each with three connections and in that sense they should be the same electrically but I can't see how one morphs into the other. But then I'm not a topologist. Also, the tetrahedron projects into a 2D plane with no overlap while the square does not.
Just lift one of the vertices out of the plane.
 
  • #24
bob012345 said:
If you get rid of a resistor even if it has no net current flowing, I think that changes the circuit.
Nope. If the current is zero, the voltage drop is zero, whether R=1Ω or 10MΩ. So you can change its value to ∞ provided you know that the current will always be zero. Same with branches that you know are 0V, those can be replaced with short circuits.
 
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  • #25
Seriously guys what does it matter from the point of view of circuit theory for the angles between the resistors and the length of resistors, given they all are 100ohm. Only if you going to view this circuit from the point of view of EM fields then it matters of course
 
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  • #26
Also note that when they say the source can be connected across any branch, that means you can pick whichever one makes it easier to draw or visualize.

Anyway, this is a better schematic without the irrelevant geometry stuff.

20220328_233643~2.jpg
 
  • #27
DaveE said:
Also note that when they say the source can be connected across any branch, that means you can pick whichever one makes it easier to draw or visualize.

Anyway, this is a better schematic without the irrelevant geometry stuff.

View attachment 299065
Well, I think the ”best” planar diagram is the one that follows from doing this:
bob012345 said:
Pick any vertex and look straight into it and what would you see?
It is the approach that conserves most of the symmetry of the setup. Of course, all of this is quite moot and not really clarifying the description of the circuit since it was adequately described in words in the OP:
Jaccobtw said:
Six 100 ohm resistors are soldered together to form a regular tetrahedron.
I think this is quite adequate.
 
  • #28
DaveE said:
Anyway, this is a better schematic without the irrelevant geometry stuff.
Way to let the OP do his own work :rolleyes:
 
  • #29
I will point out that the OP solved the problem in post #18. IMO there's an awful lot of chatter in this thread which isn't helpful.
 
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  • #30
Jaccobtw said:
I think I figured it out. I connected the terminals to the left most and right most vertices, got rid of the middle vertical resistor due to symmetry, and the rest is just adding up the resistors in series and parallel.
Nice, but can you give the answer you got for completeness of this thread?
 
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  • #31
phinds said:
Way to let the OP do his own work :rolleyes:
Yes, I get that. I thought the thread had moved beyond the OP's solution. So I was more in the mode of "train the trainers". Showing how EEs use schematics and circuit analysis to focus on the nature of lumped element networks. It truly doesn't matter at all what the physical shape of the circuit is once you've defined the lumped elements; it could be a sphere or a square, whatever. It's a network problem, not a mechanical one. This is one of the key things that many physicists don't get about circuits, they are highly contextualized to allow you to focus on tools like KVL/KCL etc.

Step 1: strip out all of the irrelevant information. Step 2: speak to other EEs in a language we all are comfortable with. This avoids the confusion you saw at the beginning with other EEs not understanding about "fading resistors" and such.

I confess that I get tired of HW problems that are trying to trick the students instead of focusing on the key analytical issues.
 
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  • #32
DaveE said:
This is one of the key things that many physicists don't get about circuits, they are highly contextualized to allow you to focus on tools like KVL/KCL etc.
I don’t think this video supports the claim that many physicists don’t understand KVL application to circuits. It shows that Walter Lewin may not understand it (and possibly anyone he taught). In fact, Lewin himself says everybody else is wrong.

DaveE said:
It truly doesn't matter at all what the physical shape of the circuit is once you've defined the lumped elements; it could be a sphere or a square, whatever. It's a network problem, not a mechanical one.
While true, drawing it one way or the other can certainly help in guiding your mind in identifying good ways of analysing a circuit.

Ultimately, the image in the OP does not seem as much a circuit diagram to me as an attempt to illustrate the construction (which was perfectly well defined in terms of descriptive text).
 
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  • #33
Orodruin said:
Walter Lewin may not understand it
Yes there's no doubt that he knows EM. What he doesn't know is this: "Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits." -- the first line of Wikipedia on Kirchhoff's laws.

Orodruin said:
Ultimately, the image in the OP does not seem as much a circuit diagram to me as an attempt to illustrate the construction (which was perfectly well defined in terms of descriptive text).
OK, but then they shouldn't have used the lumped element resistor symbol. When I see that, I assume that all of the interconnections are assumed to have zero resistance (and no induced voltage). That symbol tells me it's a network problem, not a mechanical one and thus shape doesn't matter.

It's not the concepts, physicists don't understand, it's the language.

PS: Plus Dr. Lewin doesn't know much about oscilloscope probes, but that's a different issue.
 
  • #34
DaveE said:
OK, but then they shouldn't have used the lumped element resistor symbol. When I see that, I assume that all of the interconnections are assumed to have zero resistance (and no induced voltage). That symbol tells me it's a network problem, not a mechanical one and thus shape doesn't matter.
That’s more of a language problem based on where you are coming from I think. I personally had no problems understanding what was implied from neither text nor image, but I have a different background and do not make the same assumptions. Similar to how people speaking dialects of the same language sometimes not understanding exactly what the other means. That doesn’t mean either is correct or wrong, just that they will be misunderstood in some situations when surrounded by people from the other place.
 
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  • #35
Orodruin said:
That’s more of a language problem based on where you are coming from I think. I personally had no problems understanding what was implied from neither text nor image, but I have a different background and do not make the same assumptions. Similar to how people speaking dialects of the same language sometimes not understanding exactly what the other means. That doesn’t mean either is correct or wrong, just that they will be misunderstood in some situations when surrounded by people from the other place.
Yes, exactly. The issue is, do you know which language you are speaking? Does the listener? This is precisely Lewin's problem with induced voltages and KVL.
 

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