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In a nutshell, my question is: what is the correlation between symmetry (as in symmetry in geometrical shapes) and current and/or voltage and how do I utilise it in understanding circuit analysis?

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- Thread starter Capt1801
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- #1

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In a nutshell, my question is: what is the correlation between symmetry (as in symmetry in geometrical shapes) and current and/or voltage and how do I utilise it in understanding circuit analysis?

- #2

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The point generally is that if the potential at two points in a circuit are the same then you could just put a short circuit between the two points and it would have no effect on the behavior of the circuit. This can help simplify circuits.

In a nutshell, my question is: what is the correlation between symmetry (as in symmetry in geometrical shapes) and current and/or voltage and how do I utilise it in understanding circuit analysis?

- #3

anorlunda

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We were told to note that the potential of all circuit elements lying on a line of symmetry would be the same.

It's not clear what that means, or what you are thinking about.

- In physics, we talk about the symmetries of natures, such as space translation invariance.
- If you have a bulk material, like a sheet of steel, and pass a current through it, there will be equipotential lines depending on the shape which might be symmetrical.
- In circuit analysis, we can have equal branches (as @phinds pointed out), but symmetry is not the word I would use.

- #4

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I understand that. My question is: how does symmetry fit into this? What has a line of symmetry through a network of circuit elements got to do with the potential at different points along the line?The point generally is that if the potential at two points in a circuit are the same then you could just put a short circuit between the two points and it would have no effect on the behavior of the circuit. This can help simplify circuits.

I understand that if the potential of two points is the same, current will not flow between them. But what has symmetry (as in symmetry in geometrical shapes) got anything to do with it? How do I get from 'A line divides this circuit symmetrically, i.e., it is a line of symmetry' to saying 'The potential is the same along this line'. Where does that come from?

- #5

Borek

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- #6

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It's not clear what that means, or what you are thinking about.

Can you elaborate on what you mean by symmetry?

- In physics, we talk about the symmetries of natures, such as space translation invariance.
- If you have a bulk material, like a sheet of steel, and pass a current through it, there will be equipotential lines depending on the shape which might be symmetrical.
- In circuit analysis, we can have equal branches (as @phinds pointed out), but symmetry is not the word I would use.

Sure.

The symmetry I'm referring to is the one in geometrical shapes, like a square or a rectangle.

Consider this circuit:

This circuit is symmetrical about the red line - a perpendicular bisector. I have been taught that since it is symmetrical about that line, the potential of each point along that line would be the same, and thus no current would flow through the resistor placed on that line. I am aware that it forms a balanced Wheatstone Bridge and it is therefore obvious that no current would flow through the middle resistor; but that is beside the point. My teacher simplified more complex circuits using the same principle: that no current would flow through any circuit element that lies on a line of symmetry of the circuit. I just don't understand the correlation between symmetry and this. How can I get to the conclusion that the potential of all points on the red line is the same, starting with the fact that the red line is a line of symmetry? How does symmetry fit into this? I guess it could have something to do with equipotential lines, as you said. But without anything to link the two statements, I can't possibly begin to apply this concept in questions. Of course, it might not be called 'symmetry' (that is the heading under which we were taught this little concept).

- #7

Borek

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This circuit is symmetrical about the red line - a perpendicular bisector.

There is another symmetry, horizontal one.

I have been taught that since it is symmetrical about that line, the potential of each point along that line would be the same, and thus no current would flow through the resistor placed on that line.

Doesn't sound correct to me.

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Exactly! Which is why I wanted to know the reason.Doesn't sound correct to me.

- #9

anorlunda

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See the PF Insights article https://www.physicsforums.com/posts/5872068/

Also, most circuits do not contain any kind of symmetry. That circuit in #6 is what we call a bridge. such as https://en.wikipedia.org/wiki/Bridge_circuit#/media/File:Wheatstonebridge.svg. It is a special purpose circuit specifically designed to take advantage of those balanced paths and identical resistors. So yes, you can use symmetry in a few special cases, but not in most cases.

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CWatters

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Borek

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Exactly! Which is why I wanted to know the reason.

Actually I think we can be both wrong. My first instinct was that there can be a current flowing through the vertical resistor, but I am unable to construct an example of such a circuit. Hardly a proof, still.

- #12

CWatters

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Imagine that point A is at say 10V and B is at 0V then there are two identical paths from A to B. The top path and the bottom path. Symmetry says that the voltage of any point on the top path equals the voltage on the corresponding point on the bottom path. No current flows through the vertical resistor.

- #13

jim hardy

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It's a Wheatstone Bridge.

https://www.grc.nasa.gov/www/k-12/airplane/tunwheat.html

I dislike shortcuts .

Solve the circuit using Kirchoff's Laws .

That makes it obvious why a balanced Wheatstone Bridge has no current through its measuring leg.

My advice is forget about 'symmetry' . It's a useless complication IMHO.

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Imagine that point A is at say 10V and B is at 0V then there are two identical paths from A to B. The top path and the bottom path. Symmetry says that the voltage of any point on the top path equals the voltage on the corresponding point on the bottom path. No current flows through the vertical resistor.

That makes much more sense. In fact, I think that is the concept. It has just been given the wrong title. When we were made to solve the 3D cube problem, we used what you stated above.

I dislike shortcuts .

So do I. Especially when they seem so forced, like this one. The reason we're being taught shortcuts and tricks is because we're preparing for the IIT JEE, a competitive examination for admission to undergraduate courses in IITs. Speed matters in the exam, so we're taught a lot of tricks. Most of them make perfect sense, unlike this one. Thanks for all the help!

- #15

jim hardy

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Wheatstone bridge is such a fundamental building block it'll become second nature to you.

I hope the test goes well for you.

old jim

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