# Removing connections between equipotential points in solving circuits?

• gregorspv
In summary: The circuit in the upper figure in image 3 is symmetric about a horizontal line through the middle. (Where that cuts a resistor, split that into two in series.) Consequently, all points it cuts are at the average potential, so all at the same potential.This is still true if the two horizontal lines that lie on that central line are removed, so no current flows in those and they can be removed.In summary, the circuit in the upper figure in image 3 is symmetric about a horizontal line through the middle. (Where that cuts a resistor, split that into two in series.) Consequently, all points it cuts are at the average potential, so all at the same potential. This is still true if the two horizontal
gregorspv
Homework Statement
An octahedron is made of wire, where an edge has resistance R. What is the resistance between two adjacent points?
Relevant Equations
Kirchhoff's rules, in parallel and in series resistance.
A sketch of the setup and the equivalent circuit are attached.

I believe the correct way to solve this is to redraw the circuit as shown in Fig. 3 and then remove the connections between evidently equipotential points, which reduces the problem to a familiar setup of in parallel and in series resistors.

However, the circuit may also be redrawn as shown in Fig. 4 where there are also two lines connecting equipotential points, yet if one removes them, an incorrect solution is obtained.

How do I then know, when such connections are removable? I have encountered a statement that such a connection must lie on the circuit's axis of symmetry and this seems to do the trick in this example, but I don't understand why it works that way.

#### Attachments

• IMG_20191028_110426.jpg
44.4 KB · Views: 235
• IMG_20191028_113012.jpg
32.8 KB · Views: 224
• IMG_20191028_112303.jpg
36.7 KB · Views: 220
• IMG_20191028_110513.jpg
34.5 KB · Views: 287
gregorspv said:
the circuit may also be redrawn as shown in Fig. 4
Are you sure? Looks to me that you removed two current-carrying wires.

gregorspv
Oh, I did not word that correctly. In Fig. 4 the connections between equipotential points from Fig. 2 were removed. Perhaps then equipotentiality is not a valid criterion for removal, but rather the absence of current? Would you care to explain why there is no current in the wires removed in Fig. 3?

gregorspv said:
Perhaps then equipotentiality is not a valid criterion for removal, but rather the absence of current
They're the same thing. The wires you removed in fig 3 do connect equipotential points, so carry no current. But in fig 4 it looks to me that you removed wires that do carry current, so are not connecting equipotential points. As a result the resistance increased.

I think I see what you're saying, but it still seems odd to me that there can be a voltage drop (potential difference) between two points on a perfect conductor. Or else how are the wires removed in Fig. 4 connecting points of different potential?

gregorspv said:
I think I see what you're saying, but it still seems odd to me that there can be a voltage drop (potential difference) between two points on a perfect conductor. Or else how are the wires removed in Fig. 4 connecting points of different potential?
My statement that two points being equipotential implies no current flows between them is not correct when the connection has no resistance. The test should be whether they would still be equipotential were the wire removed.

So the flaw in your method is using the equipotentiality test on zero resistance wires while they are in place; it would justify removing all of them!

gregorspv
D'oh – it had occurred to me that my reasoning would lead to removing virtually any wire, but I could not have found a way out were it not for your edifying response. Thank you!

I find it remarkable that some people just happen to get this intuitively (the solution I was provided with does not comment on this step at all), whereas I just spent the last 20 minutes or so proving that the current is indeed zero. I believe the symmetry argument still rests on the assumption that one is intuitively able to figure out the current directions in a circuit which I don't think is necessarily a trivial task (especially when dealing with different resistors and other elements).

gregorspv said:
I believe the symmetry argument still rests on the assumption that one is intuitively able to figure out the current directions in a circuit
No, it's easier than that.
The circuit in the upper figure in image 3 is symmetric about a horizontal line through the middle. (Where that cuts a resistor, split that into two in series.) Consequently, all points it cuts are at the average potential, so all at the same potential.
This is still true if the two horizontal lines that lie on that central line are removed, so no current flows in those and they can be removed.

gregorspv
Aren't 7 resistors in series and then connected in parallel with one resistor?
I must be mssing something

rude man said:
Aren't 7 resistors in series and then connected in parallel with one resistor?
I must be mssing something
No. Which figure are you looking at?

I
haruspex said:
No. Which figure are you looking at?
I confused "octagon" and "octahedron".

## 1. What is the purpose of removing connections between equipotential points in solving circuits?

Removing connections between equipotential points is important in solving circuits because it allows us to simplify the circuit and analyze it more easily. It also helps us to identify the individual components and their functions within the circuit.

## 2. When should connections between equipotential points be removed in circuit analysis?

Connections between equipotential points should be removed when solving complex circuits with multiple components. This will help to reduce the number of components and connections in the circuit, making it easier to analyze and solve.

## 3. How do you remove connections between equipotential points in circuit analysis?

To remove connections between equipotential points, you can redraw the circuit diagram by eliminating any unnecessary connections between equipotential points. You can also use Kirchhoff's laws to mathematically remove these connections in the circuit equations.

## 4. What are the benefits of removing connections between equipotential points in circuit analysis?

Removing connections between equipotential points can help to simplify the circuit and make it easier to analyze. It also allows us to identify the individual components and understand their role in the circuit. This can also help in troubleshooting and identifying any issues in the circuit.

## 5. Are there any drawbacks to removing connections between equipotential points in circuit analysis?

One potential drawback of removing connections between equipotential points is that it can sometimes lead to inaccuracies in the circuit analysis. This is because removing connections may ignore certain factors such as the resistance of the connections themselves. However, in most cases, the benefits of removing connections outweigh any potential drawbacks.

Replies
3
Views
3K
Replies
12
Views
5K
Replies
14
Views
7K
Replies
2
Views
11K
Replies
12
Views
2K
Replies
7
Views
8K
Replies
4
Views
11K
Replies
1
Views
6K
Replies
9
Views
3K
Replies
11
Views
2K