Equivalent resistance of this 6-sided circuit

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The discussion focuses on determining the equivalent resistance of a six-sided circuit using symmetry arguments. Participants explore the relationship between currents in resistors connected to a central node, concluding that currents through symmetrical resistors are equal, allowing for simplifications without removing resistors. They discuss how to split the central node into two to analyze the circuit more effectively, leading to calculations of equivalent resistances. The conversation highlights the importance of understanding current flow and symmetry in circuit analysis. Ultimately, the participants aim to simplify the circuit to find the equivalent resistance accurately.
  • #31
Sat-P said:
It has to divide equally since, maybe cuz it can't go anywhere else
@haruspex has had you number the nodes so that you could be more specific, yet you make ambiguous statements like "It has to divide equally divided", because "it can't go anywhere else"".

I'm assuming that you were saying that the current into node 2 (from resistor 1-2) has to divide equally into current through 2-0 and current through (who knows what), because current into node 2 can't go anywhere else.

Just because current "splits" at some node, it does not have to split equally.

Consider again what @haruspex is asking.

What current leaving node 0 has to be equal to the current entering node 0 via resistor 2-0?
Hint: It is not the current through resistor 3-0 .
 
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  • #32
Is current flowing through resistor 0-2 the same as the current flowing through resistor 0-5
 
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  • #33
Sat-P said:
Is it resistor 2-3
There is that "it" again.
 
  • #34
Let's try again.

Answer the question I asked at the end of Post #31. Using symmetry may help you.
SammyS said:
What current leaving node 0 has to be equal to the current entering node 0 via resistor 2-0 ?
Hint: It is not the current through resistor 3-0 .

What current leaving node 0 has to be equal to the current entering node 0 via resistor 2-0 ?
 
  • #35
Are you asking about the magnitude of the current leaving node 0 which flowed through resistor 2-0?
 
  • #36
Can someone just solve this question and tell me the answer?
 
  • #37
Sat-P said:
Are you asking about the magnitude of the current leaving node 0 which flowed through resistor 2-0?
Yes, essentially.

More specifically:
By considering symmetry, we expect that current through one of the other three resistors (those being 3-0, 4-0, or 5-0) connected to node 0 will have the same current as the resistor 3-0 has, Furthermore, if the current through resistor 3-0 flows into node 0, the the current through the resistor in question, must flow out from node 0.
 
  • #38
Sat-P said:
Is current flowing through resistor 0-2 the same as the current flowing through resistor 0-5
Yes! Do you see how that follows from symmetry?
And what about the remaining radial pair, 3-0 and 0-4?
 
  • #39
haruspex said:
Yes! Do you see how that follows from symmetry?
And what about the remaining radial pair, 3-0 and 0-4?
Oh my!
I see that we've had a bit of an editing job in Post# 32 by OP.

I never saw that reference to 0-5 .
 
  • #40
Sat-P said:
Can someone just solve this question and tell me the answer?
Perhaps you should review the Homework forum guidelines.
Yes, I could solve this problem. But, what good is that for you? Would you be able to solve it next year by yourself? The only purpose of problems like this is education. They virtually never appear IRL.
 
  • #41
haruspex said:
Can you be clearer? Number the nodes 1 to 6 anticlockwise around the hexagon starting with 1 at a. Call the centre 0.
A good idea ! (the naming in #1 is rather confusing)
1721514927929.png


And I have this mental deformation that wants to redraw circuits with components at right angles, like in school books:

1721516591535.png


Does that help you analyze your circuit :wink: ?

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  • #42
Earlier, I thought that current through resistor a0 and 0b will be the same but I was wrong, when I interpreted this mathematically I got that the current through the resistor 0b will be more than that through the resistor a0, and then I thought that current through f0 and c0 might be mirror images after passing a plain in between ab and ed, but here it is; the magnitude of current passing through them was'nt same so I couldn't just take them out, there might be another reason that the current flowing through resistor 0b has contributions of current entering from the radial resistors, so I just couldn't take this one out. Now, the current passing through resistor e0 and d0 looks symmetric, but it is not, again because of the magnitude of the current passing through it. Now, I have an idea, don't know whether it is crazy or something, but I gues if I flod the grid from cf like ab coincides ed, but I now it won't work because in the symmetry the terminals must be symmetric, so any hints what I should do next
17215436012576906450670531066776.jpg

 
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  • #43
Sat-P said:
I thought that current through resistor a0 and 0b will be the same
It must be, by symmetry.
Sat-P said:
when I interpreted this mathematically I got that the current through the resistor 0b will be less than that through the resistor a0
Please show the details.
 
  • #44
I am sorry for the misinterpretation, but the current through resistor 0b will be more i.e 0.65 i (approx) and current through resistor a0 is 0.334 I (approx). You can see the results in the image above. I've edited the above post.
 
  • #45
Sat-P said:
I am sorry for the misinterpretation, but the current through resistor 0b will be more i.e 0.65 i (approx) and current through resistor a0 is 0.334 I (approx). You can see the results in the image above. I've edited the above post.
The image is too blurry to read with any confidence, but it looks to me that you are assuming the currents along the three resistors connected to a are equal. That's not how electricity works.
Say all the resistors are 1 Ohm and the voltages are +1 at a and -1 at b.
The voltage at 0 must be 0, so the current along ab is 2A and along a0 and 0b it is 1A.
 
  • #46
haruspex said:
The image is too blurry to read with any confidence, but it looks to me that you are assuming the currents along the three resistors connected to a are equal. That's not how electricity works.
Can you please clarify why the current through the resistors connected to a should not be equal?
 
  • #47
Sat-P said:
Can you please clarify why the current through the resistors connected to a should not be equal?
See my edit to my previous post.
 
  • #48
haruspex said:
See my edit to my previous post.
How can you say that the current through ab is 2 Ampere
 
  • #49
Sat-P said:
How can you say that the current through ab is 2 Ampere
It's the law (Ohm's law, to be specific). :smile:

##\ ##
 
  • #50
BvU said:
It's the law (Ohm's law, to be specific). :smile:

##\ ##
Got it
 
  • #51
Sat-P said:
Got it
So can you solve it now using the splitting up of node 0?
 
  • #52
Even after I split the node 0ab , the other radial resistors are causing me trouble, if I can get a way to simplify them then it will be all good
 
  • #53
Sat-P said:
Even after I split the node 0ab , the other radial resistors are causing me trouble, if I can get a way to simplify them then it will be all good
In post #32 you recognised (guessed?) that, by symmetry, the current along 2-0 is the same as along 0-5. That allows you to split ##0_{2345}## into two nodes, ##0_{25}, 0_{34}##.
 
  • #54
1721562304696.png
Once you understand the consequences of the dashed red symmetry line, you can work from right to left.

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  • #55
Sir, I've tried both the methods you gave
The answers I am getting are different, please check is there something I did wrong or misinterpreted what you said

17215643568192792235745749524872.jpg
1721564378304636417938574175375.jpg
 
  • #56
What are we looking at ? (with a pain from craning our neck) ?
I can distinguish you replace the rightmost three with ##{2\over 3} R##, so the rightmost five give ##{8\over 3} R##. OK.

But why ##R/2## ?

1721576599814.png


##\ ##
 
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  • #57
Oh, I just saw the mistake. Let me do it again
 
  • #58
After solving this one I got 11R/20 so is it the answer, did I do it right
 
  • #59
Sat-P said:
After solving this one I got 11R/20 so is it the answer, did I do it right
Agrees with mine, but I didn't double check.

Checking your own work is a key skill in the physical sciences. Out in the real world they don't want to pay 2 people to do one job, and they don't want to pay one person to do the job incorrectly.
 
  • #60
Sat-P said:
After solving this one I got 11R/20 so is it the answer, did I do it right
The answer isn't really the important element. Whther you did it right can only be confirmed if you post your work.

Free tip: work it out neatly ... :wink:

1721594406599.png
 

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