Equivalent resistance of this 6-sided circuit

[Sat-P]

Homework Statement

Each resistor has resistance R, find R equivalent b/w ab

Relevant Equations

Req in series; R1 +R2 +...+Rn
In parallel ; 1/Req = 1/R1 +1/R2 +...+1/Rn

 

[haruspex]

You need to use symmetry arguments to simplify the circuit.
E.g. consider the two resistors connecting a and b to the middle node. What can you say about the currents through them?

 

[Sat-P]

I know mirror symmetry not sure which resistor to remove

 

[haruspex]

I know mirror symmetry not sure which resistor to remove

I didn't say to remove any, yet. I asked what you can say about the relationship between the currents in the two resistors connecting a and b to the centre.

 

[Sat-P]

We can conclude after placing a plain in their mid, current from a to centre and centre to b will be symmetric.

 

[Sat-P]

Since it is parallel symmetry, so it has to be true

 

[haruspex]

We can conclude after placing a plain in their mid, current from a to centre and centre to b will be symmetric.

Ok, so the current along the resistor from a to centre equals the current along the resistor from centre to b, right?
Can you see what simplification that allows you to make? Hint: it does not involve removing a resistor.

 

[Sat-P]

The resistance is also same so this means the potential across both resistors is same[ since current flow is same too], so there must be no flow across, so we can ignore this and take others in account

 

[Sat-P]

But I dont understand why there will be no current flow through resistors, even though we know current is flowing through that branch??

 

[haruspex]

The resistance is also same so this means the potential across both resistors is same[ since current flow is same too], so there must be no flow across, so we can ignore this and take others in account

The magnitudes of the potential differences are the same, but they don't cancel, they add up. If $a$ is at potential 1 and $b$ is at potential -1 then the central node is at potential 0. Current flows from a to the central node then to b.

 

[Sat-P]

So, what should I do next after knowing they bothe have same potential difference

 

[haruspex]

So, what should I do next after knowing they bothe have same potential difference

That's not the point. The point is what I wrote in post #7, that the current along the resistor from a to centre equals the current along the resistor from centre to b.
Can you see a simple change you can make to the circuit that does not change any of the current flows?

 

[Sat-P]

I can pluck that resistance R-R out of keeping it attached, and finding the Req of others and take this with it in parallel

 

[haruspex]

I can pluck that resistance R-R out of keeping it attached, and finding the Req of others and take this with it in parallel

Yes, you can cut the central node into two, one that connects the pair of resistors we've been discussing and one that connects the other four radial ones.
Can you apply the same trick again?

 

[Sat-P]

But there are still resistors attached to the centre, so I don't know how to solve this, meanwhile the R Equivalent of branch taken out is 2R and on taking this in parallel with the ab resistor we get Req 2R/3 . But how to solve the rest of it??

 

[haruspex]

But there are still resistors attached to the centre, so I don't know how to solve this, meanwhile the R Equivalent of branch taken out is 2R and on taking this in parallel with the ab resistor we get Req 2R/3 . But how to solve the rest of it??

I asked,

Can you apply the same trick again?

 

[Sat-P]

Yes, you can cut the central node into two, one that connects the pair of resistors we've been discussing and one that connects the other four radial ones.
Can you apply the same trick again?

Do you mean I can take the pairs of two out of the grid and solve them as I did with the previous one??

 

[Sat-P]

Then it will give 2R/3 and 2R/3 in series and this 2R/3 in parallel with the other two, which eventually gives 4R/9

 

[haruspex]

Do you mean I can take the pairs of two out of the grid and solve them as I did with the previous one??

Can you be clearer? Number the nodes 1 to 6 anticlockwise around the hexagon starting with 1 at a. Call the centre 0.
We have split 0 into $0_{16}$ and $0_{2345}$. What next?

 

[Sat-P]

After taking out the grid O16 , I took out O23 and O45 then after solving them

Then it will give 2R/3 and 2R/3 in series and this 2R/3 in parallel with the other two, which eventually gives 4R/9

 

[haruspex]

After taking out the grid O16 , I took out O23 and O45 then after solving them

Why can you take out $0_{23}$? Does all the current along resistor 2-0 then flow along 0-3?

 

[Sat-P]

So, how should I do it?

 

[haruspex]

So, how should I do it?

Look for another pair of radial resistors such that all the current flowing to 0 from one of them, and only that one, can be presumed then to flow to the other.

 

[Sat-P]

Ok even if I don't take it out the current in 05 is same as that in 04 so even if I take 045 out(i.e. then simplify it in grid ) , the resistance might be the same as I calculated

 

[haruspex]

the current in 05 is same as that in 04

I do not see why that should be.

 

[Sat-P]

Since we took 016 out, current goes to 20 and also from 30 but from 20 and 30 the current had two paths only, so current must divide equally

 

[haruspex]

Since we took 016 out, current goes to 20 and also from 30 but from 20 and 30 the current had two paths only, so current must divide equally

Sorry, I have no idea what you mean by "current goes to 20". There is no node called 20. There is a resistor we can call 20, but current goes through a resistor, not to it.
Please try to be clearer.

 

[Sat-P]

I meant current flowing through node 2 to node 0 through resistor 2-0

 

[haruspex]

I meant current flowing through node 2 to node 0 through resistor 2-0

Ok, so what did you mean by the current dividing equally? Between where and where, and using what symmetry?

 

[Sat-P]

It has to divide equally since, maybe cuz it can't go anywhere else

 

[SammyS]

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 .

 

[Sat-P]

Is current flowing through resistor 0-2 the same as the current flowing through resistor 0-5

 

[SammyS]

Is it resistor 2-3

There is that "it" again.

 

[SammyS]

Let's try again.

Answer the question I asked at the end of Post #31. Using symmetry may help you.

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 ?

 

[Sat-P]

Are you asking about the magnitude of the current leaving node 0 which flowed through resistor 2-0?

 

[Sat-P]

Can someone just solve this question and tell me the answer?

 

[SammyS]

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.

 

[haruspex]

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?

 

[SammyS]

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 .

 

[DaveE]

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.

 

[BvU]

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)

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

Does that help you analyze your circuit ?

$\$

 

[Sat-P]

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

 

[haruspex]

I thought that current through resistor a0 and 0b will be the same

It must be, by symmetry.

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.

 

[Sat-P]

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.

 

[haruspex]

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.

 

[Sat-P]

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?

 

[haruspex]

Can you please clarify why the current through the resistors connected to a should not be equal?

See my edit to my previous post.

 

[Sat-P]

See my edit to my previous post.

How can you say that the current through ab is 2 Ampere

 

[BvU]

How can you say that the current through ab is 2 Ampere

It's the law (Ohm's law, to be specific).

$\$

 

[Sat-P]

It's the law (Ohm's law, to be specific).

$\$

Got it