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