Equivalent resistance of 2015 points

[timetraveller123]

Homework Statement

2. Homework Equations [/B]

The Attempt at a Solution

so i was thinking i could take a look at small number of points and then generalize it but my method soon breaks down

in the diagram i have denoted the resistance of the segment beside it if there is a bracket it means how many of such branches there are in parallel to it i tried to conjecture some kind of pattern right in the next one i ran into some problems..

well for the four point situation the one on the right shows what i believe would be the equivalent transformed circuit but
1)for the case 3r looking at it there is actually two paths which has 3r the other one being the one that goes along the diagonal first then left and then back to end point
2) there also seems to be a path of 4r

here is picture of what i mean
i know such paths are not possible(can someone verify this for me it is just a gut feeling ) for at some point the current is going in the opposite direction but i don't know how to make mathematical pattern that would reject such paths
i was thinking of maybe assigning an appropriate numbering system and then imposing restrictions like the numbers need to be arranged in a non decreasing order this is a wild guess but maybe something along that line
or is my approach even valid here am i doing it totally wrong is there any other approach any help would be appreciated thanks

 

[timetraveller123]

i am sorry something went wrong when attaching
the second diagram is supposed to be
https://www.physicsforums.com/attachments/upload_2017-9-26_21-41-6-png.211750/

edit it is fixed ignore this post

 

[cnh1995]

Do you have the answer key? Is the answer 2R/2015?

 

[timetraveller123]

i don't have the answer key but how did you derive the result what method did you use

 

[jbriggs444]

An important problem solving technique is to look for symmetries and see what conclusions can be drawn from them. In this case, the points on the circuit board can be divided up into three classes: The one start point, the one end point and the 2013 other points.

Symmetry tells us that all of those "other" points must be indistinguishable. That seems to me to be a fruitful observation.

 

[timetraveller123]

An important problem solving technique is to look for symmetries and see what conclusions can be drawn from them. In this case, the points on the circuit board can be divided up into three classes: The one start point, the one end point and the 2013 other points.

Symmetry tells us that all of those "other" points must be indistinguishable. That seems to me to be a fruitful observation.

i am sorry i still don't see how the fact that they are indistinguishable is showing symetry in fact i have thought of this problem for quite long now it is not just like a infinite ladder of resistors where the symetry is obvious so help me out

 

[jbriggs444]

Do you agree that the "other" 2013 points are all just like one another?

 

[timetraveller123]

ya

 

[jbriggs444]

If those 2013 points are all just like one another, what does that say about the current flowing to each of those points?
More importantly, what does it say about the current flowing from one of those points to another?

 

[timetraveller123]

current through each point is the same and to the second question i i don't know

 

[timetraveller123]

i am not really sure if i am seeing the symetry properly what is the symetry is invariant in

 

[timetraveller123]

If those 2013 points are all just like one another, what does that say about the current flowing to each of those points?
More importantly, what does it say about the current flowing from one of those points to another?

no about this question again i am confused doesn't the current through each point depend on the paths resistance i am confused again

 

[jbriggs444]

current through each point is the same

I did not ask about current going "through" each of those other points. I asked about current going "to" each of those other points from the start point.
What can we say about that current?

and to the second question i i don't know

Does current flow from one point to the other? In which direction? That's where the agreed-upon fact that the points are indistinguishable comes in.

 

[timetraveller123]

so the current flowing to point p1 is the same as the current flowing to all points
if current goes from p1 to p2 the current going from a point to all other points is also the same is that what you mean

 

[timetraveller123]

so does that mean the problem is simplified to 2015 parallel branches each with resistance 2r hence total resistance is 2R/2015 as said before

 

[cnh1995]

i don't have the answer key but how did you derive the result what method did you use

It's a bit lengthy and kind of non-scientific.
When you have all the points connected together through some resistance, the effective resistance between two points(nodes) should be constant, no matter which two points you select. This is what the symmetry tells you.

Then I worked out the equivalent resistance when you have i) two points ii)three points and iii)4 points and using those three results, I generalized it for 'n' number of points. For n=2015, I am getting the above answer.
(I am sure it is correct since I later verified it with n=5 as well.)

 

[jbriggs444]

so the current flowing to point p1 is the same as the current flowing to all points

Right. The current going from the start point directly to anyone of the 2013 points is the same as the current going directly to any other.

if current goes from p1 to p2 the current going from a point to all other points is also the same is that what you mean

You have not explained what "p1" and "p2" are supposed to denote. I will assume that p1 is one of the 2013 points-in-the-middle and that p2 is another.

If you were to reason carefully and determine that the current was flowing directly from p1 to p2 then you could use that exact same careful argument to determine that current was flowing directly from p2 to p1. That's where symmetry comes in -- anything you can say about p1 in relation to p2 you can say with equal force about p2 in relation to p1.

To make this excruciatingly explicit, symmetry tells us that

$$I_{p1p2} = I_{p2p1}$$

But, by definition

$$I_{p1p2} = - I_{p2p1}$$

There is only one solution for ##I_{p1p2}## that is consistent with both equations.

What is that value?

 

[timetraveller123]

@jbriggs444
since the current going into each resistor is the same as the current going out of every resistor it essentially means that there is 2015 paths 2r resistance in parallel to each other does it mean what you say

 

[jbriggs444]

@jbriggs444
since the current going into each resistor is the same as the current going out of every resistor it essentially means that there is 2015 paths 2r resistance in parallel to each other does it mean what you say

2013 paths. Keep working to see why the final answer has a 2015 in it.

 

[timetraveller123]

yah 2013 paths is that right
so is the total resistance 2R/2013

 

[jbriggs444]

yah 2013 paths is that right
so is the total resistance 2R/2013

No. You forgot one path.

 

[timetraveller123]

the path of one resistance?

 

[jbriggs444]

the path of one resistance?

That's the one.

 

[timetraveller123]

in that case it is 2R/2015 ?

 

[jbriggs444]

in that case it is 2R/2015 ?

Yes. One way of seeing that without evaluating ##\frac{1}{\frac{2013}{2R}+\frac{1}{R}}## is to realize that one path of resistance R is the same as two paths of resistance 2R. So now you have the equivalent of 2013+2 paths each of resistance 2R and the result of ##\frac{2R}{2015}##

 

[timetraveller123]

@cnh1995 while jbriggs444 is elegant and simple i am not sure if i would be able to exploit such symetry by myself
so initially i tried your method but i can't generalize it how do you do it any help thanks?

 

[cnh1995]

but i can't generalize it how do you do it any help thanks?

For 2 points: equivalent resistance=R=2*(R/2)
For 3 points: equivalent resistance= 2R/3=2*(R/3)
For 4 points: equivalent resistance= R/2=2*(R/4)...and so on

So, for n points,
R_eq=2*(R/n).

 

[timetraveller123]

for 4 points how did you get r equivalent as R/2 i am getting as 3/7 R there is one R path two 2R path and one 3R path in parallel to one another

 

[cnh1995]

for 4 points how did you get r equivalent as R/2 i am getting as 3/7 R there is one R path two 2R path and one 3R path in parallel to one another

It will be better if you posted your diagram here.
Choose any two points between which you want to calculate the equivalent resistance. You'll notice a balanced Wheatstone bridge.

 

[timetraveller123]

is it that there is no potential difference across A and B

 

[jbriggs444]

is it that there is no potential difference across A and B

I assume that your two terminals are the nodes in the upper left and upper right. Yes, since there is no potential difference between A and B (by a symmetry argument), one can remove the direct connection between A and B without modifying the resistance of the network.

With that link removed, one can see that there are exactly three parallel paths. Two of length 2 and one of length 1.

Similarly for any number of nodes greater than 4. Since all of the intermediate nodes are at the same potential, one can remove all of their direct connections to one another without modifying the resistance of the network. One is left with n-1 parallel paths. n-2 of length 2 and one of length 1.

 

[timetraveller123]

i was thinking of that too and i realized that was how you derived your symetry method really smart