# Concentric Circuit

1. Mar 4, 2014

### ItsImpulse

1. The problem statement, all variables and given/known data
Find the effective resistance across this circuit from two diagonally opposite points on the outer squarehttp://aiminghigh.aimssec.org/wp-content/uploads/2012/01/7squares.gif [Broken] (I took the image off google it isn't exactly a circuit but its the correct orientation) given that the squares go on to infinity.

2. Relevant equations

3. The attempt at a solution
I tried to do current analysis which would effectively eliminate the squares in the centre past the second square but upon checking with my teacher, he said that that will not yield the correct answer.

Last edited by a moderator: May 6, 2017
2. Mar 4, 2014

### voko

Are there supposed to be two resistors per side, and the inner squares are connected in between? Are the resistors all identical?

3. Mar 4, 2014

### ItsImpulse

Sorry I forgot. Assume the wires to be of a constant material thus length of the largest square to be of resistance R. Yes they are all connected such that the corner of each square bisects the length of the larger square.

4. Mar 4, 2014

5. Mar 4, 2014

### ItsImpulse

Firstly I apologize for the bad image as I don't know how to properly photo edit things.

Now, I started off by taking using A and B as start and end points, respectively.
Upon entering A, the current will split into the two branches. At C, the current would have gone through a resistance of R/2 and at D the current would have gone through a resistance of R/2 as well. This means that potential difference across CD is 0 and so I essentially did not consider CD as part of my working.

After C, the current will split and go to E. the branch connecting to the corner of the big triangle would be a series of two 0.5R wires. the other branch would be a resistance of sqrt(2)R/2 by pythagoras theorem which is in parallel to the two 0.5R wires and will reconnect at E. Similarly, after D the same would happen and connect at F. So potential difference of EF is also 0 so I also did not consider EF as well.

From E to B and F to B, they are essentially 0.5R wires in parallel.

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6. Mar 4, 2014

### voko

You are quite correct in observing that CD and EF have no potential difference. What about CE and DF?

7. Mar 4, 2014

### ItsImpulse

As I have mentioned above, probably not clearly, I took into account that the resistance of CE and DF is simply sqrt(2)R/2 and R in parallel. This gives an effective resistance of sqrt(2)/(sqrt(2) + 2) across CE and DF

8. Mar 4, 2014

### voko

I am not entirely sure I understand what you mean by "effective resistance across CE and DF". Anyway, think about this.

Let's label the the inner square's vertices as A' (bottom left), G' (top left), B' (top right) and H' (bottom right).

Is there potential difference between A' and B'? Between G' and H'?

9. Mar 4, 2014

### ItsImpulse

I meant that consider that between node C and node E is a parallel of circuit, by which one branch has a resistance of R while the other branch has a resistance of sqrt(2)R/2.

I'm sorry, by inner square do you mean the second largest square? or the largest square in black. If its the earlier, what do you mean by bottom left and top left. (Does bottom left refer to the point at D?)

10. Mar 4, 2014

### voko

I meant the biggest black square.

11. Mar 5, 2014

### ItsImpulse

I would not think so since current does not flow in CD of EF?

12. Mar 5, 2014

### voko

A'G'B'H' and inner squares together are a conductor. Current can flow through it. So if there is a potential difference between A' and B' or between G' and H', it will.

13. Mar 5, 2014

### ItsImpulse

But then assuming the current flows from A to B and CD and EF has no potential difference, how does the current get into the black square in the first place?

14. Mar 5, 2014

### voko

Look at this circuit.

Would current flow through its extreme left to its extreme right (or vice versa)?

Observe there is no potential difference between A and B, nor between C and D.

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