# Confused as to how to simplify this circuit

1. Jan 31, 2016

### Potatochip911

1. The problem statement, all variables and given/known data

The sides of the biggest square have resistance $R$ and length $L$ (the amount of resistance in the wire is equal to the length), the smaller square is placed so that it's corners intersect midway through the sides of the larger square. Find the resistance between points X and Y

2. Relevant equations
Parallel: $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots+\frac{1}{R_n}$
Series: $R_{eq}=R_1+R_2+\cdots R_n$

3. The attempt at a solution

To start off I can see that at the top and bottom of the circuit the L/2 sections are in series so they add together giving L and then this is in parallel with the $\frac{L}{\sqrt{2}}$ section of the square. Then the equivalent resistance from that is $$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{\sqrt{2}}{R}$$, i.e. $$R_{eq}=\frac{R}{1+\sqrt{2}}$$

Then I obtain the following circuit:

At this point I'm currently stuck as I'm not sure how to simplify it further.

2. Jan 31, 2016

### ehild

Use symmetry. Connect a battery to the points X and Y. As the circuit is symmetric to the XY axis, the current from X to A must be the same as from X to B. So The potential drops by the same value from X to A as from X to B. The potential of A is the same as the potential of B, there is no potential difference between A and B, so you can remove the resistor AB, it does not influence the other currents and voltages. Do the same at points C and D, and calculate the resistance of the new circuit.

3. Jan 31, 2016

### Potatochip911

Okay I removed those resistors and then I found the total resistance for the entire circuit to be $R_{T}=R(\frac{1}{2}+\frac{1}{2+2\sqrt{2}})$

I'm trying to calculate the total resistance with another cube inside of it now, i.e.

So we have $V_a=V_b$ and $V_h=V_i$ which allows me to obtain the following circuit:

And I'm once again stuck as to how I should simplify this. I'm not sure how to deal with the diamond of resistors since they are also connected to the resistor at the top/bottom of the box they're enclosed in.

4. Jan 31, 2016

### cnh1995

What do you think about the resistor pairs
af-fh and bg-gj?

5. Jan 31, 2016

### ehild

Find more equipotential points. You can connect them with a zero-resistance wire, so they become a single node in the circuit. The colored lines represent such nodes. All four resistors between the blue and green lines are connected - how? and so are those between the green and red lines.

6. Jan 31, 2016

### Potatochip911

They're in series so they add

Ok, I've thought quite a bit about how exactly these resistors are connected and I've come to the conclusion that I should split up the top and bottom resistor in the blue-green node rectangle into two resistors with half of the original resistance. (also I've placed a R/2 resistor at the front to represent the R/2 resistors in parallel at the very start and very end of the circuit).

In this case $R_1$ is half of the parallel combination af-fh & ah which gives $R_1=\frac{1}{2}\frac{R}{1+\sqrt{2}}=\frac{R}{2+2\sqrt{2}}$ and by symmetry $R_1=R_3$, from trig the resistance of cd/ce is R/2 and since they're in parallel this gives R/4 for their combination which I've assigned as $R_2$ (the resistor in the middle). Since these resistors are all in parallel and it's the same resistors in parallel twice the resistance from the parallel combination is:
$$\frac{1}{R_{||}}=\frac{2(2+\sqrt{2})}{R}+\frac{4}{R}\Rightarrow R_{||}=\frac{R}{8+4\sqrt{2}}$$

Therefore the total resistance is $R_{T}=\frac{R}{2}+\frac{R}{8+4\sqrt{2}}$, is this correct?

7. Jan 31, 2016

Yes.