Why and how does symmetry work in circuits?

Why symmetry work in circuits? In my book there is no mention explanation as such for symmetry arguments and circuits. But there are circuits that are very difficult to solve without symmetry. Also I have heard that they save a lot of time. Can anyone please tell me why symmetry works in circuits? And what are some basic arguments that we make using symmetry while analyzing circuits? If anyone can point me to any link for the same, it would be very helpful.

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Related: physics.stackexchange.com/q/53734/2451 and links therein. –  Qmechanic Feb 13 '13 at 14:06
That was little too tough for me. An easy explanation with some examples would be nice! Thank you –  Harsh Feb 13 '13 at 14:18
Could you please give a concrete example of such a symmetry? In that case, we could explain why symmetry considerations work for your specific system. –  Frederic Brünner Feb 13 '13 at 14:54

Imagine you have an electric circuit which is in the form of a square mesh ABCDA $4\times 4$ say, and in each branch you have a resistor of value R. Then you connect a battery of emf =E at two diagonally opposite points such as A and C, say. If you draw the diagram yourself you will have a picture of it. Here is an example for a $4\times 4$ square mesh with a resistor R in each branch of the mesh. I.e. I have four resistors along each horizontal and four along each vertical line.

Sorry I could not bring my diagram, but try to imagine it or draw it for yourslef.

Once you have drawn the circuit from my description, now let us say we want to find the total resistance of this circuit, current in each resistor and voltages across each resistor and so on.

Here comes the symmetry of the circuit and solves this problem.

SYMMETRY: the ends of all resistors that are on the diagonals which are perpendicular to AC, or parallel to BD, the doted lines in my diagram, are at the same potential hence all resistors between any two pairs of successive doted lines are parallel to each other. The totals of these parallel resistors are now in series! The rest is easy.

1. The resistance between A and the first doted line is: $R/2$

2. Between the first and the second diagonals: $R/4$

3. Between the next pair of doted lines : $R/6$

4. Between the next pair of doted lines : $R/8$

The steps 1 – 4 are the same on the other side of the diagonal BD. So the total resistance is the sum of all these:

$R_T = R+R/2+R/3+R/4=2{\frac {1}{12}}R$

You can apply this method with a large number of resistors, $N\times N$ mesh say, just for fun!

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Are you sure the answer is correct? I put the 4x4 mesh into the circuit simulator ngspice and got the resistance as RT=2.136364R whereas your value is RT=1.958333R . –  Stephen Blake Feb 14 '13 at 22:22
@StephenBlake Please see edited version of my answer. I missed one brach, the fourth one, and te answer was short by 1/8. Still your simulator answer does not much mine. I hope we are talking about the same circuit here? –  JKL Feb 14 '13 at 23:07
Are there 40 resistors in your circuit? –  Stephen Blake Feb 14 '13 at 23:18
@StephenBlake That is correct. Can you make sure the simulator is given the right input please. :) –  JKL Feb 14 '13 at 23:27
One corner is grounded, and 1V is applied to the node on the opposite corner to the grounded node. I'm not sure if it's appropriate to try to post the spice deck in the comments? –  Stephen Blake Feb 14 '13 at 23:43

I think John has assumed that the voltages are constant along the diagonals, however this is not the case as the numerical example shows.

For Example:

After using LTspiceIV to create a circuit just like the one you are proposing I have been able to analyse the voltages at set points between the resistors.

Looking at your equation you are assuming that the voltage is constant across the diagonal nodes. After analyzing these nodes using the software I can assure you that they do differ.

Voltage between resistors are as follows
R2-R3 = 627.66 mV
R12-R13 =670.213 mV
R14-R15 =627.66 mV

Using this software it is then easy to calculate the total resistance
this is 2.13636 (5dp)

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I am not sure I understand what you mean by voltage between:R2-R3 and R12-R13. It will be much more clear if you could please make reference to voltages in terms of diagonal nodes perpendicular to the feeding diagonal. I am trying to see how can your simulator produce a vltage 627.213 V(?) from 1V input (?) Do you mean 627.213 mV? It would be interesting to check the voltages you get from your simulator that are across successive nodes along the main diagonal (from the bottom left to the top right), so that you could add them and see that you get 1 V. –  JKL Feb 15 '13 at 16:55