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.
 A: 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)

A: 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.  


*

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

*Between the first and the second diagonals: $R/4$

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

*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 now find anything else you want about this circuit. 
You can apply this method with a large number of resistors, $N\times N$ mesh say, just for fun!
