Circuit resembling wheatstone bridge but not exactly 
The below circuit resembles that of two individual wheatstone bridge arrangements as the ratio of resistances satisfy the conditions however we cannot come to this conclusion as potential at one end of the bridge is not same ie., Potential at S and T may or may not be same. This circuit needs tedious calculations if one goes to kirchoff's laws. But this highly symmetric circuit gives me a feel that there is some simpler approach. Could someone help me exploit the symmetry in the circuit ?
 A: This is an interesting question because it is difficult if you have the right tools for it and easy if you don’t. Setters use this kind of question to make sure that students don’t just learn clever techniques, but learn to think whether they are needed.
The “clever” way in this case would be to notice that there are triangles of resistors, remember the formula for transforming from a delta to a star configuration, apply it, and carry on from there. Necessary in the general case, but with the particular values of resistances shown here, what they are asking is not “Does she know how to apply this advanced technique?” but “Is she alert enough to notice she doesn’t need to?”
Here is an answer with no calculation at all.

*

*Throw away all the resistors except the line of three at the top. Note that they are all equal so that the potential difference across the chain including P and S is divided exactly into thirds: in this particular example, 0V, 4V, 8V, 12V.

*Throw away all the resistors except the line of three at the bottom. Note that they are all equal so that the potential difference across the chain including Q and T is divided exactly into thirds: in this particular example, 0V, 4V, 8V, 12V.

*Put together 1 and 2. Then both lines of resistors are dividing the same potential difference into thirds. So the potential at P equals the potential at Q and the potential at S equals the potential at T.

*You can put the vertical resistors in now if you like. Their value doesn’t matter because the potential at each end of each of them is identical, so no current will flow.

A: The currents in PQ and ST are null.
If you're not confident in "simplifying" electrical circuits, I leave it a solution of the circuit, exploting principle of superposition. We can define a "ring" current for every loop of the circuit, and write the current in the wire as the sum of ring currents, paying attention at the directions of the currents. As an example, following the (arbitrary) directions of the drawing, currents in PQ, PS and TS read
$i_{PQ} = i_1 - i_2, \quad i_{PS} = i_2,  \quad i_{TS} = i_2 - i_3$

A: First of all, welcome to  physics stack exchange
Instead of using the Wheatstone bridge, I suggest you have a look at how the current I1 is being divided at the common node(hint:-It's in such a way that the potential difference across both the 2ohm and 4ohm resistor is the same). That should help you decide which branches you can eliminate
