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Thanks that seems the way one can simplify the circuit and rest becomes straight forward. Just out of curiosity, can it be modified as shown in attached figure ( the central node is open in different way).
As a rule, this kind of circuits can be methodically solved with the bisection theorem.
In this case it is like this:
Imagine a voltage V symmetrically applied between the terminals A and B, i.e. V/2 at A and -V/2 at B, and split the two vertical resistors as the series of two R/2.
Take the central horizontal line that is the symmetry axis of the circuit.
By symmetry, all the points cut by this line are at the same voltage (0) and they can be virtually short-circuited to ground. So, calling Req the total equivalent resistance, we have
Req/2 = (1 || 1/2) || (1 + (1 || 1/2)) * R
Of course, the result is the same as obtained with Eduardo's method.
While the symmetry afforded here does lead to a simple solution, the problem is also solvable even if all 8 resistors are different. Just imagine imposing 1 volt between A and B. If we could calculate the total current we would know the equivalent network resistance. So we can immediately factor out the left-most resistor because it is in parallel with all the rest and its current contribution is already known. Next define unknown v1, v2, and v3 for the voltages at the 3 remaining nodes of the circuit. In terms of these unknowns we can write expressions for all the currents though all the remaining 7 resistors. Now for each of the 3 nodes, write the equation that says the current into the node is equal to the current out of the node. You now have 3 linear equations in 3 unknowns. Solve for v1, v2, and v3. Use these values to calculate the needed currents, and therefore the equivalent network resistance.
Yes tru, zorro's method is simple and easy to work out and is based on reasoning. Tunelabyguy's method seems to by analytical and general applicable to cases where all R's are not same, but appears to be laborious when easier solutions are available. Of course, it is useful as a general tool working with all possible values of R's.
Thanks again for useful suggestions and help.