The effective resistance between A and B, if resistance of each resistor is R, will be :

Solution

Physics sir · 4 steps

By symmetry the top, centre and bottom nodes share one potential, so the vertical resistors drop out.

1. Name the nodes. Call the left junction P and the right junction Q. Between them lie the top node T, the centre node C and the bottom node D. Each of T, C and D connects to P through one R and to Q through one R.
2. Use symmetry. The network between P and Q looks the same from either end, and T, C and D all lie on the vertical mirror line. So all three are at the same potential, exactly halfway between P and Q. The two vertical resistors T–C and C–D therefore carry no current and can be removed.
3. Reduce the diamond. From P to the line T, C, D there are three resistors R in parallel, giving R/3. From that line to Q there are another three in parallel, giving R/3. So R_(PQ) = R/3 + R/3 = 2R/3.

   R_(AB) = R + 2R/3 + R = 8R/3

4. So. Adding the two end resistors in series gives R_(AB) = 8R/3, which is option (a).

The idea to keep. In a network that is symmetric about a line perpendicular to AB, every node on that line is at the same potential. Resistors joining such nodes carry no current, so remove them first.

Answer from the NTA final answer key. Solution by PrepWiser.