An Enginner Cannot Reproduce This Problem

The problem can be solved by superposition.

Inject I current at point A. There are 6 branches at A and hence by symmetry, each branch will carry a current of I/6 This is the first case.

Now remove I current from point B. Using the same argument as above, the branch AB carries a current of I/6

Superpose the above cases. Hence effectively the branch AB carries a current of I/3

The resistance of AB is 1 hence the resistance of the remaining circuit is 0.5

AB and the remaining circuit are in parallel and hence the equivalent resistance is 0.333

i assume that there is a battery emf v0 connected between pts A nd B. if current supplied by the battery is I then by symmetry current in all wires will b same. since a triangle has three sides current will b 1/3rd of each line. so; v0=i0Req v0=i0/3R0 i0Req=i0/3R0 Req=R/3.

(Assume a battery is connected b/w A&B) now assume that a current of 6A enters at point A, it

gets distributed along each resistor as 1A(symmetry)....now assume at point B...6A current

comes out. thus the resistor AB will have a -ve 1A current in it...using superposition theorem the

net current in AB is 2A, the voltage drop across AB is 2V while total current in the circuit is

6A...the equivalent resistance= 2A/6V=1/3 Ohms