Find equivalent resistance between points `A` and `B` in the circuit shown. All resistances are equal to`R`.

Explanation

We will solve this using symmetry about the dotted line. From this we can deduce that the centre in a branch will be equal to the current in its mirror image. For example current in `AC=` current in `CB=i_(1)`.
This implies that even if we break the junction at `C`, the currents in the circuit will not be affected. Now, the circuit is simplified which can be easily solved using series parallel combination.
`R_(AB)=2R` in parallel to `(2R+(R )/(2))=(10)/(9)R`