Find the equivalent resistance between
(a) `A` and `B` and
(b) `A` and `C`.
.

Explanation

(a)
`V_A = (i)/(3) R - (i)/(6) R -(i)/(3) R +V = V_A`
`(V)/(i) = R_(eq) = (5R)/(6)`.
(b)
The paths `ADC` and `AGC` are equivalent and carry same current. No currents in `GF` and `DH`, hence remove these resistances.

`R_(eq) = (3R)/(4)`.