Find the equivalent resistance between the terminal points `A` and `B` in the network shown in the figure.
.

Explanation

Here resistance are neither in series nor in parallel. This problem can be solved by Kirchhoff's rules or by delta-star method.
Equivalent resistance by Kirchhoff's rules : First, connect a battery of emf `E` between points `A` and `B`, let the current supplied by battery is `i`, then find the value of `E//i` in terms of `R`.
`R_(eq) = (E)/(i)`

Here we are making use of symmetry in current distribution. `i_1` is entering in `R` between `A` and `C` and `i_1` will flow in `R` between `D` and `B`. Similarly `i - i_1` will enter in `2 R` between `A` and `D` and `i - i_1` will flow in `2 R` between `C` and `B`.
The current in middle resistance `R` between `C` and `D`
`i_1 = x + i - i_1 rArr x = 2 i_1 - i`
`Lo op ①, A to A, path ACDA`
`V_A - i_1 R-(2 i_1 - i) R + ( i -i_1) 2R = V_A`
`-5 i_1 + 3 i = 0 rArr i_1 = 3 i//5`
`Lo op ②`, A to A, path `ADBA`
`V_A - (i + i_1) 3 R - i_1 R + E = V_A rArr (7 R)/(5) = (E)/(i) = R_(eq)`
`R_(eq) = (7 R)/(5)`
Equivalent resistance by delta-star method :

Convert delta `ACD` into a star
`x = (R.2R)/(sum R = R + 2R + R) = (2 R^2)/(4 R) = (R)/(2)`
`y = (R .R)/(sum R) = (R^2)/(4 R) = (R)/(4)`
`z = (R .2R)/(sum R) = (2 R^2)/(4 R) = (R)/(2)`
We have new circuit as

`R_(eq) = (R)/(2) +(9 R)/(10) = (5R + 9R)/(10) = (14 R)/(10) = (7R)/(5)`.