Find equivalent resistance between A and B

Explanation

A potential difference V is applied between A and B . Let a total current of `x+ y` flows through the circuit . The current distribution is as shown. Equivalent resistance of the circuit is thus given by
`R_(eq)= (V)/(x+y)` ....(i)
In loop AEFBA, `-yR -yR - ( y+z) R+V=0`
or ,` V = ( 3 x + z) R` ......(ii)
In loop ACFEA,` -xR - zR+ yR + yR = 0 `
In loop ACDBFEA, `- xR - ( x-z) R - ( x -z) R + (y + z) R + 2y R = 0 `
`implies z = x-y` .....(iv)
From (ii) and (iv) `z= ( y)/(2)` .....(v)
From (i) & (ii), we have
`R _(eq) = ((3y+z)/(z+y))R`
From (v), `y= 2z` , from (ii) `, x= 3z`
`implies R_(eq ) = ((3(2z)+z)/(3z+2z))R = ( 7R)/(5)`

A current `(x+ y)` enters from A, therefore, a current `( x+y)` leaves at B
At point A, x goes to AC containing R and y goes to AE containing 2R. So, we can guess by symmetry that at B, a current flow from CB containing 2R and x from EB containing R

By junction rule at 'C' , a current y goes to CDB
`implies x-y ` goes to CE.
`implies z = x-y`