Connect a battery of emf `epsilon` between A and D.
Now the points E and F, and B and C are at the same potential, so no current flows through the arms EF and BC. The current through various arms of network of resistors.
Using Kirchhoff's loop law,
In closed loop AD`epsilon`A, `epsilon = 20y`
In closed loop ABEDA
`10 x + 20 x +10 x - 20y = 0`
or `40x = 20y or x = y/2`
If R is the resistance of network of resistors between point A and D, then `epsilon = (2x + y) R`
`= (2 xx y/2 + y)R = 2y R`
From (i) and (ii)
`20y= 2y R or R = 10 Omega`
