Find the equivalent resistance between the points A and B of the circuit shown in the fig.

Explanation

Supposing a source is connected between the terminals A and B . The current distribution is shown in fig.
At junction A , `i= i_1 + i_2`
Resistance between A and B , `R_(AB)= V/i = V/(i_1 +i_2)`
In close loop ACDA, -`5i_1 - 5i_3 + 10i_2 =0`
or `-i_1 - i_3 + 2i_2 =0" ".....(i)`
In close loop C B D E
`-10(i_1 - i_3) + 5(i_2 + i_3)+ 5i_3 =0`
or `-2i_i + i_2 + 4i_3 =0" "....(3)`
Now in close loop A C B E F A
`-5i_1 - 10(i_1 - i_3) + V= 0 ` or ` -3i_1 + 2i_3 = - V/5 " ".......(iii)`
From equations (i) and (iii) , we get
`-5i_1 + 4i_2 = - V/5" "......(iv)`
From equations (ii) and (iii),
we get `4i_1 + i_2 = (2V)/(5) " "...(v)`
Solving equations (iv) and (v) , we get
`i_1 = (9V)/(105)`
`i_2 = (6V)/(105)`
`i_3 = (V)/(35)`
`R_(AB) = (V)/(i_1 + i_2) =(V)/(((9V)/(105) + (6V)/(105)))+ 7 Omega`