Calculate the current in the various branches of `10 Omega` resistance of the network of resistance as shown in figure .
Also calculate the total resistance between A and B.

Explanation

Let the currents in the various braches of the network of resistance will be as shown in figure
Applying Kirchhoff's second rule to loop KACDEBJK
`I_(1) xx 10 +(I_(1) - I_(2)) 5 = 10`
or `15 I_(1) - 5I_(1) = 10 or 5 I_(1) - I_(2) = 2`
For loop CDGHC
`10 I_(1) + 5 I_(2) - 5 (I - I_(1)) = 0`
or `15 I_(1) + 5 I_(2) = 5I_(2) = 5I or 3 I_(1) + I_(2)=I`
For loop DEFGD
`5 (I_(1) -I_(2)) - 10(I - I_(1) + I_(2)) - 5 I_(2)= 0`
or `15 I_(1) - 20 I_(2) = 10 I`
or `3 I_(1)- 4I_(2)=2I`
Solving (i), (ii) and (iii), we have
`I_(1) = 4/7 A, I_(2) = - 2/7 A, I = 10/7 A`
Current in `10 Omega` of arm GF
`=10/7 - 4/7 +(- 2/7) = 4/7 A`
If R is the total resistance of the network between A and B, then
`R = epsilon/I = 10/(10//7) = 7 Omega`