Find the total resistance of the network shown in the figure given below :

Explanation

The given circuit is series combination of four identical units.
In one unit `2Omega and 2Omega` resistances in upper portion are in series.
So `R_1 = 2 + 2 = 4Omega`

Similarly in the lower portion
`R_2 = 3 + 3 = 6Omega`
Now `R_1 and R_2` are in parallel. Let net resistance of one unit is R.
So `1/R=1/R_1 +1/R_2 =1/4 +1/6` ,
`implies 1/R = (6+4)/24 =10/24`
`implies R = 24/10 =2.4 Omega`
Resistance of whole given circuit `= 4 xxR`
`=4 xx2.4`
`= 9.6Omega`