Equivalent resistance of the given network between points A and B is

Options

A. `31//5Omega`
B. `41//5Omega`
C. `36//5Omega`
D. `49//5Omega`

Answer

C

Explanation

In each segment of the combination `3 Omega and 2 Omega` resistance are connected in series separately.
`therefore R'=+3=6Omega and R''=2+2=4Omega`
R' and R'' are connected in parallel
`therefore "For first segment"(1)/(R_("eq"))=(1)/(6)+(1)/(4)+(2+3)/(12)=(5)/(12)`
`R_("eq1))=(12)/(5)Omega` Similarly for second and third segment
`R_("eq2")=(12)/(5)Omega and R_(eq3)=(12)/(5)Omega`
Now segment is connected in series then the total resistance of combination is
`R_("eq")=R_("eq1")+R_("eq2")+R_("eq3")=(12)/(5)+(12)/(5)+(12)/(5)=(36)/(5)Omega`