The correct combination of three resistances `1Omega, 2Omega and 3Omega` to get equivalent resistance `(11)/(5)Omega` is

Options

A. All three are combines in paralle
B. All three are combines in paralle
C. `1Omega and 2Omega` in parallel and `3Omega` is in series to both
D. `2Omega and 3Omega` in parallel and `1Omega` is in series to both

Answer

D

Explanation

According to option (d) `2Omega and 3 Omega` are combined in a parallel combination. Hence equivalent resitance
`(1)/(R')=(1)/(2)+(1)/(3)=(5)/(6) therefore R'=(6)/(5)`
R' combined in series to `1Omega`
`therefore R_("eq")=R'+1=(6)/(5)+1=(6+5)/(5)=(11)/(5)Omega`
Hence, the combination scheme in option (d) is correct.