Given three resistors of resistances `1 Omega, 2Omega and 3Omega`. How will you combine them to get an equivalent resistance of `(11)/(5) Omega` ?

Explanation

Here resistance `2Omega and 3Omega` are joined in series
`:.` Equivalent resistance
`R_(1) = (2 xx 3)/(2 +3) = (6)/(5)`
Now resistance `1Omega` is joined in series with `R_(1)`
`:. R=1 + R_(1) =1 + (6)/(5) = (11)/(5) Omega`