Determine the equivalent resistance of networks shown in figure.

Explanation

Equivalent resistance of one loop (parallel connection of 2 `Omega and 4 Omega `)
`R. = (2 xx 4)/(2 + 4) = (8)/(6) = (4)/(3) Omega`
Now, given network can be shown as follow,
`rArr ` Equivalent resistance of given network will be
R = 4 R.
= `4 xx (4)/(3)`
`= (16)/(3) Omega = 5.3 Omega`
(ii) Here five resistances each of R `Omega` are connected in series and so that equivatent resistance would be,
`R_(S) = R + R + R ` + R +R
`therefore R_(S) = 5 R Omega`