How can three resistors of resistances `2 Omega,3 Omega and 6 Omega` be connected to give a total resistance of
(a) `4 Omega (b) 1 Omega`?

Explanation

a) If we connect resistances of `3 Omega and 6 Omega` in parallel and then resistance of `2 Omega` is connected in series of the combination, the total resistance of combination is `4 Omega` as shown in fig.12.3(a) The combined resistance R. on the parallel grouping of `3 Omega and 6 Omega` resistances is given as:
`1/R.=1/3+1/6=(2+1)/6=3/6=1/2`
`implies R=2 Omega`
and total resistance of entire combination `R=2 Omega +R.=2+2=4 Omega`
(b) IF all the three resistances are joined in parallel as shown in fig.12.3(b) we have
`1/R=1/2+1/3+1/6=(3+2+1)/6=6/6=1/1 implies R=1 Omega`