Show how you would connect three resistors, each of resistance `6 Omega` so that the combination has a resistance of (i) `9 Omega (ii) 4 Omega`

Explanation

It is given here that `R_1=R_2=R_3=6 Omega`
To get a net resistance of `9 Omega` we join three resistances as shown in fig. 12.5 (a) Here resistances of parallel combination of two `6 Omega` resistors is
`1/R.=1/6+1/6=2/6=1/3implies R.=3 Omega`
`therefore Net resistance=R.+6=3+6=9 Omega`

(ii) To get a net resistance of `4 Omega` we join three resistances as shown in fig.12.5 (b) Here the combination of two resistores in series, having a combined resistances `R_0=6+6=12 Omega`, is joined in parallel with the third resistances of `6 Omega` Hence, the net resistance will be
`1/R=1/12+1/6=(1+2)/12=3/12=1/4 implies R= 4 Omega`