Show how you would connect three resistors, each of resistance 6 `omega` , so that the combination has a resistance of
4 `omega`

Explanation

To get total 4 `omega`resistance from three 6 `omega` resistors, we should connect two resistors in series and the third resistor in parallel with the resultant. The combination is given as follows :

Total resistance in series is given by `R_(12) = R_(1) + R_(2) = 6 omega + 6 omega = 12 omega` Now `R_(12)` and 6 `omega` are connected in parallel, so the net resistance is given by `rarr (1)/(R )=(1)/(R_(12))+1/6`
`rarrR=4 omega`