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

Explanation

There are three resistors of resistance `2Omega,3Omega` and `6Omega` respectively
The following circuit diagram shows the connection of the three resistors.

Here, `6Omega` and `3Omega` resistors are connected in parallel therefore, their equivalent resistance will be given by
`1/(1/6+1/3)=(6xx3)/(6+3)=2Omega`
This equivalent resistor of resistance `2Omega` is connected to a 2 `Omega` resistors in series
Therefore, the equivalent resistance of the circuit
`=2Omega+2Omega=4Omega`
Hence the total resistance of the circuit is `4Omega`
(b) The following circuit diagram shows the connection of the three resistors

All the resistors are connected in series. Therefore, their equivalent resistance wil I be given as -
`1/(1/2+1/3+1/6)=1/((3+2+1)/(6))=6/6=1Omega`
Therefore, the total resistance of the circuit is `1Omega`