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

There are three resistore os resistances `2 Omega, 3 Omega "and " 6 Omega` respectively.
(a) The following circuit diagram shows the connection of the three resistors.

Here, `6 Omega "and" 3 Omega` resistores are connnected in parallel. Therefore, their equivalent resistance will be given by `1/(1/6+1/3)(6x3)/(6+3)=2 Omega`
This equivalent resistor of resistance `2 Omega` is connected to a `2 Omega` resistor in series. Therefore, the equivalent resistance of the circuit `=2 Omega + 2 Omega=4 Omega`
Hence the total resistance of the circuit is ` 4 Omega`.
(b) The following circuit diagram shows the connection of the three resistors.

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