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) To get a total resistance of `4 Omega` from resistors of resistances `2 Omega,3 Omega and 6 Omega` , the resistors of resistances `3 Omega` and `6 Omega` should be connected in parallel. Their equivalent resistance is `((3 xx 6)/(3 + 6) Omega = 2 Omega)`.
This combination of equivalent resistance `2 Omega` should be connected in series with the resistor of resistance `2 Omega` so that equivalent resistance of three resistors becomes `2 Omega + 2 Omega = 4 Omega` as shown in (Fig. 3.34).
(b) To get a resistance of `1 Omega` from three given resistors of resistances `2 Omega,3 Omega,6 Omega`, we should connect all of them in parallel. The equivalent resistance is given by
`(1)/(R_p) = (1)/(2)+(1)/(3)+(1)/(6)= (3 + 2 + 1)/(6) = (6)/(6) = 1` i.e., `R_p = 1 Omega`.
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