How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of

(a) 4 Ω,

(b) 1 Ω?

(a) If we connect 3 Ω and 6 Ω in parallel, the equivalent resistance will be less than 3 Ω. Now, if 2 Ω is connected in series with this equivalent resistance, we should get a total resistance of 4 Ω.

So, the circuit will be as shown below:

3 Ω and 6 Ω are connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{2+1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$Ω

Hence, R_p = 2 Ω

The equivalent resistor 2 Ω is in series with the 2 Ω resistor.

R_eq= 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) Since the total resistance in this case is less than the resistance of each resistor hence, the resistors 2 Ω, 3 Ω and 6 Ω must be connected in parallel to get a total resistance of 1 Ω.

So, the circuit will be as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = \dfrac{6}{6} = 1Ω$

R_eq = 1Ω

Hence, the total resistance of the circuit is 1 Ω.