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\(\Omega \)?

(i) We can get a total resistance of 4\(\Omega \) by connecting the 2\(\Omega \) resistance in series with the parallel combination of 3\(\Omega \) and 6\(\Omega \).
So, \(R=R_1+\frac{R_2R_3}{R_2+R_3}= 2+\frac{6×3}{6+3}=4\Omega \)

(ii) We can obtain a total resistance of 1\(\Omega \) by connecting resistors of 2\(\Omega \), 3\(\Omega \) and 6\(\Omega \) in parallel.

\(\frac{1}R=\frac{1}{R_1} +\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{2} +\frac{1}{3}+\frac{1}{6}\)

\(\frac{1}R=1 \Omega \)

or R=1 \(\Omega \)