Show how you would connect three resistors,each of resistance $6 \, \Omega$,so that the combination has a resistance of $(i)$ $9 \, \Omega$,$(ii)$ $4 \, \Omega$.

(N/A) To obtain the desired equivalent resistance,we analyze different combinations of the three $6 \, \Omega$ resistors.
$(i)$ To get $9 \, \Omega$:
We connect two $6 \, \Omega$ resistors in parallel and the third $6 \, \Omega$ resistor in series with this combination.
Equivalent resistance of the parallel part: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \implies R_p = 3 \, \Omega$.
Total resistance: $R_{eq} = R_p + 6 \, \Omega = 3 \, \Omega + 6 \, \Omega = 9 \, \Omega$.
$(ii)$ To get $4 \, \Omega$:
We connect two $6 \, \Omega$ resistors in series and the third $6 \, \Omega$ resistor in parallel with this combination.
Equivalent resistance of the series part: $R_s = 6 \, \Omega + 6 \, \Omega = 12 \, \Omega$.
Total resistance: $\frac{1}{R_{eq}} = \frac{1}{R_s} + \frac{1}{6} = \frac{1}{12} + \frac{1}{6} = \frac{1+2}{12} = \frac{3}{12} = \frac{1}{4} \implies R_{eq} = 4 \, \Omega$.