$(a)$ What is the total resistance of $n$ resistors each of resistance $R$ connected in:
$(i)$ series?
$(ii)$ parallel?
$(b)$ Calculate the resultant resistance of $3$ resistors $3 \ \Omega, 4 \ \Omega$,and $12 \ \Omega$ connected in parallel.

(N/A) In a series combination,the total resistance is the sum of individual resistances:
$R_{S} = R + R + \dots + R \ (n \ \text{times}) = nR$
In a parallel combination,the reciprocal of the total resistance is the sum of the reciprocals of individual resistances:
$\frac{1}{R_{P}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} \ (n \ \text{times}) = \frac{n}{R}$
Therefore,$R_{P} = \frac{R}{n}$
$(b)$ For resistors connected in parallel,the resultant resistance $R_{P}$ is given by:
$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$
$\frac{1}{R_{P}} = \frac{1}{3} + \frac{1}{4} + \frac{1}{12}$
Taking the least common multiple $(LCM)$ of $3, 4$,and $12$,which is $12$:
$\frac{1}{R_{P}} = \frac{4 + 3 + 1}{12} = \frac{8}{12} = \frac{2}{3}$
$R_{P} = \frac{3}{2} = 1.5 \ \Omega$