(a) To get an equivalent resistance of $13.5\mathit{\Omega }$, the resistances should beconnected as shown in the figure given below :So,
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}$
$=\frac{1}{9}+\frac{1}{9}=\frac{1+1}{9}=\frac{2}{9}$
$\frac{1}{{\mathrm{R}}_{\mathrm{P}}}=\frac{2}{9}$
${\mathrm{R}}_{\mathrm{P}}=\frac{9}{2}=4.5\mathit{\Omega }$
${\mathrm{R}}_{\mathrm{S}}={\mathrm{R}}_3+4.5\mathit{\Omega }$
$=9\mathit{\Omega }+4.5\mathit{\Omega }$
$=13.5\mathit{\Omega }$
Now,
${\mathrm{R}}_{\mathrm{S}}={\mathrm{R}}_3+4.5\mathit{\Omega }$
$=9\mathit{\Omega }+4.5\mathit{\Omega }$
$=13.5\mathit{\Omega }$
(b) To get an equivalent resistance of $6\mathrm{Omega}$, the resistances should be connected as shown in the figure :
${\mathrm{R}}_{\mathrm{S}}={\mathrm{R}}_1+{\mathrm{R}}_2$
$=9+9$
$=18\mathit{\Omega }$
Now both the resistors are in parallel with each other so,
${\mathrm{R}}_{\mathrm{P}}=\frac{1}{18}+\frac{1}{9}$
$=\frac{1+2}{18}=\frac{3}{18}$
$=\frac{1}{6}\mathit{\Omega }$
So, ${\mathit{R}}_{\mathit{P}}=6\mathit{\Omega }$
OR
a() According to Joule's law of heating, the heat produced in a wire is directly proportional to
(i) square of current $\left({\mathrm{I}}^2\right)$,
(ii) resistance of wire (R),
(iii) time ( $\mathit{t}$ ) for which current is passed.
Thus, the heat produced in the wire by current $\mathit{I}$ in time ' $\mathit{t}$ ' is
or
$\mathrm{H}\propto {\mathrm{I}}^2\mathrm{R}\mathit{t}$
$\mathrm{H}={\mathrm{KI}}^2\mathrm{R}\mathit{t}$
$\mathrm{H}={\mathrm{I}}^2\mathrm{R}\mathit{t}$

(b) We know that, $\mathrm{P}=\mathrm{VI}$
$\Rightarrow$
$\mathit{I}=\frac{\mathit{P}}{\mathit{V}}$
First lamp :

${\mathit{I}}_1=\frac{{\mathit{P}}_1}{\mathit{V}}=\frac{100}{220}=0.45\mathrm{A}$
Second lamp : ${\mathit{P}}_2=60\mathrm{W},\mathit{V}=220$ volt
${\mathit{I}}_2=\frac{{\mathit{P}}_2}{\mathit{V}}=\frac{60}{220}=0.27\mathrm{A}$
So, Total current
$={\mathit{I}}_1+{\mathit{I}}_2$
$=0.45+0.27$
$=0.72\mathrm{A}$