(i) Three resistances ${\mathit{R}}_1,{\mathit{R}}_2$ and ${\mathit{R}}_3$ are connected in parallel to one another between the same two points. In this case, the potential difference across the ends of all the resistance will be the same.
$\therefore \mathrm{V}={\mathrm{V}}_1={\mathrm{V}}_2={\mathrm{V}}_3$...(i)
If the total current flowing through the circuit is $\mathit{I}$, then the current passing through ${\mathit{R}}_1$ will be ${\mathit{I}}_1$ through ${\mathit{R}}_2$ will be ${\mathit{I}}_2$ and through ${\mathit{R}}_3$ will be ${\mathit{I}}_3$.Then
$\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$...(ii)
If $\mathit{R}$ is the effective resistance of the circuit, connected across a battery of $\mathrm{V}$ volts, through which I current flows, then
$\mathit{I}=\frac{\mathit{V}}{\mathit{R}}$
Substituting the values in eq. (ii), we get
$\Rightarrow \frac{\mathit{V}}{\mathit{R}}=\frac{\mathit{V}}{{\mathit{R}}_1}+\frac{\mathit{V}}{{\mathit{R}}_2}+\frac{\mathit{V}}{{\mathit{R}}_3}$
$\Rightarrow \frac{\mathit{V}}{\mathit{R}}=\mathit{V}\left[\frac{1}{{\mathit{R}}_1}+\frac{1}{{\mathit{R}}_2}+\frac{1}{{\mathit{R}}_3}\right]$
$\Rightarrow \frac{1}{\mathit{R}}=\frac{1}{{\mathit{R}}_1}+\frac{1}{{\mathit{R}}_2}+\frac{1}{{\mathit{R}}_3}$
(ii) Given: ${\mathrm{R}}_1=4\mathit{\Omega },{\mathrm{R}}_2=4\mathit{\Omega },{\mathrm{R}}_3=8\mathit{\Omega }$ Let, resultant resistance between $\mathit{a}$ and $\mathit{c}$ be ${\mathrm{R}}^{\prime }$
Then, ${\mathit{R}}^{\prime }={\mathit{R}}_1+{\mathit{R}}_2$
(Series combination)
${\mathrm{R}}^{\prime }=4+4=8\mathit{\Omega }$
If $\mathit{R}$ is the effective resistance between $\mathit{A}$ and $\mathit{B}$, then
$\frac{1}{\mathit{R}}=\frac{1}{{\mathit{R}}^{\prime }}+\frac{1}{{\mathit{R}}_3}$
( ${\mathit{R}}^{\prime }$ and ${\mathit{R}}_3$ are in parallel combination)
$\frac{1}{\mathit{R}}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}$
$\Rightarrow \mathit{R}=4\mathit{\Omega }$