(a) It is observed that total current I is equal to the sum of separate currents.
$\mathrm{I}={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$....(i)
Let $\mathrm{R}\mathit{p}$ be the equivalent resistance of the parallel combination of resistors.So, by applying ohm's law
$\mathrm{I}=\frac{\mathrm{V}}{{\mathrm{R}}_{\mathit{p}}}$

So, now from equation (i), we have
$\frac{\mathrm{V}}{{\mathrm{R}}_{\mathit{p}}}=\frac{\mathrm{V}}{{\mathrm{R}}_1}+\frac{\mathrm{V}}{{\mathrm{R}}_2}+\frac{\mathrm{V}}{{\mathrm{R}}_3}$
and
$\frac{1}{{\mathrm{R}}_{\mathit{p}}}=\frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}+\frac{1}{{\mathrm{R}}_3}$
Hence, if $\mathit{n}$ resistors are connected in parallel, then the equivalent resistance of the circuit is given by -
$\frac{1}{{\mathit{R}}_{\mathit{e}\mathit{q}}}=\frac{1}{{\mathit{R}}_1}+\frac{1}{{\mathit{R}}_2}+\frac{1}{{\mathit{R}}_3}+........+\frac{1}{{\mathit{R}}_{\mathit{n}}}$
(b) Given, Two resistors of $12\mathit{\Omega }$ connected in parallel.

$\therefore \frac{1}{{\mathrm{R}}_1}+\frac{1}{{\mathrm{R}}_2}=\frac{1}{12}+\frac{1}{12}$


According to ohm's law V=IR
$6=\mathit{I}\times 6$
$\frac{6}{6}=\mathit{I}$

OR
(a) Given, ${\mathrm{R}}_1=20\mathit{\Omega },{\mathrm{R}}_2=4\mathit{\Omega }$
Since, in Series
$\mathit{R}={\mathit{R}}_1+{\mathit{R}}_2$
$\therefore$ Total resistance of circuit :
$\mathit{R}=20+4$
$=24\mathit{\Omega }$
(b) Current through circuit :
$\mathrm{V}=6\mathrm{V},\mathrm{R}=24\mathit{\Omega }$
According to ohm's law
$\mathrm{V}=\mathrm{IR}$
So,
$\mathit{I}=\frac{\mathit{V}}{\mathit{R}}$
$\mathit{I}=\frac{6}{24}=\frac{1}{4}=0.25$ ampere
(c) (i) Potential difference across conduction:
$\mathit{I}=\frac{1}{4},\mathit{R}=20\mathit{\Omega }$
${\mathit{V}}_1={\mathit{I}}_1$
${\mathit{V}}_1=\frac{1}{4}\times 20=5\mathrm{V}$
(ii) Potential difference across lamp
${\mathrm{V}}_2={\mathrm{IR}}_2$
$=\frac{1}{4}\times 4$
${\mathrm{V}}_2=1\mathrm{V}$
(d) Power of lamp :
$\mathrm{P}={\mathrm{I}}^2\mathrm{R}$
$={\left(\frac{1}{4}\right)}^2\times 20$
$=\frac{1}{4}\times \frac{1}{4}\times 20=\frac{5}{4}\mathrm{watt}$
or