Physics - Current Electricity Question with Solution | TestHub

To get maximum equivalent resistance all resistance must be connected in series
$\therefore {\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\dots \mathrm{n}\mathrm{R}.=\mathrm{n}\mathrm{R}$
To get minimum equivalent resistance all resistances must be connected in parallel.
$\therefore \frac{1}{{\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{i}\mathrm{n}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\dots \frac{1}{\mathrm{n}},\frac{1}{{\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{ }}}=\frac{\mathrm{n}}{\mathrm{R}}$
$\Rightarrow {\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{i}\mathrm{n}}=\frac{\mathrm{R}}{\mathrm{n}}\therefore \frac{{\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{a}\mathrm{x}}}{{\left({\mathrm{R}}_{\mathrm{e}\mathrm{q}}\right)}_{\mathrm{m}\mathrm{i}\mathrm{n}}}=\frac{\mathrm{n}\mathrm{R}}{\mathrm{R}/\mathrm{n}}={\mathrm{n}}^2$