Question

In the given circuit ‘a‘ is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be $\frac{\sqrt x}{2}$.The value of x is________.

Answer 1

$R_{net}=\frac{ma}{3}+\frac{a}{2m}$

=$a\bigg[\frac{m}{3}+\frac{1}{2m}–\frac{2}{\sqrt 6}+\frac{2}{\sqrt 6}\bigg]$

$=a\bigg[\bigg(\sqrt{ \frac{m}{3}}−\frac{1}{\sqrt 2m}\bigg)^2+\sqrt{ \frac{2}{3}}\bigg]$

This will be minimum when

$\sqrt{\frac{m}{3}}=\frac{1}{\sqrt{ 2m}}\; or$ $m=\sqrt{\frac{3}{2}}$

so $x = 3$