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

Correct Answer: 3

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

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

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

This will be minimum when

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

so \(x = 3\)