question_answer

In the arrangement shown in figure, the current through \[{{\tan }^{-1}}\left( \frac{4}{3} \right)\]resistor is

A) \[{{\cos }^{-1}}\left( \frac{4}{3} \right)\]               

B) Zero

C) \[[M{{L}^{3}}{{l}^{-1}}{{T}^{-3}}]\]                          

D) \[[{{M}^{2}}{{L}^{2}}{{l}^{-1}}{{T}^{-2}}]\]

Correct Answer: A

Solution :

The circuit may be redrawn as shown in the adjacent figure. \[C{{l}^{-}}\] \[{{\tan }^{-1}}\left[ \frac{{{p}^{2}}+PQ+{{Q}^{2}}}{PQ} \right]\] \[i=\frac{{{E}_{eq}}}{R+{{\operatorname{R}}_{eq}}}=\frac{12}{5+1}=\frac{12}{6}=2\Omega \]