2. Solved Papers
  3. Chhattisgarh PMT

Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    A wire is bent in the form of a triangle now the equivalent resistance R between its one end and (he midpoint of the side is

    A) \[\frac{5R}{12}\]                                              

    B) \[\frac{7R}{12}\]

    C) \[\frac{3R}{12}\]                                              

    D) \[\frac{R}{12}\]

    Correct Answer: A

    Solution :

    Resistances R, R and \[\frac{R}{2}\] are in series, the resultant resistance, \[R'=R+R+\frac{R}{2}=\frac{5R}{2}\] Resistances R' and \[\frac{R}{2}\] are in parallel. Their resultant resistance                 \[\frac{1}{R}=\frac{2}{5R}+\frac{2}{R}=\frac{2+10}{5R}\]                 \[R=\frac{5R}{12}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner