2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    Four resistances of 10\[\Omega \], 60\[\Omega \]., 100\[\Omega \]. and 200\[\Omega \]. respectively taken in order are used to form a Wheatstone's bridge. A 15V battery is connected to the ends of a 200\[\Omega \], resistance, the current through it will be

    A) \[7.5\times {{10}^{-5}}A\]           

    B) \[7.5\times {{10}^{-4}}A\]

    C) \[7.5\times {{10}^{-3}}A\]           

    D) \[7.5\times {{10}^{-2}}A\]

    Correct Answer: D

    Solution :

    Here, the resistance for 10\[=2\times {{10}^{-14}}\left( W/{{m}^{2}} \right)\], 60\[=\frac{{{l}_{ear}}}{{{l}_{eye}}}=\frac{{{10}^{-13}}}{2\times {{10}^{-14}}}=5\] and 100\[\mu =\frac{\Delta {{V}_{p}}}{\Delta {{V}_{g}}}\] are in series and they together are in parallel to 200\[\Rightarrow \] resistance when a potential difference of 15 V is applied across 200\[\Delta {{V}_{p}}=-\mu \times \Delta {{V}_{s}}\], then current through it is                           \[=-50\left( -20 \right)=10V\]


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