VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    A galvanometer has current range of 15 mA and voltage range 750 mV. To convert this galvanometer into an ammeter of range 25 A, the required shunt is

    A) 0.8 \[\Omega \]               

    B) 0.93\[\Omega \]

    C) 0.03 \[\Omega \]            

    D) 2.0 \[\Omega \]

    Correct Answer: C

    Solution :

    \[{{X}^{2}}=3{{R}^{2}}\]                     \[X=\sqrt{3}R\]            and  l = 25 A Using the relation                   \[e=M\frac{di}{dt}\]                   \[=0.005\times \frac{d}{dt}\left( {{i}_{0}}\sin \omega t \right)\]                   \[=0.005\times {{i}_{0}}\cos \omega t\]                \[{{e}_{\max }}=0.005\times 10\times 100\pi =5\pi \]       \[{{v}_{0}}=\frac{1}{2\pi \sqrt{LC}}\]                   \[=\frac{1}{2\pi \sqrt{1\times {{10}^{-3}}\times 0.1\times {{10}^{-6}}}}\]


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