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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    The scale of a galvanometer of resistance 100\[\Omega \] contains 25 divisions. It gives a deflection of one division on passing a current of \[4\times {{10}^{-4}}.\] The resistance in ohm to be added to it, so that it may become a voltmeter of range 2.5 volt is:

    A)  100                                       

    B)  150

    C)  250                                       

    D)  300

    Correct Answer: B

    Solution :

                     Given: Resistance of galvanometer \[G=100\Omega \] Number of division contained = 25 division Value of current \[i=4\times {{10}^{-4}}\]amp/division Value   of   \[i\]   for   25   division \[{{i}_{g}}=4\times 25\times {{10}^{-4}}={{10}^{-2}}A\] \[V=2.5\,\text{volt}\] From the relation to convert galvanometer into voltmeter the shunt resistance R is given \[R=\left[ \frac{V}{{{i}_{g}}}-G \right]\] \[R=\left( \frac{2.5}{{{10}^{-2}}}-100 \right)=250-100\] \[R=150\Omega \]


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