CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The resistance of a galvanometer is\[50\,\Omega \]. and it shows full scale deflection for a current of 1 mA. To convert it into a voltmeter to measure 1 V and as well as 10 V (Refer circuit diagram) the resistances\[{{R}_{1}}\]and\[{{R}_{2}}\]respectively are

    A)  \[950\,\Omega \]and \[9150\,\Omega \]

    B)  \[900\,\Omega \]and\[9950\,\Omega \]

    C)  \[900\,\Omega \]and\[9900\,\Omega \]

    D)  \[950\,\Omega \]and \[9000\,\Omega \]

    E)  \[950\,\Omega \]and\[9950\,\Omega \]

    Correct Answer: D

    Solution :

    Given:       \[V=1\]volt \[{{I}_{g}}=1\,mA\] \[=1\times {{10}^{-3}}A\] Resistance of a galvanometer                 \[G=50\,\Omega \] \[\therefore \]  \[{{R}_{T}}=\frac{V}{{{I}_{g}}}-G\] or            \[{{R}_{T}}=\frac{1}{{{10}^{-3}}}-50\] or            \[{{R}_{1}}=950\,\Omega \] AISO       \[{{R}_{2}}=\frac{10}{{{10}^{-3}}}-50\] \[=9950\,\Omega \] Additional resistance for attaining \[10V=9950-950\] \[=9000\,\Omega \] Hence,  \[{{R}_{1}}=950\,\Omega ,{{R}_{2}}=9000\,\Omega \]


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