CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    A galvanometer of resistance \[100\,\,\Omega \] is converted to a voltmeter of range 10 V by connecting a resistance of\[10\,k\Omega \]. The resistance required to convert the same galvanometer to an ammeter of range 1 A is

    A)  \[0.4\,\Omega \]            

    B)         \[0.3\,\,\Omega \]

    C)  \[1.2\,\Omega \]           

    D)         \[0.2\,\,\Omega \]

    E)  \[0.1\,\Omega \]

    Correct Answer: E

    Solution :

    Resistance of galvanometer, \[{{R}_{1}}=100\text{ }\Omega \] Connected resistance, \[{{R}_{2}}=10\text{ k}\Omega =10000\,\Omega \] Total resistance, \[R=10000+100\] \[=10100\,\Omega \] Current of galvanometer,                 \[{{i}_{g}}=\frac{V}{R}\]                 \[=\frac{10}{10100}\]                 \[=9.9\times {{10}^{-4}}\]                 \[\approx 10\times {{10}^{-4}}A\] Shunt,   \[S=\frac{{{i}_{g}}}{i-{{i}_{g}}}\times 100\]                 \[=\frac{{{i}_{g}}}{(1-10\times {{10}^{-4}})}\times 100\]                 \[=\frac{100}{0.99}\times 10\times {{10}^{-4}}=0.1\,\Omega \]


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