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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    A resistance of a galvanometer\[G=6\Omega \]                maximum current of 2 amp is measured by it. Then required resistance to convert it into ail ammeter reading up to 6 A, will of:

    A)  5\[5\,\Omega \]                             

    B)  \[4\,\Omega \]

    C)  \[3\,\Omega \]               

    D)         \[2\,\Omega \]

    Correct Answer: C

    Solution :

    Here: Resistance of galvanometer \[G=6\,\Omega \] Maximum   current   through   the galvanometer\[{{i}_{g}}=2A\] Current in ammeter \[i=6A\] The current through galvanometer is given by \[{{i}_{g}}=\frac{S}{S+g}i\] or            \[2=\frac{S}{S+6}\times 6\]    or            \[25+12=69\] So,          \[S=3\,\Omega \]


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