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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of \[4\times {{10}^{-4}}\]A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) 200 ohm                       

    B) 6250 ohm

    C) 6200 ohm        

    D)   250 ohm

    Correct Answer: A

    Solution :

    The full scale deflection current \[{{I}_{g}}=25\times 4\times {{10}^{-4}}A\] Let R be the resistance connected \[{{I}_{g}}=\frac{V}{R+{{R}_{g}}}\Rightarrow 25\times 4\times {{10}^{-4}}=\frac{2.5}{R+50}\] \[\Rightarrow \]\[R=200\Omega \]


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