JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    A galvanometer with its coil resistance \[25\Omega \]requires a current of 1mA for its full deflection. In order to construct an ammeter to read up to a current of 2A, the approximate value of the shunt resistance should be            [JEE Main 16-4-2018]

    A) \[2.5\times {{10}^{-2}}\Omega \]               

    B) \[1.25\times {{10}^{-3}}\Omega \]

    C) \[2.5\times {{10}^{-3}}\Omega \]               

    D) \[1.25\times {{10}^{-2}}\Omega \]  

    Correct Answer: D

    Solution :

     Formula:\[S={{G}^{*}}I/(I-Ig)\] where S= shunt resistance , (I-Ig)= current through the shunt resistance , I=Current in the circuit. Here, galvanometer resistance, G= 25 ohm ;   Ig= 0.001 A , I= 2A \[s=({{25}^{*}}0.0001)/(2-0.001)\] \[=1.25\times {{10}^{-3}}\Omega \] A shunt resistance of\[1.25\times {{10}^{-2}}\Omega \]must be connected to make the galvanometer in the range of 0 to 2A


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