JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Static EMI

  • question_answer
    The mutual inductance between a primary and secondary circuits is 0.5 H. The resistances of the primary and the secondary circuits are 20 ohms and 5 ohms respectively. To generate a current of 0.4 A in the secondary, current in the primary must be changed at the rate of  [MP PMT 1997]

    A)            4.0 A/s                                    

    B)            16.0 A/s

    C)            1.6 A/s                                    

    D)            8.0 A/s

    Correct Answer: A

    Solution :

                       \[{{e}_{2}}=M\frac{d{{i}_{1}}}{dt}\Rightarrow {{i}_{2}}{{R}_{2}}=M\frac{d{{i}_{1}}}{dt}\Rightarrow 0.4\times 5=0.5\times \frac{d{{i}_{1}}}{dt}\]            \[\Rightarrow \frac{d{{i}_{1}}}{dt}=4\ A/\sec .\]


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