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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    Two coils have self-inductance\[{{L}_{1}}=4\,mH\]and \[{{L}_{2}}=I\,mH\]respectively. The currents in the two coils are increased at the same rate. At a certain instant of time both coils are given the same power. If \[{{I}_{1}}\] and \[{{I}_{2}}\] are the current in the two coils, at that instant of time respectively, then the value of \[({{I}_{1}}/{{I}_{2}})\] is:

    A)  1/8                                       

    B)  ¼

    C)   1/2                                      

    D)  1

    Correct Answer: B

    Solution :

                     Given: Self-inductance of first coil \[{{L}_{1}}=4\,mH\] Self-inductance of second coil \[{{L}_{2}}=1\,mH\] From the relation induced emf \[e=L\frac{di}{dt}\] Hence, \[{{e}_{1}}={{L}_{1}}\frac{d{{i}_{1}}}{dt}\]and \[{{e}_{2}}={{L}_{2}}\frac{d{{i}_{2}}}{dt}\] As the power given are same \[{{e}_{1}}={{e}_{2}}\] or            \[{{L}_{1}}=\frac{d{{I}_{1}}}{dt}={{L}_{2}}\frac{d{{I}_{2}}}{dt}\] or            \[\frac{{{L}_{1}}}{{{L}_{2}}}=\frac{d{{I}_{2}}}{d{{I}_{1}}}=\frac{{{I}_{2}}}{{{I}_{1}}}\] or            \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{L}_{2}}}{{{L}_{1}}}=\frac{1}{4}\]


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