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J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • question_answer
    Two different coils have self-inductance \[{{L}_{1}}=8\,\,mH,\]\[{{L}_{2}}=2\,\,mH,\] The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same rate. At a certain instant of time, the power given to the two coils is the same. At that time the current, the induced voltage and the energy stored in the first coil are \[{{i}_{1}},\,\,{{V}_{1}}\] and \[{{W}_{1}}\] respectively. Corresponding values for the second coil at the same instant are \[{{i}_{2}},\,{{V}_{2}}\] and \[{{W}_{2}}\] respectively. Then,

    A)  \[{{W}_{2}}/{{W}_{1}}=8\]

    B)  \[{{W}_{2}}/{{W}_{1}}=1/8\]

    C)  \[{{W}_{2}}/{{W}_{1}}=4\]

    D)  \[{{W}_{2}}/{{W}_{1}}=1/4\]

    Correct Answer: C

    Solution :

    Given,  \[\frac{d{{l}_{1}}}{dt}=\frac{d{{l}_{2}}}{dt}\] \[\Rightarrow \] \[{{L}_{1}}=8\,mH\] and \[{{L}_{2}}=2\,mH\] \[\Rightarrow \] \[{{p}_{1}}={{p}_{2}}\] \[\frac{dW}{dt}=\frac{dW}{dt}\] \[{{L}_{1}}{{l}_{1}}\frac{d{{l}_{1}}}{dt}={{L}_{2}}{{l}_{2}}\frac{d{{l}_{2}}}{dt}\]\[\left[ \because \,\,\,\frac{d{{l}_{1}}}{dt}=\frac{d{{l}_{2}}}{dt},Given \right]\] \[\frac{I{{ & }_{1}}}{{{I}_{2}}}=\frac{{{L}_{2}}}{{{L}_{1}}}\] \[\Rightarrow \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{2\,mH}{8mH}\] \[\Rightarrow \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{4}\] \[\Rightarrow \]We know \[w=\frac{1}{2}L{{I}^{2}}\] \[\Rightarrow \]So, \[\frac{{{w}_{1}}}{{{w}_{2}}}=\frac{{{L}_{1}}I_{1}^{2}}{{{L}_{2}}I_{2}^{2}}=\frac{{{L}_{1}}}{{{L}_{2}}}{{\left( \frac{{{I}_{1}}}{{{I}_{2}}} \right)}^{2}}\] \[=\left( \frac{8}{2} \right){{\left( \frac{1}{4} \right)}^{2}}=4\times \frac{1}{16}=\frac{1}{4}\] \[\frac{{{w}_{2}}}{{{w}_{1}}}=4\]


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