2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    Two different coils have self-inductances \[{{L}_{1}}=8\] mH and \[{{L}_{2}}=8\] mH. The current in one coil is increased at a constant rate. The current in the second coils is also increased at the same constant rate. At a certain instant, the power given to the two coil is the same. At that instant of time, if \[{{W}_{1}}\]and \[{{W}_{2}}\]are the energies stored in the first and the second coil, respectively, the \[{{W}_{1}}:{{W}_{2}}\]is

    A) 1 : 2                

    B) 2 : 1

    C) 4 : 1                            

    D) 1 : 4

    Correct Answer: D

    Solution :

    \[E=-L\frac{dI}{dt};W=\frac{1}{2}L{{I}^{2}}\] We know that \[E=-\frac{LdI}{dt}\]As\[\frac{dI}{dt}\] is same, Hence, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}=4\] Power is constant so \[{{E}_{1}}{{I}_{1}}={{E}_{2}}{{I}_{2}};\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{I}_{2}}}{{{I}_{1}}}=4\]We know that \[W=\frac{1}{2}L{{I}^{2}};\] \[\therefore \]\[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{L}_{1}}I_{1}^{2}}{{{L}_{2}}I_{1}^{2}}=4\times \frac{1}{16}=\frac{1}{4}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner