BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer

    Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

    Two different coils have self-inductances, \[{{L}_{1}}=8\,mH\]and\[{{L}_{2}}=2\,mH\]. The current in one coil is increased at a constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power is 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  \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{1}{4}\]                                \[\frac{{{V}_{1}}}{{{V}_{2}}}=4\]  \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{1}{4}\]                         (4) \[\frac{{{i}_{1}}}{{{i}_{2}}}=4\]

    A)  1, 2 and 3 are correct

    B)  1 and 2 are correct

    C)   2 and 4 are correct

    D)  1 and 3 are correct

    Correct Answer: A

    Solution :

                     Induced emf in first coil \[={{V}_{1}}\]                 \[\therefore \]  \[{{V}_{1}}=-{{L}_{1}}\frac{di}{dt}\]                 Induced emf in second coil\[={{V}_{2}}\]                 \[\therefore \]  \[{{V}_{2}}=-{{L}_{2}}\frac{di}{dt}\]                 \[\therefore \]  \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}},\]as \[\frac{di}{dt}\]is same.                 Or           \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{8}{2}=\frac{4}{1}\]                 Hence (2) is a correct option.                 Power given to coils is same.                 \[\therefore \]  \[{{V}_{1}}{{i}_{1}}={{V}_{2}}{{i}_{2}}\]                 Or\[\left( 8\times {{10}^{-3}}\times \frac{di}{dt} \right){{i}_{1}}=\left( 2\times {{10}^{-3}}\times \frac{di}{dt} \right){{i}_{2}}\]                 \[\therefore \]  \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{1}{4}\]                 ?..(ii)                 Option (1) is correct.                 \[\therefore \]Energy stored in a coil \[=W=\frac{1}{2}L{{i}^{2}}\]                 \[\therefore \]  \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{L}_{1}}}{{{L}_{2}}}\times {{\left( \frac{{{i}_{1}}}{{{i}_{2}}} \right)}^{2}}\]                 Or           \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{8\times {{10}^{-3}}}{2\times {{10}^{-3}}}\times {{\left( \frac{1}{4} \right)}^{2}}\]                 Or           \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{1}{4}\]                 Option (3) is correct.


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