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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    An inductor (L = 0.03 H) and a resistor \[(R=0.15k\Omega )\]are connected in series to a battery of 15V EMF in a circuit shown below. The key \[{{K}_{1}}\] has been kept closed for a long time. Then at \[t=0,{{K}_{1}}\]is opened and key \[{{K}_{2}}\] is closed simultaneously. At t = 1ms, the current in the circuit will be : \[\left( {{e}^{5}}\cong 150 \right)\] [JEE Main Solved Paper-2015 ]

    A) 6.7 mA                 

    B) 0.67 mA

    C) 100 mA                                

    D) 67 mA

    Correct Answer: B

    Solution :

                    \[i=\frac{15}{0.15\times {{10}^{+3}}}\] \[15\times {{10}^{-2}}\times {{10}^{+3}}\] \[1={{10}^{-1}}{{e}^{-1\frac{Rt}{L}}}\] \[=\frac{1}{150}\]                            \[\frac{0.15\times {{10}^{3}}}{0.03}\times {{10}^{-3}}\] \[\frac{1}{1500}\] \[=1\]    \[{{10}^{-2}}\] \[=6.67\times {{10}^{-4}}\] \[=0.667\times {{10}^{-3}}\] \[=0.667mA\]


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