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JEE Main & Advanced Physics Alternating Current / प्रत्यावर्ती धारा Question Bank Self Evaluation Test - Alternating Current

  • 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 f = 1 ms, the current in the circuit will be : \[({{e}^{5}}\cong 150)\]

    A) 6.7 Ma

    B) 0.67 mA      

    C) 100 mA

    D) 67 mA     

    Correct Answer: B

    Solution :

    [b] \[I\,(0)=\frac{15\times 100}{0.15\times {{10}^{3}}}=0.1A\] \[I\,(\infty )\,=0\] \[I\,(t)\,={{[I\,(0)-I(\infty )]}_{{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}}}\] \[I\,(t)\,=0.1\,{{e}^{^{\frac{-t}{L/R}+i(\infty )}}}=0.1\,{{e}^{\frac{R}{L}}}\] \[I(t)\,=0.1\,e\frac{0.15\times 1000}{0.03}\,=067\,mA\]


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