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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Mock Test - Electromagnetic Induction

  • question_answer
    Two resistors of 10\[\Omega \] and 20\[\Omega \] and an ideal inductor of 10 H are connected to a 2 V battery as shown in figure. Key K is inserted at time t = 0. The initial (t = 0) and final (\[t\to \infty \]) currents through the battery are

    A) \[\frac{1}{15}A,\frac{1}{10}A\] 

    B) \[\frac{1}{10}A,\frac{1}{15}A\]

    C) \[\frac{2}{15}A,\frac{1}{10}A\] 

    D) \[\frac{1}{15}A,\frac{2}{25}A\]

    Correct Answer: A

    Solution :

    [a] At \[t=0\]i.e., when the key is just pressed, no current exists inside the inductor. So \[10\Omega \] and \[20\Omega \] resistors are in series and a net resistance of (10+20) = 30\[\Omega \]exists across the circuit. Hence, \[{{I}_{l}}=\frac{2}{30}=\frac{1}{15}A\] As\[t\to \infty \], the current in the inductor grows to attain a maximum value. i.e., the entire current passes through the inductor and no current passes through 10 \[\Omega \]resistor. Hence, \[{{I}_{2}}=\frac{2}{20}=\frac{1}{10}A\]


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