JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and \[5\,\Omega \]resistor. The ratio of the currents a time \[t=\,\,\infty \] and at \[t=40\]s is close to (Take\[{{e}^{2}}=7.389\]) [JEE MAIN Held on 07-01-2020 Evening]

    A) 1.06    

    B) 1.15

    C) 1.46

    D) 0.84

    Correct Answer: A

    Solution :

    [a] \[I={{I}_{0}}=\left( 1-{{e}^{\frac{tR}{L}}} \right)\] \[{{I}_{\infty }}=I\,(t=\infty )\,={{I}_{0}}\] \[{{I}_{40}}=I(t=40\,s)={{I}_{0}}\,\left( 1-e\frac{-\,40\times 5}{10\times {{10}^{-\,3}}} \right)={{I}_{0}}\,(1-{{e}^{-\,20,000}})\]\[\Rightarrow \,\frac{{{I}_{\infty }}}{{{I}_{40}}}=\frac{1}{1\,-{{e}^{-\,20,\,000}}},\] which is slightly greater than 1

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