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

  • question_answer
    A resistor 'R' and \[2\mu F\] capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed, \[(lo{{g}_{10}}2.5=0.4)\]

    A) \[1.7\times {{10}^{5}}\Omega \]

    B) \[2.7\times {{10}^{6}}\Omega \]

    C) \[3.3\times {{10}^{7}}\Omega \]

    D) \[1.3\times {{10}^{4}}\,\Omega \]

    Correct Answer: B

    Solution :

    [b] We have, \[V={{V}_{0}}\,(1\,-{{e}^{-t/RC}})\] \[\Rightarrow \,120\,=200(1-{{e}^{-t/RC}})\,\Rightarrow \,t=RC\] in \[(2.5)\,\Rightarrow \,R\] \[=2.71\times {{10}^{6}}\,\Omega \]


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