JEE Main & Advanced AIEEE Solved Paper-2011

  • 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. (\[{{\log }_{10}}2.5=0.4\])   AIEEE  Solved  Paper-2011

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

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

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

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

    Correct Answer: D

    Solution :

                 Charge on capacitor \[q={{q}_{0}}(1-{{e}^{-t/RC}})\] or \[V=\frac{{{q}_{0}}}{C}(1-{{e}^{-t/RC}})\] \[V=200(1-{{e}^{-t/RC}})\] \[120-200(1-{{e}^{-t/RC}})\] \[{{e}^{-t/RC}}=0.4\] \[\frac{-t}{RC}=\ln (0.4)\] \[\frac{t}{RC}=\ln \left( \frac{10}{4} \right)=2.303\times 0.4\] \[R=\frac{5}{2\times {{10}^{-6}}\times 2.303\times 0.4}\] \[=2.7\times {{10}^{6}}\Omega \]


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