JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • 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 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 :

    [d] 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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