A resistor 'R' and $2 \mu \mathrm{F}$ capacitor in series is connected through a switch to $200 \mathrm{~V}$…

A resistor 'R' and $2 \mu \mathrm{F}$ capacitor in series is connected through a switch to $200 \mathrm{~V}$ direct supply. Across the capacitor is a neon bulb that lights up at $120 \mathrm{~V}$. Calculate the value of $R$ to make the bulb light up $5 \mathrm{~s}$ after the switch has been closed. $\left(\log _{10} 2.5=0.4\right)$
  1. $1.7 \times 10^5 \Omega$
  2. $2.7 \times 10^6 \Omega$
  3. $3.3 \times 10^7 \Omega$
  4. $1.3 \times 10^4 \Omega$

Solution

$V_c=E\left(1-e^{-t / R c}\right)$ $1-e^{-t / R c}=\frac{120}{200}=\frac{3}{5}$ $\Rightarrow R=\frac{5}{1.84 \times 10^{-6}}=2.7 \times 10^6 \Omega$

Asked in: JEE Main 2011

Practice more Electrostatics questions on Aicharya