Search any question & find its solution
Question:
Answered & Verified by Expert
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. $\left(\log _{10} 2.5=0.4\right)$
Options:
Solution:
2990 Upvotes
Verified Answer
The correct answer is:
$2.7 \times 10^6 \Omega$
$2.7 \times 10^6 \Omega$
$V_c=E\left(1-e^{-t / R c}\right)$
$1-e^{-t / R c}=\frac{120}{200}=\frac{3}{5}$
$\Rightarrow R=\frac{5}{1.84 \times 10^{-6}}=2.7 \times 10^6 \Omega$
$1-e^{-t / R c}=\frac{120}{200}=\frac{3}{5}$
$\Rightarrow R=\frac{5}{1.84 \times 10^{-6}}=2.7 \times 10^6 \Omega$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.