Current Electricity Charging Discharging Of Capacitors Question 666
Question: 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) $
Options:
A) $ 1.3\times 10^{4}\Omega $
B) $ 1.7\times 10^{5}\Omega $
C) $ 2.7\times 10^{6}\Omega $
D) $ 3.3\times 10^{7}\Omega $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ V=200(1-{{e}^{-t/\tau }}) $
$ 120=200(1-{{e}^{-t/\tau }}) $
$ \Rightarrow {{e}^{-t/\tau }}=\frac{200-120}{200}=\frac{80}{200} $
$ \frac{t}{\tau }={\log _{e}}2.5 $
$ =2.302\times {\log _{10}}2.5=2.302\times 0.4=0.9210 $
$ 5=(0.9210)\times R\times 2\times {{10}^{-6}} $
$ \Rightarrow R=2.7\times 10^{6}\Omega $
BETA
