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. \[(lo{{g}_{10}}2.5=0.4)\]

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

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

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

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

Correct Answer: C

Solution :

[c] \[V=200(1-{{e}^{-t/\tau }})\] \[120=200(1-{{e}^{-t/\tau }})\] \[\Rightarrow {{e}^{-t/\tau }}=\frac{200-120}{200}=\frac{80}{200}\] \[\frac{t}{\tau }={{\log }_{e}}2.5\] \[=2.302\times {{\log }_{10}}2.5=2.302\times 0.4=0.9210\] \[5=(0.9210)\times R\times 2\times {{10}^{-6}}\] \[\Rightarrow R=2.7\times {{10}^{6}}\Omega \]