A resistor 'R' and `2(mu)F` capacitor in series is connected through a switch to 200 V direct supply. A cross 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. `1.7 xx 10^(5) Omega`
B. `2.7 xx 10^(6) Omega`
C. `3.3 xx 10^(7) Omega`
D. `1.3 xx 10^(4) Omega`

Answer

B

Explanation

Here, `V_(C) = 120 V V_(0) = 200 V`
so `V_(C ) = V_(0) (1 - e^(-t//RC))`
`1 - e^(-t//RC) = (V_(C ))/(V_(0)) = (120 V)/(200 V) = (3)/(5)`
or `e^(-t//RC) = 2//5`
or `log e^(-t//RC) = log_(e) 2//5`
or `(-t)/(RC) = 2.3026 log_(10) 0.4 = - 0.92` ltbgt or `R = (t)/(0.92 C) = (5)/(0.92 xx 2 xx 10^(-6)) = 2.7 xx 10^(6) Omega`