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. `1.7xx10^(5) Omega`
B. `2.7xx10^(6) Omega`
C. `3.3xx10^(7) Omega`
D. `1.3xx10^(4) Omega`

Answer

B

Explanation

As, `V = V_(0) (1 - e^(-t//RC)) :. 120 = 200 (1 - e^(-5//RC))`
or `(120)/(200) = 1 - e^(-5//RC)`
or `e^(-5//RC) = 1 - (120)/(200) = (2)/(5)` or `e^(5//RC) = (5)/(2) = 2.5`
Taking log of both the sides, we get
`(5)/(RC) log_(e) = log_(e) 2.5`
`(5)/(RC) xx1 = 2.3026 log_(10) 2.5 = 2.3026xx0.4`
`R = (5)/(2.3026xx0.4xx C) = (5)/(2.3026xx0.4xx(2xx10^(-6)))`
`= 2.7xx10^(6) Omega`