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

(b) When switch is closed the charge starts flowing in circuit .
Voltage across the capacitor in time t = 5 seconds

` V_(0) = 200 V `
`V_(C)= V_(0)(1-e^(-(t)/(RC)))`
`V_(B)= V_(0)(1-e^(-(t)/(RC)))`
`implies (120)/(200)=1-e^(-(5)/(RC))` ,
`(12)/(20)=e^(-(5)/(RC))`
`e(5)/(RC)=(20)/(8)`
`(5)/(RC) = In((20)/(8)) ` = In(2.5)
`R= (5)/(C"In"(2.5))`
`=(5)/(2xx10^(-6)[2.303"log"_(10)(2.5)])`
`R=(5)/(2xx2.303xx0.4xx10^(-6))`
`=2.7xx10^(6)Omega`