A resistor 'R' and 2 μ 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

Correct answer: B. 2.7 × 10^6 V_c=E(1-e^ -t / R c ) 1-e^ -t / R c =(120)/(200)=(3)/(5) R= 5 1.84 × 10^ -6 =2.7 × 10^6

JEE Main 2011

JEE Main 2011 · Physics · Capacitance

A resistor 'R' and 2 μ F capacitor in series is connected through a...

A resistor 'R' and

2 \mu \mathrm{F}

capacitor in series is connected through a switch to

200 \mathrm{~V}

direct supply. Across the capacitor is a neon bulb that lights up at

120 \mathrm{~V}

. Calculate the value of

R

to make the bulb light up

5 \mathrm{~s}

after the switch has been closed.

\left(\log _{10} 2.5=0.4\right)

B. $2.7 \times 10^6 \Omega$

V_c=E\left(1-e^{-t / R c}\right)

1-e^{-t / R c}=\frac{120}{200}=\frac{3}{5} \Rightarrow R=\frac{5}{1.84 \times 10^{-6}}=2.7 \times 10^6 \Omega