JEE Main2011PhysicsCapacitanceActual
A resistor 'R' and 2 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. ( ₁₀ 2.5=0.4 )
Options
- A1.7 10^5
- B2.7 10^6
- C3.3 10^7
- D1.3 10^4
Correct answer
B. 2.7 10^6
Step-by-step solution
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