NTA Abhyas JEE Main2020PhysicsCurrent ElectricityPractice
The capacitor shown in the figure is initially uncharged, the battery is ideal. The switch S is closed at time t = 0 , then the time after which the energy stored in the capacitor becomes one-fourth of the energy stored in it in steady-state is:
Options
- AR C
- BR C ln 2
- C  R C ln 4
- D2 R C
Correct answer
B. R C ln 2
Step-by-step solution
U = U m a x 4 ⇒ q 2 2 C = 1 4 Q m a x 2 2 C ⇒ q = Q m a x 2 q = C E 2 ⇒ q = C E 1 - e - t / R C C E 2 = C E 1 - e - t / R C ⇒ 1 2 = 1 - e - t / R C e - t / R C = 1 2 ⇒ t R C = ln 2 ⇒ t = R C ln 2