NTA Abhyas JEE Main2020PhysicsAlternating CurrentPractice
An inductor of inductance L = 400 mH and resistors of R 1 = 2 Ω and R 2 = 2 Ω are connected to a battery emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0 . The potential drop across L as a function of time is
Options
- A12 e -5 t   V
- B12 t e 3 t   V
- C6 1 - e - t 0.2   V
- D6   e -5 t   V
Correct answer
A. 12 e -5 t   V
Step-by-step solution
In the branch containing L and R 2 i = E R 2 1 - e - R 2 t L d i d t = E R 2 e - R 2 t L . R 2 L = E L e - R 2 t L ∴ V L = L d i d t = E e - R 2 t L = 12 e - 2 t 400 × 10 - 3 = 12    e - 5 t   V