BITSAT2016PhysicsElectromagnetic InductionActual
An inductor of inductance L=400 mH and resistors of resistance R₁=2 and R₂=2 are connected to a battery of 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 t e^ -3 t ~V
- B6 (1-e^ -t / 0.2 ) V
- C12 e ^ -5 t ~V
- D6 e ^ -5 t ~V
Correct answer
C. 12 e ^ -5 t ~V
Step-by-step solution
Growth in current in L R₂ branch when switch is closed is given by array l i= E R₂ [1-e^ -R₂ t / L ] d i d t = E R₂ R₂ L e^ -R₂ t / L = E L e^ - R₂ t L array Hence, potential drop across L = ( E L e^ -R₂ t / L ) L=E e^ -R₂ t / L =12 e^ - 2 t 400 10⁻³ =12 e ^ -5 t ~V