JEE Main2009PhysicsElectromagnetic InductionActual
An inductor of inductance L =400 mH and resistors of resistances R ₁=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
- A6 e^ -5 t V
- B12 t e^ -3 t V
- C6 (1-e^ -t / 0.2 ) V
- D12 e^ -5 t V
Correct answer
D. 12 e^ -5 t V
Step-by-step solution
aligned & I ₁= F R ₁ = 12 2 =6 ~A & E = L dl_ 2 dt + R ₂ I ₂ & I ₂= I ₀ (1- e ^ - t / t _ c ) I _ o = E R ₂ = 12 2 =6 ~A & t _ c = L R = 400 10⁻³ 2 =0.2 & I ₂=6 (1- e ^ - t / 0.2 ) aligned Potential drop across L=E-R₂ l₂=12-2 6 (1-e^ -b t )=12 e^ -5 t Directions: Question numbers 28,29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle A B C D A , as shown in the P-T diagram.