JEE Main2015PhysicsAlternating CurrentActual
An inductor ( L = 0.03   H ) and a resistor ( R = 0.15   kΩ ) are connected in series to a battery of 15   V E.M.F. in a circuit shown below. The key K 1 has been kept closed for a long time. Then at t = 0 , K 1 is opened and key K 2 is closed simultaneously. At t = 1   ms , the current in the circuit will be : Take , e 5 ≈ 150
Options
- A0 . 67   mA
- B100   mA
- C67   mA
- D6 . 7   mA
Correct answer
A. 0 . 67   mA
Step-by-step solution
Case I: K 1 is closed for long time, for a long time, the inductor acts as a conducting wire, ⇒ current in the circuit = V R = 1 5 1 5 0 ⇒ i 0 = 0.1 A Case II: K 1 is open and K 2 is closed Current in the circuit, i = i 0 e - t τ ;   τ = L R After t = 1   ms = 1 0 - 3   s ⇒ i = i 0 e - 1 0 - 3 × 1 5 0 3 × 1 0 - 2 = 0.1 e - 1 5 3 = 0.1 1 e 5 = 0.1 1 5 0   A = 0.67  mA .