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NTA Abhyas JEE Main2020PhysicsAlternating CurrentPractice

An inductor ( L = 0.03 H ) and a resistor ( R = 0.15 kΩ ) are connected in series to a battery of 15 V EMF 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 : e 5 ≅ 150

Options

  1. A0.67 mA
  2. B100 mA
  3. C67 mA
  4. D6.7 mA

Correct answer

A. 0.67 mA

Step-by-step solution

Case I: K 1 is closed for long tiime for long time, 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

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