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

Question

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, 0 . 67 mA, 100 mA, 67 mA, 6 . 7 mA

MARKS App · Accepted Answer

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 .

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