SATHEE: Electro Magnetic Induction And Alternating Currents Question 204

Electro Magnetic Induction And Alternating Currents Question 204

Question: An inductor (L=0.03 H) and a resistor $(R = 0.15 k Ω)$ are connected in series to a battery of 15V 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 f = 1 ms, the current in the circuit will be : $(e^{5} ≅ 150)$

Options:

A) 6.7 Ma

B) 0.67 mA

C) 100 mA

D) 67 mA

Show Answer

Answer:

Correct Answer: B

Solution:

$I (0) = \frac{15 \times 100}{0.15 \times 10^{3}} = 0.1 A$ $I (\infty) = 0$

$I (t) = {[I (0) - I (\infty)]}_{e^{^{\frac{- t}{L / R} + i (\infty)}}}$

$I (t) = 0.1 e^{^{\frac{- t}{L / R} + i (\infty)}} = 0.1 e^{\frac{R}{L}}$

$I (t) = 0.1 e \frac{0.15 \times 1000}{0.03} = 067 m A$

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Electro Magnetic Induction And Alternating Currents Question 204

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