Electro Magnetic Induction And Alternating Currents Question 408

Question: Two resistors of 10 $ \Omega $ and 20 $ \Omega $ and an ideal inductor of 10 H are connected to a 2 V battery as . Key K is inserted at time t = 0. The initial (t = 0) and final ( $ t\to \infty $ ) currents through the battery are

Options:

A) $ \frac{1}{15}A,\frac{1}{10}A $

B) $ \frac{1}{10}A,\frac{1}{15}A $

C) $ \frac{2}{15}A,\frac{1}{10}A $

D) $ \frac{1}{15}A,\frac{2}{25}A $

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Answer:

Correct Answer: A

Solution:

  • At $ t=0 $ i.e., when the key is just pressed, no current exists inside the inductor.

    So $ 10\Omega $ and $ 20\Omega $ resistors are in series and a net resistance of (10+20) = 30 $ \Omega $ exists across the circuit.

    Hence, $ I_{l}=\frac{2}{30}=\frac{1}{15}A $

    As $ t\to \infty $ , the current in the inductor grows to attain a maximum value. i.e., the entire current passes through the inductor and no current passes through 10 $ \Omega $ resistor.

    Hence, $ I_{2}=\frac{2}{20}=\frac{1}{10}A $



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