Electro Magnetic Induction And Alternating Currents Question 408

Question: Two resistors of 10 Ω and 20 Ω 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 ) currents through the battery are

Options:

A) 115A,110A

B) 110A,115A

C) 215A,110A

D) 115A,225A

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

    Hence, Il=230=115A

    As t , 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 Ω resistor.

    Hence, I2=220=110A



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