Electro Magnetic Induction And Alternating Currents Question 549

Question: An ideal coil of 10 henry is joined in series with a resistance of 5 ohm and a battery of 5 volt. 2 second after joining, the current flowing in ampere in the circuit will be [MP PET 1995]

Options:

A) $ {{e}^{-1}} $

B) $ (1-{{e}^{-1}}) $

C) $ (1-e) $

D) e

Show Answer

Answer:

Correct Answer: B

Solution:

From $ i=i_{0}[1-{{e}^{-Rt/L}}] $ , where $ i_{0}=\frac{5}{5}=1\ amp $
$ \therefore \ i=1( 1-{{e}^{\frac{-5\times 2}{10}}} )=(1-{{e}^{-1}})amp $



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