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 $
