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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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An ideal coil of $10H$ is connected in series with a resistance of $5 \Omega$ and a battery of $ 5V$. 2 second after the connection is made the current flowing in amperes in the circuit is

$\begin {array} {1 1} (A)\;1-e & \quad (B)\;e \\ (C)\;e^{-1} & \quad (D)\;1-e^{-1} \end {array}$

 

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1 Answer

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$I = I_0 (1 – e^{\large\frac{Rt}{L}}) = 1 – e^{-1}$
Ans : (D)
answered Feb 16, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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