logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Induction

An inductor $(L = 100\: mH)$, a resistor $(R = 100\Omega)$ and a battery $(E = 100 \: V)$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit $1\: ms$ after the circuit is

 

$\begin {array} {1 1} (A)\;1A & \quad (B)\;\large\frac{1}{e}A \\ (C)\;eA & \quad (D)\;0.1A \end {array}$

 

1 Answer

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

Related questions

...