Q.
An inductor (L = 100 mH), a resistor $(R\ = \ 100\ \Omega)$ and a battery (E = 100 V) are initially connected in series as shown in 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 short circuit is
A) $1/e\ A$
B) $e\ A$
C) 0.1 A
D) 1 A
Explanation
(a) This is a combined example of growth and decay of current in an L – R circuit.
[image:7_image54.png]
The current through circuit just before shorting the battery,
$I_{0} = \frac{E}{R} = 1\ $A
(as inductor would be shorted in steady state)
After this decay of current starts in the circuit according to the equation $I = I_{0}e^{- t/\tau}$
Where $\tau = L/R.$
[image:7_image55.png]
$I = 1 \times e^{- (1 \times 1\ \ 0^{- 3})/(1\ \ 00 \times 1\ {\ 0}^{- 3}/1\ \ 00)} = (1/e)\ $A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.