Q.
An inductor $(L = 100\ mH)$, a resistor $(R = 100\Omega)$ and a battery $(E = 100V)$ 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 short circuit is
A) $e\ A$
B) $0.1\ A$
C) $1\ A$
D) $1/e\ A$
Explanation
(d) During decay of current
$i = i_{0}e^{- \frac{Rt}{L}} = \frac{E}{R}e^{- \frac{Rt}{L}} = \frac{100}{100}e^{- \frac{100 \times 10^{- 3}}{100 \times 10^{- 3}}} = \frac{1}{e}A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.