Q. An ideal coil of 10 H is connected in series with a resistance of $5\ \Omega$ and a battery of 5 V. 2s after the connection is made, the current flowing (in ampere) in the circuit is
A) $(1 - e)$
B) e
C) $e^{- 1}$
D) $(1 - e^{- 1})$
Explanation
(d) Rise of current in L – R circuit is given by
[image:7_image59.png]
$\ \ I = I_{0}(1 - e^{- t/\tau})$
Where, $I_{0} = \frac{E}{R} = \frac{5}{5} = 1\ $A
Now, $\tau = \frac{L}{R} = \frac{10}{5} = 2\ s$
After 2s, ie, at t = 2 s
Rise of current $I = \left( 1 - e^{- 1} \right)$A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.