Q. An inductor of $2\ henry$ and a resistance of $10\ ohms$ are connected in series with a battery of $5\ volts$. The initial rate of change of current is
A) $0.5\ amp/sec$
B) $2.0\ amp/sec$
C) $2.5\ amp/sec$
D) $0.25\ amp/sec$
Explanation
(c) $i = i_{0}\left( 1 - e^{\frac{- Rt}{L}} \right) \Rightarrow \frac{di}{dt} = \frac{d}{dt}i_{0} - \frac{d}{dt}i_{0}e^{\frac{- Rt}{L}}$
$\Rightarrow \frac{di}{dt} = 0 - i_{0}\left( - \frac{R}{L} \right)e^{- \frac{Rt}{L}} = \frac{i_{0}R}{L}e^{- \frac{Rt}{L}}$
Initially, $t = 0 \Rightarrow \frac{di}{dt} = \frac{i_{0} \times R}{L} = \frac{E}{L} = \frac{5}{2} = 2.5\ amp/sec$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.