Q. An ideal coil of $10\ henry$ is joined in series with a resistance of $5\ ohm$ and a battery of $5\ volt$. 2 second after joining, the current flowing in $ampere$ in the circuit will be
A) $e^{- 1}$
B) $(1 - e^{- 1})$
C) $(1 - e)$
D) $e$
Explanation
(b) From $i = i_{0}\lbrack 1 - e^{- RT/L}\rbrack$, where $i_{0} = \frac{5}{5} = 1\ amp$
$\therefore i = 1\left( 1 - e^{\frac{- 5 \times 2}{10}} \right) = (1 - e^{- 1})amp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.