Q. Two resistance of $10\ \Omega$ and $20\ \Omega$ and an inductor of inductance $5\ H$ are connected to a battery of $2\ V$ through a key $k$ as shown in the figure. At time $t = 0,$ when the key $k$ is closed the initial current through the battery is

A) $0.2\ A$
B) $\frac{2}{15}A$
C) $\frac{1}{15}A$
D) 0
Explanation

(c) Initially the inductance will oppose the current which tries to flow through the inductance. But $10\Omega$ and $20\Omega$ can conduct. The current will be
$\frac{2V}{30\Omega} = \frac{1}{15}A$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.