Q.
In the circuit shown below, the key K is closed at t = 0. The current through the battery is
A) $\frac{VR_{1}R_{2}}{\sqrt{R_{1}^{2} + \ R_{2}^{2}}}\ at\ t\ = \ 0\ and\frac{V}{R_{2}}\ at\ t = \infty$
B) $\frac{V}{R_{2}}at\ t\ = \ 0\ and\frac{V(R_{1}{+ R}_{2})}{R_{1}R_{2}}\ at\ t = \infty$
C) $\frac{V}{R_{2}}at\ t\ = \ 0\ and\frac{VR_{1}R_{2}}{\sqrt{R_{1}^{2} + \ R_{2}^{2}}}at\ t = \infty$
D) $\frac{V(R_{1}{+ R}_{2})}{R_{1}R_{2}}at\ t\ = \ 0\ and\frac{V}{R_{2}}\ at\ t = \infty\ $
Explanation
(b) At t = 0, inductor behaves like an infinite resistance. So at
$\ \ \ \ t = 0,\ i = \frac{V}{R_{2}}$
And at $t = \infty,$ inductor behaves like a conducting wire,
$i = \frac{V}{R_{eq}} = \frac{V(R_{1} + R_{2})}{R_{1}R_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.