Q.
In the circuit shown below, the key $K$ is closed at $t = 0$. The current through the battery is
A) $\frac{V\left( R_{1} + R_{2} \right)}{R_{1}R_{2}}$ at $t = 0$ and $\frac{V}{R_{2}}$ at $t = \infty$
B) $\frac{V\left( R_{1} + R_{2} \right)}{\sqrt{R_{1}^{2}R_{2}^{2}}}$ at $t = 0$ and $\frac{V}{R_{2}}$ at $t = \infty$
C) $\frac{V}{R_{2}}$ at $t = 0$ and $\frac{V\left( R_{1} + R_{2} \right)}{R_{1}R_{2}}$ at $t = \infty$
D) $\frac{V}{R_{2}}$ at $t = 0$ and $\frac{V\left( R_{1} + R_{2} \right)}{\sqrt{R_{1}^{2}R_{2}^{2}}}$ at $t = \infty$
Explanation
(c) At $t = 0$ inductor behaves as broken wire then $i = \frac{V}{R_{2}}$
[image:6_image60.emf]
At $t = \infty$ Inductor behaves as conducting wire
$i = \frac{V}{R_{2}R_{2}/(R_{1} + R_{2})} = \frac{V(R_{1} + R_{2})}{R_{1}R_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.