Question

In the circuit shown below, the key $K$ is closed at $t = 0$. The current through the battery is

• A. $\frac{VR_{1}R_{2}}{\sqrt{R^{2}_{1}+R^{2}_{2}}}$ at $t = 0$ and $\frac{V}{R^{2}}$ at $t = \infty$
• B. $\frac{V}{R^{2}}$ at $t = 0$ and $\frac{V\left(R_{1}+R_{2}\right)}{R^{2}_{1}R^{2}_{2}}$ at $t = \infty$
• C. $\frac{V}{R^{2}}$ at $t = 0$ and $\frac{VR_{1}R_{2}}{\sqrt{R^{2}_{1}+R^{2}_{2}}}$ at $t = \infty$
• D. $\frac{V\left(R_{1}+R_{2}\right)}{R^{2}_{1}R^{2}_{2}}$ at $t = 0$ and $\frac{V}{R^{2}}$ at $t = \infty$

Answer 1

Answer: (B)

$\frac{V}{R^{2}}$ at $t = 0$ and $\frac{V\left(R_{1}+R_{2}\right)}{R^{2}_{1}R^{2}_{2}}$ at $t = \infty$

Answer 2

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 $i = \frac{V}{R_{eq}} = \frac{V\left(R_{1}+R_{2}\right)}{R^{2}_{1}R^{2}_{2}}$