In the circuit shown below, the key $\mathrm{K}$ is closed at $\mathrm{t}=0$. The current through the…

In the circuit shown below, the key $\mathrm{K}$ is closed at $\mathrm{t}=0$. The current through the battery is
  1. $\frac{\mathrm{VR}_1 \mathrm{R}_2}{\sqrt{\mathrm{R}_1^2+\mathrm{R}_2^2}}$ at $\mathrm{t}=0$ and $\frac{\mathrm{V}}{\mathrm{R}_2}$ at $\mathrm{t}=\infty$
  2. $\frac{V}{R_2}$ at $t=0$ and $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=\infty$
  3. $\frac{V}{R_2}$ at $t=0$ and $\frac{V R_1 R_2}{\sqrt{R_1^2+R_2^2}}$ at $t=\infty$
  4. $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=0$ and $\frac{V}{R_2}$ at $t=\infty$

Solution

At $\mathrm{t}=0$, 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 $ \mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{\mathrm{V}\left(\mathrm{R}_1+\mathrm{R}_2\right)}{\mathrm{R}_1 \mathrm{R}_2} $

Asked in: JEE Main 2010

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