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In the circuit shown below, the key $\mathrm{K}$ is closed at $\mathrm{t}=0$. The current through the battery is
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Verified Answer
The correct answer is:
$\frac{V}{R_2}$ at $t=0$ and $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=\infty$
$\frac{V}{R_2}$ at $t=0$ and $\frac{V\left(R_1+R_2\right)}{R_1 R_2}$ at $t=\infty$
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
$$
\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{\mathrm{V}\left(\mathrm{R}_1+\mathrm{R}_2\right)}{\mathrm{R}_1 \mathrm{R}_2}
$$
So at $t=0, i=\frac{V}{R_2}$ and at $t=\infty$, inductor behaves like a conducting wire
$$
\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {eq }}}=\frac{\mathrm{V}\left(\mathrm{R}_1+\mathrm{R}_2\right)}{\mathrm{R}_1 \mathrm{R}_2}
$$
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