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An inductor of inductance $2 \mathrm{mH}$ is connected to a $220 \mathrm{~V}, 50 \mathrm{~Hz}$ a.c. source. Let inductive reactance in the circuit is $X_1$. If a $220 \mathrm{~V}$ d.c. source replaces the a.c. source in the circuit, then the inductive reactance in the circuit is $\mathrm{X}_2 \cdot \mathrm{X}_1$ and $\mathrm{X}_2$ respectively are:
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Verified Answer
The correct answer is:
$0.628 \Omega$, zero
For AC circuits,
$\begin{aligned} X_{\mathrm{L}} & =\omega \mathrm{L} \\ \mathrm{X}_{\mathrm{L}} & =2 \pi n \mathrm{~L} \\ \mathrm{X}_{\mathrm{L}} & =0.628 \Omega \\ \text {Hence, } & \mathrm{X}_1=0.628 \Omega\end{aligned} \quad(\because \omega=2 \pi n)$
For DC circuits, $\omega=0$
$\begin{aligned}
X_{\mathrm{L}} & =\omega \mathrm{L} \\
\mathrm{X}_{\mathrm{L}} & =0 \\
\mathrm{X}_2 & =0
\end{aligned}$
Hence $\quad X_2=0$
$\begin{aligned} X_{\mathrm{L}} & =\omega \mathrm{L} \\ \mathrm{X}_{\mathrm{L}} & =2 \pi n \mathrm{~L} \\ \mathrm{X}_{\mathrm{L}} & =0.628 \Omega \\ \text {Hence, } & \mathrm{X}_1=0.628 \Omega\end{aligned} \quad(\because \omega=2 \pi n)$
For DC circuits, $\omega=0$
$\begin{aligned}
X_{\mathrm{L}} & =\omega \mathrm{L} \\
\mathrm{X}_{\mathrm{L}} & =0 \\
\mathrm{X}_2 & =0
\end{aligned}$
Hence $\quad X_2=0$
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